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Page 1: Analysis Spm Additional Mathematics
Page 2: Analysis Spm Additional Mathematics
Page 3: Analysis Spm Additional Mathematics

copyrights by :

Cerdik Publications Sdn Bhd

No. 39, Jalan Nilam 1/2, Subang Square, Subang Hi-Tech Industrial Park, Batu Tiga,

40000 Shah Alam, Selangor

Tel : 603 5637 9044Fax : 603 5637 9043

www.cerdik.com.my

Page 4: Analysis Spm Additional Mathematics

1 (a) Onetoonerelation. (b) {(1,b),(2,a),(3, c)}

2 (a)

–2

2

4

6

4

6

8

SetA SetB

(b) (i) Manytoonerelation. (ii) Objects=–2,2,4,6 Images=4,6,8

3 (a) (i) Domain={a,b,c,d} Codomain={e,f,g} (ii) Range={e,f,g} (b) Manytomanyrelation.

4 (a) Domain={–1,1,3,5} Codomain={1,9,25} (b) 9 (c) 5 (d) Range={1,9,25}

5 (a)

SetQ

SetP

7

5

3

2

8 9 12 25 49

(b) (i) 3 (ii) 25

6 (a) Manytoonerelation. (b) (i) Images=p,q Range={p,q} (ii) Noobject

7 (a) Codomain={a,b,c} Range={b} (b) 3,6,9

8 (a)

Multiple of

2

3

4

6

8

9

12

SetP SetQ

(b) {(2,6),(2,8),(2,12),(3,6), (3,9),(3,12),(4,8),(4,12)}

(c)

Set Q

Set P

12

9

8

6

2 3 4

9 (a) Notafunction. (b) Notafunction. (c) Afunction. (d) Afunction.

10 (a) Onetoonerelation. (b) Manytoonerelation. (c) Onetomanyrelation. (d) Manytomanyrelation.

11 (a) Afunction. (b) Notafunction. (c) Notafunction. (d) Afunction.

12 (a) Onetoonefunction. (b) Notonetoonefunction. (c) Notonetoonefunction. (d) Notonetoonefunction.

13 (a) Yes (b) Yes (c) No (d) Yes

14 (a) f:xa x (b) f:xax2

15 (a) (i) f(–1)= 5–2(–1) = 7 (ii) f(2)+f(–2) = [5–2(2)]+[5–2(–2)] = 1+9 = 10

(iii) g( 27 ) = 7(2

7 )+4

= 6

(iv) g( 17 )–g(– 1

7 ) = [7(1

7 )+4]–[7(– 17 )+4]

= 5–3 = 2

(b) (i) f(x) = g(x) 5–2x = 7x+4 9x = 1

x =19

(ii) f(x) = 9 5–2x = 9 2x = –4 x = –2

16 (a) f(3)+g(4)

= ( 33

+2)+(34

(4)–1) = 3+2 = 5 (b) 2f(6)–3g(8)

= 2(63

+2)–3[34

(8)–1] = 8–15 = –7 (c) f(–12)–g(–12)

= (–123

+2)–[ 34

(–12)–1] = –2+10 = 8 (d) f(–3)–g(–4)

= (– 33

+2)–[ 34

(–4)–1] = 1+4 = 5

17 (a) f(0) = –7 m(0)+n = –7 n = –7 f(3) = 2 3m+n = 2 3m–7 = 2 3m = 9 m = 3 \m+n=3+(–7)=–4 (b) f(x) = 3x–7 f(3) = 3(3)–7 = 2 (c) f(x) = 2 3x–7 = 2 3x = 9 x = 3

18 f(3) = 10 a(3)2+b = 10 9a+b = 10… 1 f(–2) = –10 a(–2)2+b = –10 4a+b = –10… 2 1 – 2 : 5a = 20 a= 4 Substitutea=4into 1 : 9(4)+b = 10 b = 10–36 = –26 \ a

b = – 4

26

= – 213

19 f(x)=3x–2 (a) f(–2) = 3(–2)–2 = 8 f(1) = 3(1)–2 = 1

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 5: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) f(x) = 8 3x–2 = 8 3x–2 = 8 or 3x–2= –8 3x = 10 3x = –6 x = 10

3 x = –2

(c) f(x) = x 3x–2 = x 3x–2 = x or 3x–2 = –x 2x = 2 4x = 2 x = 1 x = 1

220 (a) k=1 (b) 0<f(x)<3

21 (a) f(x) = x2–2x+3 f(–1) = (–1)2–2(–1)+3 = 6 \m=6 f(4) = 42–2(4)+3 = 11 \n=11 (b) 2<f(x)<11

22 (a) h(3)–h(2) = 8–4 = 4 (b) f(2)+g(12) = 9+8 = 4 (c) h(2)3h(1) = 4(4) = 16

(d)h(3)h(1)

=84

= 2

23 (a) s (b) s (c) u (d) w (e) u (f) w

24 (a) 4 (b) –3 (c) 3 (d) –1

25 (a) f=g 3x+4= x2+6 x2–3x+2= 0 (x–1)(x–2)= 0 x=1orx=2 (b) fg = gf f(x2+6) = g(3x+4) 3(x2+6)+4 = (3x+4)2+6 3x2+22 = 9x2+24x+22 6x2+24x = 0 6x(x+4) = 0 x=0orx=–4

26 (a) f 2=g 2(2x+3)+3 = 3x+6 4x+9 = 3x+6 x = –3 (b) 3(3x+6)+6 = 2x+3 9x+24 = 2x+3 7x = –21 x = –3

27 fg(a) = 8 f(a–3) = 8 (a–3)2+4 = 8 a2–6a+5 = 0 (a–1)(a–5) = 0 a=1ora=5

28 fg(–4)+2 = gf(2) f(2)+2 = g(4–2m) 4–2m+2 = 3(4–2m)+14 6–2m = 26–6m 4m = 20 m = 5

29 (a) gf(x) = x+5 4f(x)–1 = x+5 4f(x) = x+6

f(x)=x+6

4 (b) gf(x) = x

f(x)

1–f(x) = x

f(x) = x–xf(x) f(x)+xf(x) = x f(x)[1+x] = x

f(x) =x

1+x ,x ≠ –1

(c) fg(x) =2x2–13x+22 f(x–3) =2x2–13x+22 f(k) =2(k+3)2–13(k+3)+22 =2(k2+6k+9)–13k –39+22 =2k2+12k+18–13k –17 =2k2–k+1 \ f(x)=2x2–x+1 (d) fg(x)=4x2+12x f(2x+1) = 4x2+12x

f(k) = 4(k–12 )2

+12(k–12 )

= 4( k2–2k+14 )+

6(k–1) = k2–2k+1+6k–6 = k2–4k–5 f(x)=x2–4x–5

30 (a) fg(x) = 4x2+2x 2g(x)+4 = 4x2+2x 2g(x) = 4x2+2x–4 g(x) = 2x2+x–2 \g(2) = 2(2)2+2–2 = 8

(b) fg(x) = x–2

f (2x+1x ) = x–2

f(k) =1

k–2–2

f(x) =1

x–2–2

\f(4) =1

4–2–2

= –112

31 f 2(x) = 16x–15 f (ax+b) = 16x–15 a(ax+b)+b = 16x–15 a2x+ab+b = 16x–15 a2 =16 and ab+b = –15 a =±4 4b+b = –15 =4(0) 5b = –15 b = –3

\a+b = 4+(–3) = 1

32 g(2) = 5 4a+b = 5 … 1 gf(1) = –1 g(–1) = –1 a+b = –1 … 2 1 – 2 : 3a = 6 a = 2 Substitutea=2into 1 : 8+b = 5 b = –3 \a=2,b=–3

33 (a) –2 (b) f(a) = 8 a+5 = 8 a = 3

34 (a) a (b) q (c) a (d) b

35 (a) y = 3x–2 y+2 = 3x

x =y+2

3

\f –1:xa x+2

3

(b) y =5x+23x–2

3xy–2y = 5x+2 3xy–5x = 2y+2 x(3y–5) = 2y+2

x =2y+23y–5

\f –1:xa 2x+23x–5 ,x ≠

53

(c) y =3x+2

4x 4xy = 3x+2 4xy–3x= 2 x(4y–3) = 2

x =2

4y–3

\f –1:xa 2

4x–3 ,x ≠ 34

(d) y =x–14–x

4y–xy = x–1 x+xy = 4y+1 x(1+y) = 4y+1

x =4y+1y+1

\f –1:xa 4x+1x+1 ,x ≠ –1

(e) y =3x

5x–2 5xy–2y = 3x x(5y–3) = 2y

x =2y

5y–3

\f –1:xa 2x

5x–3 ,x ≠ 35

Page 6: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(f) y =2x–5

2 2y = 2x–5 2x = 2y+5

x =2y+5

2

\ f –1:xa 2x+5

2

36 (a) 7–5y = x 5y = 7–x

y =7–x

5

\ f :xa 7–x

5

(b)–5y–13y–2 = x

–5y–1 = 3xy–2x 3xy+5y = 2x–1 y(3x+5) = 2x–1

y =2x–13x+5

\ f :xa 2x–13x+5 ,x ≠ –

53

(c)y+4

3 = x

y+4 = 3x y = 3x–4 \ f :xa 3x–4

(d)3y

y–3 = x

3y = xy–3x xy–3y = 3x y(x–3) = 3x

y = 3x

x–3

\ f :xa 3x

x–3 ,x ≠ 3

(e)23 y = x

y =32 x

\ f:x a 32

x

(f)5y = x

y =5x

\ f : x a 5x

,x ≠ 0

37 (a) f(6) = 5

12+k

3 = 5

12+k = 15 k = 3

(b) y =2x+3x–3

xy–3y = 2x+3 xy–2x = 3y+3 x(y–2) = 3y+3

x =3y+3y–2

f –1(x)=3x+3x–2 ,x ≠ 2

f –1(–7) =3(–7)+3

–7–2

=–18–9

= 2

38 y =x+2x–3

xy–3y = x+2 xy –x = 3y+2 x(y–1) = 3y+2

x =3y+2y–1

f –1(x) = 3x+2x–1

,x ≠ 1

f –1(4) = 3(4)+24–1

=143

39 y = 1x +3

xy+3y = 1 xy = 1–3y

x = 1–3yy

f –1(x)=1–3xx

,x ≠ 0

f –1(m) = 3

1–3mm

= 3

1–3m = 3m 6m = 1

m =16

40 fg(x) = 1+4x

2x–3

f(x) = 1+4x2x–3

,x ≠ 32

y = 1+4x2x–3

2xy–3y = 1+4x x(2y–4) = 3y+1

x = 3y+12y–4

\f –1:xa 3x+12x–4

,x ≠ 2

41 y–15

= x

y = 5x+1 f(x) = 5x+1

3–y2

= x

3–y = 2x y = 3–2x g(x) = 3–2x fg(x) = f(3–2x) = 5(3–2x)+1 = 16–10x y = 16–10x 10x = 16–y

x = 16–y10

( fg)–1(x) = 16–x10

( fg)–1(6) = 16–610

= 1

42 g–1(x) = 1–xx

g–1(y) = x

1–yy

= x

1–y = xy xy+y = 1 y(x+1) = 1

y = 1x+1

g(x) = 1x+1

,x ≠ – 1

gf(x) = g( 1x–1 )

=

1

( 1x–1 )+1

=1

( xx–1 )

= x–1x

,x ≠ 0

h(x) = gf(x)

= x–1x

,x ≠ 0

y = x–1x

xy = x–1 1 = x–xy 1 = x(1–y)

x = 11–y

\h–1:xa 11–x

,x ≠ 1

1 (a) f:xax+4 (b) a=–1

2 (a) f(x–2) = (x–2)2–5 = x2–4x–1 (b) 5f(–2) = 5[(–2)2–5] = 5(–1) = –5

3 f(m2–7m+6)= g(m+2) m2–7m+6+2= (m+2)2–7(m+2) +6 m2–7m+8= m2+4m+4–7m –14+6 m2–7m+8= m2–3m–4 4m = 12 m = 3

4 (a) (i) Objects=3,2,–3 Range={4,9} (ii) 2 (b) Notafunctionbecausef –1isnot

onetoonefunction.

5 (a) f(2) = 1

k

2–k= 1

k = 2–k 2k = 2 k = 1

Page 7: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) y =1

x–1

xy–y = 1 xy = y+1

x =y+1

y

f –1(x) =x+1

x

f –1(–3) =–3+1

–3

=23

6 (a) f(–1) = –12

24

–p+q= –12

–2 = –p+q p–q = 2… 1 f(1) = 6

24

p+q = 6

p+q = 4… 2 1 + 2 : 2p = 6 p = 3 Substitutep=3into 1 : 3–q = 2 q = 1 \p=3,q=1

(b) f(x) = x

24

3x+1 = x

24 = 3x2+x 3x2+x–24 = 0 (3x–8)(x+3) = 0

x=83

orx=–3

7 (a) Atx-axis,y=0, 0= 2x–1

x =12

\k=12

(b) |2x–1|=3 2x–1= 3 or 2x–1= –3 2x = 4 2x = –2 x = 2 x = –1 \–1<x<2

8 y =3–x2x

2xy+x = 3 x(2y+1) = 3

x =3

2y+1

f–1(x) =3

2x+1

3

2x+1 =

3–x2x

6x = 3+5x–2x2

2x2+x–3 = 0 (2x+3)(x–1) = 0

x=–32

orx=1

9 f(mx+n) = m(mx+n)+n = m2x+mn +n

Comparewithf 2(x)=4x–9 m2= 4 m = ±2 Whenm=2, 3n = –9 n = –3 Whenm=–2, –n = –9 n = 9

10 (a) f(3x+2) = 3(3x+2)+2 f 2(x) = 9x+8

(b) f 2(2) = 9(2)+8 = 26 f 2(–1) = 9(–1)+8 = –1 Range:–1<f 2(x)<26

11 (a) gf = g(2x+1) = (2x+1)2–1 = 4x2+4x (b) 4x2+4x = x2–1 3x2+4x+1 = 0 (3x+1)(x+1) = 0

x=–13

orx=–1

12 (a) f(x)=x2andg(x)=x+1 gf = g(x2) = x2+1 \gf:xax2+1

(b) gf(3) = 32+1 = 10

13 (a) 25 (b) 2

14 (a) (i) 3 (ii) 7

(b) g(x2) = x2–2 g(k) = ( k )2–2 = k–2 \g:xax–2

15 fg(x) = x2+4x+5 [g(x)]2+1 = x2+4x+5 [g(x)]2 = x2+4x+4 = (x+2)2

g(x) = x+2 \g:xax+2

16 (a) f (1)= 3(1)+51

= 8

(b) 3x+5x

= 8

3x2–8x+5= 0 (3x–5)(x–1)= 0

x=53

orx=1

17 h(x) = gf(x) = g(2–3x) = 2–3x+4 = 6–3x y = 6–3x

x =6–y

3

h–1(x) =6–x

3

\h–1(3) =6–3

3

= 1

18 (a) f –1(y) = x

2y–13–y

= x

2y–1 = 3x–xy 2y+xy = 3x+1 y(2+x) = 3x+1

y =3x+12+x

f(x) =3x+12+x

f(x) = 2

3x+12+x

= 2

3x+1 = 4+2x x = 3

(b) gf(2) = g[ 3(2)+12+2 ]

= g(74)

= 4(74)

= 7

19 y= 2x–m

x=y+m

2

f –1(x)=12

x+m2

Comparewithf –1(x)=nx+

52

\ m=5,n=12

20 (a) f –1(y) = x

y–k

2 = x

y–k = 2x y = 2x+k f(x) = 2x+k f(–1) = 1 –2+k = 1 k = 3

(b) f (12) = 2(1

2)+3

= 4

21 (a) gf(5) = 10 g(2) = 10 4m–2 = 10 4m = 12 m = 3

(b) f ( 4x–3 ) =

4

( 4x–3 )–3

=4

( 13–3xx–3 )

=

4x–1213–3x

Comparewithax–bc–dx

.

\a=4,b=12,c=13,d=3

Page 8: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

22 (a) A(32

,–4) (b) –4<f(x)<3

(c) |2x–3|–4 = –2 |2x–3| = 2 2x–3 = 2 or 2x–3 = –2 2x = 5 2x = 1

x =52

x =12

23 (a) y = 3x–2

x =y+2

3

\f –1:x a13

x +23

(b) 3x–2 =13

x+23

9x–6 = x+2 8x = 8 x = 1 Whenx=1, y= 3(1)–2 = 1 \P(1,1)

(c) –2<f(x)<7,0<f –1(x)<3

24 (a) f 2 = f ( 1x+1 )

=1

( 1x+1 )+1

=x+1x+2

,x ≠ –2

gf = g( 1x+1 )

= 3( 1x+1 )+1

=x+4x+1

,x ≠ –1

(b)x+1x+2

=x+4x+1

x2+2x+1 = x2+6x+8 4x = –7

x = – 74

25 (a) f(x) =18x

+8x

f (x) = 0

–18x2 +8 = 0

18x2 = 8

x2 =94

x = ±32

=32

(0)

Range:32

<f(x)<30

(b) y = 8–3x

x =8–y

3

f –1(x) =8–x

3

g(8–3x) = 8–3(8–3x) = 9x–16

8–x

3 = 9x–16

8–x = 27x–48 28x = 56 x = 2

(c) h(x)=ax–b At(3,2), 2= 3a–b b = 3a–2… 1 y = ax–b

x =y+b

a

h–1(x) =x+b

a Forthepointofintersection

wherex=4,

ax–b =x+b

a a2x–ab = x+b 4a2–ab = 4+b… 2 4a2–a(3a–2) = 4+3a–2 a2–a–2 = 0 (a–2)(a+1) = 0 a = 2(0) When a=2,b= 3(2)–2 = 4

26 (a) (i) y = 5–6x

6x

= 5–y

x =6

5–y

\f –1:xa 6

5–x,x ≠ 5

(ii)6

5–x = 5–

6x

6

5–x+

6x

= 5

30

x(5–x) = 5

30 = 25x–5x2

5x2–25x+30 = 0 x2–5x+6 = 0 (x–2)(x–3) = 0 x=2orx=3

(b) (i) y =x–3

2 x = 2y+3 \g–1:xa2x+3

(ii) 2x+3 =x–3

2 4x+6 = x–3 3x = –9 x = –3

27 (a) gf (b) g–1f (c) f –1g

28 (a) f(x) = –x2+6x+1 = –(x2–6x–1) = –(x2–6x+(–3)2–(–3)2–1] = –[(x–3)2–10] = 10–(x –3)2

y

y=f(x)

xO 3

10

1

(b) Becauseforx>3,itsonetoonefunction.

(c) y = 10–(x–3)2

(x–3)2 = 10–y

x–3 = 10–y

x = 10–y +3

\f –1:xa 10–x +3

29 (a) f(–3) =–4 g(–4) = 6 –3a+5 =–4 b

–4+6 = 6 3a =9b = 12 a =3

\a=3,b=12

(b) gf(–3)=6,f –1g–1(6)=–3

(c) gf(x) = g(3x+5)

=12

3x+5+6

=12

3x+11,x ≠ –

113

30 (a) f(2) = –1 p–2q = –1… 1 g(2) = 2

6

2q–1 = 2

6 = 4q–2 4q = 8 q = 2 Substituteq=2into 1 : p–2(2)= –1 p = –1+4 = 3 \ p=3,q=2

(b) (i) f(y) = 3–2y x = 3–2y 2y = 3–x

y =3–x

2

f –1(x) =3–x

2

\f –1:y a3–y

2

(ii) gf –1(y) = g( 3–y2 )

=6

2( 3–y2 )–1

=6

2–y ,y ≠ 2

gf –1:ya6

2–y ,y ≠ 2

31 (a) (i) f 2 = f(6–3x) = 6–3(6–3x)

Page 9: Analysis Spm Additional Mathematics

= 9x–12

(ii) y = 9x–12

x =y+12

9

\( f 2)–1(x)=x+12

9

(b) y = 6–3x

x =6–y

3

f –1(x) =6–x

3

(f –1)2 = f –1( 6–x3 )

=

6–( 6–x3 )

3 =

x+129

\( f –1)2=( f 2)–1

(c)

y

x24

3O

6

12

Range:0<y<12

32 (a) y

xO 2

1y=2x–4

y=x +1

4

(b) Numberofsolutions=2

33 (a) (i) f(2) = 5 4a–b = 5… 1 f(–3) = 15 9a–b = 15… 2 2 – 1 : 5a = 10 a = 2 Substitutea=2into 1 : 4(2)–b = 5 b = 3

(ii) f(x) = 2x2–3 f(1) = 2(1)2–3 = –1

(b) 2x2–3 = x 2x2–x–3 = 0 (2x–3)(x+1) = 0

x=32

orx=–1

34 (a) f 2(x)= f ( xm–nx )

=

( xm–nx )

m–n( xm–nx )

=( xm–nx )

( m2–mnx–nxm–nx )

=

xm2–mnx–nx

=

xm2–(mn+n)x

Comparewithf 2(x)=

x4–3x

m2= 4 or mn+n = 3 m = ±2 3n = 3 = 2(0) n = 1 \m=2,n=1

(b) y =x

2–x 2y–xy = x 2y = x(1+y)

x =2y

1+y

f –1(x)=2x

1+x ,x ≠ –1

\f –1(12) =

2(12)

1+12

=

23

35 (a) f(2) = 9 4a+1 = 9 4a = 8 a = 2 gf(2) = 25 6(2)2+b = 25 24+b = 25 b = 1 \a=2,b=1

(b) g(2x2+1) = 6x2+1

g(k) = 6( k–12 )2

+1

= 6( k–12 )+1

= 3k–2 \g:xa3x–2

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 10: Analysis Spm Additional Mathematics

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-311

Quadratic Equations

1 (a) No (b) No (c) Yes (d) Yes

2 (a) 3x2 – 12x = 0, a = 3, b = –12, c = 0 (b) x2 – 25 = 0, a = 1, b = 0, c = –25 (c) x2 – 8x + 8 = 0, a = 1, b = –8,

c = 8 (d) x2 + x – 5 = 0, a = 1, b = 1,

c = –5

3 (a) –3, 8 (b) –1, 5

(c) –2

12

, 4 (d) 0, 5

4 (a) –4, 0 (b) 4, 9

(c) 7 (d) –4,

52

5 (a) –2, 12

(b) –5, –4

(c) –4, 2 (d) –3, 2

6 x2 + 9x + 2p = 0 When x = –5, (–5)2 + 9(–5) + 2p = 0 2p = 20 p = 10

7 mx2 – 5x – 3 = 0 When x = 3, m(3)2 – 5(3) – 3 = 0 9m = 18 m = 2

8 (a) ax2 + 5x = 12 When x = 2, a(2)2 + 5(2) = 12 4a = 2

a = 12

(b)

12

x2 + 5x = 12

x2 + 10x – 24 = 0 (x + 12)(x – 2) = 0 x = –12 or x = 2 \ x = –12

9 x2 + (2p – 1)x + (p2 – 3p – 4) = 0 When x = 0, p2 – 3p – 4 = 0 (p + 1)(p – 4) = 0 p = –1 or p = 4

10 x2 – 2x = 5 When x = a, a2 – 2a = 5 a(a – 2) = 5

a – 2 = 5a

1a

= a – 2

5 (shown)

11 (a) (x + 6)(x + 2) = 32 x2 + 8x – 20 = 0 (x + 10)(x – 2) = 0 x = –10 or x = 2 (b) (x + 4)2 – 3(x + 4) = 10 x2 + 8x + 16 – 3x – 12 = 10 x2 + 5x – 6 = 0 (x + 6)(x – 1) = 0 x = – 6 or x = 1 (c) 8x2 – 16 = 7x 2

x2 – 16 = 0 (x + 4)(x – 4) = 0 x = – 4 or x = 4

(d)

x + 6

x(x + 3) =

12

2x + 12 = x2 + 3x x2 + x – 12 = 0 (x + 4)(x – 3) = 0 x = – 4 or x = 3 (e) 4y2 – 12y + 9 = 6y + 1 4y2 – 18y + 8 = 0 2y2 – 9y + 4 = 0 (2y – 1)(y – 4) = 0

y = 12

or y = 4

(f) 2x2 + 3x – 2 = x2 + 16 x2 + 3x – 18 = 0 (x + 6)(x – 3) = 0 x = – 6 or x = 3

12 (a) x2 + 8x = 20 x2 + 8x + (4)2 = 20 + (4)2

(x + 4)2 = 36 x = 36 – 4 or x = – 36 – 4 = 2 = –10 \ x = 2 or x = –10 (b) 5y2 – 30y = 18

y2 – 6y = 185

y2 – 6y + (–3)2 =

185

+ (–3)2

(y – 3)2 =

635

y =

635

+ 3 or y = –

635

+ 3

= 6.55 = – 0.55 \ y = 6.55 or y = – 0.55 (c) 2y2 – 12y = 15

y2 – 6y =

152

y2 – 6y + (–3)2 =

152

+ (–3)2

(y – 3)2 =

332

y =

332

+ 3 or y = –

332

+ 3

= 7.062 = – 1.062 \ y = 7.062 or y = –1.062

(d) x2 – 4x = 21 x2 – 4x + (–2)2 = 21 + (–2)2

(x – 2)2 = 25 x = 5 + 2 or x = –5 + 2 = 7 = –3 \ x = 7 or x = –3 (e) 3y2 + 6y = 2

y2 + 2y =

23

y2 + 2y + (1)2 =

23

+ (1)2

(y + 1)2 =

53

y =

53

– 1 or y = –

53

– 1

= 0.291 = – 2.291 \ y = 0.291 or y = –2.291 (f) 5y2 – 30y = 18

y2 – 6y =

185

y2 – 6y + (–3)2 =

185

+ (–3)2

(y – 3)2 =

635

y =

635

+ 3 or y = –

635

+ 3

= 6.5496 = – 0.5496 \ y = 6.5496 or y = –0.5496

13 (a) 2x2 + 5x – 3 = 0

x =

–5 52 – 4(2)(–3)

2(2)

=

–5 494

=

–5 + 7

4 or

–5 – 7

4

=

12

or –3

\ x = 12

or x = –3

(b) x2 + 3x – 28 = 0

x =

–3 32 – 4(1)(–28)

2(1)

=

–3 1212

=

–3 + 11

2 or

–3 – 11

2 = 4 or –7 \ x = 4 or x = –7 (c) x2 – 2x – 8 = 0

x =

2 (–2)2 – 4(1)(–8)

2(1)

=

2 362

=

2 + 6

2 or

2 – 6

2

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 11: Analysis Spm Additional Mathematics

2© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

= 4 or –2 \ x = 4 or x = –2 (d) 12x2 – 7x – 12 = 0

x =

7 (–7)2 – 4(12)(–12)

2(12)

=

7 62524

=

7 + 25

24 or

7 – 25

24

= 1

13

or – 34

\ x = 1 13

or x = – 34

(e) 5x2 – 6x –8 = 0

x =

6 (–6)2 – 4(5)(–8)

2(5)

=

6 19610

=

6 + 14

10 or

6 – 14

10

= 2

or –

45

\ x = 2 or x = – 45

(f) 4x2 – 8x –5 = 0

x =

8 (–8)2 – 4(4)(–5)

2(4)

=

8 1448

=

8 + 12

8 or

8 – 12

8

= 2

12

or – 12

\ x = 2 12

or x = – 12

14 (a) x2 – 3x – 5 = 0 a + b = 3 ab = –5 (b) 2x2 + 6x = 5x + 9 2x2 + x – 9 = 0

x2 + x2

– 92

= 0

a + b = – 12

ab = –

92

(c) 2x2 + 2 = 3x 2x2 – 3x + 2 = 0

x2 – 32

x + 1 = 0

a + b = 32

ab = 1

(d)

1

x(x + 1) =

14

x2 + x – 4 = 0 a + b = –1 ab = –4

15 (a) x2 – (6 + 10)x + (6)(10) = 0 x2 – 16x + 60 = 0

(b) x2 – (–10 + 7)x + (–10)(7) = 0 x2 + 3x – 70 = 0

(c) x2 –

( 23

+ 12 )x +

( 23 )( 1

2 ) = 0

x2 –

76

x + 26

= 0

6x2 – 7x + 2 = 0 (d) x2 – (–7 – 3)x + (–7)(–3) = 0 x2 + 10x + 21 = 0

(e) x2 – ( 1

3 +

13 )x +

( 13 )( 1

3 ) = 0

x2 –

23

x + 19

= 0

9x2 – 6x + 1 = 0 (f) x2 – (0 + 4)x + (0)(4) = 0 x2 – 4x = 0

16 3x2 – 5x + 2 = 0

a + b =

53

ab =

23

(a)

1a

+ 1b

=

a + bab

=

( 53 )

( 23 )

=

52

(b)

ab

+ ba

=

a 2 + b 2

ab =

(a + b)2 – 2abab

=

( 53 )2

– 2( 2

3 )23

=

(139 )

( 23 )

=

136

(c) 2a + 2b = 2(a + b)

= 2( 5

3 )

=

103

(d)

3a

+ 3b

=

3(a + b)ab

=

3( 53 )23

=

152

17 2x2 + 4x – 7 = 0 a + b = –2

ab = –

72

(a) (a – 1) + (b – 1) = (a + b) – 2 = –2 – 2 = –4

(a – 1)(b –1) = ab – a – b + 1 = ab – (a + b) + 1

= –

72

– (–2) + 1

= –

12

\ x2 –(– 4)x +

(– 12 )

= 0

x2 + 4x –

12

= 0

2x2 + 8x – 1 = 0

(b) 2a +

2b =

2(a + b)ab

=

2(–2)

(– 72 )

=

87

( 2a)( 2

b ) =

4

ab

=

4

(– 72 )

= –

87

\ x2 –

87

x – 87

= 0

7x2 – 8x – 8 = 0 (c) 4a + 4b = 4(a + b) = 4(–2) = –8 (4a)(4b) = 16ab

= 16(– 72 )

= –56 \ x2 + 8x – 56 = 0 (d) a2 + b 2 = (a + b)2 – 2ab

= (–2)2 – 2(–

72 )

= 4 + 7 = 11 (a2)(b 2) = (ab)2

=

(– 72 )2

=

494

\ x2 – 11x +

494

= 0

4x2 – 44x + 49 = 0

(e)

a5

+ b5

=

a + b5

= –

25

(a5 )( b

5 ) =

ab25

=

(– 72 )

25

= –

750

\ x2 +

25

x –

750

= 0

50x2 + 20x – 7 = 0

Page 12: Analysis Spm Additional Mathematics

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-33

(f) a2b + ab2 = ab(a + b)

= –

72

(–2)

= 7 (a2b)(ab2) = (ab)3

=

(– 72 )3

= –

3438

\ x2 – 7x –

3438

= 0

8x2 – 56x – 343 = 0

18 x2 – 2(m + 1)x + m(m + 1) = 0 a + 2a = 2(m + 1) 3a = 2m + 2

a =

2m + 2

3 … 1

a(2a) = m(m + 1) 2a2 = m(m + 1) … 2 Substitute 1 into 2 :

2( 2m + 23 )2

= m2 + m

2(4m2 + 8m + 4) = 9m2 + 9m 8m2 + 16m + 8 = 9m2 + 9m m2 – 7m – 8 = 0 (m + 1)(m – 8) = 0 m = –1 or m = 8

19 px2 + (p – 1)x + 2p + 3 = 0

x2 + (p – 1)

p x + 2p + 3

p = 0

a + b = – (p – 1)

p ab =

2p + 3p

(a) a + (–a) = – (p – 1)

p 0 = 1 – p p = 1

(b) a( 1a) =

2p + 3p

p = 2p + 3 p = –3

20 a + (a + 2) = –(2 – h) 2a + 2 = h – 2 2a = h – 4

a = h – 4

2 … 1

a(a + 2) = h a2 + 2a = h … 2

Substitute 1 into 2 :

(h – 42 )2

+ 2(h – 4

2 ) = h

h2 – 8h + 164

+ h – 4 = h

h2 – 8h + 16 + 4h – 16 = 4h h2 – 8h = 0 h(h – 8) = 0 h = 0 or h = 8

Substitute h = 8 into 1 :

a =

8 – 42

= 2 a + 2 = 2 + 2 = 4 (a) The roots are 2 and 4. (b) h = 8

21 2(x2 + 3x + 2) = n 2x2 + 6x + 4 – n = 0 a + (a + 3) = –3 2a = –6 a = –3

a(a + 3) =

4 – n

2 … 1

Substitute a = –3 into 1 :

–3(–3 + 3) =

4 – n2

0 = 4 – n n = 4

22 x2 + 3k2 – 5kx = 1 – 2k x2 – 5kx + 3k2 + 2k – 1 = 0 a + 4a = 5k 5a = 5k a = k a(4a) = 3k2 + 2k – 1 4a2 = 3k2 + 2k – 1… 1 Substitute a = k into 1 : 4k2 = 3k2 + 2k – 1 k2 – 2k + 1 = 0 (k – 1)(k – 1) = 0 k = 1

23 x2 – 3x – 2 = 0 a + b = 3 ab = –2 x2 – 6x + p = 0

ha

+ hb

= 6

h(a + b) = 6ab 3h = –12 h = –4

( ha)( h

b ) = p

h2

ab = p

h2 = p(–2) (– 4)2 = –2p 2p = –16 p = –8 \ h = –4, p = –8

24 3x2 + 5x + 1 = 0

a + b = –

53

ab =

13

mx 2 – 4x + n = 0

a + 3 + b + 3 = 4m

a + b + 6 = 4m

– 53

+ 6 =

4m

133

=

4m

m =

1213

(a + 3)(b + 3) =

nm

ab + 3(a + b) + 9 =

nm

13

+ 3(– 53 )

+ 9 =

1312

n

133

=

1312

n

n = 4

\ m = 1213

, n = 4

25 2x2 + kx + 24 = 0

a + b = –

k2

… 1

ab = 12 … 2 Substitute a = 4 + b into 2 : b(4 + b) = 12 b 2 + 4b – 12 = 0 (b + 6)(b – 2) = 0 b = –6 or b = 2 Substitute b = –6 into 1 :

4 + b + b = –

k2

–8 = –

k2

k = 16 Substitute b = 2 into 1 :

4 + b + b = –

k2

8 = –

k2

k = –16 \ k = –16 ( 0)

26 x2 – 9x + 2k = 0 a + b = 9 … 1 ab = 2k … 2 Substitute a = 2b into 1 : 3b = 9 b = 3 Substitute b = 3 into 2 : (2b)(b) = 2k 2k = 18 k = 9

27 mx2 + 7x – n = 0

x2 +

7m

x – nm

= 0

a + b = –

7m

ab = – nm

– 4 +

12

= – 7m

– 72

= – 7m

Page 13: Analysis Spm Additional Mathematics

4© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

m = 2

– 4( 12 )

= – nm

–2 = –

n2

n = 4 \ m = 2, n = 4

28 x2 – hx – k = 0 a + b = h ab = –k –3 + 5 = h h = 2 –3(5) = –k k = 15 \ h = 2, k = 15

29 p + 1 + q + 1 = 1 p + q = –1 p = –1 – q … 1 (p + 1)(q + 1) = –12 pq + p + q + 1 = –12 … 2

Substitute 1 into 2 : q(–1 – q) – 1 – q + q + 1 = –12 q2 + q – 12 = 0 (q + 4)(q – 3) = 0 q = – 4 or q = 3 When q = – 4, p = –1 – (– 4) = 3 When q = 3, p = –1 – 3 = – 4 \ p = 3, q = – 4; p = – 4, q = 3

30 a + 3a = –p 4a = –p

a = – p4 … 1

a(3a) = q 3a2 = q … 2

Substitute 1 into 2 : 3(–

p4 )2

= q

3p2

16 = q

3p2 = 16q (shown)

31 (a) 2x2 + 5x – 3 = 0 b2 – 4ac = 52 – 4(2)(–3) = 25 + 24 = 49 ( 0) \ Two distinct roots. (b) 2 – 4x – x2 = 0 b2 – 4ac = (–4)2 – 4(–1)(2) = 16 + 8 = 24 ( 0) \ Two distinct roots. (c) x2 – 6x + 11 = 0 b2 – 4ac = (–6)2 – 4(1)(11) = 36 – 44 = –8 ( 0) \ No roots. (d) x2 + x – 4 = 0 b2 – 4ac = 12 – 4(1)(–4) = 1 + 16 = 17 ( 0) \ Two distinct roots.

(e) x2 – 6x + 9 = 0 b2 – 4ac = (–6)2 – 4(1)(9) = 36 – 36 = 0 \ Two equal roots. (f) x2 – 4x + 4 = 0 b2 – 4ac = (–4)2 – 4(1)(4) = 0 \ Two equal roots.

32 (a) 4x2 – kx + 25 = 0 (–k)2 – 4(4)(25) = 0 k2 – 400 = 0 (k + 20)(k – 20) = 0 k = –20 or k = 20 (b) (k + 1)x2 + 3k – 2(k + 3) = 0 (k + 1)x2 + k – 6 = 0 02 – 4(k + 1)(k – 6) = 0 – 4(k2 – 5k – 6) = 0 k2 – 5k – 6 = 0 (k + 1)(k – 6) = 0 k = –1 or k = 6 (c) k2x2 – (k + 2)x + 1 = 0 (–k – 2)2 – 4(k2)(1) = 0 k2 + 4k + 4 – 4k2 = 0 3k2 – 4k – 4 = 0 (3k + 2)(k – 2) = 0 k = –

23

or k = 2

(d) (4k – 14)2 – 4(2k + 3)(16k + 1) = 0 16k2 – 112k + 196 – 4(32k2 + 50k + 3) = 0 112k2 + 312k – 184 = 0 14k2 + 39k – 23 = 0 (7k + 23)(2k – 1) = 0

k = – 237

or k = 12

(e) (k – 3)x2 + 4k + 1 = 4kx – 6x (k – 3)x2 + 6x – 4kx + 4k + 1 = 0 (6 – 4k)2 – 4(k – 3)(4k + 1) = 0 36 – 48k + 16k2 – 16k2 + 44k + 12 = 0 –4k + 48 = 0 4k = 48 k = 12

33 (a) 2x2 + 4c + 3 = 0 02 – 4(2)(4c + 3) 0 –32c – 24 0 32c + 24 0

c – 34

(b) cx2 + 5x + 1 = 0 52 – 4c(1) 0 –4c –25

c

254

(c) x2 + 2cx + c2 – 5c + 7 = 0 (2c)2 – 4(1)(c2 – 5c + 7) 0 4c2 – 4(c2 – 5c + 7) 0 20c – 28 0

c 75

(d) x2 – 2x + 2 – c = 0 (–2)2 – 4(1)(2 – c) 0 4c – 4 0 c 1

34 (a) (m + 2)x2 – 2mx + m – 5 = 0 (–2m)2 – 4(m + 2)(m – 5) 0 4m2 – 4m2 + 12m + 40 0 12m + 40 0 m –

103

(b) 3x2 – 6x + m = 0 (–6)2 – 4(3)m 0 36 – 12m 0 –12m –36 m 3 (c) 5x2 + 4x + 6 – 3m = 0 16 – 4(5)(6 – 3m) 0 16 – 120 + 60m 0 60m 104

m 2615

(d) x2 – 2mx + m2 + 5m – 6 = 0 (–2m)2 – 4(1)(m2 + 5m – 6) 0 4m2 – 4(m2 + 5m – 6) 0 –20m + 24 0 –20m –24

m 65

(e) x2 + 2mx + (m – 1)(m – 3) = 0 (2m)2 – 4(1)(m – 1)(m – 3) 0 4m2 – 4(m2 – 4m + 3) 0 16m – 12 0 16m 12

m 34

35 (a) x + p = 7x – px2

px2 – 6x + p = 0 (–6)2 – 4(p)(p) = 0 36 – 4p2 = 0 4p2 = 36 p2 = 9 p = 3 \ p = –3 or p = 3 (b) x + py = 10

y = 10 – x

p

Substitute y =

10 – x

p into x2 + y2 = 10:

x2 +

(10 – xp )2

= 10

x2 + 100 – 20x + x2

p2 = 10

p2x2 + x2 – 20x + 100 = 10p2

(1 + p2)x2 – 20x + 100 – 10p2 = 0 (–20)2 – 4(1 + p2)(100 – 10p2) = 0 400 – 4(1 + p2)(100 – 10p2) = 0 400 – 4(100 + 90p2 – 10p4) = 0 –360p2 + 40p4 = 0 p2 – 9 = 0 (p + 3)(p – 3) = 0 \ p = –3 or p = 3 (c) 2y + x = p

y = p – x

2

Substitute y =

p – x

2 into y2 + 4x = 20:

( p – x2 )2

+ 4x = 20

Page 14: Analysis Spm Additional Mathematics

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-35

p2 – 2px + x2

4 + 4x = 20

x2 – 2px + p2 + 16x = 80 x2 + (16 – 2p)x + p2 – 80 = 0 (16 – 2p)2 – 4(1)(p2 – 80) = 0 256 – 64p + 4p2 – 4p2+ 320 = 0 576 – 64p = 0 64p = 576 p = 9 (d) px – 1 = x2 – 2x + 3 x2 – 2x – px + 4 = 0 x2 + (–p – 2)x + 4 = 0 (–p – 2)2 – 4(1)(4) = 0 p2 + 4p + 4 – 16 = 0 p2 + 4p – 12 = 0 (p + 6)(p – 2) = 0 p = –6 or p = 2 (e) (p – 2)y = 3x

y =

3x

p – 2

Substitute y =

3x

p – 2 into xy =

1 – x:

x( 3xp – 2)

= 1 – x

3x2 = (1 – x)(p – 2) 3x2 = p – 2 – px + 2x 3x2 + (p – 2)x + 2 – p = 0 (p – 2)2 – 4(3)(2 – p) = 0 p2 – 4p + 4 – 24 + 12p = 0 p2 + 8p – 20 = 0 (p + 10)(p – 2) = 0 p = –10 or p = 2

36 (a) x + 2y = 3

y =

3 – x

2

Substitute y =

3 – x

2 into x2 + y2 = h:

x2 +

( 3 – x2 )2

= h

x2 +

9 – 6x + x2

4 = h

4x2 + 9 – 6x + x2 = 4h 5x2 – 6x + 9 – 4h = 0 (–6)2 – 4(5)(9 – 4h) 0 36 – 180 + 80h 0 80h 144

h

95

(b) x – h = hx2 + 2hx hx2 + (2h – 1)x + h = 0 (2h – 1)2 – 4(h)(h) 0 4h2 – 4h + 1 – 4h2 0 –4h –1

h 14

(c) 2x + h = x2 – x + 1 x2 – 3x + 1 – h = 0 (–3)2 – 4(1)(1 – h) 0 9 – 4(1 – h) 0 9 – 4 + 4h 0 4h –5 h –

54

(d) 2h + 1 = x +

h2

x 2hx + x = x2 + h2

x2 – 2hx – x + h2 = 0 x2 + (–2h – 1)x + h2 = 0 (–2h – 1)2 – 4(1)(h2) 0 4h2 + 4h + 1 – 4h2 0 4h –1

h – 14

(e) 3x + h = 5 – 3x – 2x2

2x2 + 6x + h – 5 = 0 62 – 4(2)(h – 5) 0 36 – 8(h – 5) 0 36 – 8h + 40 0 –8h –76

h 192

37 (a) 2a – x = (x – 1)2 + a 2a – x = x2 – 2x + 1 + a x2 – x + 1 – a = 0 (–1)2 – 4(1)(1 – a) 0 –3 + 4a 0 4a 3

a

34

(b) ax – 3 = x + 1x

ax2 – 3x = x2 + 1 (a – 1)x2 – 3x – 1 = 0 (–3)2 – 4(a – 1)(–1) 0 9 + 4(a – 1) 0 9 + 4a – 4 0 4a –5

a – 54

(c) x + 3y = a

y =

a – x

3

Substitute y =

a – x

3 into y2 =

2x + 3:

( a – x3 )2

= 2x + 3

a2 – 2ax + x2 = 18x + 27 x2 + (–2a – 18)x + a2 – 27 = 0 (–2a – 18)2 – 4(1)(a2 – 27) 0 4a2 + 72a + 324 – 4a2 + 108 0 72a + 432 0 a –6 (d) Substitute y = ax + 6 into 2x2 – xy

= 3: 2x2 – x(ax + 6) = 3 2x2 – ax2 – 6x – 3 = 0 (2 – a)x2 – 6x – 3 = 0 (–6)2 – 4(2 – a)(–3) 0 36 + 12(2 – a) 0 36 + 24 – 12a 0 –12a –60 a 5 (e) 5x – a = x 2 + 3x + 3 x2 – 2x + 3 + a = 0 (–2)2 – 4(1)(3 + a) 0 4 – 4(3 + a) 0 –8 – 4a 0

– 4a 8 4a –8 a –2

38 kx2 + 3hx + 2h = 0 (3h)2 – 4(k)(2h) = 0 9h2 – 4k(2h) = 0 9h2 = 8hk h =

89

k

39 3cx2 – 7dx + 3c = 0 (–7d)2 – 4(3c)(3c) = 0 49d2 – 4(3c)(3c) = 0 49d2 – 36c2 = 0 36c2 = 49d2

( cd )

2

=

4936

cd

= 76

\ c : d = 7 : 6

40 (a) x2 + kx + 2k – 4 = 0 b2 – 4ac = k2 – 4(1)(2k – 4) = k2 – 8k + 16 = (k – 4)2

b2 – 4ac 0 for all values of k. Hence, the roots are real. (b) x2 + (1 – k)x – k = 0 b2 – 4ac = (1 – k)2 – 4(1)(–k) = 1 – 2k + k2 + 4k = k2 + 2k + 1 = (k + 1)2

b2 – 4ac 0 for all values of k. Hence, the roots are real.

1 4x – 53

=

x2 + 8x

4x2 – 5x = 3x2 + 24 x2 – 5x – 24 = 0 (x – 8)(x + 3) = 0 x = 8 or x = –3

2 3x2 – x – 10 = x2 – 2x 2x2 + x – 10 = 0 (2x + 5)(x – 2) = 0

x = – 52

or x = 2

3 x2 + 10x + 25 – 3x – 12 = 15 x2 + 7x – 2 = 0

x =

–7 72 – 4(1)(–2)

2(1)

=

–7 49 – 4(–2)

2

=

–7 57

2 = 0.275 or –7.275 \ x = 0.275 or x = –7.275

4 x2 + 2x – 8 = 6 x2 + 2x = 14

Page 15: Analysis Spm Additional Mathematics

6© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

x2 + 2x + (1)2 = 14 + (1)2

(x + 1)2 = 15 x = 15 – 1 or – 15 – 1 = 2.873 or – 4.873 \ x = 2.873 or x = – 4.873

5 2x2 – 4x – 3 = 0 a + b = 2

ab = –

32

a + 3 + b + 3 = a + b + 6 = 2 + 6 = 8 (a + 3)(b + 3) = ab + 3(a + b) + 9

= –

32

+ 6 + 9

=

272

\ x2 – 8x +

272

= 0

2x2 – 16x + 27 = 0

6 mx2 + (m – 5)x – 20 = 0

a + (–a) = –(m – 5)

m 0 = 5 – m m = 5

7 2x2 + px + p – 1 = 0

a + b = –

p2

ab =

p – 12

a2 + b2 = 4 (a + b)2 – 2ab = 4 (–

p2 )2

– 2( p – 1

2 ) = 4

p2

4 – p + 1 = 4

p2

4 – p = 3

p2 – 4p – 12 = 0 (p – 6)(p + 2) = 0 p = 6 or p = –2

8 3x2 – x + 2 = 0

a + b =

13

ab =

23

a2 + b 2 = (a + b)2 – 2ab

=

( 13 )2

– 2( 2

3 )

=

19

– 43

= – 119

9 2x2 – 9k = 2 – 3k

2( 12 )2

– 9k = 2 – 3k

12

– 2 = 6k

– 32

= 6k

k = – 14

10 b2 – 4ac = 121 (–9)2 – 4(2)(h) = 121 81 – 4(2)(h) = 121 –8h = 40 h = –5

11 a( 1a)

=

m – 53

3 = m –5 m = 8

12 a + 3a = – 4 4a = – 4 a = –1 a(3a) = c – 4 When a = –1, 3 = c – 4 c = 7

13 (a) Substitute x = 5 into x2 + (m + 1)x + 3m + 2: 25 + (m + 1)(5) + 3m + 2 = 0 8m + 32 = 0 8m = –32 m = – 4 (b) x2 – 3x – 10 = 0 (x – 5)(x + 2) = 0 x = 5 or x = –2 \ x = –2

14 a + 2a =

3m

3a =

3m

a =

1m

… 1

2a2 =

1

2m … 2

Substitute 1 into 2 :

2( 1m)2

=

1

2m

2

m2 =

1

2m m2 – 4m = 0 m(m – 4) = 0 m = 0 or m = 4 \ m = 4

15 4x2 – px + 25 = 0 (–p)2 – 4(4)(25) = 0 p2 – 400 = 0 (p + 20)(p – 20) = 0 p = –20 or p = 20

16 x2 – 5x + 4p – 2 = 0 a + b = 5 … 1 ab = 4p – 2 … 2 a – b = 9 … 3 1 + 3 : 2a = 14 a = 7 Substitute a = 7 into 1 : 7 + b = 5 b = –2 Substitute a = 7 and b = –2 into 2 : 7(–2) = 4p – 2

4p = –12 p = –3

17 p + q = –p 2p = –q

p = –

q2

… 1

p – q = 6 … 2 Substitute 1 into 2 :

– q2

– q = 6

–3q = 12 q = –4

When q = –4, p =

–(–4)

2 = 2 \ pq = 2(–4) = –8

18 x2 + (2p – 1)x + p2 + 3p – 4 = 0 (2p – 1)2 – 4(1)(p2 + 3p – 4) 0 4p2 – 4p + 1 – 4p2 – 12p + 16 0 17 – 16p 0 16p 17

p 1716

19 k + 3x = 5 – 3x – 2x2

2x2 + 6x + k – 5 = 0 (6)2 – 4(2)(k – 5) = 0 36 – 8(k – 5) = 0 76 – 8k = 0

k = 9 12

20 a + 3a = – m4

4a = – m4

a = – m

16 … 1

3a2 = n4

… 2

Substitute 1 into 2 :

3(– m16)2

=

n4

3m2

256 =

n4

n = 3m2

64

21 (5)2 – 4(1)(3 – c) 0 25 – 4(3 – c) 0 13 + 4c 0 4c –13

c

– 13

4

22 q2 – 4p(–9) = 0 q2 + 36p = 0

p = – 136

q2

23 x2 + (2x + k)2 = 5 x2 + 4x2 + 4kx + k2 = 5 5x2 + 4kx + k2 – 5 = 0

Page 16: Analysis Spm Additional Mathematics

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-37

(4k)2 – 4(5)(k2 – 5) = 0 16k2 – 20(k2 – 5) = 0 – 4k2 + 100 = 0 k2 = 25 k = 5

24 a + b = 2

ab =

h2

a2 + b 2 = –k (a + b)2 – 2ab = –k

(2)2 – 2( h

2 ) = –k

4 – 2( h2 )

= –k

h – k = 4 … 1 (ab)2 = 16

h2

4 = 16

h2 = 64 h = 8 ( 0) Substitute h = 8 into 1 : 8 – k = 4 k = 4 \ h = 8, k = 4

25 2c – x – (x – 1)2 = c c – x – (x2 – 2x + 1) = 0 x2 – x + 1 – c = 0 (–1)2 – 4(1)(1 – c) 0 1 – 4(1 – c) 0 –3 + 4c 0 4c 3

c

34

26 a + (a + 3k) = 8 2a = 8 – 3k

a =

8 – 3k

2 … 1

a(a + 3k) = h … 2 Substitute 1 into 2 :

(8 – 3k2 )(8 – 3k

2 + 3k)

= h

(8 – 3k2 )(8 + 3k

2 ) = h

64 – 9k2 = 4h

h =

64 – 9k2

4

27 x2 – 2mx + m2 + 5m– 6 = 0 (a) (–2m)2 – 4(1)(m2 + 5m – 6) = 0 4m2 – 4(m2 + 5m – 6) = 0 24 – 20m = 0

m = 65

(b) 24 – 20m 0 20m 24

m 65

(c) 24 – 20m 0 20m 24

m 65

28 2x2 – 5x – 4 = 0

a + b =

52

ab = –2 (a) a2 + b2 = (a + b)2 – 2ab

=

( 52 )2

– 2(–2)

=

254

+ 4

= 10 14

(b)

2ab

+ 2ba

=

2(a2 + b 2)ab

=

2(414 )

–2

= –

414

(2ab )( 2b

a ) =

4abab

= 4

\ x2 +

414

x + 4 = 0

4x2 + 41x + 16 = 0

29 3x2 + 2mx – m = 0 (a) (2m)2 – 4(3)(–m) = 0 4m2 + 12m = 0 4m(m + 3) = 0 m = 0 or m = –3 \ m = –3

(b) a( 1

a ) = – m

3 3 = –m m = –3

(c) a + 3a =

– 2m

3

4a =

– 2m

3

a = –

16

m

… 1

3a2 =

– m

3

a2 = – m

9 … 2

Substitute 1 into 2 :

m2

36 =

– m

9 m = – 4

30 (a) 2(x2 – x – 2) = p 2x2 – 2x – 4 – p = 0 a + b = 1 … 1 a – b = 5 … 2 1 + 2 : 2a = 6 a = 3 Substitute a = 3 into 1 : 3 + b = 1 b = –2

ab = – 4 – p

2

– 6 = – 4 – p

2 –12 = – 4 – p p = 8

(b) mx2 + nx – k = 0

a + b = –

nm

ab = –

km

kx2 – mx – 1 – n = 0

a + b =

mk

ab =

–1 – n

k

mk

= – nm

so, m2 = – kn

k = – m2

n … 1

and –1 – nk

= – km

– m – mn = – k2

k2 = m(1 + n) … 2

Substitute 1 into 2 :

(– m2

n )2

= m(1 + n)

m4

n2 = m(1 + n)

m3 = n2(1 + n)

31 x2 + hx + k = 0 (a) –2 + 6 = –h h = – 4 –2(6) = k k = –12 \ h = – 4, k = –12 (b) x2 – 4x – 12 – p = 0 (– 4)2 – 4(1)(–12 – p) 0 16 – 4(–12 – p) 0 64 + 4p 0 p –16

32 (a) 2x2 + kx – 12 = 0 When x = – 4, 2(– 4)2 + k(– 4) – 12 = 0 32 – 4k – 12 = 0 4k = 20 k = 5 (b) 2x2 + 5x – 12 = 0 (2x – 3)(x + 4) = 0

x = 32

or x = – 4

33 (a) 2x2 – 12x + k = 0

x2 – 6x +

k2

= 0

a + 5a = 6 6a = 6 a = 1

5a2 =

k2

k = 10 (b) 2x2 – 12x + 10 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x = 1 or x = 5 \ The roots are 1 and 5.

Page 17: Analysis Spm Additional Mathematics

8© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

37 (a)

12

[(x + 3) + (2x – 2)](x – 1) = 112

12

(3x + 1)(x – 1) = 112

3x2 – 2x – 1 = 224 3x2 – 2x – 225 = 0 (b) (3x + 25)(x – 9) = 0

x = – 253

(Reject) or x = 9

The length of QR = 2x – 2 = 2(9) – 2 = 16 cm

38 (a) (3x + 2)2 + (2x + 2)2 = (5x)2

9x2 + 12x + 4 + 4x2 + 8x + 4 = 25x2

12x2 – 20x – 8 = 0 3x2 – 5x – 2 = 0 (shown) (b) (3x + 1)(x – 2) = 0

x = –

13

(Reject) or x = 2

\ x = 2

(c) Area =

12

[3(2) + 2][2(2 + 1)]

=

12

(8)(6)

= 24 cm2

39 (a) 2x2 – 4x + 5 = 0 a + b = 2

ab =

52

(b) 2a + 1 + 2b + 1 = 2(a + b) + 2 = 2(2) + 2 = 6 (2a + 1)(2b + 1) = 4ab + 2(a + b) + 1

= 4( 5

2 ) + 2(2)

+ 1 = 15 \ x2 – 6x + 15 = 0

40 (a) 2x2 – mx + 3 = 0

a + b =

m2

ab =

32

9x2 – 52x + 4 = 0

1a2

+ 1b 2

=

529

a2 + b2

(ab)2 =

529

(a + b)2 – 2ab

(ab)2 =

529

(m2 )2

– 2( 3

2 )( 3

2 )2

=

529

m2 – 12

9 =

529

m2 = 64 m = 8 \ m = 8 (b) 2x2 – 6x – 3 = 0 a + b = 3

ab = –

32

(i) a2 + b 2 = (a + b)2 – 2ab

= (3)2 – 2(–

32 )

= 9 – 2(–

32 )

= 12 (ii) (a – b)2 = a2 – 2ab + b2

= a2 + b2 – 2ab

= 12 – 2(– 32 )

= 15 a – b = 15 \ a – b = 15 (a > b)

41 (a) ( x2 + (17 – x)2)( x2 + (17 – x)2) = 169 x2 + 289 – 34x + x2 = 169 2x2 – 34x + 120 = 0 x2 – 17x + 60 = 0(shown) (b) (x – 5)(x – 12) = 0 x = 5 or x = 12

34 (a) x2 – 4x – 1 = 2mx – 10m x2 + (– 4 – 2m)x + 10m – 1 = 0 (– 4 – 2m)2 – 4(1)(10m – 1) = 0 16 + 16m + 4m2 – 40m + 4 = 0 4m2 – 24m + 20 = 0 m2 – 6m + 5 = 0 (m – 1)(m – 5) = 0 m = 1 or m = 5 (b) 2x2 + 5x + 3 – k = 0 (5)2 – 4(2)(3 – k) 0 25 – 8(3 – k) 0 1 + 8k 0

k – 18

35 1 – c = 3x2 – 2x 3x2 – 2x + c – 1 = 0 (a) (–2)2 – 4(3)(–1) 0 4 – 12(c – 1) 0 16 – 12c 0 12c 16

c 43

(b) 16 – 12c = 0 12c = 16

c = 43

(c) 16 – 12c 0 12c 16

c 43

36 3mx2 – 7nx + 3m = 0 (a) (–7n)2 – 4(3m)(3m) = 0 49n2 – 36m2 = 0 49n2 = 36m2

m2

n 2 =

4936

(mn )2

=

4936

mn

= 76

\ m : n = 7 : 6 (b) 3(7)x2 – 7(6)x + 3(7) = 0 21x2 – 42x + 21 = 0 x2 – 2x + 1 = 0 (x – 1)(x – 1) = 0 x = 1

Page 18: Analysis Spm Additional Mathematics

1 (a) Quadraticfunction,a=1,b=–4,c=5 (b) Quadraticfunction,a=–3,b=2,c=7 (c) f(x)=x2–7x Quadraticfunction,a=1,b=–7,c=0 (d) f(x)=6+4x–x3

Notaquadraticfunction.

(e) f(x)=x–4x

+2

Notaquadraticfunction. (f) Notaquadraticfunction.

2 (a)

(b)

f(x)

xO–2 –1 1 2 3 4 5

–5

–10

–15

–20

–25

–35

3 (a) x –2 –1 0 1 2 3 4 5

f(x) –15 –3 5 9 9 5 –3 –15

f(x)

10

5

O–1–2 1 2 3 4 5x

–5

–10

–15Axisofsymmetryx=1.5

(1.5,9.5) Maximumpoint

(b) x –2 –1 0 1 2 3 4 5 6

f(x) 6 2.5 0 –1.5 –2 –1.5 0 2.5 6

f(x)

6

5

4

3

2

1

O–1–2 1 2 3 4 5 6x

–1

–2(2,-2) Minimumpoint

Axisofsymmetryx=2

4 (a) Twodistinctroots. (b) Noroots. (c) Twoequalroots. (d) Twodistinctroots.

5 (a) b2–4ac = (–7)2–4(2)(9) = 49–4(2)(9) = –23(0) ∴Noroots. (b) b2–4ac = (–8)2–4(1)(16) = 64–4(16) = 0 ∴Twoequalroots.

5

f(x)

25

20

15

10

5

O–3 –2 –1 1 2 3 4x

–5

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 19: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(c) b2–4ac = (–4)2–4(–1)(3) = 16–4(–1)(3) = 16+12 = 28(0) ∴Twodistinctroots. (d) f(x) = x2–4x+1 b2–4ac = (–4)2–4(1)(1) = 16–4 = 12(0) ∴Twodistinctroots.

6 (a) f(x)=x2+mx+2m–3 m2–4(1)(2m–3) = 0 m2–8m+12 = 0 (m–2)(m–6) = 0 m=2orm=6 (b) (–4m–2)2–4(m+2)(4m) = 0 16m2+16m+4–16m2–32m = 0 4–16m = 0

m =14

(c) f(x) = x2–4x–2mx+10m–1 = x2+(–4–2m)x+10m–1 (–4–2m)2–4(1)(10m–1)= 0 16+16m+4m2–40m+4= 0 4m2–24m+20= 0 m2–6m+5= 0 (m–1)(m–5)= 0 m=1orm=5 (d) (4m–8)2–4(m+1)(2m) = 0 16m2–64m+64–8m2–8m = 0 8m2–72m+64 = 0 m2–9m+8 = 0 (m–1)(m–8) = 0 m=1orm=8

7 (a) (–2k)2–4(1)(3k) 0 4k2–12k 0 4k(k–3) 0 k0ork3 (b) (4k)2–4(5)(k2–20) 0 16k2–20k2+400 0 400–4k2 0 k2–100 0 (k+10)(k–10) 0 –10k10 (c) (2k2)–4(1)(k2–5k+7) 0 4k2–4(k2–5k+7) 0 20k–28 0 20k 28

k 75

(d) f(x)=kx2+2kx+k–4 (2k)2–4(k)(k–4)0 4k2–4k(k–4)0 16k0 k0

8 (a) f(x) = 2x2–px2–6x–3 = (2–p)x2–6x–3 (–6)2–4(2–p)(–3)0 36+12(2–p)0 60–12p0 12p–600 12p60 p5 (b) f(x) = x2+4x2+4px+p2–5 = 5x2+4px+p2–5

(4p)2–4(5)(p2–5)0 16p2–20(p2–5)0 –4p2+1000 4p2–1000 p2–250 p–5orp5 (c) (–4p)2–4(p)(4p–5) 0 16p2–4p(4p–5) 0 20p 0 p 0 (d) f(x)=(p+2)x2–2px+p–5 (–2p)2–4(p+2)(p–5)0 4p2–4(p2–3p–10)0 12p+400

p–103

9 (5)2–4(2)(3–k) 0 25–8(3–k) 0 25–24+8k 0 1+8k 0 8k –1

k –18

(shown)

10 (2m)2–4(1)(m–1)(m–3) 0 4m2–4(m2–4m+3) 0 16m–12 0

m 34

(shown)

11 (a) Minimumvalue=5,x=3

(b) Maximumvalue=4,x=12

(c) Maximumvalue=–3,x=6

(d) Minimumvalue=–7,x=–32

(e) Minimumvalue=1,x=12

(f) Maximumvalue=23

,x=–1

12 (a) y = 2(x2–2x+52 )

= 2[x2–2x+(–1)2+52

– (–1)2] = 2[(x–1)2+

32 ]

= 2(x–1)2+3 ∴Minimumvalue=3,x=1 (b) y = –x2–2x+3 = –(x2+2x–3) = –[x2+2x+(1)2–3–(1)2] = –[(x+1)2–4] = 4–(x+1)2

∴Maximumvalue=4,x=–1 (c) y = x2+3x+4

= x2+3x+( 32 )2

+4–(32 )2

= (x+32 )2

+74

∴Minimumvalue=74

,x=–32

(d) y = 3(x2+2x–43 )

= 3[x2+2x+(1)2–43

–(1)2] = 3[(x+1)2–

73 ]

= 3(x+1)2–7 ∴Minimumvalue=–7,x=–1 (e)

y = –2(x2–52

x–12 )

= –2[x2–52

x+(– 54 )2

–12

–(– 54 )2]

= –2[(x–54 )2

–3316 ]

=338

–2(x–54 )2

∴Maximumvalue=338

,x=54

(f) y = –(x2–4x–5) = –[x2–4x+(–2)2–5–(–2)2] = –[(x–2)2–9)] = 9–(x–2)2

∴Maximumvalue=9,x=2

13 y = 4(x2+2x–94 )

= 4[x2+2x+(1)2–94

–(1)2] = 4[(x+1)2–

134 ]

= 4(x+1)2–13 ∴Minimumvalue=–13,x=–1

14 f(x) = 2(x2–3x+52 )

= 2[x2–3x+(– 32 )2

+52

–(– 32 )2]

= 2[(x–32 )2

+14 ]

= 2(x–32 )2

+12

(a) a=2,p=32

,q=12

(b) ( 32

,12 )

15 f(x) = x2+5x–6

= x2+5x+( 52 )2

–6–( 52 )2

= (x+ 52 )2

–494

∴Theleastvalue=–494

,x=–52

16 f(x) = x2–4x–k = x2–4x+(–2)2–k–(–2)2

= (x–2)2–k–4 ∴–k–4 = –9 k+4 = 9 k = 5

17 f(x) = –2x2+px+10

= –2(x2–p2

x–5) = –2[x2–

p2

x+(– p4 )2

–5–(– p4 )2]

= –2[(x–p4 )2

–5–p2

16] =

80+p2

8–2(x–

p4 )2

∴ 80+p2

8 = 18

80+p2 = 144 p2 = 64 p =±8

Page 20: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

18 f(x) = –x2+4x+12 = –(x2–4x–12) = –[x2–4x+(–2)2–12–(–2)2] = –[(x–2)2–16] = 16–(x–2)2

∴Thegreatestvalueoff(x)is16.

19 y =12

(x2+10x+25+x2–14x+49)

=12

(2x2–4x+74)

= x2–2x+37 = x2–2x+(–1)2+37–(–1)2

= (x–1)2+36 ∴Minimumvalue=36,x=1

20 f(x) = x2+6mx+144 = x2+6mx+(3m)2+144–(3m)2

= (x+3m)2+144–9m2

∴h=3mandk=144–9m2

21 (a) y = –(x2+x–6)

= –[x2+x+( 12 )2

–6–( 12 )2]

= –[(x+12 )2

–6–14 )]

=254 –(x+

12 )2

y

x2O–3

6

–12

254

Maximumpoint=(– 12

,254 )

(b) y = 2(x2+72

x–2) = 2[x2+

72 x+( 7

4 )2

–2–( 74 )2]

= 2[(x+74 )2

–8116 ]

= 2(x+74 )2

–818

y

xO

–4

–4

–818

12

–74

Minimumpoint=(– 74

,–818 )

(c) y = –(x2–4x–5) = –[x2–4x+(–2)2–5–(–2)2] = –[(x–2)2–9] = 9–(x–2)2

y

x

9

5

–1 O 2 5

Maximumpoint=(2,9)

(d) y = –3(x2–4x–53 )

= –3[x2–4x+(–2)2–53 –(–2)2]

= –3[(x–2)2–173 ]

= 17–3(x–2)2

y

x

17

O 2

5

Maximumpoint=(2,17)

22 y = –(x2+3x–12)

= –[x2+3x+( 32 )2

–12–( 32 )2]

= –[(x+32 )2

–12–94 ]

=574 –(x+

32 )2

y

x

574

O–

32

12

23 (a) y=a(x–1)2+3 At(0,5), 5 = a+3 a = 2 ∴y=2(x–1)2+3 (b) y=a(x–1)2+8 At(0,6), 6 = a+8 a = –2 y=–2(x–1)2+8 (c) y=a(x–2)2–1 At(1,0), 0 = a–1 a = 1 ∴y=(x–2)2–1 (d) y=a(x+2)2+18 At(0,10), 10 = 4a+18 4a = –8 a = –2 y=–2(x+2)2+18

24 y = 2(x2+4x+52 )

= 2[x2+4x+(2)2+52

–(2)2]

= 2[(x+2)2–32 ]

= 2(x+2)2–3 ∴Theminimumvalueis–3.

25 y = –x2–4x+5 = –(x2+4x–5) = –[x2+4x+(2)2–5–(2)2] = –[(x+2)2–9] = 9–(x+2)2

∴Themaximumvalueis9.

26 b2–4ac = 16–4(–1)(–6) = 16–24 = –8(0) Thecurveisaparabolawitha

maximumpointanddoesnotintersectthex-axis,soyisalwaysnegativeforallrealvaluesofx.

27 b2–4ac= 16–4(3)(2) = –8(0) Thecurveisaparabolawitha

minimumpointanddoesnotintersectthex-axissoyisalwayspositiveforallvalueofx.

28 y = a(x+2)2+5 7 = a+5 a = 2 y = 2(x+2)2+5

–2x

O

13

5

y

29 y = 3(x2–2x+113 )

= 3[x2–2x+(–1)2+113 –(–1)2]

= 3[(x–1)2+83 ]

= 3(x–1)2+8

y

xO 1

8

11

Turningpoint=(1,8) y-intercept=11

30 (a) y = 4(x2–3x–74 )

= 4[x2–3x+(– 32 )2

–74 –(– 3

2 )2] = 4[(x–

32 )2

–74 –

94 ]

= 4(x–32 )2

–16

Page 21: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

y

x

x-intercept

72

32

–12–16

x-intercept

y-intercept

O–7

Minimumpoint=( 32

,–16) (b) y = –2(x2–

12

x–212 )

= –2[x2–12

x+(– 14 )2

–212

–(– 14 )2]

= –2[(x–14 )2

–16916 ]

=1698

–2(x–14 )2

y

x14

72

–3 O

1698

21

x-intercept

y-intercept

x-intercept

Maximumpoint=( 14

,1698

)31 (a) 2x2–x–60 (2x+3)(x–2)0

32

2x

∴x–32

orx2

(b) 2x2–3x–20 (2x+1)(x–2)0

12

2x

∴x–12

orx2

(c) x2–x–60 (x–3)(x+2)0

–2 3

x

∴x–2orx3 (d) 2x2+5x–120 (2x–3)(x+4)0

–4 32

x

∴–4x32

(e) 2x2+7x–220 (2x+11)(x–2)0

112

2x

∴–112

x2

(f) 2x2–7x–40 (2x+1)(x–4)0

12

x4

∴x–12

orx4

(g) 4(4x2–12x+9)x2

15x2–48x+360 5x2–16x+12 0 (5x–6)(x–2) 0

65

x2

∴65

x2

(h) x2+4x+4 2x+7 x2+2x–3 0 (x–1)(x+3) 0

–3 1 x

∴x–3orx1

32 (a) x2–4x–5 = 0 (x+1)(x–5) = 0 x=–1orx=5 ∴A(–1,0);B(5,0) (b) (i) x–1orx5 (ii) –1x5

33 2x2–7x+30 (2x–1)(x–3)0

12

3x

∴x12

orx3

34 3x2–5x+1–x2

4x2–5x+10 (4x–1)(x–1)0

14

1x

∴x14

orx1

35 –1x2–4x+27 x2–4x+2 –1 x2–4x+3 0 (x–1)(x–3) 0 x1orx3

x2–4x+27 x2–4x–50 (x+1)(x–5)0 –1x5 ∴–1x1or3x5

36 (a) (x+3)(x–5)0 x2–2x–150 x2–2x15 ∴a=2,b=15 (b) (x+2)(x–4) 0 x2–2x–8 0 x2–8 2x 2x2–16 4x ∴a=–16,b=4

37 x2+xy+8 = 0

x2+x(k–x2 )+8 = 0

2x2+kx–x2+16 = 0 x2+kx+16 = 0 k2–4(1)(16) 0 k2–64 0 (k+8)(k–8) 0 k–8ork8

38 (–m–4)2–4(1)(1) 0 m2+8m+12 0 (m+6)(m+2) 0

–6 –2x

∴m–6orm–2

39 (4–k)2–4(2–3k)(2) 0 16–8k+k2–16+24k 0 k2+16k 0 k(k+16) 0 ∴–16k0

40 (c–4)2–4(1)(1) 0 c2–8c+12 0 (c–2)(c–6) 0 ∴2c6

41 x2–x(2x–p)+(2x–p)2 = 1 x2–2x2+px+4x2–4px+p2–1 = 0 3x2–3px+p2–1 = 0 (–3p)2–4(3)(p2–1) 0 9p2–12(p2–1) 0 9p2–12p2+12 0 12–3p2 0 3p2–12 0 3p2 12 p2 4 ∴p–2orp2

42 hx–9= x2+3x x2+(3–h)x+9 = 0 (3–h)2–4(1)(9)0 9–6h+h2–360 h2–6h–270 (h+3)(h–9)0 ∴h–3orh9

43 x2+(mx+2)2 = 2 x2+m2x2+4mx+2 = 0

Page 22: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(1+m2)x2+4mx+2 = 0 16m2–4(1+m2)(2)0 16m2–8–8m20 8m2–80 m21 ∴m–1orm1

44 x2+4x2–20x+25

a2 = 5

a2x2+4x2–20x+25–5a2 = 0 (a2+4)x2–20x+25–5a2 = 0 (–20)2–4(a2+4)(25–5a2) 0 400–4(25a2–5a4+100–20a2) 0 20a4–100a2+80a2 0 20a4–20a2 0 a2–1 0 (a+1)(a–1) 0 ∴–1a1

45 (–p)2–4(1)(p+3)0 p2–4(p+3)0 p2–4p–120 (p+2)(p–6)0 ∴p–2orp6(shown)

1 (a) y=a(x–1)2–18 At(–2,0), 0 = a(–2–1)2–18 0 = a(–3)2–18 9a = 18 a = 2 ∴a=2,p=1,q=18 (b) y=2(x–1)2–18 AtA,x=0, y = 2(–1)2–18 = –16 ∴A(0,–16)

2 b2–4ac0 (m–1)2–4(1)(4)0 m2–2m+1–160 m2–2m–150 (m+3)(m–5)0 ∴m–3orm5

3 (p–1)2–4(1)(–p+4)0 p2–2p+1+4p–160 p2+2p–150 (p+5)(p–3)0 ∴–5p3

4 (a) y=a(x–1)2+4 At(3,0), 0 = a(3–1)2+4 4a = –4 a = –1 ∴a=–1,p=1,q=4 (b) x=1

5 (a) x=2 (b) y=a(x–2)2+1 At(0,9), 9 = 4a+1 4a = 8 a = 2 ∴y=2(x–2)2+1

6 p=–2 At(0,3), 3 = p2+q 3 = (–2)2+q q = –1 ∴p=–2,q=–1

7 (–m–3)2–4(–1)(3–4m) 0 m2+6m+9+12–16m 0 m2–10m+21 0 (m–3)(m–7) 0 ∴3m7

8 y = –x2+4x+1 = –(x2–4x–1) = –[x2–4x+(–2)2–1–(–2)2] = –[(x–2)2–5] = 5–(x–2)2

∴Maximumvalueofyis5and x=2.

9 x2+5x+6x+6 x2+4x0 x(x+4)0

–4 0x

∴x–4orx0

10 x2–7x+100 (x–2)(x–5)0 2x5 ∴p=2,q=5

11 (a) A(–1,3) (b) y=a(x+1)2+3 At(0,1), 1= a+3 a = –2 ∴y=–2(x+1)2+3

12 (a) f(x)=x2–8x+12 Atx-axis, f(x) = 0 x2–8x+12 = 0 (x–2)(x–6) = 0 x=2orx=6 Aty-axis,x=0 f(x) = 02–8(0)+12 = 12 ∴a=2,b=6,c=12 (b) f(x)=(x–4)2–4

13 f(x)= –x2+nx+m = –(x2–nx–m)

= –[x2–nx+(– n2 )2

–m–(– n2 )2]

= –[(x–n2 )2

–m–n2

4 ] = 4m+n2

4–(x–n

2 )2 ∴ n

2= 2 and 4m+(4)2

4= 5

n = 4 4m+16 = 20 4m = 4 m = 1 ∴m=1,n=4

14

f(x)=k–(x–h)2

x(3,0)(1,0)

x=2

O

∴h=2 At(1,0), 0= k–(1–2)2

0= k–1 k = 1 ∴h=2,k=1

15 f(x)=ax2+bx+c At(0,10), 10= a(0)2+b(0)+c c = 10

f(x) = a(x2+ba

x+ ca )

= a[x2+ba

x+( b2a)2

+ ca

–( b2a)2]

= a[(x+ b2a)2

+4ac–b2

4a2 ] = a(x+ b

2a)2

+4ac–b2

4a

– b2a

= 2

–b = 4a b = –4a… 1

4ac–b2

4a = 18

40a–b2 = 72a… 2 Substitute 1 into 2 : 40a–(–4a)2 = 72a 40a–16a2 = 72a 16a2+32a = 0 16a(a+2) = 0 a=0(reject)ora=–2 Substitutea=–2into 1 : b = –4(–2) = 8 ∴a=–2,b=8,c=10

16 Minimumvalue=3 Whenx=1

f(x)

4

3

O x1

17 (a) h=1,k=–4 (b) f(x) = (x+1)2–4 = x2+2x–3 Atx-axis, x2+2x–3 = 0 (x+3)(x–1) = 0 x=–3orx=1 ∴a=–3,b=1 Aty-axis, f(x) = 02+2(0)–3 = –3 ∴a=–3,b=1,c=–3

Page 23: Analysis Spm Additional Mathematics

18 (a) p=2,q=18 Aty-axis, 10 = 18–a(0–2)2

10 = 18–4a 4a = 8 a = 2 ∴a=2,p=2,q=18 (b)

y=f(x)

f(x)

xO

–10

(2,–18)

19 (a) f(x) = x2–x+7

= x2–x+(– 12 )2

+7–(– 12 )2

= (x– 12 )2

+274

∴p= 12

,q=274

(b) x= 12

20 (k–4)2–4(1)(1) 0 k2–8k+12 0 (k–2)(k–6) 0 ∴k2ork6

21 (a) k=5 (b) y=a(x–3)2+b At(0,–10),–10=9a+b… 1 At(1,0),0=4a+b… 2 1 – 2 : –10= 5a a= –2 Substitutea=–2into 1 : –10= –18+b b = 8 ∴y=–2(x–3)2+8

22 f(x)=x2+px+2p–3 p2–4(1)(2p–3) = 0 p2–8p+12 = 0 (p–2)(p–6) = 0 p=2orp=6

23 mx+4= x2–4x+5 x2+(–4–m)x+1= 0 (–4–m)2–4(1)(1) 0 16+8m+m2–4 0 m2+8m+12 0 (m+6)(m+2) 0 ∴–6m–2

24 Minimumvalue=4whenx= 13

25 (a) p=1,q=5 f(x)=a(x–1)2+5 At(0,7), 7= a+5 a = 2 ∴a=2,p=1,q=5 (b) y=–2(x–1)2–5

26 y=a(x–1)2+4 At(3,0), 0 = a(3–1)2+4 4a = –4 a = –1 ∴y=4–(x–1)2

27 (a) (4m)2–4(m+1)(9) = 0 16m2–36(m+1) = 0 16m2–36m–36 = 0 4m2–9m–9 = 0 (4m+3)(m–3) = 0

m=– 34

orm=3

(b) mx–5= x2–1 x2–mx+4= 0 (–m)2–4(1)(4)0 m2–160 (m+4)(m–4)0 ∴m–4orm4

28 f(x) = x2–4x+3 = x2–4x+(–2)2+3–(–2)2

= (x–2)2–1 ∴p=–2,q=–1 (a) Minimumvalue=–1 (b) x=2

y

x31O

–1

3

2

29 (a) x2+x– 34

0

4x2+4x–3 0 (2x+3)(2x–1) 0

x– 32

orx 12

(b) Minimumvalue=4whenx=3

y

3

4

O x

13

30 (a) y=p–(x–2)2

At(1,0), 0 = p–(1–2)2

0 = p–1 p = 1 ∴p=1,q=2 (b) Maximumvalue=1 (c) y=1–(x–2)2

Aty-axis, y = 1–(–2)2

= –3 ∴A(0,–3) (d) y=(x–2)2–1

31 (a) y= –x2–px+7 = –(x2+px–7)

= –[x2+px+(p2 )2

–7–(p2 )2]

= –[(x+p2 )2

–7–p2

4 ] = 28+p2

4–(x+p

2 )2

∴ 28+p2

4= 16

28+p2= 64 p2= 36 p = 6(0) q = p

2

= 62

= 3 ∴p=6,q=3 (b) f(x)=16–(x+3)2

Maximumvalue=16when x=–3 (c) f(x)0 –x2–6x+70 x2+6x–70 (x+7)(x–1)0 ∴–7x1

32 f(x) = x2+hx+7

= x2+hx+(h2 )2

+7–(h2 )2

= (x+h2 )2

+7–h2

4

∴ h2

= 2

h = 4 Whenh=4,theminimumpoint

= 7–(4)2

4 = 7–4 = 3 (a) f(x)=x2+4x+7=(x+2)2+3 Aty-axis, f(x) = 02+4(0)+7 = 7 ∴A(0,7);B(–2,3) (b) h=4

33 (a) A(–2,0) Aty-axis, y = 12+p(0)–02

= 12 ∴B(0,12) ∴A(–2,0);B(0,12) (b) y = –x2+px+12 = –(x2–px–12)

= –[x2–px+(– p2 )2

–12–(– p2 )2]

= –[(x–p2 )2

–12–p4

2] = 48+p2

4–(x–p

2 )2

∴ p2

= 2

p = 4

q = 48+(4)2

4

= 16

34 (a) f(x)=2(x2– 72

x+ 32 )

= 2[x2– 72

x+(– 74 )2

+32

–(– 74 )2]

= 2[(x– 74 )2

+32

–4916 ]

= 2[(x– 74 )2

–2516 ]

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 24: Analysis Spm Additional Mathematics

= 2(x– 74 )2

–258

∴a=2,b= 74

,c=258

(b) Minimumvalue=– 258

(c)

y

x

y=| f(x)|

O 12

74

3 4

7

2583

35 (a) f(x)=16–4x–2x2

Atx-axis,16–4x–2x2 = 0 2x+4x–16 = 0 x2+2x–8 = 0 (x+4)(x–2) = 0 x=–4orx=2 ∴A(2,0);B(–4,0) (b) f(x)= –2x2–4x+16 = –2(x2+2x–8) = –2[x2+2x+(1)2–8–(1)2] = –2[(x+1)2–9] = 18–2(x+1)2

∴p=1 (c) Forf(x)=18–2(x+1)2, themaximumpointis(–1,18). Forf(x)=9–(x+1)2, themaximumpointis(–1,9).

36 f(x) =2(x2–4x+ 32 )

=2[x2–4x+(–2)2+32

–(–2)2] =2[(x–2)2– 5

2 ] =2(x–2)2–5 (a) (2,–5) (b)

y

x52O

–5–1

3

13

(c) 2x2–8x+3 k 2x2–8x+3–k 0 64–4(2)(3–k) 0 40+8k 0 8k –40 k –5

37 (a) mx–5 = 6x–6–x2

x2+(m–6)x+1 = 0 (m–6)2–4(1)(1) 0 m2–12m+32 0 (m–4)(m–8) 0 ∴m4orm8 (b) (–2)2–4(1)(p+3) = 0 4–4(p+3) = 0 4–4p–12 = 0 4p = –8 p = –2

38 (a) y=a(x+1)2–18 At(2,0), 0 = a(3)2–18 9a = 18 a = 2 ∴a=2,b=–18,c=–1 (b) A(–4,0) (c) f(x)=|2(x+1)2–18| Whenx=3,f(x)= |2(16)–18| = 14 Whenx=1,f(x)= |2(4)–18| = 10 ∴Range:0f(x)14

39 (a) A(4,0) (b) f(x)=8–a(x–2)2 At(4,0), 0 = 8–a(4) 4a = 8 a = 2 ∴a=2,p=–2,q=8 (c) f(x)=8–2(x+2)2

40 (a) f(x) = x2+2hx+(h)2+4h–(h)2

= (x+h)2+4h–h2

∴ 4h–h2 = 3 h2–4h+3 = 0 (h–1)(h–3) = 0 h=1orh=3 (b)

f(x)

x

12

43

O–1–3

(c) (–1,3)and(–3,3)

41 (a) f(x)=x2–3x+5 Aty-axis,f(x) = 02–3(0)+5 = 5 ∴P(0,5)

(b) f(x) = x2–3x+(– 32 )2

+5

–(– 32 )2

= (x– 32 )2

+114

∴Thecoordinatesofminimum

point=(32

,114 )

42 (a) f(x)=a(x–2)2–6 Aty-axis, –4 = a(0–2)2–6 4a = 2

a = 12

∴a= 12

,p=–2,q=–6

(b) x=2

(c) f(x)=12

(x+2)2–6

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 25: Analysis Spm Additional Mathematics

1 (a) y=3–x… 1 x2–3x+y2=5… 2 Substitute 1 into 2 : x2–3x+(3–x)2 = 5 x2–3x+9–6x+x2 = 5 2x2–9x+4 = 0 (2x–1)(x–4) = 0

x=12

orx=4

Substitutex=12

into 1 :

y = 3–12

= 212

Substitutex=4into 1 : y = 3–4 = –1

∴x=12

,y=212

;x=4,y=–1

(b) y=14–2x… 1 2x2–y2+6=2xy… 2 Substitute 1 into 2 : 2x2–(14–2x)2+6 =2x(14–2x) 2x2–(196–56x+4x2)+6 =28x–4x2

2x2+28x–190 =0 x2+14x–95 =0 (x+19)(x–5) =0 x=–19orx=5 Substitutex = –19into 1 : y = 14–2(–19) = 52 Substitutex = 5into 1 : y = 14–2(5) = 4 ∴x=–19,y=52;x=5,y=4

(c) y=2x+3… 1 2x2+y2–4x=39… 2 Substitute 1 into 2 : 2x2+(2x+3)2–4x = 39 2x2+4x2+12x+9–4x = 39 6x2+8x–30 = 0 3x2+4x–15 = 0 (3x–5)(x+3) = 0

x=53

orx=–3

Substitutex=53

into 1 :

y = 2(53 )+3

= 613

Substitutex=–3into 1 : y = 2(–3) + 3 = –3

∴x=53

,y=613

;x=–3,y=–3

(d) x=10–2y… 1 2y2–7y+x=0… 2 Substitute 1 into 2 : 2y2–7y+10–2y = 0 2y2–9y+10 = 0 (2y–5)(y–2) = 0

y=52

ory=2

Substitutey=52

into 1 :

x = 10–2(52 )

= 5 Substitutey=2into 1 : x = 10–2(2) = 6

∴x=5,y=52

;x=6,y=2

(e) x=4y–11… 1 y2–2x=7… 2 Substitute 1 into 2 : y2–2(4y–11) = 7 y2–8y+22 = 7 y2–8y+15 = 0 (y–3)(y–5) = 0 y=3ory=5 Substitutey=3into 1 : x = 4(3)–11 = 1 Substitutey=5into 1 : x = 4(5)–11 = 9 ∴x=1,y=3;x=9,y=5

(f) 2y=3x–1

y=3x–1

2… 1

9x2+y=7… 2 Substitute 1 into 2 :

9x2+3x–1

2 = 7

18x2+3x–1 = 14 18x2+3x–15 = 0 6x2+x–5 = 0 (6x–5)(x+1) = 0

x=56

orx=–1

Substitutex=56

into 1 :

y =

3(56)–1

2

=34

Substitutex=–1into 1 :

y =3(–1)–1

2 = –2

∴x=56

,y=34 ;x=–1,y=–2

2 (a) 3x=2y+1

x=2y+1

3… 1

4x2+9y2=15xy… 2 Substitute 1 into 2 :

4(2y+13 )2

+9y2 = 15y(2y+13 )

4(4y+4y+19 )+9y2 = 5y(2y+1)

16y2+16y+4+81y2 = 90y2+45y 7y2–29y+4 = 0 (7y–1)(y–4) = 0

y=17

ory=4

Substitutey=17

into 1 :

x =

2(17)+1

3

=37

Substitutey=4into 1 :

x =2(4)+1

3 = 3

∴x=37

,y=17

;x=3,y=4

(b) 2x=10–3y

x=10–3y

2 … 1

2y+3x=5xy… 2 Substitute 1 into 2 :

2y+3(10–3y2 ) = 5y(10–3y

2 ) 4y+30–9y = 50y–15y2

15y2–55y+30 = 0 3y2–11y+6 = 0 (3y–2)(y–3) = 0

y=23

ory=3

Substitutey=23

into 1 :

x =10–3(2

3 )2

= 4 Substitutey=3into 1 :

x =10–3(3)

2

=12

∴x=4,y=23

;x=12

,y=3

(c) y=3x–2… 1 2x2+y2=3xy… 2 Substitute 1 into 2 : 2x2+(3x–2)2 = 3x(3x–2) 2x2+9x2–12x+4 = 9x2–6x

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 26: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

2x2–6x+4 = 0 x2–3x+2 = 0 (x–1)(x–2) = 0 x=1orx=2 Substitutex=1into 1 : y = 3(1)–2 = 1 Substitutex=2into 1 : y = 3(2)–2 = 4 ∴x=1,y=1;x=2,y=4

(d) y=7–2x… 1 4y–3x=xy… 2 Substitute 1 into 2 : 4(7–2x)–3x = x(7–2x) 28–8x–3x = 7x–2x2

2x2–18x+28 = 0 x2–9x+14 = 0 (x–2)(x–7) = 0 x=2orx=7 Substitutex=2into 1 : y = 7–2(2) = 3 Substitutex=7into 1 : y = 7–2(7) = –7 ∴x=2,y=3;x=7,y=–7

(e) y=3–3x… 1 3x+2y=6xy… 2 Substitute 1 into 2 : 3x+2(3–3x) = 6x(3–3x) 3x+6–6x = 18x–18x2

18x2–21x+6 = 0 6x2–7x+2 = 0 (3x–2)(2x–1) = 0

x=23

orx=12

Substitutex=23

into 1 :

y = 3–3(23 )

= 1

Substitutex=12

into 1 :

y = 3–3(12 )

=32

∴x=23

,y=1;x=12

,y=32

(f) 3y = 2x+2

y =2x+2

3… 1

8x–9y=xy5

40x–45y=xy… 2 Substitute 1 into 2 :

40x–45( 2x+23 ) = x( 2x+2

3 ) 40x–30x–30 =

13

(2x2+2x)

3(10x–30) = 2x2+2x 2x2–28x+90 = 0 x2–14x+45 = 0 (x–5)(x–9) = 0 x=5orx=9

Substitutex=5into 1 :

y =2(5)+2

3 = 4 Substitutex=9into 1 :

y =2(9)+2

3

=203

∴x=5,y=4;x=9,y=203

3 (a) y=5–2x… 1 x2+y2=5… 2 Substitute 1 into 2 : x2+(5–2x)2 = 5 x2+25–20x+4x2 = 5 5x2–20x+20 = 0 x2–4x+4 = 0 (x–2)(x–2) = 0 x = 2 Substitutex=2into 1 : y = 5–2(2) = 1 ∴x=2,y=1

(b) 2x+3y=7

x=7–3y

2… 1

x2+xy+y2=7… 2 Substitute 1 into 2 :

( 7–3y2 )2

+y( 7–3y2 )+y2 =7

49–42y+9y2

4+

7y–3y2

2+y2 =7

49–42y+9y2+14y–6y2+4y2 =28 7y2–28y+21 =0 y2–4y+3 =0 (y–1)(y–3) =0 y=1ory=3 Substitutey=1into 1 :

x =7–3(1)

2 = 2 Substitutey=3into 1 :

x =7–3(3)

2 = –1 ∴x=2,y=1;x=–1,y=3

(c) 5x+3y=2x+y+1 2y=1–3x

y=1–3x

2… 1

3x2–y2=2x+y+1… 2 Substitute 1 into 2 :

3x2–(1–3x2 )2

= 2x+1–3x

2+1

3x2–( 1–6x+9x2

4 )= 2x+1–3x

2+1

12x2–1+6x–9x2= 8x+2–6x+4 3x2+4x–7= 0 (3x+7)(x–1)= 0

x=–73

orx=1

Substitutex=–73

into 1 :

y =1–3(– 7

3)2

= 4 Substitutex=1into 1 :

y =1–3(1)

2 = –1

∴x=–73

,y=4;x=1,y=–1

(d) 2x+y = 1 y = 1–2x… 1 4x2+12x+y2=1… 2 Substitute 1 into 2 : 4x2+12x+(1–2x)2 = 1 4x2+12x+1–4x+4x2 = 1 8x2+8x = 0 x2+x = 0 x(x+1) = 0 x=0orx=–1 Substitutex=0into 1 : y = 1–2(0) = 1 Substitutex=–1into 1 : y = 1–2(–1) = 3 ∴x=0,y=1;x=–1,y=3

(e) y=7–2x… 1 x2–xy+y2=7… 2 Substitute 1 into 2 : x2–x(7–2x)+(7–2x)2 = 7 x2–7x+2x2+49–28x+4x2 = 7 7x2–35x+42 = 0 x2–5x+6 = 0 (x–2)(x–3) = 0 x=2orx=3 Substitutex=2into 1 : y = 7–2(2) = 3 Substitutex=3into 1 : y= 7–2(3) = 1 ∴x=2,y=3;x=3,y=1

(f) 2(x+y) = 10 x+y = 5 x = 5–y… 1 x2–y+y2 = 10… 2 Substitute 1 into 2 : (5–y)2–y+y2 = 10 25–10y+y2–y+y2 = 10 2y2–11y+15 = 0 (2y–5)(y–3) = 0

y=52

ory=3

Substitutey=52

into 1 :

x = 5–52

=52

Substitutey=3into 1 : x = 5–3 = 2

∴ x=52

,y=52

;x=2,y=3

Page 27: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

4 (a) x=3–2y… 1 xy+y2=1… 2 Substitute 1 into 2 : y(3–2y)+y2 = 1 3y–2y2+y2 = 1 y2–3y+1 = 0

y =–(–3)±(–3)2–4(1)(1)

2(1)

=3±5

2

= 2.618or0.382 Substitutey=2.618into 1 : x = 3–2(2.618) = –2.236 Substitutey=0.382into 1 : x = 3–2(0.382) = 2.236 ∴x=–2.236,y=2.618; x=2.236,y=0.382

(b) y=2–x… 1 x2–3x–y=4… 2 Substitute 1 into 2 : x2–3x–(2–x)= 4 x2–2x–6= 0

x =–(–2)±(–2)2–4(1)(–6)

2(1)

=2±28

2

= 3.6458or–1.6458 Substitutex=3.6458into 1 : y = 2–3.6458 = –1.6458 Substitutex=–1.6458into 1 : y = 2–(–1.6458) = 3.6458 ∴x=3.646,y=–1.646; x=–1.646,y=3.646

(c) x=y+4 … 1 6y+x=3xy… 2 Substitute 1 into 2 : 6y+y+4 = 3y(y+4) 7y+4 = 3y2+12y 3y2+5y–4 = 0

y =–5±52–4(3)(–4)

2(3)

=–5±73

6 = 0.5907or–2.2573 Substitutey=0.5907into 1 : x = 0.5907+4 = 4.5907 Substitutey=–2.2573into 1 : x = –2.2573+4 = 1.7427 ∴x=4.591,y=0.591; x=1.743,y=–2.257

(d) x=2y+3… 1 y2+2x2=5xy… 2

Substitute 1 into 2 : y2+2(2y+3)2 =5y(2y+3) y2+2(4y2+12y+9) =10y2+15y y2+8y2+24y+18 =10y2+15y

y2–9y–18 =0

y =–(–9)±(–9)2–4(1)(–18)

2(1)

=9±153

2 = 10.6847or–1.6847 Substitutey=10.6847into 1 : x = 2(10.6847)+3 = 24.3694 Substitutey=–1.6847into 1 : x = 2(–1.6847)+3 = –0.3694 ∴x=24.369,y=10.685; x=–0.369,y=–1.685

(e) y=1–3x… 1 x2+2xy+2y2=10… 2 x2+2x(1–3x)+2(1–3x)2=10 x2+2x–6x2+2(1–6x+9x2)=10 x2+2x–6x2+2–12x+18x2=10 13x2–10x–8=0

x =–(–10)±(–10)2–4(13)(–8)

2(13)

=10±516

26 = 1.2583or–0.4891 Substitutex=1.2583into 1 : y = 1–3(1.2583) = –2.7749 Substitutex=–0.4891into 1 : y = 1–3(–0.4891) = 2.4673 ∴x=1.258,y=–2.775; x=–0.489,y=2.467

(f) x=5–3y… 1 x2+y2=6x–4y… 2 (5–3y)2+y2 = 6(5–3y)–4y 25–30y+9y2+y2 = 30–18y–4y 10y2–8y–5 = 0

y =–(–8)±(–8)2–4(10)(–5)

2(10)

=8±264

20 = 1.2124or–0.4124 Substitutey=1.2124into 1 : x = 5–3(1.2124) = 1.3628 Substitutey=–0.4124into 1 : x = 5–3(–0.4124) = 6.2372 ∴x=1.363,y=1.212; y=6.237,y=–0.412

5 (a) y=–2x–2… 1

y+2x=xy2

2y+4x=xy… 2 Substitute 1 into 2 : 2(–2x–2)+4x = x(–2x–2) –4x–4+4x = –2x2–2x 2x2+2x–4 = 0 x2+x–2 = 0 (x+2)(x–1) = 0 x=–2orx=1

Substitutex=–2into 1 : y = –2(–2)–2 = 2 Substitutex=1into 1 : y = –2(1)–2 = –4 ∴A(–2,2),B(1,–4)

(b) 2x = –3y+2

x =–3y+2

2 … 1

2xy = –1… 2 Substitute 1 into 2 :

2y(–3y+22 ) = –1

–3y2+2y = –1 3y2–2y–1 = 0 (3y+1)(y–1) = 0

y=–13

ory=1

Substitutey=–13

into 1 :

x =–3(– 1

3 )+2

2

=32

Substitutey=1into 1 :

x =–3(1)+2

2

= –12

∴A(– 12

,1),B(32

,–13 )

6 x=3y–3… 1 2x–3y=6xy… 2 Substitute 1 into 2 : 2(3y–3)–3y =6y(3y–3) 6y–6–3y =18y2–18y 18y2–21y+6 =0 6y2–7y+2 =0 (3y–2)(2y–1) =0

y=23

ory=12

Substitutey=23

into 1 :

x = 3(23 )–3

= –1

Substitutey=12

into 1 :

x = 3(12 )–3

= –32

∴(–1,23 )and(– 3

2,

12 )

7 (a) x=2y+2… 1 x2+y2= 2xy+1… 2 Substitute 1 into 2 : (2y+2)2+y2=2y(2y+2)+1 4y2+8y+4+y2=4y2+4y+1 y2+4y+3=0 (y+3)(y+1)=0 y=–3ory=–1

Page 28: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Substitutey=–3into 1 : x = 2(–3)+2 = –4 Substitutey=–1into 1 : x = 2(–1)+2 = 0 ∴(–4,–3)and(0,–1)

(b) Midpoint =[–4+02

,–3+(–1)

2 ] =(–2,–2)

8 2x=y+1 At(m,3), 2m =3+1 m =2 3x2+nxy+y2=9 At(2,3), 3(2)2+n(2)(3)+(3)2 = 9 12+6n = 0 6n = –12 n = –2 ∴m=2,n=–2 y=2x–1… 1 3x2–2xy+y2=9… 2 Substitute 1 into 2 : 3x2–2x(2x–1)+(2x–1)2 = 9 3x2–4x2+2x+4x2–4x+1 = 9 3x2–2x–8 = 0 (3x+4)(x–2) = 0

x=–43

orx=2

Substitutex=–43

into 1 :

y = 2(– 43 )–1

= –113

Theothersolutionis(– 43

,–113 ).

9 4x+4y= 48 x+y=12 x=12–y… 1 x2+y2=80… 2 Substitute 1 into 2 : (12–y)2+y2 = 80 144–24y+y2+y2 = 80 2y2–24y+64 = 0 y2–12y+32 = 0 (y–4)(y–8) = 0 y=4ory=8 Substitutey=4into 1 : x = 12–4 = 8 Substitutey=8into 1 : x = 12–8 = 4 ∴x=4,y=8

10 2x = 1–3y

x =1–3y

2 … 1

3y2 = x2+2… 2 Substitute 1 into 2 :

3y2 = (1–3y2 )2

+2

3y2 =1–6y+9y2

4 +2

12y2 = 1–6y+9y2+8 3y2+6y–9 = 0 y2+2y–3 = 0 (y+3)(y–1) = 0 y=–3ory=1 Substitutey=–3into 1 :

x =1–3(–3)

2 = 5 Substitutey=1into 1 :

x =1–3(1)

2

= –1 Thepointsofintersectionare(5,–3)

and(–1,1). ∴Distance = [1–(–3)]2+(–1–5)2

= 16+36

= 52

=7.21units

1 y=3–2x … 1 x2+y2=2 … 2 Substitute 1 into 2 : x2+(3–2x)2 = 2 x2+9–12x+4x2 = 2 5x2–12x+7 = 0 (5x–7)(x–1) = 0

x=75

orx=1

Substitutex=75

into 1 :

y = 3–2(75 )

=15

Substitutex=1into 1 : y = 3–2(1) = 1

∴x=75

,y=15

;x=1,y=1

2 x=1–3y… 1 x2–3y2=2xy… 2 Substitute 1into 2 : (1–3y)2–3y2 = 2y(1–3y) 1–6y+9y2–3y2 = 2y–6y2

12y2–8y+1 = 0 (6y–1)(2y–1) = 0

y=16

ory=12

Substitutey=16

into 1:

x = 1–3(16 )

=12

Substitutey=12

into 1:

x = 1–3(12 )

= –12

∴x=12

,y=16

;x=–12

,y=12

3 y=1–2x… 1 x2–2y2=4xy… 2 Substitute 1into 2: x2–2(1–2x)2 = 4x(1–2x) x2–2(1–4x+4x2) = 4x–8x2

x2–2+8x–8x2 = 4x–8x2

x2+4x–2 = 0

x =–4±42–4(1)(–2)

2(1)

=–4±24

2

= 0.4495or–4.4495 Substitutex=0.4495into 1 : y = 1–2(0.4495) = 0.101 Substitutex=–4.4495into 1 : y = 1–2(–4.4495) = 9.899 ∴x=0.450,y=0.101; x=–4.450,y=9.899

4 y=5–x… 1 4x2–6y=24… 2 Substitute 1 into 2 : 4x2–6(5–x) = 24 4x2–30+6x = 24 4x2+6x–54 = 0 2x2+3x–27 = 0 (2x+9)(x–3) = 0

x=–92

orx=3

Substitutex=–92

into 1 :

y = 5–(– 92 )

=192

Substitutex=3into 1 : y = 5–3 = 2

∴x=–92

,y=192

;x=3,y=2

5 2x–2y= x+y–1 x= 3y–1… 1 2x2–11y2=x+y–1… 2 Substitute 1 into 2 : 2(3y–1)2–11y2 = 3y–1+y–1 2(9y2–6y+1)–11y2= 4y–2 18y2–12y+2–11y2 = 4y–2 7y2–16y+4 = 0 (7y–2)(y–2) = 0

y=27

ory=2

Substitutey=27

into 1 :

x = 3(27 )–1

= –17

Substitutey=2into 1 : x = 3(2)–1 = 5

x=–17

,y=27

;x=5,y=2

Page 29: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

6 [2(6–3y2 )+1]2

+6(y–2)2 = 49

(7–3y)2+6(y–2)2 = 49 49–42y+9y2+6(y2–4y+4) = 49 15y2–66y+24 = 0 5y2–22y+8 = 0 (5y–2)(y–4) = 0

y=25

ory=4

Substitutey=25

intox=6–3y

2:

x =

6–3(25 )

2

=125

Substitutey=4intox=6–3y

2:

x =6–3(4)

2 = –3

∴x=125

,y=25

;x=–3,y=4

7 x–y=4 y=x–4… 1 2x2–y2=17… 2 Substitute 1 into 2 : 2x2–(x–4)2 = 17 2x2–(x2–8x+16) = 17 x2+8x–33 = 0 (x+11)(x–3) = 0 x=–11orx=3 Substitutex=–11into 1 : y = –11–4 = –15 Substitutex=3into 1 : y = 3–4 = –1 ∴x=–11,y=–15;x=3,y=–1 (–11,–15);(3,–1)

8 y=3–5x… 1 x2+y2–3x=2… 2 Substitute 1 into 2 : x2+(3–5x)2–3x = 2 x2+9–30x+25x2–3x = 2 26x2–33x+7 = 0 (26x–7)(x–1) = 0

x=726

orx=1

Substitutex=726

into 1 :

y = 3–5( 726 )

=4326

Substitutex=1into 1 : y = 3–5(1) = –2

∴x=7

26,y=

4326

;x=1,y=–2

( 726

,4326 );(1,–2)

9 (a) x=1–3y… 1 2x2+xy=y2+36… 2 2(1–3y)2+y(1–3y) = y2+36 2(1–6y+9y2)+y–3y2 = y2+36

14y2–11y–34 = 0 (14y+17)(y–2) = 0

y=–1714

ory=2

Substitutey=–1714

into 1 :

x = 1–3(– 1714 )

=6514

Substitutey=2into 1 : x = 1–3(2) = 1–6 = –5

∴M(–5,2)andN (6514

,–1714 )

(b) Midpoint

= [ –5+6514

2,

2+(– 1714 )

2 ] = (– 5

28,

1128 )

10 px+qy=2 At(1,2), p(1)+2q =2 p =2–2q… 1 qx+p2y=10 At(1,2), q(1)+2p2=10 q+2p2=10… 2 Substitute 1 into 2 : q+2(2–2q)2 = 10 q+2(4–8q+4q2) = 10 8q2–15q–2 = 0 (8q+1)(q–2) = 0

q=–18

orq=2

Substituteq=–18

into 1 :

p = 2–2(– 18 )

=94

Substituteq=2into 1 : p = 2–2(2) = –2

∴p=94

,q=–18

;p=–2,q=2

11 x=5–2y… 1 2x+y=2xy… 2 2(5–2y)+y = 2y(5–2y) 10–4y+y = 10y–4y2

4y2–13y+10 = 0 (4y–5)(y–2) = 0

y=54

ory=2

Substitutey=54

into 1 :

x = 5–2(54 )

=52

Substitutey=2into 1 : x = 5–2(2) = 1

∴P(1,2)andQ(52

,54 )

12 (a) y=–2x–5… 1 y2+(2x+3)2=10… 2 Substitute 1 into 2 : (–2x–5)2+(2x+3)2 = 10 4x2+20x+25+4x2+12x+9 = 10 8x2+32x+24 = 0 x2+4x+3 = 0 (x+3)(x+1) = 0 x=–3orx=–1 Substitutex=–3into 1 : y = –2(–3)–5 = 1 Substitutex=–1into 1 : y = –2(–1)–5 = –3 ∴A(–3,1)andB(–1,–3) (b) Midpoint

= [ –3+(–1)2

,1+(–3)

2 ] = (–2,–1)

13 y=3x–7… 1 x2+y2–xy=7… 2 Substitute 1 into 2 : x2+(3x–7)2–x(3x–7) = 7 x2+9x2–42x+49–3x2+7x = 7 7x2–35x+42 = 0 x2–5x+6 = 0 (x–2)(x–3) = 0 x=2orx=3 Substitutex=2into 1 : y = 3(2)–7 = –1 Substitutex=3into 1 : y = 3(3)–7 = 2 ∴x=2,y=–1;x=3,y=2 R(2,–1);S(3,2)

LengthofRS: [2–(–1)]2+(3–2)2

= 32+12

= 10 = 3.162units

14 xy=70… 1

π(x2 )–y = 1

227 (x

2 )–y = 1

11x7

–y = 1

y =11x7

–1… 2

Substitute 2 into 1 :

x(117

x–1) = 70

11x2–7x = 490 11x2–7x–490 = 0 (11x+70)(x–7) = 0 x = 7(0) Substitutex=7into 1 : xy = 70 7y = 70 y = 10 ∴x=7,y=10

Page 30: Analysis Spm Additional Mathematics

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-31

Indices and Logarithms

1 (a) (–3)4 = –3 –3 –3 –3 = 81 (b) 64

23 = (43)

23

= 42

= 16 (c) (–125)

13 = [(–5)3]

13

= –5 (d) (–27)

– 23 = [(–3)3]

– 23

= (–3)–2

=

1

(–3)2

=

19

(e)

(2581)3

2

=

[( 59 )2]3

2

= ( 5

9 )3

=

125729

(f)

(1258 )–

43

=

1

(1258 )4

3

=

1

[( 52 )3]4

3

=

1

(52 )4

=

16625

2 (a) 913 9

16 = 9

13 +

16

= 912

= 9 = 3 (b) 27

12 3

12 = 3

32 3

12

= 332 +

12

= 32

= 9 (c) 2–3 16

34 = 2–3 (24)

34

= 2–3 23

= 2–3 + 3

= 20

= 1 (d) 36

16 6

23 = (62)

16 6

23

= 613 +

23

= 6 (e) 16

12 64

– 23 = (42)

12 (43)

– 23

= 4 4–2

= 41 + (–2)

= 4–1

=

14

(f) 53 2512 = 53 (52)

12

= 53 + 1

= 54

= 625

3 (a) 3235 16

14 = (25)

35 (24)

14

= 23 21

= 23 – 1

= 22

= 4 (b) 25

12 125

– 23 = (52)

12 (53)

– 23

= 51 5–2

= 51 – (–2)

= 53

= 125 (c) 8

– 23 4

12 = (23)

– 23 (22)

12

= 2–2 21

= 2–2 – 1

= 2–3

=

18

(d) 27–

43 81

– 14 = (33)

– 43 (34)

– 14

= 3–4 3–1

= 3–4 – (–1)

= 3–3

=

127

(e) 8–3 2–5 = (23)–3 2–5

= 2–9 – (–5)

= 2–4

=

116

(f) 4914 7

34 = (72)

14 7

34

= 712

– 34

= 7–

14

=

1

714

4 (a) ( 1

27)– 43

81

14 3–1

= (3–3)–

43 (34)

14 3–1

= 3 4 3 3–1

= 34 + 1 – (–1)

= 36

= 729 (b) 8

23 4–1 512

– 43

= (23)23 (22)–1 (29)

– 43

= 22 2–2 2–12

= 22 – 2 – 12

= 2–12

=

1

212

=

1

4096

(c) 713 49

14 7

– 16 = 7

13 (72)

14 7

– 16

= 713 +

12 – (–

16)

= 71

= 7

(d)

823

4 12847

=

(23)23

22 (27)47

=

22

22 24

=

22

26

=

124

=

116

(e)

313 9

13

2723

=

313 (32)

13

(33)23

=

3

13 +

23

32 =

31

32

=

13

(f)

27

– 43 81

14

9 =

(33)

– 43 (34)

14

32

= 3–4 31

32

=

3–3

32

=

135

=

1

243

5 (a) (a2b–3)5 = a2 5b–3 5

= a10b–15

(b)

(a–2

b4 )2

=

a–2 2

b4 2

=

a–4

b8

=

1

a4b8

(c) 16a

– 52 4a

– 32 =

164

a–

52 – (–

32)

= 4a–1

=

4a

(d) x–8y–4 = (x–8)12(y–4)

12

= x–4y–2

=

1

x4y2

(e) (a5b–3)–2 (a4b3)3 = a–10b6 a12b9

= a–10 + 12 b6 + 9

= a2b15

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 31: Analysis Spm Additional Mathematics

2© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(f)

9h5k4

27h8k3 =

13

h5 – 8k4 – 3

=

k

3h3

(g) (2m23n

– 16)6 = 26(m

23)6(n

– 16)6

= 64m4n–1

=

64m4

n

(h) (m–4)12

n (m

23

n)9

= m–2n m6n

= m–2n – 6n

= m–8n

=

1

m8n

(i) 3 p6q–3 = (p6q–3)13

= p2q–1

=

p2

q

6 (a) ( m)3 (3 m)4 (6 m)5

= m32 +

43 –

56

= m2 (b) 3p + 1 9p 27 = 3p + 1 32p 33

= 3p + 1 + 2p – 3

= 33p – 2

(c) 23a – 1 8a + 1 16 = 23a – 1 23(a + 1) 24

= 23a – 1 – 3(a + 1) + 4

= 20

= 1 (d) 32k 9k – 2 812k – 1

= 32k 32(k – 2) 34(2k – 1)

= 32k + 2(k – 2) – 4(2k – 1)

= 3–4k

=

1

34k

(e)

4b + 1 4b – 1

16–b =

4b + 1 + (b – 1)

4–2b

=

42b

4–2b

= 42b – (–2b)

= 44b

(f)

81y + 1 92y + 1

3y – 4 27y + 2

=

34(y + 1) 32(2y + 1)

3y – 4 33(y + 2)

=

34(y + 1) – 2(2y + 1)

3y – 4 + 3(y + 2)

=

32

34y + 2

= 32 – (4y + 2)

= 3–4y

=

1

34y

7 (a) 3x – 1 – 3x + 1 = 3x

31 – 3x 31

=

y3

– 3y

= – 83

y

(b) 4(31 – x) = 4( 3

3x) =

12y

(c) 9x – 2713

x + 23

= 32x – 33(1

3 x + 23)

= 32x – 3x + 2

= 32x – 3x 9 = y2 – 9y (d) 3x + 3x + 1 + 3x + 2

= 3x + 3x 3 + 3x 9 = y + 3y + 9y = 13y

(e) 32x + 1 + 33x – 1 = 32x 3 +

33x

3

= 3y2 + 13

y3

(f) 3x – 912

x + 1 = 3x – 3

2(12

x + 1) = 3x – 3x + 2

= 3x – 3x 9 = y – 9y = –8y

8 (27k2)3 – h( 13k )h

= (33k2)3 – h(3k)–h

= 39 – 3hk6 – 2h(3–hk–h) = 39 – 3h –hk6 – 2h – h

= 39 – 4hk6 – 3h

9 (a) 4n – 22n + 1 + 5(4n + 1) = 22n – 22n 21 + 5[22(n + 1)] = 22n – 2 22n + 5(22n 22) = 22n – 2 22n + 20 22n

= 19 22n

\ 19 22n is divisible by 19 for all positive integers of n.

(b) 2n – 1 + 2n + 1 + 2n

= 2n 2–1 + 2n 21 + 2n

=

12

2n + 2 2n + 2n

=

72

2n

\

72

2n is divisible by 7 for

all positive integers of n. (c) 4(2n + 2) + 2n + 1 – 3(2n) = 4(2n 22) + 2n 2 1 – 3 2n

= 16 2n + 2 2n – 3 2n

= 15 2n

\ 15 2n is divisible by 15 for all positive integers of n.

(d) 3n + 3n + 1 + 3n + 2

= 3n + 3n 31 + 3n 32

= 3n + 3 3n + 9 3n

= 13 3n

\ 13 3n is divisible by 13 for all positive integers of n.

10 (a) Let 3a = 2b = 6c = k, then 3 2 = 6 k

1a k

1b = k

1c

k1a +

1b = k

1c

a + b

ab =

1c

c =

aba + b

(b) Let 3x = 4y = 12z = k then 3 4 = 12

k1x k

1y = k

1z

k1x +

1y = k

1z

x + yxy =

1z

xz + yz = xy xy – xz = yz x(y – z) = yz

x = yz

y – z

11 (a) 53 = 125 ⇔ log5125 = 3 (b) 73 = 343 ⇔ log7343 = 3 (c) 100 = 1 ⇔ log101 = 0

(d) 4–2 = 1

16 ⇔ log4( 1

16) = –2

(e) 3–1 = 1

3 ⇔

log3(1

3) = –1

(f) 63 = 216 ⇔ log6216 = 3

12 (a) log39 = 2 ⇔ 9 = 32 (b) log81 = 0 ⇔ 1 = 80

(c) log101000 = 3 ⇔ 1000 = 103

(d) log2(12) = –1 ⇔

12

= 2–1

(e) log6( 1

36) = –2 ⇔ 136

= 6–2

(f) log216 = 4 ⇔ 16 = 24

13 (a) log40.25 = x 0.25 = 4x

4x =

14

4x = 4–1

\ x = –1 (b) log5 5 = x 5

12 = 5x

\ x =

12

(c) log9x = – 1

2

x = 9–

12

=

1

912

\ x = 13

(d) logx27 = 3 27 = x3

x3 = 33

\ x = 3 (e) log2(3x + 1) = 4 3x + 1 = 24

3x + 1 = 16 3x = 15 \ x = 5 (f) logx + 181 = 2 81 = (x + 1)2

92 = (x + 1)2

x + 1 = 9 \ x = 8 (g) logx(5x – 6) = 2 5x – 6 = x2

x2 – 5x + 6 = 0 (x – 2)(x – 3) = 0 x = 2 or x = 3

Page 32: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(h) log335 = x

35 = 3x

\ x = 5

14 (a) log264 = x 64 = 2x

26 = 2x

x = 6 \ log264 = 6 (b) log41 = x 1 = 4x

40 = 4x

x = 0 \ log41 = 0 (c) log366 = x 6 = 36x

61 = 62x

2x = 1

x =

12

\ log366 = 12

(d) log124 = x

4 =

(12)x

22 = 2–x

x = –2 \ log1

24 = –2

(e) log80.25 = x 0.25 = 8x

14

= 8x

2–2 = 23x

x = – 2

3

\ log80.25 = – 23

(f) log 2 = x 2 = 2

x

212 = 2

x2

x = 1 \ log 2 = 1 (g) log100.001 = x 0.001 = 10x

10–3 = 10x

x = –3 \ log100.01 = –3 (h) log24

–1 = x 4–1 = 2x

2–2 = 2x

x = –2 \ log24

–1 = –2

(i) log5

125

= x

125

= 5x

5–2 = 5x

x = –2

\ log5 125

= –2

15 (a) log245 = log2(9 5) = log29 + log25 = 2 log23 + log25 = 2p + q

(b) log210 = log2(2 5) = log22 + log25 = 1 + q (c) log275 = log2(3 25) = log23 + 2 log25 = p + 2q

16 (a) log10x2y = 2 log10x + log10y

= 2m + n (b) log10 10xy3

=

12

(log1010 + log10x + 3 log10y)

=

12

(1 + m + 3n)

=

1 + m + 3n

2

(c) log10(100 x

y2 ) = log10100 + log10 x – log10y

2

= 2 log1010 + 12

log10x – 2 log10y

= 2 + 12

m – 2n

(d) log10(10y

x ) = log1010 + log10y – log10x = 1 + n – m

17 (a) log320 = log3(4 5) = 2 log32 + log35 = 2(0.631) + 1.465 = 2.727 (b) log315 = log3(3 5) = log33 + log35 = 1 + 1.465 = 2.465

(c) log3 2 =

12

log32

=

12

(0.631)

= 0.3155

(d) log32.5 = log3(5

2) = log35 – log32 = 1.465 – 0.631 = 0.834

(e) log33 1

3 = log3

103

= log310 – log33 = log3(2 5) – log33 = log32 + log35 – log33 = 0.631 + 1.465 – 1 = 1.096

(f) log3

18

= log31 – log38

= log31 – log323

= log31 – 3 log32 = 0 – 3(0.631) = –1.893

18 (a) loga9loga5a =

2 loga3loga5 + logaa

=

2(0.254)0.721 + 1

= 0.2952

(b) loga(3a2) = loga3 + 2 logaa = 0.254 + 2 = 2.254

(c) loga(25

3a) = 2 loga5 – (loga3 + logaa) = 2(0.721) – (0.254 + 1) = 0.188

19 (a) log2(818 )

+ 2 log2(23)

– 2 log2(34)

= log2[(818 )(4

9)( 916) ]

= log28 = log22

3

= 3 log22 = 3

(b) log48 – log42 = log4(82)

= log44 = 1

(c) log636 + log7(1

7) = log66

2 + log77–1

= 2 log66 + (–log77) = 2 – 1 = 1 (d) log3(3 3) + log3 3

= log33 +

12

log33 + 12

log33

= 1 +

12

+ 12

= 2

(e) 2 log105 – log102 + log10( 4

35) +

log1070

= log10[(25)( 4

35)(70)

2 ] = log10100 = 2 log1010 = 2 (f) log32 + log34 – log372

= log3[(2)(4)

72 ]

= log3

19

= log33–2

= –2

20 (a) 2 logm5 – 3 logm2 = logm25 – logm8

= logm 258

(b) loga7 + loga 7 = loga(7 7) = loga7

1 + 12

=

32

loga7

(c) 2 logp5 + logp4 – 2 logp3 = logp25 + logp4 – logp9

Page 33: Analysis Spm Additional Mathematics

4© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

= logp((25)(4)9 )

= logp(1009 )

(d) 4 log10(3

2) + log10( 8

75) – 2 log10(3

5)

= log10[(8116)( 8

75)( 925) ]

= log10(32)

(e) 2 + log25 = 2 log22 + log25 = log2(4 5) = log220 (f) 3 – 3 log102 = 3 log1010 – 3 log102 = log10103 – log102

3

= log101000 – log108

= log10(1000

8 ) = log10125

21 (a) log47 = log107log104

= 1.4037 (b) log312 =

log1012log103

= 2.2619

(c) log128 =

log108

log10 12

= –3 (d) log4(5.2) =

log105.2log104

= 1.1893

(e) log2.56.5 = log106.5log102.5

= 2.0428 (f) log5p =

log10p

log105 = 0.7113

22 (a) log85 log57 log78 =

log105log108

log107log105

log108log107

= 1

(b)

1

logaabc +

1

logbabc +

1

logcabc = logabca + logabcb + logabcc = logabcabc = 1 (c) 4 log35 2 log53 =

4 log55log53 2 log53

= 8 (d) logba logcb logac

= log10alog10b

log10blog10c

log10clog10a

= 1

23 (a) log224 = log2(8 3) = log28 + log23

= log223 + log23

= 3 log22 + log23 = 3 + 1.585 = 4.585

(b) log824 = log224log28

= log224log22

3

=

4.5853

= 1.528

24 (a) log3a3 = 3 log3a

= 3m

(b) log3(1

a) = log3a

–1

= –log3a = –m (c) log9a =

log3alog39

= log3alog33

2

=

m2

25 (a) log2mn = log2m + log2n = p + q

(b) log4(mn ) =

log2(mn )

log24

=

log2m – log2n

2 log22

= p – q

2

(c) logm4n =

log24nlog2m

=

2 log22 + log2n

log2m

= 2 + q

p

26 (a) log4(1a)

= log41 – log4a

= 0 – b = –b

(b) log28a =

log48alog42

=

log48 + log4a

log42

=

log4(4 2) + log4a

log42

=

log44 + log42 + log4a

log42

=

1 + 12

+ b

12

=

32

+ b

12

= 3 + 2b

27 (a) log464h = log464 + log4h = log44

3 + log4h = 3 log44 + log4h = 3 + k

(b) log8h =

log4hlog48

=

log4h

log4(4 2)

=

log4h

log44 + log42

=

k

1 + 12

=

23

k

(c) log28h3 =

log48h3

log42

=

log48 + 3 log4h

log42

=

log4(4 2) + 3 log4h

log42

=

log44 + log42 + 3 log4h

log42

=

1 + 12

+ 3k

12

= 3 + 6k

28 logy81 = log981log9y

=

log99

2

log9y

= 2x

29 m = 2a ⇒ log2m = a n = 2b ⇒ log2n = b

(a) log2(m3n

8 ) = log2m

3 + log2n – log28 = 3 log2m + log2n – 3 log22 = 3a + b – 3

(b) log8m + log4n =

log2mlog28

+ log2nlog24

=

log2mlog22

3 +

log2nlog22

2

= log2m

3 +

log2n

2

=

a3

+ b2

=

2a + 3b

6

30 log1pk =

logpk

logp1p

=

logpklogp p

–1

= –h

31 (a) 4x = 8 22x = 23

Page 34: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

\ 2x = 3 x =

32

(b) 22x – 1 = 32 22x – 1 = 25

\ 2x – 1 = 5 2x = 6 x = 3 (c) 4x + 1 = 0.25

4x + 1 =

14

4x + 1 = 4–1

\ x + 1 = –1 x = –2

(d) ( 2 )3x =

18

232

x = 2–3

\

32

x = –3

x = –2 (e) 9x = ( 3 )x + 2

32x = 3x + 2

2

\ 2x =

x + 2

2 4x = x + 2 3x = 2

x =

23

(f) 16x =

12

24x = 2–1

\ 4x = –1

x = – 14

(g) 54 + x = (0.2)x

54 + x =

(15)x

54 + x = 5–x

\ 4 + x = –x 2x = –4 x = –2 (h) 34x = 27x + 3

34x = 33(x + 3)

\ 4x = 3x + 9 x = 9

32 (a) 9x – 1 = (13)4x – 1

32(x – 1) = 3–1(4x – 1)

\ 2x – 2 = –4x + 1 6x = 3 x =

12

(b) 32x + 1 = 9x – 2

32x + 1

2 = 32(x – 2)

\

2x + 1

2 = 2x – 4

2x + 1 = 4x – 8 2x = 9

x =

92

(c)

(14)x – 1

= 3 23x + 1

2–2(x – 1) = 23x + 1

3

\ –2x + 2 =

3x + 1

3 –6x + 6 = 3x + 1 9x = 5

x = 59

(d) 34x – 5 = 3 34x + 5

34x – 5

2 = 34x + 5

3

\

4x – 5

2 =

4x + 5

3 12x – 15 = 8x + 10 4x = 25

x = 254

(e)

53x

25x + 1 =

1

125

53x

52(x + 1) = 5–3

53x – 2(x + 1) = 5–3

\ 3x – 2x – 2 = –3 x = –1

(f)

(12)2x + 1

=

24x – 1

128

2–(2x + 1) =

24x – 1

27

2–2x – 1 = 24x – 8

2

\ –2x – 1 =

4x – 8

2 – 4x – 2 = 4x – 8 8x = 6

x =

34

33 (a) 22x + 1 + 2x – 3 = 0 21 22x + 2x – 3 = 0 Let y = 2x

2y2 + y – 3 = 0 (2y + 3)(y – 1) = 0

y = –

32

or y = 1

When y = –

32

2x = –

32

(inadmissible)

When y = 1 2x = 1 2x = 20

\ x = 0 (b) 22x + 3 + 2x + 3 = 1 + 2x

23 22x + 23 2x = 1 + 2x

8 22x + 8 2x = 1 + 2x

Let y = 2x

8y2 + 8y = 1 + y 8y2 + 7y – 1 = 0 (8y – 1)(y + 1) = 0

y =

18

or y = –1

When y =

18

,

When y = –1

2x = –1

2x =

18

(inadmissible)

2x = 2–3

\ x = –3

(c) 41 – x + 23 – x = 12

22(1 – x) +

82x

= 12

22 – 2x + 82x

= 12

4

22x +

82x

= 12

Let y = 2x

4y2

+ 8y

= 12

4y + 8y2 = 12y3

12y3 – 8y2 – 4y = 0 4y(3y2 – 2y – 1) = 0 4y(3y + 1)(y – 1) = 0

y = 0 or y = –

13

or y = 1

When y = 0 2x = 0 (inadmissible)

When y = –

13

2x = –

13

(inadmissible)

When y = 1 2x = 1 2x = 20

\ x = 0 (d) 6(9x) + 3x = 2 6 32x + 3x = 2 Let y = 3x

6y2 + y – 2 = 0 (3y + 2)(2y – 1) = 0

y = –

23

or y = 12

When y = – 23

3x = –

23

(inadmissible)

When y =

12

3x =

12

x log103 = log10

12

\ x =

log10 12

log103 x = – 0.631 (e) 52x + 5x + 1 = 6 52x + 5 5x = 6 Let y = 5x

y2 + 5y – 6 = 0 (y + 6)(y – 1) = 0 y = –6 or y = 1 When y = –6 5x = –6 (inadmissible) When y = 1 5x = 1 5x = 50

\ x = 0 (f) 9x = 4(3x) – 3 32x = 4(3x) – 3 Let y = 3x

y2 – 4y + 3 = 0 (y – 1)(y – 3) = 0 y = 1 or y = 3

Page 35: Analysis Spm Additional Mathematics

6© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

When y = 1 When y = 3 3x = 1 3x = 3 3x = 30 3x = 31

x = 0 x = 1 \ x = 0 or x = 1 (g) 32x + 1 + 9 = 3x + 3 + 3x

3 32x + 9 = 33 3x + 3x

3 32x + 9 = 27 3x + 3x

Let y = 3x

3y2 + 9 = 27y + y 3y2 – 28y + 9 = 0 (3y – 1)(y – 9) = 0

y =

13

or y = 9

When y =

13

When y = 9

3x = 9

3x =

13

3x = 32

3x = 3–1 \ x = 2

\ x = –1 \ x = –1 or x = 2 (h) 5(2x) = 2(4x) + 2 5 2x = 2 22x + 2 Let y = 2x

5y = 2y2 + 2 2y2 – 5y + 2 = 0 (2y – 1)(y – 2) = 0

y = 12

or y = 2

When y =

12

When y = 2

2x = 2–1 2x = 2

\ x = –1 \ x = 1

\ x = –1 or x = 1

34 (a) 8x – 1 = 4y

23(x – 1) = 22y

3x – 3 = 2y 3x – 2y = 3 … 1 27x = 3y + 3

33x = 3y + 3

3x = y + 3 3x – y = 3 … 2 2 – 1 : y = 0 Substitute y = 0 into 1 : 3x – 2(0) = 3 3x = 3 x = 1 \ x = 1, y = 0 (b) 3x 92y = 27 3x 34y = 33

3x + 4y = 33

x + 4y = 3 … 1

2x 4–y =

18

2x 2–2y = 2–3

2x – 2y = 2–3

x – 2y = –3 … 2 1 – 2 : 6y = 6 y = 1 Substitute y = 1 into 1 : x + 4(1) = 3 x = –1 \ x = –1, y = 1

(c) 7x – y = 49 7x – y = 72

x – y = 2 … 1 7x + y = 343 7x + y = 73

x + y = 3 … 2 1 + 2 : 2x = 5

x =

52

Substitute x =

52

into 1 :

52

– y = 2

y =

52

– 2

y =

12

\ x = 52

, y = 12

(d) 3x 81y = 27 3x 34y = 33

3x + 4y = 33

x + 4y = 3 … 1

2x 8y =

116

2x 23y = 2– 4

2x + 3y = 2– 4

x + 3y = – 4 … 2 1 – 2 : y = 7 Substitute y = 7 into 1 : x + 4(7) = 3 x + 28 = 3 x = –25 \ x = –25, y = 7 (e) 52x + y = 625 52x + y = 54

2x + y = 4 … 1

24x – 2y =

116

24x – 2y = 2–4

4x – 2y = –4 2x – y = –2 … 2 1 + 2 : 4x = 2

x =

12

Substitute x =

12

into 1 :

2(1

2) + y = 4

1 + y = 4 y = 3

\ x = 12

, y = 3

(f) 8x = 2y + 1

23x = 2y + 1

3x – y = 1 … 1 5y = 25x + 1

5y = 52(x + 1)

y = 2x + 2 2x – y = –2 … 2 1 – 2 : x = 3

Substitute x = 3 into 1 : 3(3) – y = 1 y = 9 – 1 = 8 \ x = 3, y = 8

35 y = mxn – 5 7 = m(2)n – 5 m(2)n = 12 … 1 22 = m(3)n – 5 m(3)n = 27 … 2

1 2 :

(23)n

=

1227

(23)n

=

49

(23)n

=

(23)2

\ n = 2 Substitute n = 2 into 1 : m(2)2 = 12 4m = 12 \ m = 3 \ m = 3, n = 2

36 (a) 0.1x = 0.25

x log100.1 = 5 log100.2 x = 3.4949 (b) 2x 3x = 5x + 1

6x = 5x + 1

x log106 = (x + 1) log105 x log106 = x log105 + log105 x (log106 – log105) = log105 x = 8.8275 (c) 3x + 1 = 7 (x + 1) log103 = log107 x + 1 = 1.7712 x = 0.7712 (d) 4x = 9(5x)

(45)x

= 9

x log10(4

5) = log109

x = –9.8467 (e) 2x = 3x – 2

x log102 = (x – 2) log103 x log102 = x log103 – 2 log103 x (log103 – log102) = 2 log103 x = 5.419 (f) 3x + 1 = 4x

(x + 1) log103 = x log104 x log103 + log103 = x log104 x (log104 – log103) = log103 x = 3.819

37 (a) 5 logm6 – logm96 = 4

logm( 65

96) = 4

m4 = 81 m4 = 34

\ m = 3 (b) log1025 + log10x – log10(x – 1) = 2

log10( 25xx – 1)

= 2

Page 36: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

25xx – 1

= 100

25x = 100x – 100 75x = 100 x =

43

(c) log104 + 2 log10p = 2 log104p2 = 2 4p2 = 100 p2 = 25 p = 25 = 5 \ p = 5 (–5 is rejected) (d) 2 log103 + log102x = log10(3x + 1) log10(3

2 2x) = log10(3x + 1) log1018x = log10(3x + 1) 18x = 3x + 1 15x = 1

x =

115

(e) log2(2x + 5) – 2 log22x = 2

log2

2x + 5(2x)2

= 2

2x + 5

4x2 = 4

2x + 5 = 16x2

16x2 – 2x – 5 = 0 (8x – 5)(2x + 1) = 0

x = 58

or x = – 12

(f) log10(x + 6) = log10(3x – 2) x + 6 = 3x – 2 2x = 8 x = 4

38 (a) log5x = 4 logx5 + 3

log5x = 4(log55

log5x ) + 3

log5x =

4

log5x + 3

Let y = log5x

y = 4y + 3

y2 = 4 + 3y y2 – 3y – 4 = 0 (y + 1)(y – 4) = 0 y = –1 or y = 4 When y = –1 When y = 4 log5x = –1 log5x = 4 x = 5–1 x = 54

= 625

= 15

\ x = 15

or x = 625

(b) log3x + log9x = 6

log3x +

log3x

2 log33 = 6

log3x +

12

log3x = 6

log3x x = 6 x

32 = 36

x 32 = 729

x = 72923

= (36)23

= 34

= 81 (c) log4(6 – x) – log28 = log93 log4(6 – x) – log22

3 = log9912

log4(6 – x) – 3 =

12

log4(6 – x) =

72

6 – x = 472

6 – x = (22)72

6 – x = 128 x = –122 (d) log5x – log25(x + 10) =

12

log5x – log5(x + 10)

log525 =

12

log5x –

12

log5(x + 10) = 12

log5 x

x + 10 =

12

x

x + 10 = 5

12

x = ( 5)( x + 10) x2 = 5(x + 10) x2 – 5x – 50 = 0 (x + 5)(x – 10) = 0 x = –5 or x = 10 \ x = 10 (e) log3x + 2 = 3 logx3

log3x + 2 =

3

log3x Let y = log3x

y + 2 =

3y

y2 + 2y – 3 = 0 (y + 3)(y – 1) = 0 y = –3 or y = 1 When y = –3 When y = 1 log3x = –3 log3x = 1 x = 3–3 x = 3

=

127

\ x =

127

or x = 3

(f) 4 log4x = 9 logx4

4 log4x =

9

log4x Let y = log4x

4y =

9y

y2 =

94

y = 3

2

When y =

32

When y = – 32

log4x =

32

log4x = – 32

x = 432 x = 4

– 32

= (2 2 )32 = (22)

– 32

= 8 = 18

\ x = 8 or x = 18

39 (a) log10y = 2 – log10x log10xy = 2 xy = 100

y = 100

x (b) 2 log10xy = 2 + log10(x + 1)

+ log10y

log10

x2y2

y(x + 1) = 2

x2y

x + 1 = 100

y = 100(x + 1)

x2

(c) 3 + log2(x + y) = log2(x – 2y)

3 = log2(x – 2yx + y )

8 =

x – 2yx + y

8x + 8y = x – 2y 10y = –7x

y = – 710

x

(d) 3 + log10x = 2 log10y

log10

y2

x = 3

y2

x = 1000

y2 = 1000x y = 1000x

40 (a) 2x(4y) = 128 2x 22y = 27

x + 2y = 7 … 1 log10(4x – y) = log102 + log105 log10(4x – y) = log10(2 5) log10(4x – y) = log1010 4x – y = 10 … 2 2 2: 8x – 2y = 20 … 3 1 + 3 : 9x = 27 x = 3 Substitute x = 3 into 1 : 3 + 2y = 7 2y = 4 y = 2 \ x = 3 , y = 2 (b) log5(3x – y) + log56 = log524 log56(3x – y) = log524 6(3x – y) = 24 3x – y = 4 … 1 75x 73y = 1 75x – 3y = 70

5x – 3y = 0 … 2 1 3: 9x – 3y = 12 … 3 3 – 2 : 4x = 12 x = 3 Substitute x = 3 into 1 : 3(3) – y = 4 y = 9 – 4 = 5 \ x = 3, y = 5

Page 37: Analysis Spm Additional Mathematics

8© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

1 8x – 1 = 322x + 5

(23)x – 1 = (25)2x + 5

3x – 3 = 10x + 25 7x = –28 x = –4

2 (1

3)435x + 1 = 81

3– 4(35x + 1

2 ) = 34

3– 4 +

5x + 12 = 34

– 4 + 5x + 1

2 = 4

5x + 12

= 8

5x + 1 = 16 5x = 15 x = 3

3 2n + 1 2n – 2

41 – n =

22n – 1

22(1 – n)

= 22n – 1 – 2(1 – n)

= 24n – 3

4

5a + 1 – 5a – 1

6 53a =

5 5a – 5a

51

6 53a

=

(245 5a)

6 53a

=

45

5–2a

\ Compare with m 5an

we have m =

45

and n = –2.

5

(6423x3)5

(1654x

53)3

=

[(26)23x3]5

[(24)54x

53]3

=

(24x3)5

(25x53)3

=

220x15

215x5

= 25x10

= (2x2)5

\ Compare with (2xm)n we have m = 2 and n = 5.

6 92x + 4 = (1

3)–(3x + 3)

32(2x + 4)

2 = 3–1(–3x – 3)

2x + 4 = 3x + 3 x = 1

7 23(x – 1) =

12

32

23(x – 1) = 2–1 252

23(x – 1) = 232

3x – 3 =

32

3x =

92

x =

32

8 Let 2x = 3y = 6z = k 2 3 = 6

k1x k

1y = k

1z

k1x +

1y = k

1z

1x +

1y =

1z

x + y

xy = 1z

z = xy

x + y (shown)

9 4x + 1 + 7(2x) = 2 22(x + 1) + 7(2x) = 2 22x + 2 + 7(2x) = 2 4 22x + 7(2x) = 2 Let y = 2x

4y2 + 7y = 2 4y2 + 7y – 2 = 0 (4y – 1)(y + 2) = 0

y =

14

or y = –2

When y =

14

2x =

14

2x = 2–2

x = –2 When y = –2 2x = –2 (inadmissible) \ x = –2

10 3235 16

14 = (25)

35 (24)

14

= 23 21

= 22

= 4

11

a32b

23

c34

=

43227

23

1634

=

(22)32(33)

23

(24)34

=

2332

23

= 32

= 9

12 2n + 1 + 4(2n + 2) – 3(2n) = 2 2n + 4(4 2n) – 3(2n) = 2 2n + 16 2n – 3 2n

= 15 2n

\ 15 2n is divisible by 15 for all positive integers of n.

13 52x + y = 625 52x + y = 54

2x + y = 4 … 1 22(2x – y) = 2–4

4x – 2y = –4 2x – y = –2 … 2 1 + 2 : 4x = 2

x =

12

Substitute x = 12

into 1 :

2(1

2) + y = 4

1 + y = 4 y = 3

\ x = 12

, y = 3

14 2x 4x – 1 = 82x – 1

2x 22(x – 1) = 23(2x – 1)

23x – 2 = 26x – 3

3x – 2 = 6x – 3 3x = 1

x =

13

15 4x + 2 32x = 576 (16 4x)(9x) = 576 4x9x = 36 36x = 36 x = 1

16

( 964)1

2

(38)–1

(2764)2

3

=

38

83

916

=

169

17 y = nxm – 4 8 = n(2)m – 4 12 = n(2)m … 1 71 = n(5)m – 4 75 = n(5)m … 2

1 2 :

425

= (2

5)m

(25)2

=

(25)m

m = 2 Substitute m = 2 into 1 : 12 = n(2)2

n = 3 \ m = 2, n = 3

18

(–2k)3(2k)– 23

(16k4)13

=

–(2k)3(2k)– 23

(2k)43

= –(2k)3 –

23 –

43

= –2k

19 log7x = 9 logx7

log7x =

9 log77log7x

Let y = log7x

y =

9y

y2 = 9 y = 3

Page 38: Analysis Spm Additional Mathematics

9© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

When y = 3 When y = –3 log7x = 3 log7x = –3 x = 343 x = 7–3

=

1343

\ x = 343 or x =

1343

20 3 log7x – 2 = log7y 3 log7x – log7y = 2

log7(x3

y ) = 2

x3

y = 49

y =

x3

49

21 2 logx4 + log4x = 3

2 logx4 +

1

logx4 = 3

Let y = logx4

2y + 1y = 3

2y2 – 3y + 1 = 0 (2y – 1)(y – 1) = 0

y =

12

or y = 1

When y =

12

When y = 1

logx4 = 1

logx4 =

12

x = 4

x12 = 4

x = 16 \ x = 16 or x = 4

22 logmp2q = 8 2 logmp + logmq = 8 … 1

logm q2

p = 6

–logmp + 2 logmq = 6 … 2 2 2: –2 logmp + 4 logmq = 12 … 3 1 + 3 : 5 logmq = 20 logmq = 4 Substitute logmq = 4 into 1 : 2 logmp + 4 = 8 2 logmp = 4 logmp = 2 \ logmq = 4, logmp = 2

23 log4mn = 10 log4m + log4n = 10 … 1

2 log8m = 3 log8n

2 log4mlog48 =

3 log4nlog48

2 log4m – 3 log4n = 0 … 2 1 2: 2 log4m + 2 log4n = 20 … 3 3 – 2 : 5 log4n = 20 log4n = 4 n = 44

= 256 Substitute n = 256 into 1 : log4m + log4256 = 10 log4m + 4 log44 = 10 log4m = 6

m = 46

= 4096 \ m = 4096, n = 256

24 log82x + log2x =

53

log22x3 log22

+ log2x =

53

log22x + 3 log2x = 5 log2(2x)(x3) = 5 2x4 = 32 x4 = 16 x = 2

25 logp 27 p125

= logp(33 p12

53 )

= 3 logp 3 + 12

logp p –

3 logp 5

= 3x +

12

– 3y

26 log3y +

12

log3x = 2 log3z

log3y x = log3z2

y x = z2

x =

z2

y

x = ( z2

y )2

= z4

y2

27 log45 log56 log67 log78

= log25log24

log26log25

log27log26

log28log27

=

log28log24

=

3 log222 log22

=

32

28 log927 log381 =

log327log39

log381

=

32

4

= 6

29 log5x = 4 logx5

log5x =

4log5x

Let y = log5x

y =

4y

y2 = 4 y = 2 When y = 2 When y = –2 log5x = 2 log5x = –2 x = 25

x =

125

\ x = 25 or x =

125

30 log2x – 2 = log4(x – 4)

log2x – 2 = log2(x – 4)

2 log22 2 log2x – 4 = log2(x – 4)

log2( x2

x – 4) = 4

x2

x – 4 = 16

x2 = 16x – 64 x2 – 16x + 64 = 0 (x – 8)(x – 8) = 0 x = 8

31 log10(2x + 7) = 1 + log10x

log10 2x + 7

x = 1

2x + 7

x = 10

2x + 7 = 10x 8x = 7

x = 78

32 (a) log4(1n)

= log41 – log4n

= –m (b) log28n = log28 + log2n = 3 log22 +

log4nlog42

= 3 + 2m

33 log28 + log3 19

+ log2 12

= 3 log22 – 2 log33 – log22 = 3 – 2 – 1 = 0

34 log1045 = log10(9 5) = log109 + log105 = 2 log103 + log105 = 2(0.477) + 0.699 = 1.653

35 log264 p = 6 log22

6p = 6 6p log22 = 6 6p = 6 p = 1

36 log38 = 3 log22log23

= 3p

37 logh10h = logh10 + loghh

=

1

log10h + 1

=

1k

+ 1

=

1 + k

k

38 logm(5m2

27 ) = logm5 + 2 logmm – 3 logm3

= q + 2 – 3p

Page 39: Analysis Spm Additional Mathematics

10© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

39 log20.3 = log2

310

= log2[3 (5 2)] = log23 – (log25 + log22) = 1.585 – 2.322 – 1 = –1.737

40 2 log2(23)

+ log2(818 )

– 2 log2(34)

= log2

(49)(81

8 )916

= log28 = 3 log22 = 3

41 log98 = a

3 log22log29 = a

log29 = 3a

log43 = log93log94

= 1

2 log94

=

1

(2 log24log29 )

=

1

(4 log22log29 )

=

1

(4a3 )

=

34a

42 uv = 25 v = logu25

v = 2 log55log5u

log5u = 2v

Page 40: Analysis Spm Additional Mathematics

BC = (1–1)2+[2–(–1)]2

= 9

= 3units

CA = (1–5)2+(2–2)2

= 16 = 4units Perimeter = 5+3+4 = 12units

7 (a) P(x,0),M(5,2)andN(–1,4) MP =NP

22+(5–x)2 = 42+(–1–x)2

4+25–10x+x2 = 16–8y+y2+1 29–10x = 17+2x 12x = 12 x = 1 ∴P(1,0) (b) Q(0,y),M(5,2)andN(–1,4) MQ=NQ

(2–y)2+52 = (4–y)2+(–1–0)2

4–4y+y2+25 = 16–8y+y2+1 29–4y = 17–8y 4y = –12 y = –3 ∴Q(0,–3)

8 (–3+h2

,4+k

2)=(5,7)

–3+h

2 = 5 and

4+k2

= 7

–3+h = 10 4+k = 14 h = 13 k = 10 ∴h=13,k=10

9m2+n2

2 = 5

m2+n2 = 10… 1

m+n

2 = 1

m+n = 2 m = 2–n… 2 Substitute 2 into 1 : (2–n)2+n2 = 10 4–4n+n2+n2 = 10 2n2–4n–6 = 0 n2–2n–3 = 0 (n+1)(n–3) = 0 n=–1orn=3 Substituten=–1into 2 : m = 2–(–1) = 3 Substituten=3into 2 : m = 2–3 = –1 ∴m=3,n=–1;m=–1,n=3

1 (–4h)2+(–3h)2 = 10 25h2 = 100 h2 = 4 h = ±2

2 AB = (–1–3)2+(–3)2

= 25 = 5units

BC = [–1–(–1)]2+(–5)2

= 25 = 5units

CD = [(3–(–1)]2+[–2–(–5)]2

= 25 = 5units

DA = (3–3)2+(–2–3)2

= 25 = 5units AB=BC=CD=DA ∴ABCDisarhombus.

3 RP=SP (2–y)2+(3–x)2

= (–2–y)2+(4–x)2

4–4y+y2+9–6x+x2

= 4+4y+y2+16–8x+x2

13–6x–4y=20–8x+4y 2x–8y–7=0

4 AB = (6–2)2+(6–1)2

= 41units

BC = (1–6)2+(10–6)2

= 41units

AC = (1–2)2+(10–1)2

= 82 AB=BC=2AC ∴A,BandCaretheverticesofa

right-angledtriangle.

5 (a) PQ = [3–(–4)]2+(3–2)2

= 50units

QR = (3–4)2+[3–(–4)]2

= 50units

(b) PR = (–4–4)2+[2–(–4)]2

= 100 = 10units PQ=QR ≠ PR ∴PQRisanisoscelestriangle.

6 AB = (1–5)2+(–1–2)2

= 25 = 5units

10p+8

2 = 4 and

5+q2

= 3

p+8 = 8 5+q = 6 p = 0 q = 1 ∴p=0,q=1

11 (a) MidpointofAC

= [6+(–4)2

, 4+62 ]

= (1,5) (b) MidpointofAC=midpointofBD LetDbe(x,y),

( x+82

,y+7

2 )=(1,5)

x+8

2 = 1 and

y+72

= 5

x+8 = 2 y+7 = 10 x = –6 y = 3 ∴D(–6,3)

12 MidpointofAC

= (–1+22

,4+5

2 ) = (1

2,

92 )

MidpointofBD

= (–4+52

,6+3

2 ) = (1

2,

92 )

Thesamepoint( 12

,92 )isthe

midpointofACandBD,sothelinesACandBDmustbisectoneanother.

13 BD=AC

(p+(–7)2

,6+q

2 )=(4+(–2)2

,–4+4

2 )

p–72

= 1 and6+q

2=0

p–7 = 2 q =–6 p = 9 ∴p=9,q=–6

14 y=1–2x… 1 y2+(2x+3)2=10… 2 Substitute 1 into 2 : (1–2x)2+(2x+3)2=10 1–4x+4x2+4x2+12x+9 =10 8x2+8x+10 =10 x2+x =0 x(x+1) =0 x=0orx=–1 Whenx=0,y = 1–2(0) = 1 Whenx=–1,y = 1–2(–1) = 3 A(0,1);B(–1,3)

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 41: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

∴MidpointofAB

= (0+(–1)2

,1+3

2 ) = (– 1

2,2)

15 (a) A = ( –2+42

,7+9

2 ) = (1,8)

B = (6+42

,1+9

2 ) = (5,5)

(b) AB = (5–8)2+(5–1)2

= 25 = 5units

PQ = (1–7)2+[6–(–2)]2

= 100 = 10units

∴AB=12

PQ

16 (a) P = (2(3)+3(–2)2+3

,2(3)+3(–2)

2+3 ) = (0,0)

(b) P=(2(4)+3(–1)2+3

,2(–2)+3(3)

2+3 ) =(1,1)

17 (a) M= ( 2+02

,–3+1

2 ) = (1,–1)

(b) (1,–1)=( 3+x2

,0+4

2 )

3+x2

=10+y

2 = –1

x =–1 y = –2 S(–1,–2)

18 (a) P = (3(4)+1(–4)3+1

,3(3)+1(–1)

3+1 ) = (2,2)

(b) P=(1(1)+2(–2)1+2

,1(–2)+2(4)

1+2 ) =(–1,2)

19 (a)2x+1(2)

3= 0

2x+2 = 0 2x = –2 x = –1 and

2y+1(4)

3= –2

2y+4 = –6 2y = –10 y = –5 ∴B(–1,–5)

(b)2x+5(3)

7= 1

2x+15 = 7 2x = –8 x = –4 and

2y+5(–4)

7 = –2

2y–20 = –14 2y = 6

y = 3 ∴B(–4,3)

20 (a)7m+(–3n)

m+n = 1

7m–3n = m+n 6m = 4n

mn

=46

=23

∴m:n=2:3

(b) k =2(3)+3(–2)

2+3

=6–6

5

= 0

21 (a)3m+8nm+n

= 5

3m+8n = 5m+5n 3n = 2m

mn

=32

∴m:n=3:2

(b) k =3(4)+2(–1)

3+2

=105

=2

22 (a) Area =12 |

5 1 –6 5| 2 6 –7 2

=12

(30–7–12–2+ 36+35)

= 40unit2

(b) Area =12 |

0 4 6 –50| 0 1 5 3 0

=12

(20+18–6+25)

= 28.5unit2

23 (a) A =12 |

t 2t –2 t |–3 3 –1 –3

=12

(3t–2t+6+6t+6+t)

=12

(8t+12)

= 4t+6 (b) 4t+6 = 14 4t = 8 t = 2

2412 |

5 2 8 5|10 1 r 10=±24

12

(5+2r+80–20–8–5r)=±24

12

(57–3r)=±24

12

(57–3r) = 24

57–3r = 48 3r = 9 r = 3 or

12

(57–3r) = –24

57–3r = –48 3r = 105 r = 35 ∴r=3orr=35

2512 |

1 m 4 1| 2 3 5 2= ±6

12

(3+5m+8–2m–12–5)= ±6

12

(3m–6)= ±6

12

(3m–6) = 6

3m–6 = 12 3m = 18 m = 6 or

12

(3m–6) = –6

3m–6 = –12 3m = –6 m = –2 ∴m=6orm=–2

26 (a) AreaofABC

=12 |

4 3 –4 4| 1 5 3 1

=12

(20+9–4–3+20–12)

=12

(30)

= 15unit2

AreaofACD

=12 |

4 –4–2 4| 1 3 –3 1

=12

(12+12–2+4+6+12)

=12

(44)

= 22unit2

(b) AreaofABCD = AreaofABC+Areaof ACD = 15+22 = 37unit2

2712 |

–1 0 2 –1|–5 –2 k –5 = 0

2–10+4+k = 0 k–4 = 0 k = 4

28 (a) M= ( –3+112

,1+(–3)

2 ) = (4,–1) LetSbe(x,y),

x+4

2 =4 and

y+92

= –1

x+4 =8 y+9 = –2 x =4 x = –11 ∴S(4,–11) (b) AreaofPQRS

=12 |

–3 4 11 4 –3| 1 –11–3 9 1

=12

(33–12+99+4–4+121 +12+27)

=12

(280)

= 140unit2

29m2+4–2m

6–m = 1

m2–2m+4 = 6–m

Page 42: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

m2–m–2 = 0 (m+1)(m–2) = 0 m=–1orm=2

30–6–(–k)

k–2 = –3

–6+k = –3k+6 4k = 12 k = 3

31h–24–2

=2–(–1)

2–1

h–2

2 = 3

h–2 = 6 h = 8

32 (a) y–(–3)=3(x–2) y+3=3x–6 y =3x–9 (b) y–4= –2[x–(–1)] y–4= –2(x+1) y–4= –2x–2 y+2x = 2

(c) y–6 =13

(x–5)

3y–18 = x–5 3y–x = 13

(d) y–7 = –25

[x–(–5)]

y–7 = –25

x–2

y = –25

x+5

33k–2

–3–2 = –4

k–2 = 20 k = 22

34 (a) m =3–15–1–5

= 2 y–15 = 2(x–5) y–15 = 2x–10 y = 2x+5 (b) Atx-axis, y = 0 2x = –5

x = –52

∴A(– 52

,0) Aty-axis, x = 0 y = 5 ∴B(0,5)

AB = 52+(52)2

= 1254

= 5.59units

35 m =5–26–3

= 1 y–2= 1(x–3) y–2= x–3 x–y = 1 … 1 x+2y = –5… 2 1 – 2 : –3y = 6 y = –2

Substitutey=–2into 1 : x–(–2)= 1 x = –1 Thepointofintersectionis(–1,–2).

36 2x+y=2… 1 3x+6y=3… 2 1 6: 12x+6y = 12… 3 3 – 2 : 9x = 9 x = 1 Substitutex=1into 1 : 2(1)+y = 2 y = 0 Thepointofintersectionis(1,0). x+2y+h=0 At(1,0), 1+2(0)+h = 0 h = –1

37 x+y=–4… 1 3x+y=–2… 2 2 – 1 : 2x = 2 x = 1 Substitutex=1into 1 : 1+y = –4 y = –5 Twopointsare(0,0)and(1,–5). Theequationis:

yx

=–5–01–0

y = –5x

38 3x–y=–1… 1 2x+3y=–8… 2 1 3: 9x–3y = –3… 3 2 + 3 : 11x = –11 x = –1 Substitutex=–1into 1 : –3–y = –1 y = –2 Theequationis: y–(–2)= –2[x–(–1)] y+2= –2(x+1) y+2= –2x–2 y+2x = –4

39 (a) A(–1,–1)andC(4,3) Theequationis:

y–(–1)x–(–1)

=3–(–1)4–(–1)

y+1x+1

=45

5y+5= 4x+4 5y–4x = –1 B(3,–2)andD(0,2) Theequationis:

y–(–2)

x–3 =

2–(–2)0–3

y+2x–3

= –43

3y+6 = –4x+12 3y+4x = 6 (b) 5y–4x=–1 … 1 3y+4x=6 … 2 1 + 2 : 8y = 5

y =58

Substitutey=58

into 1 :

5(58 )–4x = –1

4x =258

+1

=338

x =3332

∴Thepointofintersectionis

(3332

,58 ).

40 (a) y–x=1… 1 3y–x=–1… 2 2 – 1 :2y = –2 y = –1 Substitutey=–1into 1 : –1–x = 1 x = –2 ∴P(–2,–1) TheequationofABis: y+1 = 2(x+2) y+1 = 2x+4 y = 2x+3

∴A(0,3)andB(– 32

,0) (b)

–2m+n(0)m+n

= –32

–4m = –3m–3n m = 3n

mn

= 3

∴m:n=3:1

ABBP

=31

41 (a) C= ( 2(6)+1(–3)2+1

,2(6)+1(0)

2+1 ) = (3,4)

(b)y

x+3 =

6–06–(–3)

9y = 6x+18 9y–6x = 18 3y–2x = 6 (c) Q(0,6)andC(3,4) Theequationis:

y–6

x =

4–63

3y–18 = –2x 3y+2x = 18

42 (a)y–2

x–(–1) =

–3–24–(–1)

y–2x+1

= –55

5y–10 = –5x–5 5y+5x = 5 y+x = 1

(b) y+x=1… 1 y–2x=–5… 2 1 – 2 : 3x = 6 x = 2 Substitutex=2into 1 : y+2 = 1 y = –1 ∴P(2,–1)

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�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(c)4m+n(–1)

m+n = 2

4m–n = 2m+2n 2m = 3n

mn

=32

∴AP:PB=3:2

43 (a) 2x = y+6 y = 2x–6 m

1 = 2

y = 2x+4 m

2 = 2

∴Parallel (b) x+y = 3 y = –x+3 m

1 = –1

2x+2y = 5 2y = –2x+5

y = –x+52

m2 = –1

∴Parallel

(c)x2

+y3

= 1

3x+2y = 6 2y = –3x+6

y = –32

x+3

m1 = –

32

2y+3x = 8 2y = –3x+8

y = –32

x+4

m2 = –

32

∴Parallel (d) 2x–5y+4 = 0 5y = 2x+4

y =25

x+45

m1 =

25

2y = 5x+6

y =52

x+3

m2 =

52

∴Notparallel

44 (a) m=4,(–1,2) y–2 = 4(x+1) y–2 = 4x+4 y = 4x+6

(b) m=–23

,(–1,–1)

y+1 = –23

(x+1)

3y+3 = –2x–2 3y+2x = –5

45 (a) m=–12

,(2,3)

y–3 = –12

(x–2)

2y–6 = –x+2 2y+x = 8

(b) m=12

,(–6,1)

y–1 =12

(x+6)

y–1 =12

x+3

y =12

x+4

(c) m = –43

,(4,–1)

y+1 = –43

(x–4)

3y+3 = –4x+16 3y+4x = 13

(d) m=–23

,(–3,–5)

y+5 = –23

(x+3)

y+5 = –23

x–2

y = –23

x–7

46 (a)5–(–3)

p+8 =

23

24 = 2p+16 2p = 8 p = 4

(b) m=23

,(4,5)

y–5 =23

(x–4)

3y–15 = 2x–8 3y–2x = 7

47 (a) y–x=–2… 1 y+2x=1… 2 2 – 1 : 3x = 3 x = 1 Substitutex=1into 1 : y–1 = –2 y = –1 ∴A(1,–1) (b) m=3,A(1,–1) Theequationis: y+1 = 3(x–1) y+1 = 3x–3 y = 3x–4

48 (a) x+y = 5 x–y = 3 y = –x+5 y = x–3 m

1 = –1 m

2 = 1

∴Perpendicular (b) 2x+y = 1 y = –2x+1 m

1 = –2

x+2y = 4 2y = –x+4

y = –12

x+2

m2 = –

12

∴Notperpendicular (c) y = 3x+6 x = 3y+7 m

1= 3 3y = x–7

y =13

x–73

m2 =

13

∴Notperpendicular

(d)x3

–y5

= 1

5x–3y = 15 3y = 5x–15

y =53

x–5

m1 =

53

5y+3x = 10 5y = –3x+10

y = –35

x+2

m2 = –

35

∴Perpendicular

49 (a) m=–1,(0,0) Theequationis: y=–x (b) m=2,(3,–2) Theequationis: y+2= 2(x–3) y+2= 2x–6 y = 2x–8 (c) m=–2,(–4,–5) Theequationis: y+5 = –2(x+4) y+5 = –2x–8 y+2x+13 = 0

(d) m=–12

,(5,6)

Theequationis:

y–6 = –12

(x–5)

2y–12 = –x+5 2y+x = 17

50 (a) m=–12

y=–12

x+n

At(4,–5), –5 = –12

(4)+n

n = –5+2 = –3

∴m=–12

,n=–3

(b)k3

= –23

k = –2

51 (a) m1 =

3–5–2–4

=13

m2 = –3

Midpoint = (4+(–2)2

,5+3

2 ) = (1,4) Theequationofperpendicular

bisectoris: y–4 = –3(x–1) y–4 = –3x+3 y+3x = 7

(b) m1

=4–(–8)–2–3

= –125

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�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

m2 =

512

Midpoint = (3+(–2)2

,–8+4

2 ) = ( 1

2,–2)

Theequationofperpendicularbisectoris:

y+2 =512 (x–

12 )

12y+24 = 5x–52

24y+48 = 10x–5 24y–10x+53 = 0

(c) m1 =

8–74–3

= 1 m

2 = –1

Midpoint = (3+42

,7+8

2 ) = ( 7

2,

152 )

Theequationofperpendicularbisectoris:

y–152

= –1(x–72 )

y–152

= –x+72

2y–15 = –2x+7 2y+2x = 22 y+x = 11

(d) m1 =

6–3–1–2

= –1 m

2 = 1

Midpoint = (2+(–1)2

,3+6

2 ) = ( 1

2,

92 )

Theequationofperpendicularbisectoris:

y–92

= 1(x–12 )

2y–9 = 2x–1 2y–2x = 8 y–x = 4

52 (a) mAB

=4–21–3

= –1 m

CD = 1

MidpointofAB

= (1+32

,4+2

2 ) = (2,3) Theequationis: y–3 = 1(x–2) y = x+1 (b) At(4,t), t = 4+1 = 5 (c) LetDbe(0,y),

(0+42

,y+5

2 ) = (2,3)

y+5

2 = 3

y+5 = 6 y = 1 ∴D(0,1)

53 (a) m=–13

,A(2,–3)

Theequationis:

y+3 = –13

(x–2)

3y+9 = –x+2 3y+x = –7 (b) 3y+x=–7… 1 y–3x=1 … 2 1 3 : 9y+3x = –21… 3 2 + 3 : 10y = –20 y = –2 Substitutey=–2into 1 : –6+x = –7 x = –1 ∴P(–1,–2)

54 (a) m =2–4

2–(–1)

= –23

Theequationis:

y–9 = –23

(x+2)

3y–27 = –2x–4 3y+2x = 23

(b) m=32

,A(–1,4)

Theequationis:

y–4 =32

(x+1)

2y–8 = 3x+3 2y–3x = 11 (c) 3y+2x= 23… 1 2y–3x = 11… 2 1 3: 9y+6x = 69… 3 2 2: 4y–6x = 22… 4 3 + 4 : 13y = 91 y = 7 Substitutey=7into 1 : 3(7)+2x = 23 21+2x = 23 2x = 2 x = 1 ∴D(1,7)

55 (a) (x+1)2+(y+3)2 = 3 x2+2x+1+y2+6y+9 = 9 x2+y2+2x+6y+1 = 0

(b) x2+y2 = 3 x2+y2 = 9 x2+y2–9 = 0

(c) (x–2)2+(y–5)2 = 3 x2–4x+4+y2–10y+25 = 9 x2+y2–4x–10y+20 = 0

(d) (x–1)2+(y+2)2 = 3 x2–2x+1+y2+4y+4 = 9 x2+y2–2x+4y–4 = 0

56 (a) (x+1)2+y2= (x–1)2+y2

x2+2x+1+y2= x2–2x+1+y2

2x+1= –2x+1 4x = 0 x = 0

(b) (x+1)2+(y–1)2

= (x–1)2+(y+1)2

x2+2x+1+y2–2y+1 = x2–2x+1+y2+2y+1

2x–2y+2 = –2x+2y+2 4x–4y = 0 x–y = 0

(c) (x–2)2+(y–4)2

= (x–8)2+(y–6)2 x2–4x+4+y2–8y+16 = x2–16x+64+y2–12y+36 20–4x–8y = 100–16x–12y 12x+4y–80 = 0 3x+y–20 = 0

(d) x2+(y–2)2 = (x–2)2+y2

x2+y2–4y+4 = x2–4x+4+y2

4–4y = 4–4x 4x–4y = 0 x–y = 0

57 (a) 2PA = PB

2 (x+4)2+y2 = (x–4)2+y2

4(x2+8x+16+y2)= x2–8x+16+y2

4x2+4y2+32x+64= x2+y2–8x+16 3x2+3y2+40x+48= 0

(b) 3PA=2PB 3 (x+4)2+y2 =2 (x–4)2+y2

9(x2+8x+16+y2) =2(x2–8x+16+y2)9x2+9y2+72x+144 =2x2+2y2–16x+327x2+7y2+88x+112 =0

(c) PA2+PB2 = 10

( (x+4)2+y2 )2+( (x–4)2+y2 )2

= 10

x2+8x+16+y2+x2–8x+16+y2 = 10 2x2+2y2+22 = 0 x2+y2+11 = 0

(d) 3PA= PB

3 (x+4)2+y2 = (x–4)2+y2

9(x2+8x+16+y2)= x2–8x+16+y2

9x2+72x+144+9y2= x2–8x+16+y2

8x2+8y2+80x+128= 0 x2+y2+10x+16= 0

58 PA = x (x–3)2+(y–2)2 = x x2–6x+9+y2–4y+4 = x2

y2–6x–4y+13 = 0

59 PA=3PB

(x–2)2+y2 =3 (x+4)2+y2

x2–4x+4+y2=9(x2+8x+16+y2) x2–4x+4+y2=9x2+72x+144+9y2

8x2+8y2+76x+140=0 2x2+2y2+19x+35=0

60 PA=x+3

(x–3)2+y2 = x+3 x2–6x+9+y2= (x+3)2

x2–6x+9+y2= x2+6x+9 y2= 12x

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�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

61 3OP =2PA 3 x2+y2 = 2 (x–2)2+y2

9(x2+y2)= 4(x2–4x+4+y2) 9x2+9y2= 4x2–16x+16+4y2

5x2+5y2+16x–16= 0

62 (x–4)2+y2 = x x2–8x+16+y2 = x2

y2–8x+16 = 0

1h2+k2

2 = 5

h2+k2 = 10… 1

h+k

2 = 1

h+k = 2 h = 2–k… 2 Substitute 2 into 1 : (2–k)2+k2 = 10 4–4k+k2+k2 = 10 2k2–4k–6 = 0 k2–2k–3 = 0 (k+1)(k–3) = 0 k=–1ork=3 Substitutek=–1into 2 : h = 2–(–1) = 3 Substitutek=3into 2 : h= 2–3 = –1 ∴h=3,k=–1;h=–1,k=3

2 (p+(–7)2

,6+q

2 ) =(4+(–2)

2,

–4+42 )

p–7

2 = 1 and

6+q2

= 0

p–7 = 2 q = –6 p = 9 ∴p=9,q=–6

3 Area=6unit2

12

2 5 –1 2 | 3 r 4 3 | = ±6

12

(2r+20–3–15+r–8)= ±6

12

(3r–6)= ±6

3r–6 = 12 or 3r–6 = –12 3r = 18 3r = –6 r = 6 r = –2

4 y=5–2x… 1 x2+y2=10… 2 x2+(5–2x)2= 10 x2+25–20x+4x2 = 10 5x2–20x+15 = 0 x2–4x+3 = 0 (x–1)(x–3) = 0 x=1orx=3 Substitutex=1into 1 : y = 5–2(1) = 3

Substitutex=3into 1 : y = 5–2(3) = –1 P(1,3)andQ(3,–1) ∴MidpointofPQ

= (1+32

,3+(–1)

2 ) = (2,1)

512

0 2 4m 0 | 0 3m 6 0 | = 0

12–12m2 = 0 12m2 = 12 m2 = 1 m = ±1

6 mBC

=–12

B(0,3)

3–00–

= –12

= 6

7 (a)3t–(t+2)

t2–t =

t–32t

2t(2t–2) = (t–3)(t2–t) 4t2–4t = t3–t2–3t2+3t t3–8t2+7t = 0 t2–8t+7 = 0 (t–1)(t–7) = 0 t=1ort=7

(b) ( 3tt2 )( t–3–(t+2)

2t–t )=–1

( 3t )(–5

t )= –1

t2= 15 t = ± 15

8 x=y+6… 1 y2=8x … 2 Substitute 1 into 2 : y2= 8(y+6) y2–8y–48= 0 (y+4)(y–12)= 0 y=–4ory=12 Substitutey=–4into 1 : x = –4+6 = 2 Substitutey=12into 1 : x = 12+6 = 18 A(2,–4)andB(18,12)

∴AB = [12–(–4)]2+(18–2)2

= 512 = 22.627units

9 Midpoint = (–1+72

,8+(–2)

2 ) = (3,3) Theequationofthelinethrough(3,3) andparalleltotheline2x+5y=9:

y–3 = –25

(x–3)

5y–15 = –2x+6 5y+2x = 21

10 x–2y=1… 1 x+3y=6… 2 2 – 1 : 5y = 5 y = 1 Substitutey=1into 1 : x–2(1) = 1 x = 3 ∴Thepointofintersectionis(3,1). 3x+4y = 8 4y = –3x+8

y = –34

x+2

∴m=–34

Therequiredequationis:

y–1 = –34

(x–3)

4y–4 = –3x+9 4y+3x = 13

11 (a) MidpointofAB

= ( –6+(–2)2

,1+7

2 ) = (–4,4)

mAB

=7–1

–2–(–6)

=32

mPQ

=–23

TheequationofthelinePQis:

y–4 = –23

(x+4)

3y–12 = –2x–8 3y+2x = 4 (b) 3y+2x=4 Aty-axis,x=0 3y+2(0)= 4 3y = 4

y =43

∴P(0,43 )

Atx-axis,y=0 3(0)+2x = 4 2x = 4 x = 2 ∴Q(2,0)

12 ( –3–32–k )(–3–1

2–10 ) = –1

( –62–k )( 1

2 ) = –1

–3 = –1(2–k) –3 = k–2 k = –1

13 (a) mAB

mAC

= (5–(–3)3–(–1) )(5–2

3–9 ) = 2(– 1

2 ) = –1 Sincem

ABm

AC=–1,theangle

BACistherightangleandhenceABCisaright-angledtriangle.

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�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) AreaofABC

=12

3 –1 9 3 | 5 –3 2 5 |

=12

(–9–2+45+5+27–6)

=12

(60)

= 30unit2

14 Midpoint = ( –4+82

,6+(–2)

2 ) = (2,2)

Gradient =–2–6

8–(–4)

= –23

Sogradientoftheperpendicular

bisector=32

.

Equationoftheperpendicularbisectoris:

y–2 =32

(x–2)

2y–4 = 3x–6 2y–3x = –2

15 (a) 2x+3y =7 … 1 3x–4y =2 … 2 1 4: 8x+12y = 28… 3 2 3: 9x–12y = 6 … 4 3 + 4 : 17x = 34 x = 2 Substitutex=2into 1 : 2(2)+3y = 7 3y = 3 y = 1 ∴A(2,1) Hence,theequationoftheline

throughA(2,1)andpassesthroughB(5,7)is

y–1x–2

=7–15–2

y–1= 2(x–2) y–1= 2x–4 y = 2x–3

(b) GradientofAB =7–15–2

= 2 Sothegradientofthe

perpendicularline=–12

.

TheequationoftheperpendicularlinethroughA(2,1)is:

y–1 = –12

(x–2)

2y–2 = –x+2 2y+x = 4

16 (a) AreaofABC

=12

1 7 1 1 |–2 6 2 –2|

=12

(6+14–2+14–6–2)

=12

(24)

= 12unit2

(b) TheequationofthelineAC:

y+2x–1

=6–(–2)

7–1 3y+6 = 4x–4 3y–4x = –10… 1

mBD

=–34

andB(1,2)

TheequationofthelineBD:

y–2= –34

(x–1)

4y–8= –3x+3 4y+3x=11… 2 1 3: 9y–12x = –30… 3 2 4: 16y+12x = 44… 4 3 + 4 : 25y = 14

y = 1425

Substitutey=1425

into 1 :

3(1425 )–4x = –10

4x =4225

+10

=29225

=7325

∴D(7325

,1425 )

17 C = (3+92

,5+1

2 ) = (6,3)

mAB

=5–13–9

so mCD

=32

= –23

TheequationoftheperpendicularbisectorofABis:

y–3 =32

(x–6)

2y–6 = 3x–18 2y–3x = –12… 1 TheequationofthelineODis:

y–0= –23

(x–0)

3y= –2x 3y+2x = 0… 2 1 2: 4y–6x = –24… 3 2 3: 9y+6x = 0… 4 3 + 4 : 13y = –24

y = –2413

Substitutey=–2413

into 1 :

2(– 2413 )–3x = –12

3x =10813

x =3613

∴D(3613

,–2413 )

18 (2k)2+k2 = (2k+1)2+(k–3)2 4k2+k2 = 4k2+4k+1+k2–6k+9 –2k+10 = 0 –2k = –10 k = 5

19 (a)x

2a+

y3a

=1

(b)142a

+(–93a ) = 1

7a

–3a

= 1

4a

= 1

a = 4

20 MidpointofAB = (0+82

,5+7

2 ) = (4,6)

mAB

=7–58–0

=14

Sogradientofthebisector=–4 EquationofthebisectorofABis: y–6 = –4(x–4) y–6 = –4x+16 y+4x = 22… 1 MidpointofBC=(6,4)

mBC

=7–18–4

=32

Sogradientofthebisector=–23

EquationofthebisectorBCis:

y–4 =–23

(x–6)

3y–12 = –2x+12 3y+2x= 24… 2 2 2: 6y+4x=48… 3 3 – 1 : 5y = 26

y =265

Substitutey=265

into 1 :

265

+4x = 22

4x =845

x =215

∴Thepointofintersectionis(215

,265 ).

21 LetRbe(x,y), PR=QR

(x+3)2+(y–1)2

= (x+1)2+(y–7)2 x2+6x+9+y2–2y+1 = x2+2x+1+y2–14y+49 6x–2y+10 = 2x–14y+50 4x+12y = 40 x+3y = 10… 1 y–2x = –6… 2 1 2: 6y+2x = 20… 3 2 + 3 : 7y = 14 y = 2 Substitutey=2into 1 : x+3(2) = 10 x+6 = 10 x = 4 ∴R(4,2)

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�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

22 mPQ

=–1

10–3–

= –1

10– = –3+ + = 13… 1

MidpointofPQ=(+32

, +102 )

+102

= +32

+5

–+72

= 5

–+7 = 10 – = 3… 2 1 + 2 : 2 = 16 = 8 Substitute=8into 1 : +8 = 13 = 5 ∴=5,=8

23 (x–2)2+(y–3)2 = (x–1)2+y2 x2–4x+4+y2–6y+9 =x2–2x+1+y2

13–4x–6y = 1–2x 2x+6y–12 = 0 x+3y–6 = 0

24 (a) mPQ

=3andR(–2,3) TheequationofthelinePQ: y–3 = 3(x+2) y–3 = 3x+6 y = 3x+9 (b) y=3x+9 Atx-axis, y = 0 0 = 3x+9 –3x = 9 x = –3 ∴P(–3,0) Aty-axis, x = 0 y = 3(0)+9 = 9 ∴Q(0,9)

(c)m(0)+n(–3)

m+n = –2

–3n = –2m–2n 2m = n

mn

=12

∴PR:RQ=1:2

25 (a) LetCbe(0,y), m

BC = –5

y–10–1

= –5

y–1 = 5 y = 6 ∴C(0,6)

(b) mBD

=7–1–8–1

= –69

= –23

mAC

=32

TheequationofAC:

y–6=32

(x–0)

2y–12= 3x 2y–3x = 12 (c) 2y–3x=12 Atx-axis,y=0 2(0)–3x = 12 x = –4 ∴A(–4,0)

M = (–4+02

,0+6

2 ) = (–2,3)

(d) Area =12

–4 1 0 –8 –4 | 0 1 6 7 0 |

=12

(–4+6+48+28)

= 39unit2

26 (a) mAB

=4–(–4)–2–(–6)

=84

= 2

mBC

= –12

TheequationofthelineBC:

y–4 = –12

(x+2)

2y–8 = –x–2 2y+x = 6 (b) 2y+x=6 Aty-axis, x=0 2y = 6 y = 3 ∴D(0,3) TheequationofthelineAB: y+4 = 2(x+6) y+4 = 2x+12 y = 2x+8 Atx-axis,y=0 0 = 2x+8 2x = –8 x = –4 ∴E(–4,0)

mAC

=mED

=0–3–4–0

=34

TheequationofAC:

y+4 =34

(x+6)

4y+16 = 3x+18 4y–3x = 2 2y+x = 6… 1 4y–3x = 2… 2 1 3: 6y+3x =18… 3 2 + 3 : 10y =20 y =2 Substitutey=2into 1 : 2(2)+x = 6 x = 2 ∴C(2,2)

(c) Area =12

–6 2 0 –4 –6 |–4 2 3 0 –4|

=12

(–12+6+16+8+12)

=12

(30)

= 15unit2

27 (a) mAB

=12

andA(–3,–1)

TheequationofAB:

y+1 =12

(x+3)

2y+2 = x+3 2y–x = 1 m

AD=–2andA(–3,–1)

TheequationofAD: y+1 = –2(x+3) y+1 = –2x–6 y+2x = –7 (b) 2y–x= 6… 1 y+2x= –7… 2 1 2: 4y–2x =12… 3 2 + 3 : 5y = 5 y = 1 Substitutey=1into 1 : 2(1)–x = 6 x = –4 ∴D(–4,1) AreaofACD = 7.5

12

–3 x –4 –3 |–1 y 1 –1| = 7.5

12

(–3y+x+4+x+4y+3) = 7.5

2x+y+7 = 15 2x+y=8… 1 2y–x=6… 2 2 2: –2x+4y = 12… 3 1 + 3 : 5y = 20 y = 4 Substitutey=4into 1 : 2x+4 = 8 2x = 4 x = 2 ∴C(2,4) LetBbe(x,y),

(–4+x2

,1+y

2 )=(–3+22

,–1+4

2 )

–4+x2

= –12

and1+y

2 =

32

–4+x = –1 1+y =3 x = 3 y =2 ∴B(3,2)

28 (a) M = (–3+52

,–3+(–1)

2 ) = (1,–2)

mCD

=–1–(–3)5–(–3)

=28

=14

mAM

= –4

Page 48: Analysis Spm Additional Mathematics

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

TheequationofperpendicularbisectorofAM:

y+2 = –4(x–1) y+2 = –4x+4 y+4x = 2 Aty-axis,x=0 y+4(0) = 2 y = 2 ∴A(0,2)

(b) mAB

=14

andA(0,2)

TheequationofAB:

y–2 =14

(x–0)

4y–8 = x 4y–x = 8 m

BC=–4andC(5,–1)

TheequationofBC: y+1 = –4(x–5) y+1 = –4x+20 y+4x = 19 (c) 4y–x=8… 1 y+4x=19… 2 1 4: 16y–4x = 32… 3 2 + 3 : 17y = 51 y = 3 Substitutey=3into 1 : 4(3)–x = 8 x = 12–8 = 4 ∴B(4,3) (d) Area

=12

0 –3 5 4 0 | 2 –3 –1 3 2 |

=12

(3+15+8+6+15+4)

= 2512

unit2

29 (a) mAB

=5–3

–1–(–5)

=24

=12

mDE

=–2 MidpointofAB,

D = (–5+(–1)2

,3+5

2 ) = (–3,4) TheequationofDE: y–4 = –2(x+3) y–4 = –2x–6 y+2x = –2

mAC

=–3–3

3–(–5)

= –68

= –34

∴mBF

=43

TheequationofBF:

y–5 =43

(x+1)

3y–15 = 4x+4 3y–4x = 19

(b) y+2x =–2… 1 3y–4x =19… 2 1 2: 2y+4x =–4… 3 2 + 3 : 5y =15 y =3 Substitutey=3into 1 : 3+2x = –2 2x = –5

x = –52

∴G(– 52

,3)30 (a) m

AB = –3m

4–1–1–1

= –3m

–32

= –3m

m =12

(b) mAC

= 3m

= 3(12 )

=32

TheequationofAC:

y–1 =32

(x–1)

2y–2 = 3x–3 2y–3x = –1… 1 m

BC = m

=12

TheequationofBC:

y–4 =12

(x+1)

2y–8 = x+1 2y–x = 9… 2 2 – 1 : 2x = 10 x = 5 Substitutex=5into 1 : 2y–3(5) = –1 2y–15 = –1 2y = 14 y = 7 ∴C(5,7)

(c) AB = (4–1)2+(–1–1)2

= 13

AC = (5–1)2+(7–1)2

= 52

= 413

= 2 13 ∴AC=2AB

31 (a) mPQ

=7–34–6

= –42

= –2 (b) m

AB=–2andR(3,1)

TheequationofAB: y–1 = –2(x–3) y–1 = –2x+6 y+2x = 7 (c) m

AC = m

RP

=3–16–3

=23

Thegradientoftheperpendicularbisectoris–

32

.

TheequationoftheperpendicularbisectorofAC:

y–7 = –32

(x–4)

2y–14 = –3x+12 2y+3x = 26

32 (a) y–x= 2… 1 y+3x = 10… 2 2 – 1 : 4x = 8 x = 2 Substitutex=2into 1 : y–2 = 2 y = 4 ∴A(2,4) LetCbe(x,y),

( 2+x2

,4+y

2 )=(312

,712 )

2+x

2=

72

and4+y

2=

152

x = 5 y = 11 ∴C(5,11) (b) m

BC=1andC(5,11)

TheequationofBC: y–11= x–5 y–x = 6

33 (a) LetDbe(x,y),

x+9

2= –1 and

y+112

=7

x+9= –2 y+11= 14 x = –11 y = 3 ∴D(–11,3)

(b) mAC

=12–7

–3–(–1)

= –52

TheequationofAC:

y–7= –52

(x+1)

2y–14= –5x–5 2y+5x = 9 (c) LetCbe(x,y),

–3+x

2 = –1

–3+x = –2 x = 1 and

12+y

2 = 7

12+y = 14 y = 2 ∴C(1,2)

AC = (1+3)2+(2–12)2

= 116 = 10.77units

(d) Area =12

–3 –11 1 9 –3 |12 3 2 11 12|

Page 49: Analysis Spm Additional Mathematics

�0© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

=12

(–9–22+11+108

+132–3–18+33)

=12

(232)

= 116unit2

34 (a) mAB

=8–22–4

= –62

= –3

mBC

=13

andB(2,8)

TheequationofBC:

y–8 =13

(x–2)

3y–24 = x–2 3y–x = 22 (b) 3y–x=22… 1 y–x=–2… 2 1 – 2 : 2y = 24 y = 12 Substitutey=12into 1 : 3(12)–x = 22 36–x = 22 x = 14 ∴C(14,12) MidpointofAC

= ( 4+142

,2+12

2 ) = (9,7) LetDbe(x,y),

2+x

2 = 9 and

8+y2

= 7

2+x = 18 8+y = 14 x = 16 y = 6 ∴D(16,6) (c) TheareaofABCD

=12

4 16 14 2 4 | 2 6 12 8 2|

=12

(24+192+112+4–32–

84–24–32)

=12

(160)

= 80unit2

35 (a) 3y–5x=–34…1 4y–x=0… 2 2 5: 20y–5x = 0… 3 3 – 1 : 17y = 34 y = 2 Substitutey=2into 1 : 3(2)–5x = –34 6–5x = –34 5x = 40 x = 8 ∴C(8,2) LetBbethepoint(x,y),

( x2

,y2 )=(3+8

2,

5+22 )

x2

=112

andy2

=72

x = 11 y = 7 ∴B(11,7) m

AD=–4andA(3,5)

TheequationofAD: y–5=–4(x–3) y–5=–4x+12 y+4x=17… 1 4y–x=0… 2 2 4:16y–4x = 0… 3 1 + 3 : 17y = 17 y = 1 Substitutey=1into 1 : 1+4x = 17 4x = 16 x = 4 ∴D(4,1) (b) TheareaofOABC

=12

0 8 11 3 0 | 0 2 7 5 0 |

=12

(56+55–22–21)

=12

(68)

= 34unit2

36 (a) LetAbe(x,y),

x2+y2 = 20 x2+y2=20… 1 y=2x… 2 Substitute 2 into 1 : x2+4x2 = 20 5x2 = 20 x2 = 4 x = ±2 Substitutex=2into 2 : y = 2(2) = 4 ∴A(2,4)

mAB

=–12

andA(2,4)

TheequationofAB:

y–4 = –12

(x–2)

2y–8 = –x+2 2y+x = 10 Aty-axis, x = 0 2y = 10 y = 5 ∴B(0,5) m

BC=2andB(0,5)

TheequationofBC: y–5 = 2(x–0) y–2x = 5… 1 y+3x = 0… 2 2 – 1 : 5x = –5 x = –1 Substitutex=–1into 1 : y–2(–1) = 5 y+2 = 5 y = 3 ∴C(–1,3)

(b) Area =12

0 2 0 –1 0 | 0 4 5 3 0|

=12

(10+5)

= 7.5unit2

37 (a) AB = (6–2)2+(6–3)2

= 25

= 5units

BC = (6–6)2+(6–1)2

= 25 = 5units SinceAB=BC,ABCisan

isoscelestriangle.

(b) m = (2+62

,3+1

2 ) = (4,2)

mCD

=mBA

=6–36–2

=34

TheequationofCD:

y–1 =34

(x–6)

4y–4 = 3x–18 4y–3x = –14… 1

mAC

=1–36–2

= –12

SomBD

=2 TheequationofBD: y–6=2(x–6) y–6=2x–12 y–2x =–6… 2 2 4: 4y–8x = –24… 3 1 – 3 : 5x = 10 x = 2 Substitutex=2into 1 : 4y–3(2)= –14 4y = –8 y = –2 ∴D(2,–2) (c) DP=2MP

(x2–2)2+(y+2)2 =2 (x–4)2+(y–2)2 x2–4x+4+y2+4y+4= 4(x2–8x+16+ y2–4y+4)3x2+3y2–28x–20y+72=0

38 (a) 3x–y=–9… 1 2x+y=–1… 2 1 + 2 : 5x = –10 x = –2 Substitutex=–2into 1 : 3(–2)–y = –9 –6–y = –9 y = 3 ∴P(–2,3)

mPO

=–32

andO(0,0)

TheequationofPO:

y=–32

x

(b) mPC

=–13

andP(–2,3)

TheequationofPC:

y–3 = –13

(x+2)

3y–9 = –x–2 3y+x = 7

Page 50: Analysis Spm Additional Mathematics

1

Statistics

1 (a) Mean = 34812

= 29 Mode = 25

Median =

25 + 31

2 = 28

(b) Mean =

4215

= 2.8 Mode = 3 Median = 3

(c) Mean =

11011

= 10 Mode = 12 Median = 10

(d) Mean =

1418

= 17.625 Mode = 22

Median =

17 + 18

2 = 17.5

2 (a) Mean =

3120

= 1.55 Mode = 0

Median =

1 + 2

2 = 1.5

(b) Mean =

5025

= 2 Mode = 1 Median = 2

(c) Mean =

7828

= 2.786 Mode = 1

Median =

2 + 3

2 = 2.5

3 (a) x– = 8

55 + x

8 = 8

x = 64 – 55 = 9 (b) x– = 6

36 + y

7 = 6

36 + y = 42 y = 42 – 36 = 6

4 x– = 6

800 + 8x140 + x

= 6

800 + 8x = 840 + 6x 2x = 40 x = 20

5 (a) Modal class = 15 – 19

Mean =

68540

= 17.125

Median = 14.5 +

[20 – 1412 ]5

= 17 (b) Modal class = 161 – 165

Mean =

16 024

98 = 163.51

Median = 160.5 +

[49 – 2540 ]5

= 163.5 (c) Modal class = 50 – 59

Mean =

266550

= 53.3

Median = 49.5 + [25 – 19

16 ]10

= 53.25 (d) Modal class = 53 – 55

Mean =

164130

= 54.7

Median =

52.5 +

[15 – 98 ]3

= 54.75

6 x– = 68.25

1077 + 72x

x + 16 = 68.25

1077 + 72x = 68.25x + 1092 3.75x = 15 x = 4

7 a + b + 18 = 50 a + b = 32 b = 32 – a … 1 Median = 58

55.5 +

[25 – (b + 8)a ]10 = 58 … 2

Substitute 1 into 2 :

25 – (32 – a + 8)a = 0.25

a – 15 = 0.25a 0.75a = 15 a = 20 Substitute a = 20 into 1 : b = 32 – 20 = 12

8 (a)Number of apples

Mass (g)54.5 60.5 66.5 72.563.8

14

12

10

8

6

4

2

0

Mode = 63.8 (b)

Number of people

14

12

10

8

6

4

2

0 42.5 48.5 54.5 60.550

Age(years)

Mode = 50 (c)Number of pupils

16

14

12

10

8

6

4

2

0

14.25

Arrival time(minutes)6.5 11.5 16.5 21.5

Mode = 14.25

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 51: Analysis Spm Additional Mathematics

2© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(d)Number of workers

12

10

8

6

4

2

038.5 43.5 48.5 53.5

45.75

Amount spent (RM)

Mode = 45.75 9 (a) Cumulative frequency

40

35

30

25

20

15

10

5

024.5 44.529.5 49.534.5 54.539.5 59.5

43.75

Length (mm)

m

Median = 43.75 (b)

Cumulative frequency

m

100

90

80

70

60

50

40

30

20

10

0.5 40.510.5 50.520.5 60.530.5 70.5 80.533.5

Mark

Median = 33.5

10 (a) Mean = 5510

= 5.5 Mode = 6 Median = 6 (b) (i) New mean = 5.5 + 1 = 6.5 New mode = 6 + 1 = 7 New median = 6 + 1 = 7 (ii) New mean = 5.5 – 2 = 3.5 New mode = 6 – 2 = 4 New median = 6 – 2 = 4

(iii) New mean = 5.5(3) = 16.5 New mode = 6(3) = 18 New median = 6(3) = 18

(iv) New mean = 5.5

12

= 11

New mode = 6

12

= 12

New median = 6

12

= 12 11 (a) Range = 32 – 15 = 17 Interquartile range = 29 – 16 = 13 (b) Range = 9 – 2 = 7 Interquartile range = 7 – 3.5 = 3.5 (c) Range = 16 – 7 = 9 Interquartile range = 15 – 9 = 6 (d) Range = 8 – 2 = 6 Interquartile range = 7 – 3 = 4

12 (a) Range = 5 – 1 = 4 Interquartile range = 4 – 1 = 3 (b) Range = 6 – 1 = 5 Interquartile range = 5 – 2 = 3 (c) Range = 6 – 0 = 6 Interquartile range = 4.5 – 2 = 2.5 (d) Range = 8 – 4 = 4 Interquartile range = 8 – 4 = 4

13 (a) Range = 28 – 3 = 25

Q1 = 0.5 +

[10 – 010 ]5

= 5.5

Q3 = 10.5 +

[30 – 246 ]5

= 15.5 Interquartile range = 15.5 – 5.5 = 10 (b) Range = 75.5 – 5.5 = 70

Q1 = 30.5 +

[40 – 1968 ]10

= 30.5 + 3.088 = 33.5880

Page 52: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Q3 = 40.5 +

[120 – 8751 ]10

= 40.5 + 6.471 = 46.971 Interquartile range = 46.971 – 33.588 = 13.383 (c) Range = 70.5 – 42.5 = 28

Q1 = 48.5 +

[25 – 817 ]4

= 52.5

Q3 = 60.5 +

[75 – 6916 ]4

= 62 Interquartile range = 62 – 52.5 = 9.5 (d) Range = 65.5 – 10.5 = 55

Q1 = 40.5 +

[14 – 1315 ]20

= 41.83

Q3 = 50.5 +

[42 – 2820 ]10

= 57.5 Interquartile range = 57.5 – 41.83 = 15.67

14 (a) x– = 8010

= 8

σ2 =

66810

– (8)2

= 2.8 σ = 2.8 = 1.673

(b) x– =

255

= 5

σ2 =

1355

– (5)2

= 2 σ = 2 = 1.414

(c) x– =

7212

= 6

σ2 =

49412

– (6)2

= 5.167 σ = 5.167 = 2.273

(d) x– =

488

= 6

σ2 =

3008

– (6)2

= 1.5 σ = 1.5 = 1.225

15 (a) x– = 14020

= 7

σ2 =

42010

– (6)2

= 6 σ = 6 = 2.449 (a) x– = 6 + 2 = 8 σ2 = 6 σ = 2.449 (b) x– = 6(3) = 18 σ2 = 32 6 = 54 σ = 3 2.449 = 7.347

(c) x– =

62

= 3

σ2 =

622

= 1.5

σ =

2.449

2 = 1.2245

18 Range = 35 – 23 = 12 Interquartile range = 31 – 25.5 = 5.5 (a) New range = 12 New interquartile range = 5.5 (b) New range = 12 New interquartile range = 5.5

(c) New range =

12

12

= 6 New interquartile range

=

12

5.5

= 2.75

(d) New range =

124

= 3

New interquartile range =

5.54

= 1.375 19 (a) x– = 4

Σx5

= 4

Σx = 20

20 + x6

= 4

20 + x = 24 x = 4 (b) σ2 = 2

Σx2

5 – (4)2 = 2

Σx2 = 90 The new variance,

=

90 + 42

6 – (4)2

= 1.667

σ2 =

101620

– (7)2

= 1.8 σ = 1.8 = 1.342

(b) x– =

26440

= 6.6

σ2 =

181440

– (6.6)2

= 1.79 σ = 1.79 = 1.338

(c) x– =

24040

= 6

σ2 =

1512

40 – (6)2

= 1.8 σ = 1.8 = 1.342

(d) x– =

3015

= 2

σ2 =

8015

– (2)2

= 1.333 σ = 1.333 = 1.155

16 (a) x– =

274040

= 68.5

σ2 =

188 980

40 – (68.5)2

= 32.25 σ = 32.25 = 5.679

(b) x– =

1560

30 = 52

σ2 =

81 750

30 – (52)2

= 21 σ = 21 = 4.583

(c) x– =

43020

= 21.5

σ2 =

10 340

20 – (21.5)2

= 54.75 σ = 54.75 = 7.399

(d) x– =

85550

= 17.1

σ2 =

17 155

50 – (17.1)2

= 50.69 σ = 50.69 = 7.12

17 x– =

6010

= 6

Page 53: Analysis Spm Additional Mathematics

4© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

20 (a)

Cumulative frequency

Q1

Q3

200

180

160

140

120

100

80

60

40

20

0.5 12.53.5 15.5 24.56.5 18.5 27.59.5 21.5 30.518.2 24.94

Life span(weeks)

Interquartile range = 24.94 – 18.2 = 6.74 (b)

Cumulative frequency

Q1

Q3

100

90

80

70

60

50

40

30

20

10

19.5 39.524.5 44.5 59.529.5 49.534.5 54.531 42.5

Age (years)

Interquartile range = 42.5 – 31 = 11.5

(b) m =

3 + 4

2 = 3.5

4 m = 14.5 + (15 – 10

10 )5

= 14.5 + 2.5 = 17

5 σ = 3 12

49 + y2

4 – (11 + y

4 )2

= 3

12

49 + y2

4 – (121 + 22y + y2

16 ) = 12.25

196 + 4y2 – 121 – 22y – y2 = 196 3y2 – 22y – 121 = 0 (3y + 11)(y – 11) = 0

y = –

113

or y = 11

\ y = 11

6 (a) x– = 8

7x + 44

= 8

7x + 4 = 32 7x = 28 x = 4 (b) When x = 4 the set of data is 2, 7,

7 and 16.

σ2 =

3584

– (8)2

= 25.5

7 (a) x– = 9

2m + 184

= 9

2m + 18 = 36 2m = 18 m = 9 σ = 5

164 + (9 – n)2 + (9 + n)2

4 – (9)2

= 5

164 + 81 – 18n + n2 + 81 + 18n + n2

4 – 81

= 25

2n2 + 326 = 424 2n2 = 98 n2 = 49 n = 7 \ m = 9, n = 7

8 σ2 = Σ(x – x–)2

N

=

15020

= 7.5

9 (a) Median mark = 34.5 (b) Q1 = 27.5 Q3 = 42.5 \ Interquartile range = 42.5 – 27.5 = 15 (c) 400 – 275 = 125 students

1

28 + x4

= 10

28 + x = 40 x = 12 \ The fourth number is 12.

2 (a) RM250

(b)

1270 + x

6 = 240

1270 + x = 1440 x = 170 \ Mukhriz’s weekly wage is

RM170.

3 (a)

Number of goals scored 0 1 2 3 4 5

Number of matches 2 1 3 4 7 3

0

0

Page 54: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

10 x– = 3.5

6 + 3k + 8 + 20 + 6k + 10

= 3.5

3k + 40 = 3.5k + 35 0.5k = 5 k = 10

11 (a) x– = 357

= 5

σ2 =

2877

– (5)2

= 16 (b) 16m2

12 Q1 = 39.5 + (7 – 3

5 )5

= 43.5

Q3 = 49.5 +

(21 – 186 )5

= 52 \ Interquartile range = 52 – 43.5 = 8.5

13 (a) x– = 1086

= 18

(b) σ2 =

21846

– (18)2

= 40

14 New mean = 4(2) – 3 = 5 New variance = 22(1.5) = 6

15 (a) x– = 20m

5 = 4m σ2 = 8

90m2

5 – (4m)2 = 8

90m2 – 80m2

5 = 8

10m2 = 40 m2 = 4 m = 2 \ m = 2

16 σ2 = 4n

505

– ( m )2 = 4n

10 – m = 4n m = 10 – 4n

17 (a) Student A : x– = 205

= 4

σ =

905

– (4)2

= 2 = 1.414

Student B : x– = 205

= 4

σ =

985

– (4)2

= 3.6 = 1.897

22 (a) Number of goals scored 0 1 2 3 4 5 6

Number of matches 2 3 4 5 2 1 3

(b) (i) m = 3 (ii) Interquartile range = 3

(iii) x– =

3 + 8 + 15 + 8 + 5 + 18

20

=

5720

= 2.85

σ =

22920

– (2.85)2

= 3.3275 = 1.824

23 (a) (i) m = 8 (ii) Interquartile range = 13 – 3.5 = 9.5

(iii)

x– =

729

= 8

σ2 =

7929

– (8)2

= 24

(b)

72 + x10

= 8

72 + x = 80 x = 8

24 (a) Mode = 2

(b) m =

2 + 32

= 2.5 (c) Range = 6 – 0 = 6 (d) x– =

5220

= 2.6

σ =

19220

– (2.6)2

= 2.84 = 1.685

(b) Student A

18 (a) m = 4

(b) x– = 7820

= 3.9

σ =

35620

– (3.9)2

= 2.59 = 1.609

19 (a) k = 9, l = 5 (b) Estimated mode = 12.5

20 (a) x = 5 (b)

x + 112

x

x + 11 2x –x –11 x 11 \ x = 10 (c) x– = 2

x + 12 + 6 + 12x + 11

= 2

x + 30 = 2x + 22 x = 8

21 (a)Number of cars

6

5

4

3

2

1

060.5 70.5 80.5 90.5 100.5 110.5

Speed (km h–1)92.5

(b) Estimated mode = 92.5

25 (a) Frequency

10

8

6

4

2

050.5 60.5 70.5 80.5 90.5 100.5

76.5110.5

Mass (g)

Mode = 76.5

(b) m = 70.5 +

(18 – 1110 )10

= 77.5

Page 55: Analysis Spm Additional Mathematics

6© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(c) x– =

281836

= 78.278

26 Number of books Number of students

1 – 3 5

4 – 6 7

7 – 9 8

10 – 12 10

13 – 15 6

16 – 18 4

(a) Q1 = 3.5 +

(10 – 57 )3

= 5.643

Q3 = 9.5 +

(30 – 2010 )3

= 12.5 \ Interquartile range = 12.5 – 5.643 = 6.857

(b) x– = 37140

= 9.275

σ =

424940

– (9.275)2

= 20.1994 = 4.494

27 (a) Frequency

12

10

8

6

4

2

00.5 5.5 10.5 15.5 20.5 25.5

9.530.5

Time(minutes)

Mode = 9.5

(b) x– =

32530

= 10.833

σ2 =

445530

– (10.833)2

= 31.146

(c) m = 5.5 +

(15 – 412 )5

= 10.083

28 (a) Range = 52.5 – 36.5 = 16 Modal class = 47 – 50

(b)Number of workers

14

12

10

8

6

4

2

034.5 38.5 42.5 46.5 50.5 54.5

48.7

Number ofhours

Mode = 48.7

(c) x– =

186040

= 46.5

σ =

87 418

40 – (46.5)2

= 23.2 = 4.817

29 (a) (i) Median mark = 29.5 (ii) Interquartile range = 37.5 – 21.5 = 16 (b) 40 marks

30 (a)Cumulative frequency

Q3

Q1

m

40

3532.5

30

25

20

15

10

5

25.5 30.5 35.5 40.5 45.5 50.5 55.5 60.536 46 4841.75

Length (mm)

(i) m = 41.75 (ii) Interquartile range = 46 – 36 = 10 (iii) The number of leaves which are longer than 48 mm = 40 – 33 = 7 leaves (b) (i) p = 9, q = 3

0

Page 56: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(ii)

Frequency

12

10

8

6

4

2

025.5 30.5 35.5 40.5 45.5 50.5 55.5 60.5

42.5Length (mm)

Mode = 42.5

31 (a) x– =

168080

= 21

(b) Score 5 10 15 20 25 30 35

Number of students 0 4 14 34 64 76 80

(c)

Number of students

m

8076

70

60

50

40

30

20

10

5.5 10.5 15.5 20.5 25.5 30.5 35.521.5

Score

(i) Median score = 21.5 (ii) The percentage of students whose scores are 31 and above

=

80 – 7680

100

= 5% 32 (a) (i) Σx = 3m + 112

(ii) x– = 50

3m + 112

5 = 50

3m + 112 = 250 3m = 138 m = 46

(b) σ2 =

12 652

5 – (50)2

= 30.4

New variance =

30.422

= 7.6

33 (a) (i) x– = 6

Σx5

= 6

Σx = 30

(ii) σ = 2

12

Σx2

5 – (6)2

= 2

12

Σx2

5 – 36 = 6.25

Σx2 = 211.25 (b) (i) New mean = (6 – 1)(2) = 5(2) = 10

(ii) New standard deviation = 2

12

(2)

= 5

34 (a) (i) x– = 8

Σx50

= 8

Σx = 400 (ii) σ = 3

Σx2

50 – (8)2

= 3

Σx2

50 – 64 = 9

Σx2 = 3650

(b) σ =

3650 + 82

51 – (8)2

= 8.824 = 2.97

35 (a) x– = 12010

= 12

σ2 =

165010

– (12)2

= 21

(b) (i)

120 – k

9 = 11

120 – k = 99 k = 21

(ii) σ2 =

1650 – 212

9 – (11)2

= 13.33

0

Page 57: Analysis Spm Additional Mathematics

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-31

1 (a) 18° =

18°180°

p

=

110

p radian

(b) 45° =

45°180°

p

=

14

p radian

(c) 80° =

80°180°

p

=

49

p radians

(d) 150° = 150°180°

p

=

56

p radians

(e) 225° = 225°180°

p

=

54

p radians

(f) 300° = 300°180°

p

=

53

p radians

2 (a) 20° =

20°180°

p

= 0.349 radian

(b) 42.5° = 42.5°180°

p

= 0.742 radian

(c) 75° =

75°180°

p

= 1.309 radians (d) 142° 25 =

142° 25

180° p

= 2.486 radians

(e) 250° = 250°180°

p

= 4.363 radians

(f) 320° 15 = 320° 15

180° p

= 5.589 radians

3 (a) 0.72 rad = 0.72

p 180°

= 41° 15 (b) 1.5 rad =

1.5p 180°

= 85° 57

(c) 2.425 rad = 2.425

p 180°

= 138° 57

(d) 56 p rad =

56

p

p 180°

= 150°

(e) 45 p rad =

45

p

p 180°

= 144°

(f) 74 p rad =

74

p

p 180°

= 315°

4 (a) s = 8(1.6) = 12.8 cm

(b) s = 20(290

180 p)

= 20(5.061) = 101.22 cm (c) s = 10(0.65) = 6.5 cm

(d) s = 50(3

4 p)

= 117.81 cm

5 (a) r =

120.8

= 15 cm

(b) r =

222.2

= 10 cm

(c) r =

91.003

= 8.97 cm

(d) r =

353.491

= 10.03 cm

6 (a) q = 154

= 3.75 radians

(b) q =

87

= 1.143 radians

(c) q = 1512

= 1.25 radians

(d) q =

918

= 0.5 radian

7 p = rq + 2r sin q2

(a) p = 9(1.047) + 2(9)(sin 30°) = 9.423 + 9 = 18.423 cm (b) p = 10(2.5) + 2(10)(sin 1.25 rad) = 25 + 18.98 = 43.98 cm

(c) p = 15(106° 16

180 p)

+ 2(15)

(sin 53° 8) = 27.821 + 24.001 = 51.822 cm

(d) p = 8(1.5) + 2(8)(sin 0.75 rad) = 12 + 10.906 = 22.906 cm

8 (a) ∠POQ =

710

= 0.7 rad

PR10

= sin 0.7 rad

PR = 10(sin 0.7 rad) = 6.442 cm

OR10

= cos 0.7 rad

OR = 10(cos 0.7 rad) = 7.648 cm QR = 10 – 7.648 = 2.352 cm \ Perimeter of the shaded region = 7 + 6.442 + 2.352 = 15.794 cm

(b) sPQ = 6(p

2) = 9.425 cm PQ = 62 + 62

= 72 = 8.485 cm \ Perimeter of the shaded region = 9.425 + 8.485 = 17.91 cm (c)

3 cm

P

SO q

1.25 cm

sin q = 1.25

3 q = 24° 37 \ Perimeter of the shaded region

= 3(49° 14

180° p)

+ 2.5

= 3(0.859) + 2.5 = 5.077 cm (d) sPQ = 8(0.6) = 4.8 cm

PR8

= tan 0.6 rad

PR = 8(tan 0.6 rad) = 5.473 cm OR = 5.4732 + 82

= 9.693 cm RQ = 9.693 – 8 = 1.693 cm \ Perimeter of the shaded region = 4.8 + 5.473 + 1.693 = 11.966 cm

Circular Measures

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 58: Analysis Spm Additional Mathematics

2© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

9

8 cm

R

C

O30°

8

OR = tan 30°

OR =

8

tan 30° = 13.856 cm BR = AP = 24 – 13.856 = 10.144 cm

sPQR = 8(240°

180° p)

= 8(4.189) = 33.512 cm

sAQB = 24(p

3) = 25.133 cm \ Perimeter of the shaded region = 2(10.144) + 33.512 + 25.133 = 78.933 cm

10

OBBD

=

21

2BD = 8 BD = 4 cm sAB = 8(0.5) = 4 cm sCD = 12(0.5) = 6 cm \ Perimeter of the shaded region = 2(4) + 4 + 6 = 18 cm

11

20AC

= tan 0.5 rad

AC =

20

tan 0.5 rad = 36.61 cm

20OC

= sin 0.5 rad

OC =

20

sin 0.5 rad = 41.717 cm BC = 41.717 – 20 = 21.717 cm

∠AOC = p –

p2

– 0.5

= 1.071 rad sAB = 20(1.071) = 21.42 cm \ Perimeter of the shaded region = 36.61 + 21.717 + 21.42 = 79.747 cm

12

4 cm

5 cm

9 cm

A

E

D C

B

13 cm

CD = 132 – 52

= 144 = 12 cm

cos ∠BAE =

513

∠BAE = 67° 23 = 1.176 rad ∠ABC = 2p – p – 1.176 = 1.966 rad \ Perimeter of the shaded region = 12 + 9(1.176) + 4(1.966) = 30.448 cm

13 (a) ∠AOB = 1020

=

12

rad

= 0.5 radian (b) sBC = 20(p – 0.5) = 52.832 cm

14 (a) A =

12

(5)2(2.4)

= 30 cm2

(b) A =

12

(8)2(p3)

= 33.51 cm2

(c) A =

12

(20)2(300180

p) = 1047.198 cm2

(d) A =

12

(10)2(109

p) = 174.533 cm2

15 (a)

12

r2(2) = 64

r2 = 64 r = 8 cm (b)

12

r2(4.4) = 55

r2 = 25 r = 5 cm

(c)

12

r2(2p3 )

= 12p

r2 = 36 r = 6 cm

(d)

12

r2(3.5) = 175

r2 = 100 r = 10 cm

16 (a)

12

(5)2(q) = 10

25q = 20

q =

2025

= 45

radian

(b)

12

(8)2(q) = 24p

32q = 24p

q =

34

p

= 2.356 radians

(c)

12

(10)2(q) = 30

50q = 30

q =

35

radian

(d)

12

(6)2(q) = 27

18q = 27 q = 1.5 radians

17 (a) q = 1210

= 1.2 rad

Area =

12

(102)(1.2) – 12

(102)

(sin 1.2 rad) = 60 – 46.602 = 13.398 cm2

(b) q =

512

p rad

Area =

12

(12)2( 512

p) – 12

(12)2

(sin

512

p rad)

= 94.248 – 69.547 = 24.701 cm2

(c)

Area =

12

(10)2( 25

p) – 12

(10)2

(sin 25

p rad) = 62.832 – 47.553 = 15.279 cm2

(d)

12 cm 13 cm

D

C

Oq

sin q =

1213

q = 67° 23 = 1.176 rad ∠AOC = 2 1.176 = 2.352 rad

Area =

12

(13)2(2.352) – 12

(13)2(sin 2.352 rad) = 198.744 – 60.001 = 138.743 cm2

18 Area of the shaded segment

=

12

r2(1.8) – 12

r2(sin 1.8 rad)

= 0.9 r2 – 0.4869 r2

= 0.4131 r2

Area of circle = pr2

The ratio of the shaded segment to

the circle

=

0.4131r2

pr2

=

0.41313.1416

=

4131

31 416 = 81 : 616

19 ∠AOB = 26

180°

= 60°

= 13

p rad

Page 59: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

sAB = 4p

r(13

p) = 4p

r = 12 cm

∠BOC =

36

180°

= 90°

=

12

p rad

∠COD =

16

180°

= 30°

=

16

p rad

Area of shaded segment BC

=

12

(12)2(12

p) – 12

(12)2 sin 90°

= 36p – 72 Area of shaded segment CD

=

12

(12)2(16

p) – 12

(12)2 sin 30°

= 12p – 36 \ Area of the shaded region = 36p – 72 + (12p – 36) = (48p – 108) cm2

20 (a) Area of DOPR = 12

(5)(12)

= 30 cm2

Area of sector OPQ

=

12

(5)2(1.176)

= 14.7 cm2

Area of the shaded region = 30 – 14.7 = 15.3 cm2

(b) Area of sector OPQ

=

12

(14.142)2(14

p) = 78.538 cm2

Area of DORQ =

12

(10)(10)

= 50 cm2

Area of the shaded region = 78.538 – 50 = 28.538 cm2

21 AP10

= tan 0.6 rad

AP = 6.841 cm

Area of OAPB = 2[1

2 (10)(6.841)]

= 68.41 cm2

Area of sector OAB =

12

(10)2(1.2)

= 60 cm2

\ Area of the shaded region = 68.41 – 60 = 8.41 cm2

22

Area of semicircle = 12

(6.5)2(p)

= 66.366 cm2

Area of DPQR = 12

(5)(12)

= 30 cm2

\ Area of the shaded region = 66.366 – 30 = 36.366 cm2

23 (a) sPQ = 12 8q = 12

q = 32

radians

(b) Area of minor sector TOR

=

12

(4)2(32)

= 12 cm2

Area of circle = p(4)2

= 16p Area of the shaded region = (16p – 12) cm2

1 (a) tan ∠AOX =

86

=

43

∠AOX = 53° 8 = 0.927 rad

(b) Area of DAOX =

12

(6)(8)

= 24 cm2

Area of sector AOB

=

12

(6)2(0.927)

= 16.686 cm2

\ Area of the shaded region = 24 – 16.686 = 7.314 cm2

2

10 cm6 cm

qO

A

B

sin q =

610

=

35

q = 36° 52 ∠AOB = 73° 44 = 1.287 rad Area of the shaded segment

=

12

(10)2(1.287) – 12

(10)2 sin 73° 44

= 64.35 – 47.998 = 16.352 cm2

\ Area of the shaded region = p(10)2 – 16.352 = 314.159 – 16.352 = 297.81 cm2

3 (a) sACB = 15(53

p) = 78.54 cm

(b) Area of the shaded region

=

12

(15)2(p3 )

– 12

(15)2 sin p3

= 117.81 – 97.428 = 20.38 cm2

4 (a) sAB = 2p 12q = 2p

q =

p6

radian

\ ∠AOB =

p6

radian

(b) Area of the shaded region

=

12

(12)2(p6 )

– 12

(12)2 sin p6

= 37.699 – 36 = 1.699 cm2

5 Area of DOAT = 60

12

(10)(AT) = 60

AT = 12 cm

tan ∠AOT =

1210

= 1.2 ∠AOT = 50° 12 = 0.876 rad \ Area of the sector AOB

=

12

(10)2(0.876)

= 43.8 cm2

6 (a) sAB = 10(0.8) = 8 cm (b) Area of the shaded region

=

12

(10)2(0.8) – 12

(5)(10)

sin 0.8 rad = 40 – 17.934 = 22.07 cm 2

7 (a) sAB = 6(0.5) = 3 cm

(b)

BC6

= tan 0.5 rad

BC = 6 tan 0.5 rad = 3.278 cm \ Area of the shaded region

=

12

(6)(3.278) – 12

(6)2(0.5)

= 9.834 – 9 = 0.834 cm2

8 (a)

12 cm

9 cm

Q

O

T

tan ∠OTQ =

912

=

34

∠OTQ = 36° 52 ∠PTQ = 2(36° 52) = 73° 44 = 1.287 radians

Page 60: Analysis Spm Additional Mathematics

4© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) Area of the shaded region

= 2[1

2 (9)(12)]

– 12

(9)2(1.855)

= 108 – 75.128 = 32.872 cm2

9 2r + rq = 15 r(2 + q) = 15

r =

15

2 + q … 1

12

r2q = 9 … 2

Substitute 1 into 2 :

12 ( 15

2 + q)2q = 9

225q = 18(4 + 4q + q 2) 225q = 72 + 72q + 18q 2

18q 2 – 153q + 72 = 0 2q 2 – 17q + 8 = 0 (2q – 1)(q – 8) = 0

q =

12

or q = 8 (rejected)

Substitute q =

12

into 1 :

r =

15

2 + 12

= 6

r = 6, q =

12

radian

10 sAB = 10(0.6) = 6 cm

BP10

= tan 0.3 rad

BP = 10 tan 0.3 rad = 3.093 cm \ Perimeter of the shaded region = 2(3.093) + 6 = 12.186 cm

11 (a) sACB = 24 6q = 24 q = 4 radians (b) Area of minor sector AOB

=

12

(6)2(2p – 4)

= 41.097 cm2

12 (a) ∠CAO = 45°

=

p4

rad

sCD = 8.485(p

4) = 6.664 cm (b) Area of the shaded region

=

12

(8.485)2(p4)

– 12

(6)(6)

= 28.27 – 18 = 10.27 cm2

13 Let r be the radians of sector COD

12

r2(0.5) – 12

(6)2(0.5) = 16

12

r2(0.5) = 25

r2 = 100 r = 10 0

sAB = 6(0.5) = 3 cm sCD = 10(0.5) = 5 cm \ Perimeter of the shaded region = 2(4) + 3 + 5 = 16 cm

14 (a)

5 cm4 cm

B

OT

sin ∠BOT =

45

∠BOT = 53° 8 \ q = 53° 8 = 0.927 radian (b) Area of the minor segment BCD

=

12

(5)2(1.854) – 12

(5)2

sin 1.854 rad = 23.175 – 12.002 = 11.173 cm2

15 (a) sAB = 6(0.9) = 5.4 cm (b) Area of the shaded region

=

12

(6)2(0.9) – 12

(3)2sin 0.9 rad

= 16.2 – 3.525 = 12.675 cm2

16 (a) 2q + 6q + 8 = 12 8q = 4

q = 12

radian

(b) Area of the shaded region

=

12

(6)2(12) –

12

(2)2(12)

= 9 – 1 = 8 cm2

17 (a)

13 cm 12 cm

q

sin q =

1213

q = 67° 23 \ ∠AOC = 2(67° 23) = 134° 46 = 2.352 radians (b) sABC = 13(2.352) = 30.576 cm

(c) Area of the shaded region

=

p(13)2 –

12

(13)2 sin 134° 46

= 530.93 – 59.99 = 470.94 cm2

18 (a) Let r be the length of OQ

13

r + 13

(r + 2) + 4 = 10

23

r + 23

= 6

2r + 2 = 18 2r = 16 r = 8 \ OQ = 8 cm (b) Area of the shaded region

=

12

(10)2(13) –

12

(8)2(13)

= 16.67 – 10.67 = 6 cm2

19 (a) q = 45°

=

p4

radian

(b) sCD = 14.142 (1

4 p)

= 11.107 cm sACB = 10 p = 31.416 cm

\ Perimeter of the shaded region = 31.416 + 11.107 + 14.142

+ 5.858 = 62.52 cm

(c) Area of semicircle =

12

p(10)2

= 157.08 cm2

Area of sector CBD

=

12

(14.142)2(14

p) = 78.54 cm2

\ Area of the shaded region = 157.08 – 78.54 = 78.54 cm2

20 (a)

12

(3k)2(34)

– 12

(2k)2(34)

= 30

278

k2 – 128

k2 = 30

158

k2 = 30

k2 = 16 k = 4 0

(b) sCD = 12 (3

4) = 9 cm

sAB = 8

(34)

= 6 cm \ The difference in length = 9 – 6 = 3 cm

21 (a)

12

(8)2q = p (2)2

32q = 4p

q =

18

p rad \ ∠AOB = p

8 radian

Page 61: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) sAB = 2 (p

8) = 0.785 cm

sCD =

8 (p

8) = 3.412 cm \ Perimeter of the shaded

region ABDC = 0.785 + 3.142 + 12 = 15.927 cm

22 (a)

sABsBC =

45

10q

10 (p2

– q)

= 45

5q = 4(p

2 – q)

9q = 2 p

q =

29

p rad

(b) sAB = 10(29

p) = 6.981 cm

10 cmB

C

O

T

CT10

= sin 0.436 rad

CT = 10 sin 0.436 rad = 4.223 cm BC = 8.446 \ Perimeter of the shaded region = 20 + 6.981 + 8.446 = 35.427 cm

23 (a) sBC = 10(1.2) = 12 cm (b) (i) Area of sector COB

=

12

(10)2(1.2)

= 60 cm2

(ii) ∠CAO =

1.22

= 0.6 radC

T

O

10 cm

0.9708 rad

CT10

= sin 0.9708 rad

CT = 10 sin 0.9708 = 8.2534 cm CA = 2(8.2534) = 16.507 cm

Area of sector CAD

=

12

(16.507)2(0.6)

= 81.744 cm2

(c) Area of DAOC

=

12

(10)2(sin 1.9416)

= 46.602 cm2

Area of COD = 81.744 – 46.602 = 35.142 \ Area of the shaded region = 60 – 35.142 = 24.86 cm2

24 (a) Area of sector LOM

=

12

(8)2(1.25)

= 40 cm2

(b)

LN8

= tan 1.25 rad

LN = 8 tan 1.25 = 24.077 cm

Area of DLON =

12

(8)(24.077)

= 96.308 cm2

\ Area of the shaded region = 96.308 – 40 = 56.308 cm2

25 (a) ∠PRQ = p – 2(0.9) = 1.3416 rad Area of the shaded region

=

12

(10)(10) sin 1.3416 rad – 12

(10)2(0.9) = 48.692 – 45 = 3.69 cm2

(b) sRS = 10(0.9) = 9 cm

P T

R

10 cm

0.9 rad

PT10

= cos 0.9 rad

PT = 10 cos 0.9 = 6.216 cm PQ = 2(6.216) = 12.432 cm Perimeter of the shaded region = 9 + 10 + (12.432 – 10) = 21.432 cm

26 (a) 12q = 18

q = 1812

=

32

radians

(b) S

U

6 cm32

rad

O

SU6

= sin 32

rad

SU = 6 sin

32

= 5.985 cm \ RT = 5.985 cm (c)

OT

R

5.98512

sin ∠ROT =

5.985

12 ∠ROT = 29° 55 = 0.522 radian (d) Area of sector OQR

=

12

(12)2(0.522)

= 37.584 cm2

Area of DORT

=

12

(10.401)(5.985)

= 31.125 cm2

\ Area of the shaded region = 37.584 – 31.125 = 6.459 cm2

27 (a)

cos ∠POQ =

610

=

35

∠POQ = 53° 8 = 0.927 radian (b) sAB = 3(0.927) = 2.781 cm PQ = 102 – 62

= 8 cm \ Perimeter of the shaded region = 7 + 8 + 3 + 2.781 = 20.781 cm (c) Area of the shaded region

=

12

(6)(8) – 12

(3)2(0.927)

= 24 – 4.172 = 19.828 cm2

28 (a)

15 cm

1.1 rad

A

F O

AF15

= sin 1.1 rad

AF = 15 sin 1.1 rad = 13.368 cm

Page 62: Analysis Spm Additional Mathematics

6© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

AC = 2(13.368) = 26.736 cm (b)

F

A

D

20 cm13.368

sin ∠ADF =

13.368

20 ∠ADF = 41° 57 ∠ADC = 2(41° 57) = 83° 54 = 1.464 radians (c) sABC = 15(2.2) = 33 cm sAEC = 20(1.464) = 29.28 cm \ Perimeter of the shaded region = 33 + 29.28 = 62.28 cm Area of segment AEC

=

12

(20)2(1.464) – 12

(20)2

sin 1.464 rad = 292.80 – 198.86 = 93.94 cm2

Area of segment ABC

=

12

(15)2(2.2) – 12

(15)2 sin 2.2

= 247.50 – 90.956 = 156.544 cm2

Area of the shaded region = 156.544 – 93.94 = 62.604 cm2

29 (a)

12

(20)2(0.8)

12

(20 + r)2(0.8) – 12

(20)2(0.8)

= 45

5(160) = 4(0.4r2 + 16r) 1.6r2 + 64r – 800 = 0 r2 + 40r – 500 = 0 (r + 50)(r – 10) = 0 r = 10 0 (b) sPQ = 20(0.8) = 16 cm sRS = 30(0.8) = 24 \ Perimeter of region A = 16 + 24 + 20 = 60 cm

30 (a)

OC16

= cos 1.5 rad

OC = 16(cos 1.5 rad) = 1.132 cm \ BC = 16 – 1.132 = 14.87 cm (b) AC = 162 – 1.1322

= 15.96 cm sAB = 16(1.5) = 24 cm \ Perimeter of the shaded region = 24 + 15.96 + 14.87 = 54.83 cm Area of sector OAB

=

12

(16)2(1.5)

= 192 cm Area of DAOC

=

12

(1.132)(15.96)

= 9.033 cm2

\ Area of the shaded region = 192 – 9.033 = 182.97 cm2

Page 63: Analysis Spm Additional Mathematics

Differentiation

1 (a) limx→–1(x2–1

x+1 )

= lim

x→–1

(x+1)(x–1)(x+1)

= –1–1 = –2

(b) lim

x→3(x2–9

x–3 )

= limx→3

(x+3)(x–3)(x–3)

= limx→3

(x+3)

= 3+3 = 6

(c) lim

x→4( x2–x–12x2–11x+28)

= lim

x→4(x+3)(x–4)(x–4)(x–7)

= lim

x→4(x+3)(x–7)

=

4+34–7

= – 7

3

(d) lim

x→2(x2–2xx2–4 )

= lim

x→2

x(x–2)(x+2)(x–2)

= lim

x→2( xx+2)

=

24

=

12

2 (a) limx→∞(x+1

x–1)

= limx→∞

( x

x +

1x

xx

–1x)

= limx→∞

( 1+

1x

1–1x

)

=

1+01–0

= 1

(b) lim

x→∞(2x+1x )

= limx→∞

( 2x

x +1x

xx

)

= limx→∞(2+

1x)

= 2

(c) lim

x→∞( xx+5 )

= limx→∞

( x

xxx

+5x)

= limx→∞

( 1

1+5x )

=

11+0

= 1

(d) lim

x→∞( 3–4x6–7x )

= limx→∞

( 3

x –4

6x

–7)

=

47

(e) lim

x→∞(x2+7xx2–4 )

= limx→∞

( x2

x2+7x

x2

x2

x2–4x2

)

= limx→∞

(1+7x

1– 4x2

) = 1

(f) lim

x → ∞(2x3–5x2+4x–63–8x3 )

=

( 2x3

x3–5x2

x3 +4xx3 – 6

x3

3x3

–8x3

x3)

= limx→∞

(2–5x

+ 4x2

– 6

x3

3x3 –8 )

=

– 2

8

= – 1

4

3 (a) limx→ 0( x

x–1)=

0

0–1 = 0

(b) lim

x→ 0(x+1x–1)

=

0+10–1

= –1

(c) lim

x→ 0(x+31–x)

=

0+31–0

= 3

(d) lim

x→ 0( 23+x)

=

23+0

=

23

4 (a) limx→2

(x2–3x+1) = 22–3(2)+1 = –1

(b) lim

x→–3( 3x2)

=

3(–3)2

=

39

=

13

(c) lim

x→∞(4+5x2)

= 4+0

=4

(d) limx→1

3x(x–2) = 3(–1) = –3

(e) lim

x→3( 4–xx+2 )

=

4–33+2

= 1

5 (f) lim

x→4(x–2)(x–3)

= (4–2)(4–3) = 2

5 (a) mAB =

9–43–2

= 5

(b)

dydx

= 4

6 (a) y = 3x–4 y+δy = 3(x+δx)–4 δy = 3x+3δx–4–(3x–4) δy = 3δx

δyδx

= 3

dydx

=limδx→0

δyδx

= 3

(b) y = 2x2+3 y+δy = 2(x+δx)2+3 = 2[x2+2xδx+(δx)2]+3 δy = 2x2+4xδx+2(δx)2+ 3–(2x2+3) δy = 4xδx+2(δx)2

δyδx

= 4x+2δx

dydx

=limδx→0

δyδx

= 4x

(c) y =

1x

y+δy =

1

x+δx

δy =

1

x+δx–1x

δy =

x–(x+δx)x(x+δx)

δy =

δxx(x+δx)

δyδx

=

1x(x+δx)

δyδx

=

limδx→0

δyδx

= – 1x2

�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

Page 64: Analysis Spm Additional Mathematics

�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

(d) y =

12

x2

y+δy =

12

(x+δx)2

=

12

[x2+2xδx+(δx)2]

=

12

x2+xδx+12

(δx)2

–12

x2

δy = xδx+

12

(δx)2

δyδx

=x+δx2

δyδx

= limδx→0

δyδx

= x (e) y = 2x3

y+δy = 2(x+δx)3

= 2(x+δx)(x+δx)2

= 2(x+δx)[x2+2xδx+ (δx)2]

= 2[x3+2x2δx+x(δx)2+ x2δx+2x(δx)2+(δx)3] δy = 2x3+4x2δx+2x(δx)2

+2x2δx+4x(δx)2+ 2(δx)3–2x3

δyδx

= 4x2+2xδx+2x2+

4xδx+2(δx)2

= 6x2+6xδx+2(δx)2

δyδx

= limδx→0

δyδx

= 6x2

(f) y =

3x

–2

y+δy =

3

x+δx –2

δy =

3

x+δx –2–

(3x

–2)

=

3x+δx

–3x

=

3x–3(x+δx)

x(x+δx)

=

3δxx(x+δx)

δyδx

=

3x(x+δx)

δyδx

=

limδx→0

δyδx

= – 3x2

(g) y = 2x2–3x+2 y+δy = 2(x+δx)2–3(x+δx)

+2 = 2[x2+2xδx+(δx)2]

–3x–3δx+2 δy = 4xδx+2(δx)2–3δx

δyδx

= 4x+2δx–3

δyδx

= limδx→0

δyδx

= 4x–3

(h) y =

x

x+2

y+δy =

x+δx

x+δx+2

δy =

x+δx

x+δx+2 –

xx+2

=

(x+2)(x+δx)–x(x+δx+2)

(x+2)(x+δx+2)

=

x2+xδx+2x+2δx–x2–xδx–2x

(x+2)(x+δx+2)

=

2δx

(x+2)(x+δx+2)

δyδx

=

2(x+2)(x+δx+2)

δyδx

=

limδx→0

δyδx

= 2

(x+2)2

7 (a)

dydx

=6x+5

(b)

dydx

=5x4–21x2+6

(c) y = 9x2–4x–2

dydx

= 18x+8x–3

=

18x+

8x3

(d) y = 4x+3x–1

dydx

= 4–3x–2

= 4–

3x2

(e) y = 3x+4x12–3

dydx

= 3+2x– 1

2

= 3+

2

x

(f) y = 7x2+3x–x12

dydx

= 14x+3–12

x– 1

2

= 14x+3–

1

2 x

(g) y = 4x2+

23

x– 1

2

+1

dydx

= 8x–13

x– 3

2

= 8x–

1

3x32

(h) y =

x3

–3x–1–5x–2

dydx

=

13

+3x–2+10x–3

= 1

3 +

3x2

+10x3

(i) y =

32

x12

+6x

– 12

dydx

=

34

x– 1

2

–3x

– 32

=

3

4 x–

3

x32

(j) y = 2x4+

12

x2–1x

+9

= 2x4+

12

x2–x–1+9

dydx

= 8x3+x+x–2

= 8x3+x+1x2

8 (a) y =

2x2+6x

x = 2x+6

dydx

= 2

(b) y =

4x3–2x–3

8x

=

12

x2–14

–38

x–1

dydx

= x+38

x–2

= x+

38x2

(c) y =

3x2–4xx

= 3x32–4x

12

dydx

=

92

x12

–2x

– 12

=

92

x–

2

x

(d) y =

x3+x

x2

= x+x–1

dydx

= 1–x–2

= 1–1x2

(e) y = x +

1

x

= x12+x

– 12

=

12

x– 1

2

–12

x– 3

2

=

1

2 x–

1

2x32

(f) y =

6x2– x+3

2x

= 3x–

12

x– 1

2

+

32

x–1

dydx

=

3+14

x– 3

2

+

–32

x–2

= 3+

1

4x32

32x2

(g) y =

x+3 xx

= x12 +x

– 16

dydx

=

12

x– 1

2–16

x– 7

6

=

1

2 x–

1

6x76

(h) y =

2+ x

x2

= 2x–2+x– 3

2

dydx

= –4x–3–

32

x– 5

2

= –

4x3

3

2x52

Page 65: Analysis Spm Additional Mathematics

�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

(i) y =

x2+2x

= x32+2x

– 12

dydx

=

32

x12

–x

– 32

=

32

x–

1

x32

(j) y =

x2+3

x2

= 1+3x–2

dydx

= –6x–3

= – 6

x3

9 (a) y = (3x+1)2

= 9x2+6x+1

dydx

= 18x+6

(b) y =

(x2+2x)2

= x4+4x+4x–2

dydx

= 4x3+4–8x–3

= 4x3+4–8x3

(c) y = (x+1)(2x–3) = 2x2–x–3

dydx

= 4x–1

(d) y = x( x –4)

= x32–4x

dydx

=

32

x12

–4

= 3

2x

–4

(e) y = (1+ x )(1– x ) = 1–x

dydx

= –1

(f) y =

( x +

3x )2

= x+6+9x–1

dydx

=

1–9x–2

= 1–9x2

(g) y = (x2+3)2

= x4+6x2+9

dydx

= 4x3+12x

(h) y =

(2+1x2)2

= 4+4x–2+x–4

dydx

=

–8x–3–4x–5

=

– 8

x3 –4x5

(i) y = (5x+

13x)2

= 25x2+

103

+19

x–2

dydx

= 50x–29

x–3

= 50x–

29x3

(j) y = (2–5 x )2

= 4–20x12+25x

dydx

= –10x– 1

2+25

= 25–

10

x

10 (a)

dydx

=4x–5

Whenx=2,

dydx

= 4(2)–5

=3

(b) y = 2x12–x

32

dydx

= x– 1

2

–32

x12

=

1

x –32

x

Whenx=9,

dydx

=

13

–32

(3)

= – 256

(c) y = 3x2–2x–1+4

dydx

= 6x+2x–2

= 6x+

2x2

Whenx=1,

dydx

= 6(1)+

212

= 8 (d) y = 6x2–11x+4

dydx

= 12x–11

Whenx=3,

dydx

= 12(3)–11 =25

(e) y = 1–9x –1

dydx

= 9x–2

=

9x2

Whenx=–3,

dydx

=

9(–3)2

= 1

(f) y = 4x+

3x

+

1x2

dydx

=4–

3x2

2x3

Whenx=–1,

dydx

=

4–

3(–1)2

2(–1)3

= 4–3+2 = 3

(g) y =

2x2–x–3

x = 2x–1–3x–1

dydx

= 2+3x–2

= 2+

3x2

Whenx=

12

,dydx

=2+

3

(12 )2

=2+12 =14

(h) y = x32+x

12+4x

– 12

dydx

=

32

x12

+

12

x– 1

2 –2x

– 32

=

32

x +

1

2 x

2

x32

Whenx=4,

dydx

= 3+14

–28

=3

(i) y = 2x12+3

dydx

=

x– 1

2

=

1

x

Whenx=9,

dydx

=13

(j) y = 4x2+4x+1

dydx

=8x+4

Whenx=

14

,dydx

= 8(14)

+4

=6

11 (a) y = (x3+2)(x2–1)

dydx

= 3x2(x2–1)+2x(x3+2)

= 3x4–3x2+2x4+4x = 5x4–3x2+4x (b) y = (3+2 x)(3–2 x)

dydx

= x– 1

2(3–2 x)–x– 1

2(3+2 x)

= 3x– 1

2–2x– 1

2 + 12–3x

– 12

–2x– 1

2 + 12

= –4 (c) y = (x+2)(x2+3x)

dydx

= x2+3x+(2x+3)(x+2)

= x2+3x+2x2+7x+6 = 3x2+10x+6 (d) y = (x2+4)(2x2–x+6) = 2x(2x2–x+6)+(4x–1) (x2+4) = 4x3–2x2+12x+4x3+16x

–x2–4 = 8x3–3x2+28x–4

(e) y = (3x3+2x)(x2+

1x

+4)

dydx

= (9x2+2)(x2+1x

+4)+

(2x–

1x2 )(3x3+2x)

= 9x4+9x+36x2+2x2+

2x

+8+6x4+4x2–3x–

2x

= 15x4+42x2+6x+8 (f) y = 5x2(4– x)

dydx

=

10x(4– x)–

1

2 x (5x2)

= 40x–10x

32

–52

x32

= 40x–252

x32

(g) y = x( x –6)

dydx

= x –6+

1

2 x (x)

Page 66: Analysis Spm Additional Mathematics

�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

= x –6+

x2

=

12

x–6

(h) y =

(x+1x)(1–

1x)

dydx

= (1– 1

x2)(1–1x)

+1x2

(x+1x)

= 1–

1x

–1x2

+1x3

+1x

+1x3

= 1–1x2

+

2x3

(i) y = x (x+2)

dydx

=

12 x

(x+2)+

x

=

12

x+

1x

+ x

=

32

x+

1

x

(j) y =

( x+

1x )(x+

1x )

dydx

= ( 1

2 x –

1

2x32 )(x+

1x )

+

(1–1x2 )( x

+

1x )

=

12

x+

1

2x32

12 x

1

2x52

+

x +1x

1

x32

1

x52

=

32

x–

1

2x32

+

12 x

3

2x52

12 (a) y =

2x+53x–4

dydx

=

2(3x–4)–3(2x+5)(3x–4)2

=

6x–8–6x–15

(3x–4)2

=

– 23

(3x–4)2

(b) y =

3

5–x

dydx

=

0(5–x)–(–1)(3)(5–x)2

=

3(5–x)2

(c) y =

x2

2x–1

dydx

=

2x(2x–1)–2(x2)(2x–1)2

=

4x2–2x–2x2

(2x–1)2

=

2x2–2x(2x–1)2

=

2x(x–1)(2x–1)2

(d) y =

3x2+5xx

dydx

=

(6x+5) x –

12 x

(3x2+5x)x

=

6x32+5 x –3

2x

32–

52

x

x

=

92

x32

+

52

x

x

=

92

x +

5

2 x

(e) y =

2x–x4

x2

dydx

=

(2–4x3)(x2)–2x(2x–x4)x4

=

2x2–4x5–4x2+2x5

x4

=

–2x2–2x5

x4

=

– 2

x2 –2x

(f) y =

x2–x+1

2x+3

dydx

=

(2x–1)(2x+3)–2(x2–x+1)(2x+3)2

=

4x2+4x–3–2x2+2x–2

(2x+3)2

=

2x2+6x–5

(2x+3)2

(g) y =

xx+1

dydx

=

12 x

(x+1)– x

(x+1)2

=

12

x +1

2 x – x

(x+1)2

=

12 x

–12

x

(x+1)2

=

1–x2 x (x+1)2

(h) y =

x2–1x2+1

dydx

=

2x(x2+1)–2x(x2–1)(x2+1)2

=

2x3+2x–2x3+2x

(x2+1)2

=

4x

(x2+1)2

(i) y =

12

x3+2x

dydx

=

0–(3x2+2)(12)(x3+2x)2

=

– 12(3x2+2)

(x3+2x)2

(j) y =

4x+1x2+3

dydx

=

4(x2+3)–2x(4x+1)(x2+3)2

=

4x2+12–8x2–2x

(x2+3)2

=

12–2x–4x2

(x2+3)2

=

2(6–x–2x2)

(x2+3)2

13 (a) y = (1+3x)12

dydx

=

12

(1+3x)– 1

2(3)

=

32 1+3x

(b) y = (5x2+2)13

dydx

=

13

(5x2+2)– 2

3(10x)

=

10x

3(5x2+2)23

(c) y = (x+2)4

dydx

= 4(x+2)3

(d) y = (2–3x2)5

dydx

= 5(2–3x2)4(–6x)

= –30x(2–3x2)4

(e) y =

(14

x+3)8

dydx

= 8(14

x+3)7(14)

= 2(14

x+3)7

(f) y = (x2+1)–1

dydx

= –(x2+1)–2(2x)

=

–2x

(x2+1)2

(g) y = 3(3–4x)–3

dydx

= –9(3–4x)–4(–4)

=

36

(3–4x)4

(h) y = 4(2–x)–2

dydx

= –8(2–x)–3(–1)

=

8

(2–x)3

(i) y = (2– x )6

dydx

= 6(2– x )5(–1

2 x )

=

3(2– x )5

x

(j) y = 2(x2–3)12

dydx

=(x2–3)– 1

2(2x)

=

2xx2–3

14 (a) y = x 1+x2

dydx

=

1+x2+12

(1+x2)– 1

2(2x)(x)

= 1+x2 +

x2

1+x2

=

1+2x2

1+x2

(b) y = x(2x–1)3

dydx

= (2x–1)3+3(2x–1)2(2)(x)

= (2x–1)3+6x(2x–1)2

Page 67: Analysis Spm Additional Mathematics

�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

= (2x–1)2[(2x–1)+6x] = (2x–1)2(8x–1)

(c) y = x (1–x)2

dydx

=

12 x (1–x)2+2(1–x)

(–1)( x )

=

(1–x)2

2 x –2 x (1–x)

=

(1–x)2–4x(1–x)2 x

=

1–2x +x2–4x+4x2

2 x

=

5x2–6x+12 x

(d) y = x2 1–2x2

dydx

= 2x 1–2x2

+

12

(1–2x2)– 1

2

(–4x)(x2)

= 2x 1–2x2

2x3

1–2x2

=

2x(1–2x2)–2x3

1–2x2

=

2x–6x3

1–2x2

=

2x(1–3x2)1–2x2

(e) y =

x1–x

dydx

=

1–x –12

(1–x)– 1

2

(–1)(x)( 1–x)2

=

1–x +

x2 1–x

1–x =

2(1–x)+x2 1–x (1–x)

=2–x

2(1–x)32

(f) y =

2x2x+1

dydx

=

2 2x+1–12

(2x+1)– 1

2

(2)(2x)

2x+1

=

2 2x+1–

2x2x+1

(2x+1)

=

2(2x+1)–2x(2x+1)( 2x+1)

=

2x+2

(2x+1)32

=

2(x+1)

(2x+1)32

(g) y = (x–9) x–3

dydx

=

x–3+

12

(x–3)– 1

2(x–9)

= x–3 +

x–92 x–3

=

2(x–3)+x–92 x–3

=

3x–152 x–3

=

3(x–5)2 x–3

(h) y = x 9+3x

dydx

=

9+3x +12

(9+3x)– 1

2(3)(x)

= 9+3x +

3x2 9+3x

=

2(9+3x)+3x2 9+3x

=

18+9x2 9+3x

=

9(2+x)2 9+3x

15 (a) y = (2x–1)4

dydx

= 4(2x–1)3(2)

= 8(2x–1)3

Whenx=1,

dydx

=8

(b) y = 5–2x

dydx

=12

(5–2x)– 1

2(–2)

=

15–2x

Whenx=

12

,dydx

=–1

2

(c) y =

1

2x–3

dydx

=

–2(2x–3)2

Whenx=12

, dydx

=

–2

[2(12 )–3]2

=

–2

4=

–1

2

(d) y =

3x2–85–2x

dydx

=

6x(5–2x)–(–2)(3x2–8)(5–2x)2

=

30x–12x2+6x2–16

(5–2x)2

=

30x–6x2–16

(5–2x)2

Whenx=2,

dydx

=

30(2)–6(2)2–16[5–2(2)]2

=20

(e) y =

3x2

1–4x2

dydx

=

6x(1–4x2)–(–8x)(3x2)(1–4x2)2

=

6x–24x3+24x3

(1–4x2)2

=

6x

(1–4x2)2

Whenx=1,

dydx

=

69

=

23

(f) y =

x2+4

x2

dydx

=

2x(x2)–2x(x2+4)x4

=

2x3–2x3–8x

x4

= – 8

x3

Wheny=5, 5=x2+4

x2

4x2= 4 x2= 1 x = 1

Whenx=1,

dydx

=–8

Whenx=–1,

dydx

=8

(g) y = x (2–x)

dydx

=

12

x– 1

2(2–x)+(–1)( x )

=

2–x2 x

x

=

2–x–2x

2 x

=

2–3x2 x

Whenx=9,

dydx

=

2–3(9)2 9

=

–25

6

(h) y =

5–x2x

dydx

=

–2x–2(5–x)4x2

=

104x2

=

– 5

2x2

Wheny=2,2= 5–x

2x 4x = 5–x 5x = 5 x = 1

Whenx=1,

dydx

=–5

2

16 (a) y=3x2–4x+5 Whenx = –1, y =3(–1)2–4(–1)+5 = 12

dydx

=6x–4

Atx=–1,

dydx

=–10

Equationofthetangent: y–12 = –10(x+1) y–12 = –10x–10 y+10x = 2 (b) y = x2–3x+4

dydx

=2x–3

Atx=3,

dydx

=3

Equationofthetangent: y–4 = 3(x–3) y–4 = 3x–9 y = 3x–5

(c) y=x+

2x

Whenx=–2,y = –2+

2

–2 =–3

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dydx

=1–

2x2

Atx=–2,

dydx

= 1–

2(–2)2

=

12

Equationofthetangent:

y+3 =

12

(x+2)

y+3 =

12

x+1

y =

12

x–2

(d) y=

x–1x+1

Whenx=1,y =

1–11+1

= 0

dydx

=

x+1–(x–1)(x+1)2

=

2

(x+1)2

Atx=1,

dydx

=12

Equationofthetangent:

y =

12

(x–1)

2y = x–1 2y–x = –1 (e) y = 1+4x–x2

dydx

=

12

(1+4x–x2)– 1

2(4–2x)

=

2–x1+4x–x2

Atx=3,

dydx

=

2–31+4(3)–32

=

–12

Equationofthetangent:

y–2=

–12

(x–3)

2y–4= –x+3 2y+x =7

(f) y=

x2+5x+1

Whenx=1,y =

62

= 3

dydx

=

2x(x+1)–(x2+5)(x+1)2

=

2x2+2x–x2–5

(x+1)2

=

x2+2x–5

(x+1)2

Atx=1,

dydx

=

–24

=

–12

Equationofthetangent:

y–3=

–12

(x–1)

2y–6= –x+1 2y+x = 7

17 (a) y=2(x–1)3

Whenx=

12

,y = 2(12

–1)3

=2(–

18 )

=

–14

dydx

=6(x–1)2

Atx=

12

,dydx

=64

=32

Equationofthenormal:

y+

14

=

–23 (x–

12 )

y+14

= –23

x+13

y =

–23

x+

112

(b) y=x 1–2x Whenx=–4,y = –4 1–2(–4) = –4(3) = –12

dydx

= 1–2x +12

(1–2x)–1

2

(–2)(x)

= 1–2x –

x1–2x

=

(1–2x)–x1–2x

=

1–3x1–2x

Atx=–4,

dydx

=

1–3(–4)1–2(–4)

=

133

Equationofthenormal:

y+12=

313

(x+4)

13y+156= –3x–12 13y+3x = –168

(c) y=

x

x+1

Whenx=2,y=

23

dydx

=

x+1–x(x+1)2

=

1

(x+1)2

Atx=2,

dydx

=19

Equationofthenormal:

y–

23

= –9(x–2)

y–

23

= –9x+18

3y–2= –27x+54 3y+27x = 56

(d) y=2+

1x

Whenx=1,y=3

dydx

=–1x2

Atx=1,

dydx

=–1

Equationofthenormal: y–3= 1(x–1) y–3= x–1

y–x=2

(e) y=7x–

6x

Whenx=3,y = 7(3)–

63

=21–2 =19

dydx

=7+

6x2

Atx=3,

dydx

= 7+23

=

233

Equationofthenormal:

y–19=

323

(x–3)

23y–437= –3x+9 23y+3x = 446

(f) y=

x–2

2x+1 Whenx=2,y=0

dydx

=

2x+1–2(x–2)(2x+1)2

=

5

(2x+1)2

Atx=2,

dydx

=

525

=

15

Equationofthenormal: y = –5(x–2) y = –5x+10 y+5x = 10

18 (a) y = 8x+

12x2

= 8x+

12

x–2

dydx

= 8–x–3

= 8–

1x3

Forturningpoint,

dydx

= 0

8–

1x3

=0

8=

1x3

x3=

18

x=

12

Whenx=

12

,y = 8(12)

+

1

2(12)2

=4+2 =6

Theturningpointis

(12

,6).

(b) y = 4x+x–1

dydx

=4–x–2

=4–

1x2

4–

1x2

= 0

4=

1x2

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x2 =

14

x = ± 1

2

Whenx=

12

,y = 4(12 )

+

1

(12)

= 2+2 = 4

Whenx=

12

,

y=4(– 12)

+

1

(–12)

= –2–2 = –4

(12

,4)and

(–12

,–4) (c) y=x3–6x2+9x

dydx

= 0

3x2–12x+9= 0 x2–4x+3= 0 (x–1)(x–3)= 0 x=1orx=3 Whenx=1,y=13–6(1)2+9(1) =4 Whenx=3,y=33–6(3)2+9(3) =0 ∴(1,4)and(3,0) (d) y = x(x–2)2

= x(x2–4x+4) =x3–4x2+4x

dydx

=0

3x2–8x+4=0 (3x–2)(x–2)=0

x=

23

orx=2

Whenx=

23

,y = (2

3 )3

–4(2

3 )2

+4(2

3 )

=

3227

Whenx=2,y=23–4(2)2+4(2) =0

(23 ,

3227 )

and(2,0)

(e) y = 4–x2–16x–2

dydx

=0

–2x+

32x3

= 0

32x3

= 2x

16=x4

x=2

Whenx=2,y = 4–(2)2–

16(2)2

=–4 ∴(2,–4) (f) y = x2–2x4

dydx

= 0

2x–8x3 = 0

2x = 8x3

1x2

= 4

x2 =

14

x = ± 1

2

Whenx=

12

,y

=(1

2)2

–2(1

2)4

=

14

–2( 116)

=

18

Whenx=–

12

,y = (–

12)2

–2(– 1

2)4

=

18

–2( 116)

=

18

∴(1

2 ,18)

and(–

12

,18)

19 (a) y = x(x–6)2

=x(x2–12x+36) =x3–12x2+36x

dydx

= 0

3x2–24x+36= 0 x2–8x+12= 0 (x–2)(x–6)= 0 x=2orx=6 Whenx= 2, y= 23–12(2)2+36(2) = 8–48+72 =32 Whenx= 6, y=63–12(6)2+36(6) =216–432+216 =0

d2ydx2

= 6x–24

Whenx=2,

d2ydx2

= 6(2)–24

=–120 ∴(2,32)isamaximumpoint.

Whenx=6,

d2ydx2

=6(6)–24

=36–24 =120 ∴(6,0)isaminimumpoint. (b) y = x+16x–1

dydx

= 1–16x–2

= 1–

16x2

dydx

=0

1–

16x2

=0

1=

16x2

x2 = 16 x =±4

Whenx=4,y =4+

164

=4+4 =8

Whenx=–4,y= –4+

16

(–4) = –8

d2ydx2

=32x –3

=

32x3

Whenx=4,

d2ydx2

=

3264

=

12

0

∴(4,8)isaminimumpoint.

Whenx=–4,

d2ydx2

=

–3264

=

–12

0

(–4,–8)isamaximumpoint. (c) y = x3–2x2

dydx

= 3x2–4x

dydx

= 0

x(3x–4)= 0

x=0orx=

43

Whenx=0,y=0

Whenx=4

3,y=

(43)3

–2(4

3)2

=

6427

–329

=

–3227

d2ydx2

=6x–4

Whenx=0,

d2ydx2

=–40

∴(0,0)isamaximumpoint.

Whenx=

43

,d2ydx2

= 6(43)

–4

=8–4 =40

(43

,–3227)isaminimumpoint.

(d) y = 4x+9x–1

dydx

=4–

9x2

dydx

= 0

4–

9x2

=0

4 =

9x2

x2 =

94

x =± 3

2

Whenx=

32

,y=4(32)

+

9

(32)

= 6+6 =12

Whenx =

–32

,

y = 4(–

32)

+

9

(– 32)

= –6–6 = –12

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d2ydx2

=

18x3

Whenx=

32

,d2ydx2

=

18

(32)3

=

163

0

∴(3

2 ,12)

isaminimumpoint.

Whenx =

–32

,d2ydx2

=

18

(– 32)3

=

163

0

∴(– 3

2,–12)

isamaximumpoint.

(e) y =

x2–6x+9

x = x–6+9x–1

dydx

= 1–9x–2

dydx

= 0

1–

9x2

= 0

1=

9x2

x2=9 x=±3 Whenx=3,y=0 Whenx=–3,y=–12

d2ydx2

=18x3

Whenx=3,

d2ydx2

=1827

=

23

0

∴(3,0)isaminimumpoint.

Whenx=–3,

d2ydx2

=

–1827

=

– 2

30

∴(–3,–12)isamaximumpoint.

(f) y =

x2

x+1

dydx

=

2x(x+1)–x2

(x+1)2

=

x2+2x(x+1)2

dydx

= 0

x2+2x(x+1)2

=0

x(x+2)= 0 x=0orx=–2 Whenx=0,y=0 Whenx=–2,y=–4

d2ydx2

=

(2x+2)(x+1)2–2(x+1)(x2+2x)

(x+1)4

=

(2x+2)(x2+2x+1)–2(x3+2x2+x2+2x)

(x+1)4

=

2x3+4x2+2x+2x2+4x+2–2x3–4x2–2x2–4x

(x+1)4

=

2x+2(x+1)4

Whenx=0,

d2ydx2

=20

∴(0,0)isaminimumpoint.

Whenx=–2,

d2ydx2

=

2(–2)+2(–2+1)4

=

–2(–1)4

=–20 ∴(–2,–4)isamaximumpoint.

20 y = ax2+bx+c

dydx

= 0

2ax+b= 0 Atx=2, 4a+b=0… 1 Atthepoint(0,10), 10 = a(0)2+b(0)+c c = 10 Atthepoint(2,18), 18=a(2)2+b(2)+10 4a+2b= 8 2a+b = 4… 2

1 – 2 : 2a = –4 a = –2 Substitutea=–2into 1 : 4(–2)+b = 0 b=8 ∴a=–2,b=8,c=10

21 (a)

dydx

= 0

3x2–12= 0 3x2= 12 x2=4 x = ±2

(b)

dydx

=3x2–12

Atx=3,

dydx

=3(3)2–12 =27–12 =15 Equationofthetangent: y+9= 15(x–3) y+9= 15x–45 y=15x–54

22 y=mx+nx2

Atthepoint(3,5),

5= 3m+

n9

27m+n = 45… 1

dydx

=m–2nx3

Atx=3,

dydx

= 0

m–

2n27

=0

27m–2n = 0… 2

1 – 2 : 3n =45 n=15 Substituten=15into 2 : 27m–2(15)= 0

27m = 30

m=

109

∴m=109

,n=15

23 (a) 4(3x)+4(x)+4(h)= 1200 16x+4h =1200 4x+h =300 h= 300–4x Volumeofthebox, V =3x2h =3x2(300–4x) =900x2–12x3

=12x2(75–x)(shown) (b) ForthemaximumvalueofV,

dVdx

= 0

1800x–36x2= 0 36x(50–x)=0 Sincex0,x= 50

24 (a)

(24 – h) cm

R

A

B

24 cm

7 cm

cmx2

Fromthediagram,

724

=

( x2 )

24–h 7(24–h)=12x

24–h =

127

x

h =24–

127

x

TheareaofΔPQR,

A =

12

xh

=

12

x (24–127

x)

= 12x–

67

x2

=

67

x (14–x)(shown)

(b) ForamaximumvalueofA,

dAdx

=0

12–

127

x=0

127

x=12

x=7

25 (a) sAB = 40–2r rθ=40–2r θ=

40–2r

r (b) Areaofsector,

A =

12

r2θ

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�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

=

12

r2(40–2rr )

=20r–r2

=r(20–r)(shown) (c) ForamaximumvalueofA,

dAdr

= 0

20–2r=0 2r=20 r=10

andso,

d2Adr2

=–20

Hence,forthearea,Atobemaximum,r=10andthemaximumareais20(10)–(10)2=100cm2.

26 (a) y = 2x2+4x

dydx

=4x–4x2

Whenx=–2,

dydx

=

4(–2)–

4(–2)2

=–8–1 x =–9 Bythechainrule,

dydt

=

dydx

dxdt

=–9(3) =–27

(b) y =

2

(2x–3)3

dydx

=

12(2x–3)4

Whenx=2,

dydx

=–12

Bythechainrule,

dydt

=

dydx

dxdt

=–12(3) =–36 (c) y =(3x–5)4

dydt

=12(3x–5)3

Whenx=

43

,

dydx

= 12[3(43 )

–5]3

= –12 Bythechainrule,

dydt

=

dydx

dxdt

=–12(3) =–36 (d) y = 3x2–5

dydx

=6x

Whenx=

– 1

3,dydx

=6(– 13 )

= –2 Bythechainrule,

dydt

=

dydx

dxdt

=–2(3) =–6

27 (a) y = 2x2+5x+2

dydx

= 4x+5

Whenx=–1,

dydx

=1

Bythechainrule,

dydt

=

dydx

dxdt

2=1

dxdt

dxdt

=2

(b) y =x(x–4)

dydx

=2x–4

Whenx=3,

dydx

=2

Bythechainrule,

dydt

=

dydx

dxdt

2 = 2

dxdt

dxdt

= 1

(c) y = 2x +3

dydx

=

12x +3

Whenx=3,

dydx

=13

Bythechainrule,

dydt

=

dydx

dxdt

2 =

13

dxdt

dxdt

=6

(d) y=

x

x+2

Aty=

13

,13

=

xx+2

x+2= 3x 2x = 2 x = 1

dydx

=

2(x+2)2

Whenx=1,

dydx

=29

Bythechainrule,

dydt

=

dydx

dxdt

2 =

29

dxdt

dxdt

= 9

28 LetAbetheareaandrtheradius. ThenA=πr2.

Bythechainrule,dA

dt =

dAdr

drdt

= 2πr(4) =8πr

(a) Whenr=2,

dAdt

= 8π(2)

= 16πcm2s–1

(b) WhenA=9π,πr2= 9π r2= 9

r = 3(r0)

∴ dA

dt = 8π(3)

= 24πcm2s–1

29 (a) y = x2+8x

dydx

= 2x–8x2

(b) Bythechainrule,

dydt

=

dydx

dxdt

dydt

=(2x–

8x2) dx

dt

If

dydt

=6whenx=2,

6= 2

dxdt

dxdt

=3unitss–1

30 LetAbethetotalsurfaceareaandxthelengthofitsside.

ThenA=6x2

Bythechainrule,

dAdt

=

dAdx

dxdt

=12x(3) =36x

(asdxdt

=3cms–1) Whenvolume=64,x3=64 x =4

∴dA

dt =36(4)

= 144cm2s–1

31 LetVbethevolumeandrtheradius,

thenV=

43

πr3.

Bythechainrule,

dVdt

=

dVdr

drdt

8= 4πr2

drdt

drdt

=

84πr2

=

2

πr2(asdVdt

=8cm3s–1)

WhenA=16π, 4πr2= 16π r2= 4 r = 2(r0)

drdt

=

2π(2)2

=

1

2π cms–1

32 LetAbetheareaandxthelengthofitsside.ThenA=x2.

Bythechainrule,

dAdt

=

dAdx

dxdt

12=2x

dxdt

dxdt

=

6x (as

dAdt

=12cm2s–1) WhenA=9,x2= 9 x = 3(x0)

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dxdt

=

63

= 2cms–1

33 (a) A =

12

xy

=

12

x(4x–x2)

=

12

(4x2–x3)

(b) Ifxincreasesatarateof2units

persecond,thendxdt

=2.

dAdt

=

dAdx

dxdt

=

12

(8x–3x2)(2)

=8x–3x2

=x(8–3x) (i) Whenx = 2,

dAdt

= 2[8–3(2)]

=4unit2s–1

(ii) Whenx=3

dAdt

= 2[8–3(3)]

= 2(–1) =–2unit2s–1

34 (a)

r

x

10

50

Fromthediagram,

r

10=

x

50

r =

x5

Thevolumeof thewater in thecontaineris:

V =

13

πr2x

=

13

π(x5)2

x

=

13

π( x2

25)x

=

175

πx3(shown)

(b)

dVdt

=

dVdx

dxdt

18 =

( 125

πx2)dxdt

dxdt

=

450πx2

Whenx=3,dx

dt =

450

π(3)2

=

50π

cms–1

35 (a) y = 4x3–7x2

dydx

=12x2–14x

Atthepoint(2,4),

dydx

=12(2)2–14(2) =20 (b) Wheny=4,x =2 δy =4.05–4 =0.05

dydx

=20

Thenδy

dydx

δx

0.0520δx δx0.0025

36 (a) y = x3+1

dydx

=

3x2

2 x3+1

(b) Whenx=2,δx = 2.02–2 =0.02

and

dydx

= 2

Thenδy

dydx

δx

2(0.02) 0.04

37 (a) y = 2x3–7x2+15

dydx

= 6x2–14x

(b) Whenx=2,δx = 2.01–2 =0.01

and

dydx

= 6(2)2–14(2)

=24–28 =–4

Thenδy

dydx

δx

–4(0.01) –0.04

38 (a) y =

13x+1

dydx

=

–3(3x+1)2

Whenx=3,

dydx

=– 3

100 (b) Theapproximatechangeiny,

δy

dydx

δx

3100

(p)

– 3p

100

39 y =

3x+2x+2

dydx

=

3(x+2)–(3x+2)(x+2)2

=

4

(x+2)2

Whenx=2,δx = 2+p–2 =p

anddy

dx =

14

Thenδy

dydx

δx

14

(p)

p4

40 (a) y = 3x–9x

dydx

=3+9x2

(b) Whenx=3,δx = 3+p–3 =p

and

dydx

= 3+

9(3)2

=4

Thenδy

dydx

δx

4p

41 V =43

πr3

dVdr

= 4πr2

Whenr=4,δr = 3.8–4 =–0.2

and

dVdr

=4π(4)2

=64π

ThenδV

dVdr

δr

64π(–0.2) –12.8π

42 T = 2πl

10

=

2π10

(l)12

dTdl

=

12(

2π10 )l–1

2

=

π10l

Whenl=2.5,δl = 2.6–2.5 = 0.1

and

dTdl

=

π10(2.5)

=

π5

ThenδT

dTdl

δl

π5

(0.1)

0.02π

43 y =

1x

= x– 1

2

dydx

=

–12

x– 3

2

=

–1

2x32

(a) Whenx=100, y =

1

10 δx = 100.5–100 =0.5

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��©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

dydx

=

– 12000

Then

1

100.5

1

10+

(– 12000)

(0.5) 0.09975

(b) Whenx=49,y =

17

δx = 49.2–49 =0.2

dydx

= –

1686

Then

1

49.2

17

+(– 1

686)(0.2)

0.14257

(c) Whenx=25,y =

15

δx = 24.05–25 = –0.95

dydx

= –

1250

Then

124.05

15

+(– 1

250) (–0.95) 0.2038

(d) Whenx=9,y =

13

δx = 8.96–9 =–0.04

dydx

= –

154

Then

18.96

13

+(– 1

54) (–0.04) 0.33407

44 y = x– 1

3

dydx

= –13

x– 4

3

= –

1

3x43

Whenx=8,y =

12

δx =8.95–8 =0.95

dydx

= –

148

Then

13

8.95

12

+(– 1

48)(0.95)

0.4802

45 (a) y = (x+1)2

= x2+2x+1

dydx

= 2x+2

d2ydx2

= 2

(b) y = x+

1x

= x+x–1

dydx

= 1–x–2

= 1–

1x2

d2ydx2

=

2x3

(c) y = x–5

dydx

=

12

(x–5)– 1

2

=

12 x–5

d2ydx2

= –

1

4(x–5)32

(d) y =

1

x+1

dydx

= –

1(x+1)2

d2ydx2

=

2(x+1)3

(e) y =

1x

dydx

= –

1

2x32

d2ydx2

=

3

4x52

(f) y =

x

x–1

dydx

=

(x–1)–x(x–1)2

= –

1

(x–1)2

d2ydx2

=

2(x–1)3

46 (a) y = x3–6x2+5

dydx

= 3x2–12x

d2ydx2

= 0

6x–12= 0 6x= 12 x = 2 (b) y = x2–27x–2

dydx

= 2x+54x–3

d2ydx2

= 0

2–

162x4

= 0

2=

162x4

x4= 81 x = 3 (c) y = (x–2)3

dydx

= 3(x–2)2

d2ydx2

= 0

6(x–2)= 0 x = 2 (d) y = 3x2–x3

dydx

=

6x–3x2

d2ydx2

= 0

6–6x = 0 x = 1

47 y = x3–6x2+4

dydx

= 3x2–12x

d2ydx2

= 0

6x–12= 0 x = 2

dydx

= 3(2)2–12(2)

= 12–24 = –12

48 y = x4–108x

dydx

=0

4x3–108= 0 4x3= 108 x3=27 x= 3

d2ydx2

=12x2

Whenx=3,

d2ydx2 = 12(3)2

= 108

49 y = x3–2x2+3x+4

dydx

= 3x2–4x+3

d2ydx2

=6x–4

Whenx=1,

dydx

= 3(1)2–4(1)+3

=2

Whenx=1,

d2ydx2

= 6(1)–4

=2

50 y =

1–x2

x

dydx

=

–2x(x)–(1–x2)x2

=

–x2–1

x2

d 2ydx2

=

–2x(x2)–2x(–x2–1)x4

=

2xx4

=

2x3

xd2y

dx2 +2dy

dx+2= x(2

x3)+2(–x2–1x2 )+2

=

2x2

+2(–1–1x2)

+2

=

2x2

–2–2x2

+2

= 0(shown)

51 (a) y =

1x+1

dydx

= –1

1(x+1)2

= –1

1= (x+1)2

x2+2x =0 x(x+2)=0 x=0orx=–2

Whenx=0,y =

1

0+1 =1

Whenx=–2,y =

1

–2+1 = –1

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∴Thecoordinatesare(0,1)and(–2,–1).

(b)

d2ydx2

=

2(x+1)3

Atthepoint(1,1),

d2ydx2

=

2(1+1)3

=

28

=

14

1 (a) y = x12+x–

12

dydx

=

12

x– 12

–12

x– 32

=

12 x

1

2x32

(b) y =

(x+1)2

x

=

x2+2x+1

x = x+2+x–1

dydx

= 1–x–2

= 1–1x2

2 y = 8x–1–6x12

dydx

= –8x–2–3x– 12

= –

8x2

3

x

Atx=4,

dydx

= –

8(4)2

3

4

= –

12

–32

= –2

3 y=ax

+bx2

Atthepoint(3,6),6=a

3+b(3)2

a+27b = 18… 1

and

dydx

= 7

–ax2

+2bx = 7

Atthepoint(3,6),–

a9

+6b=7

–a+54b=63… 2

1 + 2 : 81b = 81 b = 1 Substituteb=1into 1 : a+27(1)= 18 a = –9 ∴a=–9,b=1

4 y =

x2

x–2

dydx

=

2x(x–2)–x2

(x–2)2

=

2x2–4x–x2

(x–2)2

=

x2–4x(x–2)2

Forturningpoints,

dydx

= 0

x2–4x(x–2)2

=0

x(x–4)= 0 x=0orx = 4 Whenx=0,y=0

Whenx=4,y =

42

4–2 =8 ∴Theturningpointsare(0,0)and(4,8)

5 y =

3x2x–3

dydx

= –94

3(2x–3)–2(3x)

(2x–3)2 =

–94

–9

(2x–3)2 =

–94

(2x–3)2= 4 4x2–12x+9=4 4x2–12x+5=0 (2x–1)(2x–5)=0

x=12

orx=52

6 y=2x+1x2+2

Aty-axis,x =0

y=

2(0)+102+2

=

12

Wheny=

12

,12

=2x+1x2+2

x2+2= 4x+2 x2–4x=0 x(x–4)=0 x=0orx=4

dydx

=

2(x2+2)–2x(2x+1)(x2+2)2

=

2x2+4–4x2–2x

(x2+2)2

=

4–2x –2x2

(x2+2)2

Whenx=0,

dydx

=

4–2(0)–2(0)2

(02+2)2

=1

Whenx=4,

dydx

=

4–2(4)–2(4)2

(42+2)2

= –

36

324

= – 19

7 (a) y = (x+6)7(x–9)8

dydx

= (x+6)6(x–9)7[7(x–9) +8(x+6)] = (x+6)6(x–9)7(7x–63 +8x+48) = (x+6)6(x–9)7(15x–15) = 15(x–1)(x+6)6(x–9)7

(b)

dydx

= 0

15(x–1)(x+6)6(x–9)7= 0 ∴x=1,x=–6,x=9

8 y = (x2+

2x )8

dydx

=

8(x2+2x )7(2x– 2

x2)

Whenx=1,dydx

= 8(3)7(0)

=0

9 y = (2x–3)3

dydx

= 6

3(2x–3)2(2)= 6 6(2x–3)2=6 4x2–12x+9=1 4x2–12x+8=0 x2–3x+2=0 (x–1)(x–2)=0 x=1orx=2 Whenx=1,y=[2(1)–3]3

=–1 Whenx=2,y=[2(2)–3]2

=1 ∴(1,–1)and(2,1)

10 y = (2x+3)3(x–4)

dydx

= 3(2x+3)2(2)(x–4)+(2x+3)3

= 6(x–4)(2x+3)2+(2x+3)3

= (2x+3)2[6(x–4)+(2x+3)] = (2x+3)2(8x–21)

Comparewith

dydx

=(2x+3)2(mx+n)

∴m=8,n=–21

11 (a) y = (1+x)(1+ x )

dydx

= 1+ x + 12 x

(1+x)

=

2 x +2x+1+x

2 x

=1+

3x

2 x +

1

2 x

=1+

32

x+

1

2 x (b) y = x(1+3x)5

dydx

= (1+3x)5+5(1+3x)4(3)(x)

= (1+3x)5+15x(1+3x)4

= (1+3x)4[(1+3x)+15x] = (1+3x)4(18x+1)

12 (a) y = 4x+x–1

dydx

= 4–x–2

= 4–1x2

(b) y=4x+

1x

Atthepointwherex=2,

y=4(2)+

12

=

172

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��©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

Atx=2,

dydx

= 4–122

=

154

Equationofthetangent:

y–

172

=

154

(x–2)

4y–34= 15x–30 4y–15x=4

13 (a) y=3x2–8x+7

Foraminimumpoint,

dydx

= 0

6x–8=0 6x=8

x =

43

(b) Atx=2,y = 3(2)2–8(2)+7 = 3 and dy

dx = 6(2)–8

= 4 Equationofthenormal:

y–3= –

14

(x–2)

4y–12= –x+2 4y+x = 14

14 y = x2–x+3

dydx

= –3

2x–1 = –3 2x = –2 x = –1 Whenx=–1,y =(–1)2–(–1)+3 =5 Atthepoint(–1,5),3(5)–(–1) = c c = 16

15 (a) dydx

= 2

2x–2 = 2 2x =4 x =2 Whenx=2,y= 22–2(2)+2 = 4–4+2 = 2 ∴A(2,2) (b) Atthepoint(2,2), 2= 2(2)+c c = 2–4 c =–2

16 (a) y =

2x–4x+2

dydx

=

2(x+2)–(2x–4)(x+2)2

=

8

(x+2)2

Atx=–6,

dydx

=

8(–6+2)2

=

8

16

=

12

EquationofthenormalatA: y–4= –2(x+6) y–4= –2x–12 y+2x = –8

(b) y=

2x–4x+2

… 1

y=–2x–8… 2

Substitute 2 into 1 :

–2x–8=

2x–4x+2

–2x2–12x–16= 2x–4 2x2+14x+12= 0 x2+7x+6= 0 (x+6)(x+1)= 0 x=–6orx = –1

Substitutex=–1into 2 : y= –2(–1)–8 =2–8 =–6 ∴B(–1,–6)

17 (a) y =

12x+3

dydx

= –

2(2x+3)2

Comparewith

dydx

=

k(2x+3)2

∴k = –2 (b) Atthepoint(–1,1),

dydx

=

– 2[2(–1)+3]2

= –2 Equationofthenormal:

y–1=

12

(x+1)

2y–2= x+1 2y–x = 3

18 (a) y =

2x–8x+2

dydx

=

2(x+2)–(2x–8)(x+2)2

=

12

(x+2)2

Comparewith

dydx

=

k(x+2)2

∴ k = 12 (b) Atx-axis,y=0

0 =

2x–8x+2

2x =8 x = 4 ∴P(4,0) Aty-axis,x=0

y =

2(0)–8

0+2 = –4 ∴Q(0,–4) AtpointP(4,0),

dydx

=

12(4+2)2

=

1236

=

13

EquationofthenormalatP: y= –3(x–4) y = –3x+12 y+3x = 12 Aty-axis,x= 0

y+3(0)=12 y = 12 ∴R(0,12) ∴ThelengthofRQ = 12+4 = 16units

19 (a) y=x+6x–2

Atx-axis,y=0

0=

x+6x–2

x= –6 ∴A(–6,0) Aty-axis,x=0

y = –

62

= –3 ∴B(0,–3)

(b)

dydx

=

x–2–(x+6)(x–2)2

= –

8

(x–2)2

AtpointB(0,–3),

dydx

= –

8(–2)2

= –2 EquationofthenormalatB:

y+3 =

12

x

y =

12

x–3

Atx-axis,y=0

12

x = 3

x = 6 ∴C(6,0)

∴MidpointofBC=

(62

,–32 )

=

(3,–32 )

20 y =

13 x

= x– 1

3

dydx

= –13

x– 43

= –

1

3x43

Whenx=8, y =

13 8

=

12

δx = 7.9–8 = –0.1

and

dydx

= –

1

3(8)43

= –

1

48

Then

1

3 7.9

12

+(–

148)(–0.1)

0.5021

21 y = x3

dydx

= 3x2

(a) Whenx=1,δx = 1.05–1 = 0.05

and

dydx

= 3(1)2

= 3

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��©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

Then δy

dydx

δx

3(0.05) 0.15 (b) Wheny=27,x3= 27 x = 3 δy= 26.8–27 = –0.2

and

dydx

= 3(3)2

= 27

Then δy

dydx

δx

–0.227δx δx–0.00741

22 y =

3x2

x+1

dydx

=

6x(x+1)–3x2

(x+1)2

=

3x2+6x(x+1)2

=

3x(x+2)(x+1)2

Bythechainrule,

dydt

=

dydx

dxdt

=

[3x(x+2)(x+1)2 ]dx

dt

If

dydt

=4whenx=2

4 =

( 83 )dx

dt

dxdt

=

128

=

32

unitspersecond

23 LetAbetheareaandrtheradius. ThenA=πr2

Bythechainrule,

dAdt

=

dAdr

drdt

5=

2πr

drdt

drdt

=

52πr

=

2.5πr

Whenthecircumference= 40cm 2πr =40

r=

402π

=

20π

∴ drdt

=

2.5

π( 20π )

=0.125cms–1

24 LetVbe thevolumeof sphereandrtheradius,then

V =43

πr3

dVdr

= 4πr2

Whenr=10,δr = 9.98–10

= –0.02

and

dVdr

= 4π(10)2

=400π

δV

dVdr

δr

400π(–0.02) –25.133

25 (a) y = x+5x–2

dydx

= 1–10x–3

= 1–

10x3

Whenx=4,

dydx

= 1–

10(4)3

=

2732

(b)

dydt

=

dydx

dxdt

2.7 =

(2732)dx

dt

dxdt

= 3.2unitspersecond

26 (a) y =

51–3x

dydx

=

15(1–3x)2

(b) Whenx=2, δx = 2+p–2 = p

and

dydx

=

1525

=

35

Then δy

dydx

δx

35

p

27 limx→∞( x2+7x

x2–5 )=

limx→∞(1+

7x

1–5x2

)

=

1+01–0

=1

28 limx→2( x2–2x

x2–4 )=lim

x→2[ x(x–2)(x+2)(x–2)]

=lim

x→2( xx+2)

=

2

2+2

=

24

=

12

29 (a) Forturningpoint,dydx

= 0

6x2+2px = 0 Atthepoint(–3,19), 54–6p =0 6p=54 p=9 Atthepoint(–3,19), 19 =2(–3)3+9(–3)2+q

19=27+q q=–8 p=9,q=–8

(b)

d2ydx2

=12x+18

Whenx=–3,

d2ydx2

=12(–3)+18

=–180 (–3,19)isamaximumpoint

30 (a) y = x3–3x2+4

dydx

= 3x2–6x

(b) Forturningpoints,

dydx

= 0

3x2–6x=0 3x(x–2)=0 x=0orx=2 Whenx=0,y=4 ∴B(0,4) Whenx=2,y=23–3(2)2+4 =0 ∴A(2,0)

31 (a) A(2,3)andB(1,0)

mAB =

–3–1

=3 3px2–3= 3 AtthepointA(2,3), 12p–3= 3 12p =6

p =

12

y=

12

x3–3x+q

AtthepointA(2,3),

3=

12

(2)3–3(2)+q

3= 4–6+q q = 5

∴p=12

,q=5

(b) TheequationofthenormalatA:

y–3= –1

3(x–2)

3y–9= –x+2 3y+x = 11 (c) Atx-axis,y=0 3(0)+x = 11 x = 11 ∴C(11,0)

32 (a) mAB = –124

=–3 2x–9=–3 2x=6 x=3 Whenx=3,y = 32–9(3)+18 =0 ∴P(3,0) (b) EquationofthenormalatP:

y =

13

(x–3)

3y = x–3

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��©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

(c) y=x2–9x+18… 1

x=3y+3… 2

Substitute 2 into 1 :

y=(3y+3)2–9(3y+3)+18 y=9y2+18y+9–27y–27+18 9y2–10y = 0 y(9y–10)= 0

y=0ory=

109

Substitutey=

109

into

2 :

x=3(10

9 )+3

=

193

∴Q(193

,109 )

33 (a) y =x2–3x+4

dydx

=2x–3

AtthepointP(1,2),

dydx

= 2(1)–3

=–1 EquationofthetangentatP: y–2 = –1(x–1) y–2 =–x+1 y+x= 3 (b) AtthepointQ(3,4),

dydx

= 2(3)–3

= 3 EquationofthenormalatQ(3,4):

y–4= –

13

(x–3)

3y–12= –x+3 3y+x = 15 (c) y+x =3… 1

3y+x = 15… 2

2 – 1 : 2y = 12 y = 6 Substitutey=6into 1 : 6+x = 3 x=–3 ∴R(–3,6)

34 (a) y = ax+bx

dydx

= a–

bx2

AtP(3,5),

dydx

=

13

a–

b9

=

13

9a–b=3… 1

and 5=3a+

b3

9a+b = 15 … 2

1 + 2 :18a=18 a=1 Substitutea=1into 1 : 9–b = 3 b=6 ∴a=1,b=6

(b) y= x+

6x

xy=x2+6… 1

y=14–3x … 2

Substitute 2 into 1 : x(14–3x)= x2+6 14x–3x2= x2+6 4x2–14x+6=0 2x2–7x+3=0 (2x–1)(x–3)=0

x=

12

orx=3

Substitutex=

12

into

2 :

y=14–3(1

2)

=

252

∴Q(12

,252 )

35 (a) A(1,2)andB(0,–1)

mAB =

–3–1

=3 AtA(1,2),3ax2–6= 3 3a–6=3 3a=9 a=3 and 2= 3(1)3–6(1)+b 2=–3+b b=5 ∴a=3,b=5 (b) LetCbe(x,0),

mAC = –

13

2

1–x =

–13

6= –1+x x=7 ∴C(7,0) (c) AreaofΔABC

=

12 12

0–1

70

12

=

12

(–1+14+7)

=

12

(20)

=10unit2

36 V=

43

πr3

dVdr

= 4πr2

(a) Whenr=16,δr= 15.9–16 =–0.1

and

dVdr

=4π(16)2

=1024π

Then

dV

dVdr

δr

1024π(–0.1) –102.4π (b) Bythechainrule,

dVdt

=dVdr

drdt

1000 = 4πr2

drdt

drdt

=

10004πr2

Whenr=5,

dAdt

=

1000100π

=

10π

cms–1

37 (a) V =

43

πr3

dVdr

= 4πr2

Bythechainrule,

dVdt

=

dVdr

drdt

dVdt

= 4πr2(0.25)

Whenr=4,

dVdt

= 64π(0.25)

= 16πcm3s–1

(b) A = 4πr2

dAdr

= 8πr

Bythechainrule,

dAdt

=

dAdr

drdt

=8πr(0.25)

Whenr=4,

dAdt

= 32π(0.25)

=8πcm2s–1

38 (a)

r

x

5

10

Fromthediagram,

rx

=

510

r =

12

x

Thevolumeofwaterinthecone,

V=

13

πr2x

=

13

π(12

x)2

x

=

112

πx3(shown)

(b) V =

1

12 πx3

dVdx

=

14

πx2

Whenx=4,δx = 4.05–4 =0.05

and

dVdx

=

14

π(4)2

=4π

Then δV=

dVdx

δx

=4π(0.05) =0.2πcm3

Page 78: Analysis Spm Additional Mathematics

��©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

39 (a) V =

13

πx3

dVdx

= πx2

Bythechainrule,

dVdt

=

dVdx

dxdt

25=πx2

dxdt

dxdt

=

25πx2

Whenx=5,

dxdt

=

2525π

=

cms–1

(b) A = πx2

dAdx

=2πx

Bythechainrule,

dAdt

=

dAdx

dxdt

= 2πx(1

π ) = 2x

Whenx=5,

dAdt

= 2(5)

=10cm2s–1

40 (a) V =2x2+16x

dVdx

=4x+16

Bythechainrule,

dVdt

=

dVdx

dxdt

12=(4x+16)

dxdt

Whenx=2,

dxdt

=

1224

=

12

cms–1

(b) 12=(4x+16)(0.2) 4x+16= 60 4x =44 x=11cm

41 (a) y =

c(x2+1)3

y =c(x2+1)–3

dydx

=–3c(x2+1)–4(2x)

= –

6cx

(x2+1)4

(b) Whenx=1,δx = 1+p–1 = p

and

dydx

= –6c16

=

–3c

8

Then δy

dydx

δx

– 3

4p

–3c

8(p)

c2

42 (a) 50x+2y = 480 25x+y=240 y=240–25x

A =24xy+

12

(24x)(5x)

=24x(240–25x)+60x2

=5760x–600x2+60x2

=5760x–540x2(shown) (b) FormaximumvalueofA,

dAdx

= 0

5760–1080x =0

x= 16

3

andso,

d2Adx2

=–1080

∴Aismaximum.

Whenx=

163

,

y=240–25(16

3 )

=106 23

Hence,themaximumarea,

A=5760(16

3 )–540(16

3 )2

=15360cm2

43 (a)

C

y

R

Q10 – x

45°

Fromthediagram,

y10–x

= tan45°

y =10–x AreaofPQRS, A=2xy =2x(10–x)(shown) (b) ForamaximumvalueofA,

dAdx

= 0

20–4x = 0 4x =20 x=5

andso

d2Adx2

=–40

Hence,themaximumareais2(5)(10–5)=50cm2.

44 (a) 2y+2x+πx = 200 2y=200–2x–πx

y=100–x–

12

πx

A= 2xy+

12

πx2

=2x(100–x–

12

πx)+12

πx2

=200x–2x2–πx2+

12

πx2

=200x–2x2–12

πx2(shown)

(b) ForamaximumvalueofA,

dAdx

= 0

200–4x–πx = 0 x(4+π)=200

x=

200

4+π =28

andso

d2Adx2

=–7.1420

Hencethemaximumarea,

A= 200(28)–2(28)2–

12

π(28)2

=5600–1568–1232 =2800cm2

45 (a) 2r+rθ = 16 rθ=16–2r

θ =

16–2r

r

A =

12

r2θ

=

12

r2(16–2rr )

=

12

r(16–2r)

= 8r–r2

=r(8–r)(shown)

(b)

dAdr

= 8–2r

dAdt

=

dAdr

drdt

=(8–2r)(0.05)

Whenr=3,

dAdt

= 2(0.05)

=0.1cm2s–1

46 (a) s =rθ

dsdθ

=r

dsdt

=

dsdθ

dθdt

2=r dθ

dt

Whenr=16,2=16 dθ

dt

dθdt

=18

radianpersecond

(b) A =

12

r2θ

dAdθ

=

12

r2

dAdt

=

dAdθ

dθdt

=

12

r2dθdt

Whenr=16,

dAdt

=

12

(16)2( 18 )

=16cm2s–1

Page 79: Analysis Spm Additional Mathematics

10 Solution of Triangles

1

Booster Zone

1 (a) xsin 40°

= 6sin 15°

x = 6 sin 40°sin 15°

= 14.9 cm

(b) xsin 40°

= 12

sin 120°

x = 12 sin 40°sin 120°

= 8.91 cm

(c) 8.5sin 123°

= 5.5sin θ

sin θ = 5.5 sin 123°

8.5 = 0.5427 θ = 32° 52'

(d) 7sin 110°

= 4sin ∠QPR

sin ∠QPR = 4 sin 110°

7 = 0.5370 ∠QPR = 32° 29' θ = 180° – (110° + 32° 29') = 37° 31'

(e) 10sin 160°

= 5

sin ∠PRQ

sin ∠PRQ = 5 sin 160°

10 = 0.1710 ∠PRQ = 9° 51' θ = 180° – (160° + 9° 51') = 10° 9'

(f) xsin 75°

= 9

sin 70°

x = 9 sin 75°sin 70°

= 9.25 cm

2 (a) 5sin 25°

= 8sin θ

sin θ = 8 sin 25°

5 = 0.6762 θ = 42° 33', 137° 27' ∴ θ = 137° 27'

(b) 8

sin 10° = 15

sin θ

sin θ = 15 sin 10°8

= 0.3256 θ = 19°, 161° ∴ θ = 161°

(c) 16sin 20°

= 20sin θ

sin θ = 20 sin 20°

16 = 0.4275 θ = 25° 19', 154° 41'

∴ θ = 154° 41'

(d) 5sin 8°

= 9

sin θ

sin θ = 9 sin 8°5

= 0.2505 θ = 14° 30', 165° 30'

∴ θ = 165° 30'

3 (a)

5

sin 25° =

9sin ∠ABC

sin ∠ABC = 9 sin 25°

5 = 0.7607 ∠ABC = 49° 32' ∴ ∠AB'C = 180° – 49° 32' = 130° 28'

(b)

12sin ∠ABC

= 6sin 15°

sin ∠ABC = 12 sin 15°6

= 0.5176 ∠ABC = 31° 10', 148° 50' ∴ ∠AB'C = 148° 50'

(c)

7sin 10°

= 16sin ∠ABC

sin ∠ABC = 16 sin 10°7

= 0.3969 ∠ABC = 23° 23', 156° 37' ∴ ∠AB'C = 156° 37'

(d)

8

sin 20° =

14sin ∠ABC

sin ∠ABC = 14 sin 20°

8 = 0.5985 ∠ABC = 36° 46', 143° 14' ∴ ∠AB'C = 143° 14'

4 (a) 10sin 130°

= 6

sin ∠RQS

sin ∠RQS = 6 sin 130°10

= 0.4596 ∠RQS = 27° 22'

(b) PRsin 42° 22'

= 10

sin 115°

PR = 10 sin 42° 22'

sin 115° = 7.44 ∴PS = 7.44 – 6 = 1.44 cm

5 (a) 12sin 85°

= 6

sin ∠QRS

sin ∠QRS = 6 sin 85°

12 = 0.4981 ∠QRS = 29° 52' ∴ ∠RQS = 180° – (85° + 29° 52') = 65° 8'

(b) PQsin 95°

= 6sin 50°

PQ = 6 sin 95°sin 50°

= 7.8 cm

6 (a) x2 = 82 + 102 – 2(8)(10) cos 160° = 314.351

x = 314.351 = 17.73 cm

(b) x2 = 42 + 92 – 2(4)(9) cos 110° = 121.625

x = 121.625 = 11.03 cm

(c) x2 = 72 + 92 – 2(7)(9) cos 115° = 183.25

x = 183.25 = 13.54 cm

(d) x2 = 52 + 122 – 2(5)(12) cos 135° = 253.853 x = 253.853 = 15.93 cm

A

9 cm 5 cm

C B' B25°

A

16 cm

7 cm

C

B'

B

10°

A14 cm 8 cm

CB' B

20°

A

12 cm

6 cm

C

B'

B

15°

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 80: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

7 (a) 172 = 82 + 122 – 2(8)(12) cos θ

cos θ = 82 + 122 – 172

2(8)(12) = –0.4219 θ = 114° 57'

(b) 162 = 142 + 202 – 2(14)(20) cos θ cos θ = 142 + 202 – 162

2(14)(20) = 0.6071 θ = 52° 37'

(c) 72 = 62 + 92 – 2(6)(9) cos θ

cos θ = 62 + 92 – 72

2(6)(9) = 0.6296 θ = 50° 59'

(d) 202 = 72 + 182 – 2(7)(18) cos θ

cos θ = 72 + 182 – 202

2(7)(18) = –0.1071 θ = 96° 9'

8 (a) BD2 = 62 + 102 – 2(6)(10) cos 50° = 58.865

BD = 58.865 = 7.67 cm

(b) 92 = 52 + 7.672 – 2(5)(7.67) cos ∠BDC

cos ∠BDC = 52 + 7.672 – 92

2(5)(7.67) = 0.0369 ∠BDC = 87° 54', 92° 6' ∴ ∠BDC = 92° 6'

9 (a) QSsin 145°

= 4sin 15°

QS = 4 sin 145°

sin 15° = 8.87 cm

(b) 8.872 = 82 + 122 – 2(8)(12) cos ∠QRS

cos ∠QRS = 82 + 122 – 8.872

2(8)(12) = 0.6736 ∠QRS = 47° 39'

10 (a) AC2 = 22 + 52 – 2(2)(5) cos 110° = 35.84

AC = 35.84 = 5.99 cm

(b) 5.99sin ∠ADC

= 8

sin 80°

sin ∠ADC = 5.99 sin 80°

8 = 0.7374 ∠ADC = 47° 30' ∴ ∠ACD = 180° – (80° + 47° 30') = 52° 30'

11 (a) A = 12

(4)(9) sin 130°

= 13.79 cm2

(b) A = 12

(3)(4.5) sin 70°

= 6.34 cm2

(c) A = 12

(5)(6.2) sin 33°

= 8.44 cm2

(d) 9sin 54°

= 10sin ∠PQR

sin ∠PQR = 10 sin 54°9

= 0.8989 ∠PQR = 64° 1' ∠QPR = 180° – (54° + 64° 1') = 61° 59'

A = 12

(9)(10) sin 61° 59'

= 39.73 cm2

(e) 22 = 42 + 52 – 2(4)(5) cos ∠PRQ

cos ∠PRQ = 42 + 52 – 22

2(4)(5) = 0.925 ∠PRQ = 22° 20'

A = 12

(4)(5) sin 22° 20'

= 3.8 cm2

(f) 122 = 62 + 92 – 2(6)(9) cos ∠ABC

cos ∠ABC = 62 + 92 – 122

2(6)(9) = –0.25 ∠ABC = 104° 29'

A = 12

(6)(9) sin 104° 29'

= 26.14 cm2

12 (a)

16.162 = 102 + 172 – 2(10)(17) cos θ

cos θ = 102 + 172 – 16.162

2(10)(17) = 0.3760 θ = 67° 55'

A = 12

(10)(17) sin 67° 55'

= 78.76 cm2

(b)

132 = 52 + 13.422 – 2(5)(13.42) cos θ

cos θ = 52 + 13.422 – 132

2(5)(13.42) = 0.2690 θ = 74° 23'

A = 12

(5)(13.42) sin 74° 23'

= 32.31 cm2

13 (a)

9.612 = 12.52 + 172 – 2(12.5)(17) cos θ

cos θ = 12.52 + 172 – 9.612

2(12.5)(17) = 0.8303 θ = 33° 52'

(b) Area of ΔACT

= 12

(12.5)(17) sin 33° 52'

= 59.18 cm2

14 (a)

tan ∠VCW = 125

= 2.4 ∠VCW = 67° 23'

(b) Area of ΔAVC = 12

(10)(13)

sin 67° 23' = 60 cm2

SPM Appraisal Zone

1 (a) 6sin 85°

= 4sin ∠ADC

sin ∠ADC = 4 sin 85°6

= 0.6641 ∠ADC = 41° 37' ∴ ∠CAD = 180° – (85° + 41° 37') = 53° 23'

(b) 42 = 22 + 32 – 2(2)(3) cos ∠ABC

cos ∠ABC = 22 + 32 – 42

2(2)(3) = – 0.25 ∠ABC = 104° 29'

(c) Area of ΔABC

= 12

(2)(3) sin 104° 29'

= 2.91 cm2

Area of ΔCAD

= 12

(4)(6) sin 53° 23'

= 9.63 cm2

∴ Area of ABCD = 2.91 + 9.63 = 12.54 cm2

16.16 cm

17 cm θ

10 cm

5 cm

13 cm

θ

13.42 cm

9.61 cm

17 cmθ

12.5 cm

A

T

C

13

5

12

W

V

C

Page 81: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

2 (a) QS2 = 72 + 92 – 2(7)(9) cos 65° = 76.75 QS = 8.76 cm

(b) 8.76sin 110°

= 5sin ∠QSR

sin ∠QSR = 5 sin 110°

8.76 = 0.5364 ∠QSR = 32° 26'

(c) Area of ΔQPS = 12

(7)(9) sin 65°

= 28.55 cm2

Area of ΔRQS

= 12

(8.761)(5) sin 37° 34'

= 13.35 cm2

∴ Area of PQRS = 28.55 + 13.35 = 41.9 cm2

3 (a) PR2 = 62 + 82 – 2(6)(8) cos 55° = 44.937 PR = 6.7 cm

(b) 6.7sin 60°

= 5

sin ∠RPS

sin ∠RPS = 5 sin 60°6.7

= 0.6463 ∠RPS = 40° 15' ∴ ∠PRS = 180° – (60° + 40° 15') = 79° 45'

(c) Area of Δ PQR = 12

(6)(8) sin 55°

= 19.66 cm2

Area of ΔPRS

= 12

(5)(6.7) sin 79° 45'

= 16.48 cm2

∴ Area of PQRS = 19.66 + 16.48 = 36.14 cm2

(d)

6.7sin 55°

= 8sin ∠PRQ

sin ∠PRQ = 8 sin 55°

6.7 = 0.9781 ∠PRQ = 77° 59' ∴ ∠PR'Q = 180° – 77° 59' = 102° 1'

4 (a) (i)

tan θ = 8

17.09 = 0.4681 θ = 25° 5'

(ii)

tan θ = 86

= 1.3333 θ = 53° 8'

(b)

162 = 12.812 + 12.812 – 2(12.81)(12.81) cos ∠ATB

cos ∠ATB

= 12.812 + 12.812 – 162

2(12.81)(12.81) = 0.22 ∠ATB = 77° 17'

(c) Area of ΔATB

= 12

(12.81)(12.81) sin 77° 17'

= 80 cm2

5 (a) (i) BD2 = 42 + 72 – 2(4)(7) cos 60° = 37 BD = 6.08 cm

(ii) 6.08

sin 30° =

8sin ∠ADB

sin ∠ADB = 8 sin 30°

6.08 = 0.6579 ∠ADB = 41° 8'

(iii) Area of ΔABD

= 12

(6.08)(8) sin 108° 52'

= 23.01 cm2

Area of ΔBCD

= 12

(4)(7) sin 60°

= 12.12 cm2

∴ Area of ABCD = 23.01 + 12.12 = 35.13 cm2

(b) (i)

(ii) ∠AD'B = 180° – 41° 8' = 138° 52'

6 (a) 12

(12)(20) sin ∠BAD = 60

sin ∠BAD = 60120

= 0.5 ∠BAD = 30°

(b) BD2 = 122 + 202 – 2(12)(20) cos 30° = 128.308 BD = 11.33 cm

(c) 11.33sin 20°

= 28

sin ∠BDC

sin ∠BDC = 28 sin 20°

11.33 = 0.8452 ∠BDC = 57° 41' ∴ ∠BDC = 180° – 57° 41' = 122° 19'

(d) Area of ΔCBD

= 12

(11.33)(28) sin 37° 41'

= 96.96 cm2

∴ Area of ABCD = 60 + 96.96 = 156.96 cm2

7 (a) (i) QSsin 30°

= 10sin 105°

QS = 10 sin 30°sin 105°

= 5.18 cm

(ii) 5.182 = 922 + 112 – 2(9)(11) cos ∠QPS

cos ∠QPS = 92 + 112 – 5.182

2(9)(11) = 0.8847 ∠QPS = 27° 48'

(iii) Area of ΔQPS

= 12

(9)(11) sin 27° 48'

= 23.09 cm2

Area of ΔRQS

= 12

(5.18)(10) sin 45°

= 18.3 cm2

∴ Area of PQRS = 23.09 + 18.3 = 41.39 cm2

(b) (i)

6 cm

R'8 cm

R

Q

P

17.09 cm

8 cm

C

F

6 cm

8 cm

θ

12.81 cm

16 cm B

T

A

8 cm

6.08 cm

A

D

B30°

D'

41° 8'

9 cm

5.18 cm

Q

P

SQ'

27° 48'

Page 82: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(ii) 5.18sin 27° 48'

= 9

sin ∠PQS

sin ∠PQS = 9 sin 27° 48' 5.18

= 0.8103 ∠PQS = 54° 7' ∠PQ'S = 180° – 54° 7' = 125° 53' ∠PSQ = 180° – (27° 48'

+ 125° 53') = 26° 19'

Area of ΔPQ'S

= 12

(5.18)(9) sin 26° 19'

= 10.33 cm2

8 (a) CD2 = 182 + 242 – 2(18)(24) cos 30° = 151.754 CD = 12.32 cm

(b) 12.322 = 19.312 + 252 – 2(19.31)(25) cos ∠CAD

cos ∠CAD = 19.312 + 252 – 12.322

2(19.31)(25) = 0.8763 ∠CAD = 28° 48'

(c) Area of ΔCAD

= 12

(19.31)(25) sin 28° 48'

= 116.28 cm2

9 (a) (i) PR2 = 92 + 182 – 2(9)(18) cos 120°

= 567 PR = 23.81 cm

(ii) 6

sin 12° = 23.81

sin ∠RSP

sin ∠RSP = 23.81 sin 12°6

= 0.8251 ∠RSP = 55° 36'

(iii) Area of ΔPQR

= 12

(9)(18) sin 120°

= 70.15 cm2

Area of ΔPRS

= 12

(6)(23.81) sin 112° 24'

= 66.04 cm2

∴ Area of PQRS = 70.15 + 66.04 = 136.19 cm2

(b) (i)

(ii) PS'

sin 43° 36' =

6sin 12°

PS' = 6 sin 43° 36'sin 12°

= 19.9 cm

10 (a) 12

(8)(12) sin ∠PQS = 24

sin ∠PQS = 0.5 ∠PQS = 30°

(b) PS2 = 82 + 122 – 2(8)(12) cos 30° = 41.723 PS = 6.46 cm

(c) 13sin 20°

= 8sin ∠QRS

sin ∠QRS = 8 sin 20° 13

= 0.2105 ∠QRS = 12° 9'

(d) Area of ΔQSR

= 12

(8)(13) sin 147° 51'

= 27.67 cm2

∴ Area of PQRS = 24 + 27.67 = 51.67 cm2

6 cm

23.81 cm

R

P

S

12°

124° 24'

S'55° 36'

Page 83: Analysis Spm Additional Mathematics

11 Index Number

1

Booster Zone

1 I10/08

= 13 23012 600

× 100 = 105

2 I09/07

= 45643260

× 100 = 140

3 (a) I06/05

= 8401200

× 100

= 70

(b) I10/08

= 1.801.20

× 100

= 150

(c) I04/08

= 26002000

× 100

= 130

(d) I09/06

= 2.101.50

× 100

= 140

4 (a) 420P

0

× 100 = 105

P0 = 420 × 100

105

= RM400

P

1

1200 × 100 = 87.5

P1 = 87.5 × 1200

100

= RM1050

(b) 560P

0

× 100 = 112

P0 = 560 × 100

112

= RM500

P

1

60 × 100 = 125

P1 = 125 × 60

100

= RM75

(c) 29P

0

× 100 = 145

P0 = 29 × 100

145

= RM20

P

1

80 × 100 = 90

P1 = 90 × 80

100

= RM72

5 p = 1110

× 100 = 110

7.8q

× 100 = 104

q = 7.80 × 100104

= 7.50

r

5.00 × 100 = 112

r = 112 × 5.00100

= 5.60

(s + 1)

s × 100 = 105

s + 1 = 1.05s 0.05s = 1 s = 20 ∴ p = 110, q = 7.50, r = 5.60, s = 20

6 P

08

P00

÷ P

05

P00

= 1.261.20

P

08

P05

= 1.05

p = 1.05 × 100 = 105

1.32 ÷ P

05

P00

= 1.10

P

05

P00

= 1.321.10

= 1.2 q = 1.2 × 100 = 120

P

08

P00

÷ 1.25 = 1.16

P

08

P00

= 1.16 × 1.25

= 1.45 r = 1.45 × 100 = 145 ∴ p = 105, q = 120, r = 145

7 x = 2424

× 100 = 100

y24

× 100 = 125

y = 125 × 24100

= 30

z = 3624

× 100

= 150 ∴ x = 100, y = 30, z = 150

8 (a) –I =

115(6) + 120(5) + 150(4)15

= 189015

= 126

(b) –I =

125(2) + 110(5) + 105(3)10

= 111510

= 111.5

9 (a) –I

09/07 =

120(4) + 110(3) + 105(2) + 125(1)

10

= 114510

= 114.5

(b) –I

09/07 =

115(4) + 104(6) + 105(8) + 110(2)

20

= 214420

= 107.2

(c) –I

09/07 =

105(2) + 115.5(4) + 108(3) + 115(6)

15 =

168615

= 112.4

10 (a)

128(2) + 110(6) + m + 85(4)

13 = 105

m + 1256 = 1365 m = 109

(b)

122(4) + 120m + 132(5) + 86(2)

m + 11 = 110

1320 = 110m + 1210 110m = 110 m = 1

(c)

120(m – 2) + 112m + 115(2)

2m = 115

232m – 10 = 230m 2m = 10 m = 5

(d)

150(m + 5) + 130(2) + 90m

+ 110(7)2m + 14

= 125

1780 + 240m = 250m + 1750 10m = 30 m = 3

11 (a)

115(2) + 8m + 125(4) + 105(6)

20 = 116

1360 + 8m = 2320 8m = 960 m = 120

(b) –I

10/05 = 116 × 1.05

= 121.8

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 84: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

12 (a) –I

07/04 =

110(3) + 105(4) + 115(2) + 140(1)

10

= 112010

= 112

(b) –I

09/04 =

120(3) + 125(4) + 115(2) + 130(1)

10

= 122010

= 122

(c) –I

09/07 =

122112

× 100

= 108.93

13 (a) p = 18.0012.00

× 100

= 150

(b) 28q

× 100 = 140

q = 28 × 100140

= 20

(c) r

7.20 × 100 = 125

r = 125 × 7.20100

= 9

(d)

150(7) + 140(3) + 125(6) + 110s

s + 16 = 133

2220 + 110s = 133s + 2128 23s = 92 s = 4

14 (a) A : 110 × 1.05 = 115.5 B : 112 × 1.1 = 123.2 C : 124 × 0.95 = 117.8 D : 115

(b) –I

10/06 =

115.5(2) + 123.2(5) + 117.8(5) + 115(2)

14

= 166614

= 119

(c) P

10

2500 × 100 = 119

P10

= 119 × 2500100

= RM2975

SPM Appraisal Zone

1 (a) x

1.20 × 100 = 125

x = 125 × 1.20100

= 1.50

(b) y + 0.30

y × 100 = 120

y + 0.3 = 1.2y 0.2y = 0.3 y = 1.50

∴ y = 1.50 and z = y + 0.30 = 1.50 + 0.30 = 1.80

(c) (i) P

09

5 × 100 = 133

P09

= 133 × 5100

= RM6.65

(ii)

125(4) + 150(7) + 120(6) + 130m

m + 17 = 133

2270 + 130m = 133m + 2261 3m = 9 m = 3

2 (a) x = 0.900.60

× 100

= 150

y

1.20 × 100 = 125

y = 125 × 1.20100

= 1.50

0.90

z × 100 = 112.5

z = 0.90 × 100112.5

= 0.80 ∴ x = 150, y = 1.50, z = 0.80

(b) (i) –I

10/07 =

150(3) + 125(6) + 120(4) + 112.5(2)

15

= 190515

= 127

(ii) 4.70P

07

× 100 = 127

P07

= 4.70 × 100127

= RM3.70

(c) –I

11/07 = 127 × 1.1

= 139.7

3 (a) (i) 0.50P

06

× 100 = 125

P06

= 0.50 × 100125

= RM0.40

(ii) P

08

P06

= 1.4

P

06

P04

= 1.1

P

08

P04

= 1.4 × 1.1

= 1.54 I

08/04 = 1.54 × 100

= 154

(b) (i) –I

08/06 = 124

140(2) + 4x + 105(3) + 125(1)

10 = 124

4x + 720 = 1240 4x = 520 x = 130

(ii) P

08

2.50 × 100 = 124

P08

= 124 × 2.50100

= RM3.10

4 (a) 1.80

x × 100 = 150

x = 1.80 × 100150

= 1.20

y

0.80 × 100 = 175

y = 175 × 0.80100

= 1.40

z = 1.201.00

× 100

= 120 ∴ x = 1.20, y = 1.40, z = 120

(b) –I

07/05 =

130(5) + 150(3) + 175(2) + 125(4) + 120(1)

15

= 207015

= 138

(c) Let a = new weightage of banana b = new weightage of

pineapple 6 + 5 + 2 + a + b = 20 13 + a + b = 20 a + b = 7 ... 1

–I = 137

130(6) + 150(5) + 175(2) + 125a + 120b

20 = 137

1880 + 125a + 120b = 2740 125a + 120b = 860 25a + 24b = 172 ... 2 1 × 24 : 24a + 24b = 168 ... 3 2 – 3 : a = 4

Substitute a = 4 into 1 : 4 + b = 7 b = 3 ∴ a = 4, b = 3

5 (a) p = 4.804.00

× 100

= 120

q

2.50 × 100 = 112

q = 112 × 2.50100

= 2.80

2.75

r × 100 = 125

r = 2.20 ∴ p = 120, q = 2.80, r = 2.20

Page 85: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) –I

08/06 =

120(4) + 112(3) + 125(1) + 150(2)

10

= 124110

= 124.1

(c) P

08

2000 × 100 = 124.1

P08

= 124.1 × 2000100

= RM2482

(d) I10/06

= 124.1 × 1.2 = 148.92

6 (a) (i) P

04

8.50 × 100 = 120

P04

= 120 × 8.50100

= RM10.20

(ii) 16.50

P00

× 100 = 132

P00

= 16.50 × 100132

= RM12.50

(b) (i) U : 1.501.20

× 100 = 125

V : 1.321.10

× 100 = 120

W : 1.471.05

× 100 = 140

X : 1.041.30

× 100 = 80

∴ U : 125, V : 120, W : 140, X : 80

(ii)

125(2) + 120(3) + 140(4) + 80m

m + 9 = 125

1170 + 80m = 125m + 1125 45m = 45 m = 1

7 (a) x = 6050

× 100

= 120

y

42 × 100 = 140

y = 140 × 42100

= 58.8 ∴ x = 120, y = 58.8

(b) –I

10/08 = 137

120(200) + 140(700) + 150zz + 900

= 137

122 000 + 150z = 137z + 123 300 13z = 1300 z = 100

(c) P

10

4100 × 100 = 137

P10

= 137 × 4100100

= RM5617

8 (a) –I

09/07 =

120(4) + 200(2) + 150(1) + 128(3)

10

= 141410

= 141.4

(b) P

09

1000 × 100 = 141.4

P09

= 141.4 × 1000100

= RM1414

(c) (i) –I

10/07 = 141.4 × 1.1

= 155.54

(ii) –I

10/09 = 110

9 (a)

Ingredient

Price index in

2010 (2008 = 100)

Weightage

Noodles 120 5

Sweet potatoes

105 3

Bean curds 102 1

Bean sprouts

104 2

Onions 110 4

–I

10/08 =

120(5) + 105(3) + 102(1) + 104(2) + 110(4)

15

= 166515

= 111

(b) (i) P

10

1.50 × 100 = 120

P10

= RM1.80

(ii) 2.20P

08

× 100 = 105

P08

= RM2.10

(c) P

10

4.50 × 100 = 111

P10

= 111 × 4.50100

= RM5

10 (a) (i)Item I08/07

Coffee 105

Milk 150

Sugar 125

Cream 150

(ii) –I

08/07 =

105(4) + 150(3) + 125(2) + 150(1)

10

= 127010

= 127

(b) (i) –I

10/08 = 127

(ii) P

10

2500 × 100 = 127

P10

= 127 × 2500100

= RM3175

Page 86: Analysis Spm Additional Mathematics

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-31

SPM-Cloned Questions

Chapter 1: Functions 1 (a) 7

(b) 2, 3 2 h(x) = 1—

2x – m

Let y = 1—

2x – m

1—

2x = y + m

x = 2y + 2m∴h–1(x) = 2x + 2mCompare with h–1(x) = nx + 6∴ 2m = 6 and n = 2 m = 3∴ m = 3, n = 2

3 h(x) = 3x – 5Let y = 3x – 5 3x = y + 5

x = y + 5——–

3

∴ h–1(x) = x + 5——–

3 ∴ gh–1(x) = g� x + 5

——–3 �

= 2� x + 5

——–3 � – 1

= 2x + 10 – 3—————

3 =

2x + 7———

3 4 (a) k(x) =

3——–4 – x

Let y = 3

——–4 – x

4y – xy = 3 xy = 4y – 3

x = 4y – 3———

y

∴ k –1(x) = 4x – 3———

x, x ≠ 0

(b) k –1� 1—2 � =

4� 1—2 � – 3

————1—2

=

–1—–1—2

= –2

5 (a) Many to one relation.(b) f : x x2 – 1

6 (a) Let y = m(x) Then y = 2x – 5 2x = y + 5 x =

y + 5——–

2

∴ m–1(x) = x + 5——–2

m–1(–3) = –3 + 5———

2 = 1(b) nm(x) = n(2x – 5) = (2x – 5)2 + 3(2x – 5) – 4 = 4x2 – 20x + 25 + 6x – 15 – 4 = 4x2 – 14x + 6

7 (a) 3 and 7(b) Domain = {3, 5, 7}

8 (a) Let y = f(x) Then y = 2x – 3 2x = y + 3

x = y + 3——–

2

∴ f –1(x) = x + 3——–

2 (b) f –1g(x) = f –1� x—

4 + 1�

=

x—4

+ 1 + 3————–

2 =

� x + 16———4 �

————2

= x + 16———8

(c) hg(x) = 1—2

x + 5

h� x—4

+ 1� = 1—2

x + 5

Let k = x—4

+ 1 x—

4 = k – 1

x = 4k – 4 h(k) = 1—

2(4k – 4) + 5

= 2k – 2 + 5 = 2k + 3 ∴ h(x) = 2x + 3

Chapter 2: Quadratic Equations

1 Sum of roots: p + q + q – 3 = 2 p + 2q = 5 p = 5 – 2q … 1Product of roots: (p + q)(q – 3) = –24 pq – 3p + q2 – 3q = –24 … 2

Substitute 1 into 2 : q(5 – 2q) – 3(5 – 2q) + q2 – 3q = –24 5q – 2q2 – 15 + 6q + q2 – 3q = –24 q2 – 8q – 9 = 0 (q + 1)(q – 9) = 0 q = –1 or q = 9Substitute q = –1 into 1 : p = 5 – 2(–1) = 7

Substitute q = 9 into 1 : p = 5 – 2(9) = –13∴ p = 7, q = –1; p = –13, q = 9

2 2x2 + 4x + p = 0Sum of roots: α + 3α = –2 4α = –2 α = – 1—

2 … 1

Product of roots: α(3α) =

p—2

3α2 = p—2

… 2

Substitute 1 into 2 :

3�– 1—2 �

2

= p—2

3—4

= p—2

p = 6—4

= 3—2

∴ The roots are – 1—2

and 3�– 1—2 � = – 3—

2and p = 3—

2.

3 x2 + 2k + 10 = x – 3kx x2 + (3k – 1)x + 2k + 10 = 0For two distinct roots, b2 – 4ac � 0 (3k – 1)2 – 4(1)(2k + 10) � 0 9k2 – 6k + 1 – 8k – 40 � 0 9k2 – 14k – 39 � 0 (9k + 13)(k – 3) � 0

∴ k � – 13—–

9 or k � 3

4 x2 + 6x + a = 2x + 1 x2 + 4x + a – 1 = 0For the line not to intersect the curve, b2 – 4ac � 0 42 – 4(1)(a – 1) � 0 16 – 4a + 4 � 0 20 – 4a � 0 4a � 20 a � 5

5 (a) 4x2 – 11x + 6 = 0 (4x – 3)(x – 2) = 0

k3 13– —–

9

ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

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2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

x = 3—4

or x = 2 (b) kx2 + px = 9 kx2 + px – 9 = 0 For two equal roots, b2 – 4ac = 0 p2 – 4(k)(–9) = 0 p2 + 36k = 0

6 x2 + x – k = 0Since –2 is one root of the equation, (–2)2 + (–2) – k = 0 k = 4 – 2 = 2

7 x2 – 2x = p – 2 x2 – 2x + 2 – p = 0For two different roots, b2 – 4ac � 0 (–2)2 – 4(1)(2 – p) � 0 4 – 8 + 4p � 0 4p � 4 p � 1

8 px2 – px + 2q = 2 + 4x px2 + (–p – 4)x + 2q – 2 = 0(a) Sum of roots:

1—p + q = p + 4——–

p 1 + pq = p + 4 pq = p + 3 q =

p + 3——–

p … 1

Product of roots:

1—p (q) = 2q – 2———p

q = 2q – 2 q = 2 Substitute q = 2 into 1 :

2 = p + 3——–p

2p = p + 3 p = 3 ∴ p = 3, q = 2(b) Sum of roots: p + (–3q) = 3 – 3(2) = –3 Product of roots: p(–3q) = 3[–3(2)] = –18 The quadratic equation with roots

p and –3q is x2 – (–3)x + (–18) = 0 x2 + 3x – 18 = 0

Chapter 3: Quadratic Functions

1 f(x) = –3x2 – 6x + 8

= –3�x2 + 2x – 8—3 �

= –3�x2 + 2x + � 2—2 �

2

– 8—3

– � 2—2 �

2

� = –3�(x + 1)2 – 11—–

3 � = 11 – 3(x + 1)2

∴ The maximum value is 11 and the axis of symmetry is x = –1.

2 (5x + 4)(x – 1) � 2(x – 1) 5x2 – x – 4 � 2x – 2 5x2 – 3x – 2 � 0 (5x + 2)(x – 1) � 0

∴ x � – 2—

5 or x � 1

3 (a) k = 1

(b) x = 1(c) (1, –4)

4 (a) p = –2(b) q = 4(c) f(x) = a(x – 2)2 + 4 At point (0, 2), 2 = a(–2)2 + 4 4a = –2

a = – 1—2

5 (a) a = 4(b) x = –4

6 f(x) = x2 + 2x – a2

= x2 + 2x + (1)2 – a2 – (1)2

= (x + 1)2 – a2 – 1∴ –a2 – 1 = –10 a2 = 9 a = ±3

7 (a) f(x) = x2 – 2px + 2p2 + 9 = x2 – 2px + (–p)2 + 2p2 + 9 – (–p)2

= (x – p)2 + p2 + 9 ∴ p2 + 9 = h2 + 6p h2 = p2 – 6p + 9 = (p – 3)2

h = p – 3 (shown)(b) p = h2 + 1 … 1 h = p – 3 … 2 Substitute 2 into 1 : p = (p – 3)2 + 1 p = p2 – 6p + 10 p2 – 7p + 10 = 0 (p – 2)(p – 5) = 0 p = 2 or p = 5 Substitute p = 2 into 2 : h = 2 – 3 = –1 Substitute p = 5 into 2 : h = 5 – 3 = 2 ∴ p = 2, h = –1; p = 5, h = 2

8 (a) At f(x)-axis, x = 0 f(x) = 02 – k(0) + 13 = 13 ∴ A(0, 13)

(b) f(x) = x2 – kx + 13

= x2 – kx + �– k—2 �

2

+ 13 – �– k—2 �

2

= �x – k—

2 �2

+ 13 – k2

—4

∴ k—2

= 3 and p = 13 – k2

—4

k = 6 = 13 – 36—–4

= 13 – 9 = 4 ∴ k = 6, p = 4

Chapter 4: Simultaneous Equations

1 1—2

y = 1 – x y = 2 – 2x … 1 y2 = 2x + 10 … 2

Substitute 1 into 2 : (2 – 2x)2 = 2x + 10 4 – 8x + 4x2 = 2x + 10 4x2 – 10x – 6 = 0 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 x = – 1—

2 or x = 3

Substitute x = – 1—

2 into 1 :

y = 2 – 2�– 1—2 �

= 3Substitute x = 3 into 1 : y = 2 – 2(3) = –4

∴ �– 1—2

, 3�; (3, –4)

2 y – 3x = 7 y = 3x + 7 … 1

x2 + y2 – xy = 7 … 2

Substitute 1 into 2 : x2 + (3x + 7)2 – x(3x + 7) = 7 x2 + 9x2 + 42x + 49 – 3x2 – 7x = 7 7x2 + 35x + 42 = 0 x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = –3 or x = –2Substitute x = –3 into 1 : y = 3(–3) + 7 = –9 + 7 = –2Substitute x = –2 into 1 : y = 3(–2) + 7 = –6 + 7 = 1∴ x = –3, y = –2; x = –2, y = 1

3 e – f = 3 f = e – 3 … 1 e2 – 4f = 10 … 2Substitute 1 into 2 : e2 – 4(e – 3) = 10 e2 – 4e + 2 = 0

e = ————————–2(1)

4 ± (–4)2 – 4(1)(2)

x1 2– —

5

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3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

= ———2

4 ± 8

= 3.414 or 0.586Substitute e = 3.414 into 1 : f = 3.414 – 3 = 0.414Substitute e = 0.586 into 1 : f = 0.586 – 3 = –2.414∴ e = 3.414, f = 0.414; e = 0.586,

f = –2.414 4 x + 1 —

2y = 2

x = 2 – 1 —

2y … 1

y2 = 4x … 2

Substitute 1 into 2 :

y2 = 4�2 – 1 —2

y� y2 = 8 – 2y y2 + 2y – 8 = 0 (y + 4)(y – 2) = 0y = –4 or y = 2

Substitute y = –4 into 1 : x = 2 – 1 —

2(–4)

= 2 + 2 = 4

Substitute y = 2 into 1 : x = 2 – 1 —

2(2)

= 2 – 1 = 1

∴ x = 4, y = –4; x = 1, y = 2

5 2x + y = 3 y = 3 – 2x … 1 2x2 + y2 + xy = 6 … 2

Substitute 1 into 2 : 2x2 + (3 – 2x)2 + x(3 – 2x) = 6 2x2 + 9 – 12x + 4x2 + 3x – 2x2 = 6 4x2 – 9x + 3 = 0 x = ————————–

2(4)9 ± (–9)2 – 4(4)(3)

= ———

89 ± 33

= 1.843 or 0.407Substitute x = 1.843 into 1 : y = 3 – 2(1.843) = –0.686Substitute x = 0.407 into 1 : y = 3 – 2(0.407) = 2.186∴ x = 1.843, y = –0.686; x = 0.407,

y = 2.186

6 x + 2y = 10 x = 10 – 2y … 1 2y2 – 7y + x – 1 = 0 … 2

Substitute 1 into 2 : 2y2 – 7y + 10 – 2y – 1 = 0 2y2 – 9y + 9 = 0 (2y – 3)(y – 3) = 0

y = 3 —2

or y = 3 Substitute y = 3 —

2 into 1 :

x = 10 – 2� 3 —

2 � = 7Substitute y = 3 into 1 : x = 10 – 2(3) = 4 ∴ x = 7, y = 3 —

2; x = 4, y = 3

7 2y – x = 1

x = 2y – 1 … 1 x2 + xy = 26 … 2 (2y – 1)2 + y(2y – 1) = 26 4y2 – 4y + 1 + 2y2 – y = 26 6y2 – 5y – 25 = 0 (3y + 5)(2y – 5) = 0

y = – 5 —3

or y = 5 —2

Substitute y = – 5 —3

into 1 : x = 2�– 5 —

3 � – 1

= – 13 —–3

Substitute y = 5 —2

into 1 : x = 2� 5 —

2 � –1 = 4

∴ x = – 13 —–3

, y = – 5 —3

; x = 4, y = 5 —2

8 k – 2p = –3 k = 2p – 3 … 1 p + pk – 4k = 0 … 2

Substitute 1 into 2 : p + p(2p – 3) – 4(2p – 3) = 0 p + 2p2 – 3p – 8p + 12 = 0 2p2 – 10p + 12 = 0 p2 – 5p + 6 = 0 (p – 2)(p – 3) = 0 p = 2 or p = 3Substitute p = 2 into 1 : k = 2(2) – 3 = 1Substitute p = 3 into 1 : k = 2(3) – 3 = 3∴ k = 1, p = 2; k = 3, p = 3

Chapter 5: Indices and Logarithms

1 log3R – log

9T = 2

log3R –

log3T

——–log

39

= 2 log

3R – 1 —

2 log

3T = 2

log

3

R —–T

= 2

R —–T

= 9 R = 9 T

2 9x = � 3 � x + 6

32x = 3x + 6 ——–

2

∴ 2x = x + 6 ——–2

4x = x + 6 3x = 6 x = 2

3 log

36.25 = log

3

25 —–4

= log325 – log

34

= 2 log35 – 2 log

32

= 2n – 2m 4 log

h � 81h ——16 � = log

h81 + log

hh – log

h16

= 4 logh3 + 1 – 2 log

h4

= 4p + 1 – 2q

5 log8 �4m —–n � =

log2�4m —–n �

————log

28

= log

24 + log

2m – log

2n

—————————log

28

= 2 log

22 + log

2m – log

2n

—————————–3 log

22

= 2 + x – y

————–3

6 4x

——2x – 3

= 82–x

22x

——2x – 3

= (23)2–x

22x – (x – 3) = 26 – 3x

∴ 2x – (x – 3) = 6 – 3x x + 3 = 6 – 3x 4x = 3

x = 3—4

7 2n–7 × 8n = 512

2n–7 × 23n = 29

2n–7 + 3n = 29

∴ n – 7 + 3n = 9 4n = 16 n = 4

8 log81

p – log3q = 0

log3p

——–log

381

= log3q

log3p

———4 log

33

= log3q

1—4

log3p = log

3q

log

3p

1—4 = log

3q

∴ p1—4 = q

p = q4

Chapter 6: Coordinate Geometry

1

A(h, 3h)

1

B(p, 2t)

C(4p, 5t)

2

Page 89: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

2h + 4p———–

3 = p

2h + 4p = 3p p = –2h … 1 6h + 5t

———–3

= 2t 6h + 5t = 6t 6h = t

h = t—6

… 2 Substitute 2 into 1 : p = –2� t—

6 � = – 1—

3t

2 x

—2

+ y

—3

= 1 3x + 2y = 6 2y = –3x + 6 y = – 3—

2x + 3

The gradient of this line is – 3—

2.

3y = 2x + 6 y = 2—

3x + 2

The gradient is 2—

3.

Since �– 3—

2 �� 2—3 � = –1 so these two lines

are perpendicular to each other.

3 x—4

– y—2

= 1

The gradient of this line,

m1 = – �– 2—

4 � = 1—

2 and the coordinates of K is (4, 0).∴ Equation of the perpendicular line

through K is y – 0 = –2(x – 4) y = –2x + 8

4 Equation of the perpendicular line CD through C is y – 2 = 1—

2(x – 0)

2y – 4 = x 2y – x = 4 … 1 y + 2x = 7 … 21 × 2 : 4y – 2x = 8 … 32 + 3 : 5y = 15 y = 3Substitute y = 3 into 1 : 2(3) – x = 4 x = 6 – 4 = 2∴ D(2, 3)

5 a =

2(–2) + 1(13)——————

3 = 9—

3 = 3

b = 2(0) + 1(15)—————–

3

= 15—–3

= 5∴ Q(3, 5)

6 (a) The equation of AB: y – 2 = 1—

2(x + 2)

2y – 4 = x + 2 2y – x = 6 At x-axis, y = 0 2(0) – x = 6 x = –6 ∴ A(–6, 0)(b)

3(–6) + 2x—————5

= –2

–18 + 2x = –10 2x = 8 x = 4 3(0) + 2y————–

5 = 2

2y = 10 y = 5 ∴ C(4, 5)(c) Equation of the perpendicular line

through C: y – 5 = –2(x – 4) y – 5 = –2x + 8 y + 2x = 13

7 (a) QP = 5 units y2 + (x – 5)2 = 5 y2 + x2 – 10x + 25 = 25 x2 + y2 – 10x = 0(b) (i) x2 + y2 – 10x = 0

At M(2, k), 4 + k2 – 10(2) = 0 k2 – 16 = 0 k2 = 16 k = ±4 ∴ k = 4 (ii) Let N be (x, y),

� 2 + x——–2

, k + y——–

2 � = (5, 0) 2 + x——–

2 = 5 and k + y——–

2 = 0

2 + x = 10 4 + y = 0 x = 8 y = –4 ∴ N(8, –4)

8 (a) (i) � 1—k �(–2) = –1

k = 2 (ii) 2y – x = 10 … 1 y + 2x = 10 … 2

1 × 2: 4y – 2x = 20 … 3 2 + 3 : 5y = 30 y = 6 Substitute y = 6 into 1 : 2(6) – x = 10 x = 2 ∴ Q(2, 6)

(b) (i) 2y = x + 10 At y-axis, x = 0 2y = 10 y = 5 ∴ D(0, 5)

1(x) + 3(2)————–

4 = 0

x + 6 = 0 x = –6 and

1(y) + 3(6)————–

4 = 5

y + 18 = 20 y = 2 ∴ R(–6, 2) (ii) m = 2 – 0——–

–6 – 0 = – 1—

3

y – 0 = – 1—3

(x – 0)

y = – 1—3

x

(c) 2EQ = ER

2 (y – 6)2 + (x – 2)2

= (y – 2)2 + (x + 6)2

4(y2 – 12y + 36 + x2 – 4x + 4) = y2 – 4y + 4 + x2 + 12x + 36 4y2 – 48y + 4x2 – 16x + 160 = y2 – 4y + x2 + 12x + 40 3x2 + 3y2 – 28x – 44y + 120 = 0

Chapter 7: Statistics 1

Score f x fx x2 fx2

21–25 1 23 23 529 529

26–30 2 28 56 784 1568

31–35 9 33 297 1089 9801

36–40 11 38 418 1444 15 884

41–45 14 43 602 1849 25 886

46–50 3 48 144 2304 6912

Σf=40 Σfx=1540 Σfx2=60 580

–(a) x = Σfx

——Σf

= 1540——

40 = 38.5(b) Median class = 36 – 40

m = L + �

1—2

N – F————

fm

�C

2

A(–6, 0)

3

B(–2, 2)

C(x, y)

3

1

R(x, y)

D(0, 5)

Q(2, 6)

Page 90: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

= 35.5 + �

1—2

(40) – 12—————–

11�5

= 39.14 (c) σ =

–Σfx2

——Σf

– (x)2

= 60 580———

40 – (38.5)2

= 5.679

2 81—–

3 – � k �2 = 2h

27 – k = 4h2

k = 27 – 4h2

– 3 (a) x = 8

Σx—–

6 = 8

Σx = 48 (b) 48 + m———

7 = 9

48 + m = 63 m = 63 – 48 = 15

4 (a) σ2 = 21, Σx2 = 1275, N = 15

1275——15

– (–x)2 = 21 (–x)2 = 85 – 21 = 64 –x = 64 = 8

(b) Σx—–15

= 8

Σx = 120

5 (a) N = 8, Σx = 120, Σx2 = 2448 –x = Σx—–

N –x = 120—–

8 = 15 σ =

Σx2

—–N

– (–x)2

=

2448——

8 – 152

= 81 = 9

(b) (i)

120 + m———–

9 = 14

120 + m = 126 m = 6 ∴ The number that being

added to the set is 6. (ii) σ =

2448 + 36————–

9 – 142

= 80 = 8.944

6 (a) Age (years)

Number of patients

1 – 10 8

11 – 20 6

21 – 30 5

31 – 40 7

41 – 50 10

Median class = 21 – 30 Median = L + �

N—2

– F———

fm

�C

= 20.5 + �18 – 14———

5 �10 = 20.5 + 8 = 28.5

(b) –x =

8(5.5) + 6(15.5) + 5(25.5) + 7(35.5) + 10(45.5)

———————————36

= 968——36

= 26.89

x f x2 fx2

5.5 8 30.25 242

15.5 6 240.25 1441.5

25.5 5 650.25 3251.25

35.5 7 1260.25 8821.75

45.5 10 2070.25 20 702.5

Σf=36 Σfx2=34 459

σ =

34 459———

36– 26.892

= 234.122 = 15.3

7 (a) Median class = 21–30 Median = 24.5

20.5 +

k + 24�———�2 – 12�——————�

1010 = 24.5

k + 24�———�2 – 12 = 4

k + 24———

2 = 16

k + 24 = 32 k = 8(b)

Mode time = 26.5

(c) From the histogram the mode time is 26.5. If the time of each athlete is decreased by 2, so the new mode is 26.5 – 2 = 24.5

8 (a) Mark Number of students

0 – 9 6

10 – 19 8

20 – 29 12

30 – 39 9

40 – 49 5

(b) Q1 = 9.5 + � 10 – 6———

8 �10

= 14.5 Q

3 = 29.5 + � 30 – 26———–

9 �10

= 33.94 ∴ Interquartile range = 33.94 – 14.5 = 19.44

Chapter 8: Circular Measures

1 Let the radius of a sector COD be r cm, 2r + 2 = 10 2r = 8 r = 4 cm s

CD = 2

4θ = 2

θ = 1—2

radian 2 (a) θ = 360° – 280°

= 80°

= 80° × π——180°

= 1.396 radians(b) s

XY = 20

1.396r = 20 r = 20——–

1.396 = 14.33 cm

3 (a) sQT

= 3 4θ = 3 θ = 0.75 radian ∴ ∠QST = 0.75 radian(b) Area of semicircle PQR = 1—

2(3)2(π)

= 14.14 cm2

Area of sector QST

= 1—2

(4)2(0.75) = 6 cm2

∴ Area of the shaded region = 14.14 – 6 = 8.14 cm2

4 (a) sQR

= 4(2.1) = 8.4 cm

Number of athletes

10

8

6

4

2

0 0.5 10.5 20.5 30.5 40.5 50.5

Time (minutes)26.5

Page 91: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) Area of sector QOR =

1—2

(4)2(2.1)

= 16.8 cm2

Area of ∆POS

= 1—2

(2)(2) sin �2.1 × 180°——π � = 1.73 cm2

Area of the shaded region = 16.8 – 1.73 = 15.07 cm2

5 (a) sQR

= 12(1.18) = 14.16 cm(b) Area of sector OQR = 1—

2(12)2(1.18)

= 84.96 cm2

Area of ∆OPS

= 1—2

(6)(6) sin �1.18 × 180°——π � = 16.64 Area of the shaded region = 84.96 – 16.64 = 68.32 cm2

6 (a) OP = 3—4

(12) = 9 cm

cos θ = 9—–12

= 0.75 θ = 41° 24' = 0.723 radian

(b) RP = 122 – 92

= 7.937 cm

Area of the shaded region = 1—

2 (12)2(0.723) – 1—

2(9)(7.937)

= 52.056 – 35.717 = 16.34 cm2

7 (a) Area of sector POQ = 1—

2(5)2(1.8)

= 22.5 cm2

(b) sPQ

= 5(1.8) = 9 cm s

PS = 8(3.142 – 1.8)

= 10.736 cm ∴ Perimeter of the shaded region = 10.736 + 9 + (8 – 5) = 22.74 cm(c) Area of sector PTS

= 1—2

(8)2(1.342)

= 42.94 cm2

Area of ∆QOT

= 1—2

(5)(3) sin �1.342 × 180°——π � = 7.31 cm2

∴ Area of the shaded region = 42.94 – 22.5 – 7.31 = 13.13 cm2

8 (a)

sin θ = 3—5

= 0.6 θ = 36° 52' ∠AOB = 0.644 radian(b) s

AB = 10(0.644)

= 6.44 cm

sDE

= 3� π—2 �

= 4.71 cm ∴ Perimeter of the shaded region = 6.44 + 5 + 4.71 + (10 – 7) = 19.15 cm(c) Area of sector OAB =

1—2

(10)2(0.644) = 32.2 cm2

Area of ∆OCD = 1—

2(4)(3)

= 6 cm2

Area of quadrant DCE

= 1—2

(3)2� π—2 �

= 7.07 cm2

Area of the shaded region = 32.2 – 6 – 7.07 = 19.13 cm2

Chapter 9: Differentiation 1 y = (3x – 4)3

dy

—–dx

= 3(3x – 4)2(3)

= 9(3x – 4)2

When x = 2,

dy—–dx

= 9[3(2) – 4]2 = 36The rate of change of y is given by

dy

—–dt

= dy

—–dx

× dx

—–dt

18 = 36 � dx—–dt �

dx

—–dt

= 18—–36

= 0.5 unit s–1

∴ x is increasing at a rate of 0.5 unit s–1.

2 h(x) = 2� 1—

2x – 4�

–2

h'(x) = –4� 1—2

x – 4�–3

� 1—2 �

= –2� 1—2

x – 4�–3

h"(x) = 6� 1—2

x – 4�–4

� 1—2 �

= 3————

� 1—2

x – 4�4

h"(4) =

3—————

� 1—2

(4) – 4�4

=

3——(–2)4

= 3—–

16 3 y = 3—

5u4

= 3—

5 (5x – 2)4

dy

—–dx

= 12—–5

(5x – 2)3(5)

= 12(5x – 2)3

4 (a) y = 2x2 + 3x – 2

dy—–dx

= 4x + 3

When x =

1—2

, dy

—–dx

= 4� 1—2 � + 3

= 5(b) When x = 2, δx = 2 + p – 2 = p

dy—–dx

= 4(2) + 3

= 11 Then, δy ≈

dy—–dx

× δx ≈ 11(p) ≈ 11p ∴ The approximate change in y is

11p.

5 Let V be the volume of a packet of butter. Then, V = x3

By the chain rule, dV—–dt

= dV—–dx

× dx—–dt

= 3x2 × dx—–dt

When x = 4, –12 = 3(4)2 dx—–dt

dx—–dt

= – 12—–48

= –0.25 cm s–1

∴ Rate of change of x is –0.25 cm s–1. 6 y =

1—2

x + 3 The gradient of this line is

1—2

.

So the gradient of the tangent at A(1, 4) is 2x – kx2 = –2 2(1) – k(1)2 = –2 2 – k = –2 k = 4

R

12

θP O9

3 5

θ

D

C O

Page 92: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

7 (a) At the turning point (k, 2),

dy

—–dx

= 0

1 – 4—x2

= 0 1 – 4—

k2 = 0

1 =

4—k2

k2 = 4 k = ±2 ∴ k = 2

(b) dy

—–dx

= 1 – 4x–2

d 2y—–dx2

= 8x–3

=

8—x3

At (2, 2),

d 2y——dx2 � 0.

∴ (2, 2) is a minimum point.

8 y = 3 – 2x –1

dy

—–dx

= 2x –2

= 2—

x2 When x = 2, dy—–

dx = 2—–

(2)2

= 1—

2 So the gradient of the tangent at

A(2, 2) is 1—2

and its equation is y – 2 = 1—

2 (x – 2)

2y – 4 = x – 2 2y – x = 2

Chapter 10: Solution of Triangles

1 (a) BD2 = 32 + 42 – 2(3)(4) cos α = 25 – 24 cos α(b) BD2 = 12 + 22 – 2(1)(2) cos β = 5 – 4 cos β α + β = 180° β = 180° – αSubstitute β = 180° – α into 5 – 4 cos β:5 – 4 cos (180° – α)= 5 – 4 [cos 180° cos α + sin 180° sin α]= 5 – 4 [–1(cos α) + 0(sin α)]= 5 – 4 (– cos α)= 5 + 4 cos α 25 – 24 cos α = 5 + 4 cos α 28 cos α = 20

cos α = 20—–28

= 5—7

2

QS———sin 40°

= 4———sin 55°

QS = 4 sin 40°————sin 55°

= 3.14 cm

PS2 = 32 + 3.142 – 2(3)(3.14) cos 95° = 20.5

PS = 20.5 = 4.53 cm

3 (a) PV = 62 + 82

= 100 = 10 cm

QR = 172 – 82

= 225 = 15 cm

PR = 62 + 152

= 261 = 16.16 cm16.162 = 102 + 172 – 2(10)(17) cos ∠PVR 340 cos ∠PVR = 127.85 cos ∠PVR = 0.3760 ∠PVR = 67° 55'(b) Area of PVR

= 1—2

(10)(17) sin 67° 55'

= 78.76 cm2

4 (a) (i) AC 2 = 62 + 102 – 2(6)(10) cos 130°

= 213.135

AC = 213.135 = 14.6 cm (ii) 16———

sin 30° = 14.6————–

sin ∠ABC sin ∠ABC =

14.6 sin 30°—————16

= 0.4563 ∠ABC = 27° 9' (iii) Area of quadrilateral ABCD = 1—

2(6)(10) sin 130° +

1—2

(14.6)(16) sin 122° 51'

= 22.981 + 98.123 = 121.104 cm2

(b) (i)

(ii) 14.6———–sin 130°

= 10————–sin ∠ACD

sin ∠ACD = 10 sin 130°—————14.6

= 0.5247 ∠ACD = 31° 39' ∴AC'D = 180° – 31° 39' = 148° 21'

5 (a) AB2 = 52 + 72 – 2(5)(7) cos 130° = 118.995 AB = 118.995 = 10.9 cm

(b)

10.9————–

sin ∠AD2B

= 5.5———sin 30°

sin ∠AD

2B =

10.9 sin 30°—————

5.5

∠AD2B = 82° 17'

∠AD1B = 180° – 82° 17'

= 97° 43'(c) (i) ∠AD

2B = 82° 17'

∠ABD2 = 180° – 30° – 82° 17'

= 67° 43'

AD2————–

sin 67° 43' = 10.9————–

sin 82° 17' AD

2 = 10.18 cm

(ii) Area of ∆ACB

= 1—2

(5)(7) sin 130°

= 13.41 cm2

Area of ∆ABD2

= 1—2

(10.9)(5.5) sin 67° 43'

= 27.74 cm2

Area of quadrilateral ACBD = 13.41 + 27.74 = 41.15 cm2

6 (a) Area = 10 cm2

1—2

(5)(6) sin ∠BAD = 10 sin ∠BAD = 10—–

15 = 0.6667 ∠BAD = 41° 48'(b) BD2 = 52 + 62 – 2(5)(6) cos 41° 48' = 16.271

BD = 16.271 = 4.03 cm (c)

BC———sin 80°

= 4.03———

sin 65° BC = 4.03 sin 80°—————–

sin 65° = 4.38 cm(d) Area of quadrilateral ABCD = 1—

2(5)(6) sin 41° 48' +

1—2

(4.03)(4.38) sin 35° = 9.998 + 5.062 = 15.06 cm2

7 (a) (i) 72 = 52 + 82 – 2(5)(8) cos ∠BDC 80 cos ∠BDC = 40

cos ∠BDC = 1—2

∠BDC = 60°

D2

5.5 cm30°

5.5 cm

7 cm130°

D1

A

C

B

5 cm

A C'

10 cm6 cm

C

D

Page 93: Analysis Spm Additional Mathematics

8 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(ii) AD———sin 40°

= 12———sin 60°

AD = 12 sin 40°————–

sin 60° = 8.907 cm (iii) Area of ∆ABD = 1—

2(5)(8.907) sin 120°

= 19.28 cm2

(b) (i)

(ii) Area of ∆AD'E

= 1—2

(8.907)(12) sin 20°

= 18.28 cm2

8 (a)

QS———sin 61°

= 6.8

———sin 80°

QS =

6.8 sin 61°————–

sin 80° = 6.04 cm(b) ∠PSQ = 180° – (61° + 80°) = 39° RS2 = (2.2)2 + (6.04)2 – 2(2.2)(6.04)

cos 39° = 20.668

RS = 20.668 = 4.55 cm

(c) sin ∠QRS————

6.04 =

sin 39°———

4.55 sin ∠QRS =

6.04 sin 39°—————

4.55 = 0.8354 ∠QRS = 56° 39', 123° 21' ∴ ∠QRS = 123° 21' (d) Area of ∆QRS

= 1—2

(2.2)(4.55) sin 123° 21'

= 4.18 cm2

Chapter 11: Index Number 1 I

– = 115

120(m – 2) + 112m + 115(2)———————————–

2m = 115

232m – 10 = 230m 2m = 10 m = 5

2 (a) I–

= 113

110(6) + 120p + 80(3)

+ 140(5) + 90(2)—————————

p + 16 = 113

120p + 1780 = 113p + 1808 7p = 28 p = 4

(b) I09/07

= 90

P

09——6.50

× 100 = 90 P

09 = 90 × 6.50————

100 = RM5.85

3 (a) x——

2.50 × 100 = 112

x = 112 × 2.50————–

100 = 2.80

(b) I–

10/08 = 113

105(2) + 112(5) + 120y——————————–

y + 7 = 113

770 + 120y = 113y + 791 7y = 21 y = 3

4 (a) (i) x——0.90

× 100 = 150 x = 150 × 0.90————–

100 = 1.35 (ii) y = 1.80——

1.50 × 100

= 120 (iii) 3.85——z × 100 = 110 z = 3.85 × 100—————

110 = 3.50

(b) I–

05/00 =

150(3) + 125(6) + 120(4) + 160(10) + 110(7)

——————————–30

= 4050——30

= 135

(c) P

05——420

× 100 = 135

P05

= 135 × 420————–

100 = 567 ∴ The total monthly cost for the

year 2005 is RM567.

(d) I–

10/00 = 135 × 1.1

= 148.5 5 (a) I

07/04 = 140

4.20——

m × 100 = 140

m = 4.20 × 100————–140

= 3.00(b) I

07/04 = 150

n + 1——–n

× 100 = 150

100n + 100 = 150n 50n = 100 n = 2 ∴ n = 2, p = 3

(c) (i) I–

07/04 = 130

P

07——6.50

× 100 = 130

P07

= 130 × 6.50————–100

= RM8.45

(ii)

140(2) + 125(4) + 150(3) + 110q———————

q + 9 = 130

1230 + 110q = 130q + 1170 20q = 60 q = 3

6 (a) 1.40——x × 100 = 175 x = 1.40 × 100————–

175 = 0.80 y = 3.00——

2.50 × 100

= 120

(b) I

–10/09

=

150(90) + 175(45) + 140(108) + 120(36) + 125(81)

————————————360

= 50 940———360

= 141.5

(c) I–

11/09 = 141.5 × 1.2

= 169.8

P11—–

25 × 100 = 169.8

P

11 = 169.8 × 25————–

100 = 42.45 ∴ The production cost for the year

2011 is RM42.45.

7 (a) h = 1.65——1.20

× 100 = 137.5 0.80——

k × 100 = 160

k = 0.80 × 100————–160

= 0.50

(b) I

–10/08

=

120(20) + 137.5(40) + 160(30) + 150(10)

—————————100

= 14 200———100

= 142(c) (i) I

–12/08

= 142 × 1.1 = 156.2 (ii)

P12——

1.60 × 100 = 156.2

P

12 = 156.2 × 1.60—————

100 = 2.50 ∴ The price of a muffi n in the

year 2012 is RM2.50. 8 (a) (i) x = 1.80——

2.40 × 100

= 75

(ii) y

——3.20

× 100 = 150

y = 150 × 3.20————–100

= 4.80

A

60°

8.907 cm

D'40°

D E

12 cm

Page 94: Analysis Spm Additional Mathematics

9© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(iii) 7.15——

z × 100 = 110

z =

7.15 × 100————–

110 = 6.50

(b) I–

10/08 = 117.5

75(4) + 112(5) + 150(7) + 110m

———————m + 16

= 117.5

1910 + 110m = 117.5m + 1880 7.5m = 30 m = 4 (c)

P10——

5240 × 100 = 117.5

P

10 = 117.5 × 5240—————–

100 = 6157 ∴ The total expenditure in the year

2010 is RM6157.

(d) I–

11/08 = 140

I–

10/08 = 112

∴ I

11/10 =

140—–112

× 100 = 125

Page 95: Analysis Spm Additional Mathematics

11

Paper 1 1 (a) f : x → x2

(b) q = 16

2 (a)

(b) –4 � f –1(x) � 6

3 f 2(x) = gf(1)

4(4x – 5) – 5 = 5

2(1)–3 16x – 25 = –1 16x = 24 x = 1.5

4 (a) f(–4) = 6

–4–2 = –1

(b) y = 6

x – 2 xy – 2y = 6

xy = 2y + 6

x = 2y + 6

y

f –1(x) = 2x + 6

x ∴ a = 2, b = 6

5 3x(2x – 4) = x – 5 6x2 – 12x = x – 5

6x2 – 13x + 5 = 0

x = 13± (–13)2 – 4(6)(5) 2(6)

= 13± 49

12

= 13 + 712

or 13 – 7

12

= 53

or 12

6 x2 – [3 + �– 12 �]x + (3)�– 1

2 � = 0

x2 – 52

x – 32

= 0

2x2 = 5x + 3∴ p = 5, q = 3

7 y = 2k+ x

8 … 1

2y2 = k + x … 2

Substitute 1 into 2 :

2�2k + x8 �

2

= k + x

2� 4k2 + 4kx + x2

64 � = k + x

x2 + 4kx + 4k2 = 32k + 32x x2 + (4k – 32)x + 4k2 – 32k = 0

b2 – 4ac = 0 (4k – 32)2 – 4(1)(4k2 – 32k) = 0 16k2 – 256k + 1024 – 16k2 + 128k = 0 1024 – 128k = 0 128k = 1024 k = 8

8 (a) f(x) = –x2 + 4x + 6 = –(x2 – 4x – 6) = –[x2 – 4x + (–2)2 – (–2)2 – 6] = –[(x – 2)2 – 10] = 10 – (x – 2)2

∴ h = –2, k = 10

(b) The maximum value is 10.

9 (a) f(x) = a(x + 1)2 + 6 At the point (0, 4),

4 = a + 6 a = –2 f(x) = –2(x + 1)2 + 6 ∴ a = –2, p = 1, q = 6

(b) f(x) = –2(x – 1)2 + 6

10 2x2 – 4 � 7x 2x2 – 7x – 4 � 0

(2x + 1)(x – 4) � 0

∴ – 12

� x � 4

11

�64

2––3 x

3––2 �5

�165––4 x

2––3 �3

= [�26� 2––

3 x3––2 ]5

[�24� 5––4 x

2––3 ]3

= 220 x15––2

215 x2

= 25 x11––2

= �2x11––10�5

∴ m = 1110

, n = 5

12 8x

2y = 64

23x–y = 26

3x – y = 6 … 1

34x × �19 �

y–1

= 81

34x × (3–2)y–1 = 34

34x – 2y + 2 = 34

4x – 2y + 2 = 4 4x – 2y = 2 2x – y = 1 … 2 1 – 2 : x = 5

Substitute x = 5 into 1 : 3(5) – y = 6 y = 15 – 6 = 9 ∴ x = 5, y = 9

13

log4� x

y � = log

2 � x

y �log

24

= log

2x – log

2y

2

= h – k2

14 (a) AC = (10 – 2)2 + (10 – 4)2

= 64 + 36

= 100

= 10 units

(b) Area of ΔABC

= 12

⎪ 4 10 2 4 ⎪⎪ 2 10 6 2 ⎪

= 12

(40 + 60 + 4 – 20 – 20 – 24)

= 12

(40)

= 20 unit2

15 2y = 4x + 7

y = 2x + 72

The gradient of the line is 2.

Midpoint = �–2 + 62

, 7 + (–9)2 �

= (2, –1)

Then the required equation is y + 1 = 2(x – 2) y + 1 = 2x – 4 y = 2x – 5

16 (a) Let P(x, y) be the moving point, AP = BP

(y–0)2 + (x–1)2 = (y – 7)2 + (x – 6)2

y2 + x2 – 2x + 1 = y2 – 14y + 49 + x2 – 12x + 36

y

O

y = f –1(x)

x

y = x(3, 6)

(6, 3)2

2–4

–4

x4

1—2

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 96: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

1 – 2x = 85 – 12x – 14y 10x + 14y – 84 = 0 5x + 7y – 42 = 0

(b) At y-axis, x = 0 5(0) + 7y – 42 = 0 7y = 42 y = 6

∴ C(0,6)

17 M = 40.5 + �15 – 139 �20

= 40.5 + 4.44 = 44.94

18 (a) Range = 15 – 3 = 12

(b) x = 405

= 8

σ = 4045

–(8)2

= 16.8 = 4.099

19 (a) The new mean = 4(9) + 2 = 38

(b) The new variance = 42�3 12 �

= 56

20 (a) tan ∠AOC = 106

= 53

∠AOC = 59° 2' = 1.03 radians

(b) sAB

= 6(1.03) = 6.18 cm

OC = 62 + 102

= 11.662

∴ Perimeter of the shaded region = 6.18 + 10 + (11.662 – 6) = 21.842 cm

21 (a) sAB

= 10�8π5 �

= 16π cm/50.265 cm

(b) Area of sector AOB

= 12

(10)2 �2π5 �

= 20π = 62.832 cm2

Area of ΔAOB

= 12

(10)2 sin �2π5 �

= 47.553 cm2

∴ Area of the shaded region = 62.832 – 47.553 = 15.28 cm2

22 lim x(x + 10) =

lim x + 10 x→0 x(x2 + 2) x→0 x2 + 2

= 102

= 5

23 (a) y = 116

(3x–4)5

dydx

= 516

(3x – 4)4 (3)

= 1516

(3x – 4)4

(b) When x = 2,

dydx

= 1516

[3(2) – 4]4

= 15

and y = 116

[3(2) – 4]5

= 2

∴ Equation of the normal:

y – 2 = – 115

(x – 2)

15y – 30 = –x + 2 15y + x = 32

24 y = x– 1–

3

dydx

= – 13

x– 4–

3

= –1

3x4–3

When x = 8, δx = 8.2 – 8 = 0.2

y = 12

and dydx

= – 148

∴ 13 8.2

= 12

+ �– 148 � (0.2)

= 0.4958

25 y = x + 5x–2

dydx

= 1 – 10x3

By the chain rule,dydt

= dydx

× dxdt

1.5 = �1 – 10x3 � dx

dt

When x = 2, 1.5 = –0.25 dxdt

dxdt

= –6

∴ x changes at a rate of –6 units s–1.

Paper 2Section A/Bahagian A

1 3y + 2x = 14 3y = 14 – 2x

y = 14 – 2x

3 … 1

y = 4x

… 2

Substitute 1 into 2 :

14 – 2x3

= 4x

14x – 2x2 = 12

2x2 – 14x + 12 = 0 x2 – 7x + 6 = 0 (x – 1)(x – 6) = 0 x = 1 or x = 6

Substitute x = 1 into 2 : y = 4

Substitute x = 6 into 2 :

y = 46

= 23

∴ x = 1, y = 4; x = 6, y = 23

∴ P(1, 4); Q�6, 23 �

2 (a) f(–1) = 2 g(2) = 3

8(–1) + m = 2 n

2 + 6 = 3

m = 10

n = 24 ∴ m = 10, n = 24

(b) f(x) = 8x + 10 y = 8x + 10

x = y – 108

f –1(y) = y – 10

8

(c) gf(x) = g(8x + 10)

= 248x + 10 + 6

= 248x + 16

= 248(x + 2)

= 3x + 2

, x ≠ – 2

3 (a) (i) a = 1 (ii) (1, –4) (iii) x = 1

(b)

4 (a) LN10

= tan 1.2

LN = 10 tan 1.2 = 25.72 cm

ON = 102 + (25.72)2

= 27.6 cm s

LM = 10(1.2)

= 12 cm

∴ Perimeter of the shaded region = 25.72 + 12 + (27.6 – 10)= 55.32 cm

(b) Area of ΔLON

= 12

(10)(25.72)

= 128.6 cm2

x

y

y = –f(x)

3

3

O–1

Page 97: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Area of section LOM

= 12

(10)2(1.2)

= 60 cm2

∴ Area of the shaded region = 128.6 – 60 = 68.6 cm2

5 (a) (i) x = ΣxN

= 12010

= 12

(ii) σ2 = Σx2

N – (x)2

= 160010

– (12)2

= 16

(b) (i) 120 – x9

= 12

x = 12

(ii) σ = 1600 – (12)2

9 – (12)2

= 17.778 = 4.216

6 (a) y = x2 – 3x + 4

dydx

= 2x – 3

At point A(1, 2), dydx

= 2(1) – 3

= –1

∴ Equation of the tangenty – 2 = –1(x – 1)y – 2 = –x + 1y + x = 3

(b) At point B(3, 4), dydx

= 2(3) – 3

= 3

∴ Equation of the normal:

y – 4 = – 13

(x – 3)

3y – 12 = –x + 3 3y + x = 15

(c) y + x = 3 … 1 3y + x = 15 … 2 2 – 1 : 2y = 12

y = 6

Substitute y = 6 into 1 : 6 + x = 3 x = –3

∴ C(–3, 6)

Section B/Bahagian B

7 (a) 12

(x – 1)[2(x – 1) + x + 3] = 112

(x – 1)(3x + 1) = 224 3x2 – 2x – 1 = 224 3x2 – 2x – 225 = 0 (shown)

(b) x = 2 + (–2)2 – 4(3)(–225)2(3)

= 2 + 27046

= 2 + 526

or 2 – 52

6

= 9 or –253

(c) (i) AB = 2(9 – 1) = 16 cm

(ii)

tan ∠DAE = 84

= 2 ∠DAE = 63° 26' ∴ ∠BAD = 63° 26'

8 (a) 4x

2y = 2

22x–y = 21

2x – y = 1 … 1

log10

(2x + 2y) = 1 2x + 2y = 10 x + y = 5 … 2

1 + 2 : 3x = 6 x = 2

Substitute x = 2 into 1 : 2(2) – y = 1 y = 3 ∴ x = 2, y = 3

(b) (i) log4 � 1

m� = log4 1 – log

4 m

= –n

(ii) log28m =

log4 8m

log42

= log

48 + log

4 m

log42

= 3 log

42 + log

4 m

log42

= 3�1

2� + n

12

= 3 + 2n

(c) (i) By January 2012 ⇒ 4 years RM50 000(1.0425)4

= RM59 057.39

(ii) 50 000(1.0425)t � 80 000 t log

10 1.0425 � log

10 1.6

t � 11.29 ∴ 12 years. It will be in the

year 2020.

9 (a) At y-axis, x = 0 y – 2(0) = 2 y = 2

∴ P(0, 2)

D

A E

8 cm

4 cm

Let S be (x, 0),

mPS

= – 12

0 – 2x – 0

= –12

x = 4 ∴ S(4, 0) ∴ P(0, 2); S(4, 0)

(b) Equation of QR:

y – 5 = –12

(x – 9)

2y – 10 = –x + 9 2y + x = 19

(c) y – 2x = 2 … 1 2y + x = 19 … 2 2 × 2: 4y + 2x = 38 … 3 1 + 3 : 5y = 40

y = 8 Substitute y = 8 into 1 :

8 – 2x = 2 2x = 6 x = 3

∴ Q(3, 8)

(d) Area of PQRS

= 12

⎪ 0 4 9 3 0 ⎪⎪ 2 0 5 8 2 ⎪

= 12

(20 + 72 + 6 – 8 – 15)

= 12

(75)

= 37.5 unit2

10 (a)

(i) Median time = 13.75 (ii) Interquartile range = 17.5 – 10.5 = 7 (iii) The number of teenagers who

spent more than 20 hours = 100 – 87 = 13 teenagers

(b) x =

3(6) + 8(18) + 13(39) + 18(25) + 23(9) + 28(3)

100

= 1410

100 = 14.1

100

90

80

70

60

50

40

30

20

10

00.5 5.5 10.5 15.5 20.5 25.5 30.5 13.75 17.5

Cumulative frequency

Time (hours)

Q3

Q1

Median

87

Page 98: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

11 (a) (i) Volume of water in he hemispherical bowl

= 23

π(6)3

= 144π ∴ dV

dt = 144π

4 = 36π cm3 s–1

(ii)

rh

= 816

r = 12

h

Volume of water in the cone,

V = 13

πr2h

= 13

π�12

h�2h

= 112

πh3

dVdh

= 14

πh2

By the chain rule,

dVdt

= dVdh

× dhdt

36π = 14

πh2 × dhdt

When h = 8,

36π = 16π dhdt

∴ dhdt

= 2.25 cm s–1

(b) (i) 36x + 2y = 120 2y = 120 – 36x y = 60 – 18x

Area = 16xy + 12

(16x)(6x)

= 16x(60 – 18x) + 48x2

= 960x – 288x2 + 48x2

= 960x – 240x2

= 240x(4 – x) (shown)

(ii) For a maximum value of A,

dAdx

= 0

960 – 480x = 0 x = 2

and so d2Adx2

= –480 � 0

Hence, for the area, A to be maximum, x = 2 and

y = 60 – 18(2) = 24 ∴ x = 2, y = 24

Section C/Bahagian C

12 (a) (i) 15sin 82°

= 8sin ∠BAD

sin ∠BAD = 8 sin 82°15

= 0.5281 ∠BAD = 31° 53'

(ii) 82 = 52 + 72 – 2(5)(7) cos ∠ BCD

cos ∠BCD = 52 + 72 – 82

2(5)(7) = 0.1429 ∠BCD = 81° 47'

(iii) Area of ΔABD

= 12

(8)(15) sin 66° 7'

= 54.86 cm2

Area of ΔBCD

= 12

(5)(7) sin 81° 47'

= 17.32 cm2

∴ Area of ABCD = 54.86 + 17.32 = 72.18 cm2

(b) (i)

(ii) 8sin 81° 47'

= 7sin ∠BDC

sin ∠BDC = 7 sin 81° 47'8

= 0.8660 ∠BDC = 60° ∴ ∠BD'C = 180° – 60° = 120°

13 (a) (i) BC2 = 72 + 12.22 – 2(7)(12.2) cos 60°

= 112.44

BC = 112.44 = 10.604 cm

(ii) 10.604sin 120°

= 3.5

sin ∠CBD

sin ∠CBD = 3.5 sin 120°

10.604 = 0.2858 ∠CBD = 16° 37'

(b) (i)

(ii) ∠ BCD = 180° – 120° – 16° 37' = 43° 23'

Area of ΔBCD

= 12

(3.5)(10.604) sin 43° 23'

= 12.75 cm2

∠CDC' = 180° – 2(43° 23') = 93° 14'

Area of ΔCDC'

= 12

(3.5)(3.5) sin 93° 14'

= 6.12 cm2

∴ Area of ΔBC'D = 12.75 – 6.12 = 6.63 cm2

14 (a) x

1.50 × 100 = 120

x = 120 × 1.50

100 = 1.80

4.20y

× 100 = 105

y = 4.20 × 100105

= 4

z = 7.005.00

× 100

= 140 ∴ x = 1.80, y = 4, z = 140

(b) –I

10/08 =

125(4) + 120(8) + 112(6) + 105(2) + 140(9)

29

= 360229

= 124.21

(c) P

10

520 × 100 = 124.21

P10

= 124.21 × 520100

= RM645.89

(d) –I

12/08 = 124.21 × 0.9

= 111.79

15 (a) x = 0.500.40

× 100

= 125

2.20

y × 100 = 110

y = 2.20 × 100110

= 2.00

z

4.00 × 100 = 90

z = 90 × 4.00100

= 3.60 ∴ x = 125, y = 2.00, z = 3.60

(b) (i) –I

11/09 =

125(9) + 110(2) + 150(10) + 90(15)

36

= 419536

= 116.53

(ii) P

11

1.20 × 100 = 116.53

P11

= 116.53 × 1.20100

= RM1.40

(iii) –I

12/09 = 116.53 × 1.2

= 139.84

h

r

8

16

B

C'

C

D

3.5 cm

B

CD

D'

5 cm

7 cm

Page 99: Analysis Spm Additional Mathematics

12 Progressions

1

Booster Zone

1 (a) T1 = 1—

2 [4(1) – 3]

= 1—2

T2 = 1—

2[4(2) – 3]

= 5—2

(b) d = 5—2

– 1—2

= 2

2 T4 = 27

a + 3d = 27 6 + 3d = 27 3d = 21 d = 7∴ p = 6 + 7 = 13and q = 13 + 7 = 20∴p = 13, q = 20

3 (a) a + 4d = 28 … 1 a + 19d = 103 … 2 2 – 1 : 15d = 75 d = 5 Substitute d = 5 into 1 : a + 4(5) = 28 a = 28 – 20 = 8 ∴ a = 8, d = 5(b) T

10 = a + 9d

= 8 + 9(5) = 53

4 (a) k + 3 – (k–1) = 2k + 1 – (k + 3) 4 = k – 2 k = 6(b) The fi rst three terms are 5, 9 and

13 So a = 5 and d = 4 ∴ T

12 = a + 11d

= 5 + 11(4) = 49

5 (a) d = 12 – 7 = 5(b) T

8 = a + 7d

= 7 + 7(5) = 42

6 4(2a + 7d) = 24 2a + 7d = 6 … 1

9(2a + 7d) = 90 2a + 17d = 10 … 2

2 – 1 : 10d = 4

d = 2—5

Substitute d = 2—5

into 1 :

2a + 7� 2—5 � = 6

2a = 16—–5

a = 8—5

∴ T15

= a + 14d

= 8—5

+ 14� 2—5 �

= 36—–5

7 a + 5d = 27 … 1

a + 13d = 59 … 2

2 – 1 : 8d = 32 d = 4Substitute d = 4 into 1 : a + 5(4) = 27 a = 7∴ S

10 = 5[2(7) + 9(4)]

= 5(50) = 250

8 (a) n—2

(4 + 104) = 1134 54n = 1134 n = 21

(b) 21—–2

[2(4) + 20d] = 1134

8 + 20d = 108 20d = 100 d = 5

9 S10

= S15

– S10

2S10

= S15

2[5(24 + 9d)] = 15—–2

(24 + 14d)

240 + 90d = 180 + 105d 15d = 60 d = 4

∴ S15

= 15—–2

[24 + 14(4)]

= 15—–2

(80) = 600

10 3—2

(2a + 2d) = 21 2a + 2d = 14 … 1

S6 – S

3 = 57

3(2a + 5d) = 57 + 21 2a + 5d = 26 … 2

2 – 1 : 3d = 12 d = 4Substitute d = 4 into 1 : 2a + 2(4) = 14 2a = 6 a = 3∴ a = 3, d = 4

11 2(2a + 3d) = 62 2a + 3d = 31 … 1 5(2a + 9d) = 365 2a + 9d = 73 … 22 – 1 :

6d = 42 d = 7Substitute d = 7 into 1 : 2a + 3(7) = 31 2a = 10 a = 5∴ T

6 = a + 5d

= 5 + 5(7) = 40

12 (a) T1 = S

1 = 2(1)2 + 3(1)

= 5(b) T

2 = S

2 – T

1

= 2(2)2 + 3(2) – 5 = 14 – 5 = 9 ∴ d = 9 – 5 = 4

13 3(2a + 5d) = 96 2a + 5d = 32 … 1

S10

= 1—3

S20

5(2a + 9d) = 1—3

[10(2a + 19d)] 15(2a + 9d) = 10(2a + 19d) 30a + 135d = 20a + 190d 10a – 55d = 0 2a – 11d = 0 … 21 – 2 :

16d = 32 d = 2Substitute d = 2 into 1 : 2a + 5(2) = 32 2a = 22 a = 11∴ T

10 = a + 9d

= 11 + 9(2) = 29

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 100: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

14 100, 104 , 117 , ..., 494 , 500

So 104 + (n–1) 13 = 494 13(n–1) = 30 n – 1 = 30 n = 31

15 (a) S500

= 500——2

(1 + 500)

= 125 250

(b) 1, 5 , 10 , ..., 495 , 500

So, 5 + (n – 1) 5 = 500 5(n – 1) = 495 n – 1 = 99 n = 100

S100

= 100—–2

[2(5) + 99(5)]

= 100—–2

(505) = 25 250∴ The sum of all integers between

1 and 500 which are not the multiples of 5.

= 125 250 – 25 250= 100 000

16 a + 3d = 9 … 1 2(2a + 3d) = 21 2a + 3d = 10.5 … 22 – 1 : a = 1.5

Substitute a = 1.5 into 1 : 1.5 + 3d = 9 3d = 7.5 d = 2.5∴ S

8 = 4 [2(1.5) + 7(2.5)]

= 4(20.5) = 82

17 Sn � 230

n—2

[2(5) + (n–1) 4] � 230

n—2

(4n + 6) � 230

2n2 + 3n – 230 � 0 (2n + 23)(n – 10) � 0

So, n � 10∴ The least value of n is 11

18 (a) n—2

(8 + 52) = 360

30n = 360 n = 12

(b) 12—–2

[2(8) + 11d] = 360 16 + 11d = 60 11d = 44 d = 4 ∴ Angles of the other sides = 12°, 16°, 20°, 24°, 28°, 32°,

36°, 40°, 44°, 48°

19 (a) n—2

(1.2 + 3.6) = 12

2.4 n = 12 n = 5

(b) 5—2

[2(1.2) + 4d] = 12

5—2

(2.4 + 4d) = 12

2.4 + 4d = 4.8 4d = 2.4 d = 0.6 ∴ Length of the other sides = 1.8 cm, 2.4 cm, 3 cm

20 (a) a = 100 and d = –5 ∴ T

10 = 100 + 9(–5)

= 55

(b) S10

= 10—–2

[2(100) + 9(–5)]

= 5(155) = 775

21 T

2—–T

1

= 52x

—–5x

= 5x

T3—–

T2

= 53x

—–52x

= 5x

So, the sequence is a geometric progression with common ratio 5x.

22 27 + 27r + 27r2 = 21 27r2 + 27r + 6 = 0 9r2 + 9r + 2 = 0 (3r + 1)(3r + 2) = 0

r = – 1—3

or r = – 2—3

When r = – 1—3

,

x = 27�– 1—3 �

= –9

y = –9�– 1—3 �

= 3

When r = – 2—3

,

x = 27�– 2—3 �

= –18

y = –18�– 2—3 �

= 12∴ x = –9, y = 3 or x = –18, y = 12

23 (a) x——–x – 4

= 5x – 12———–x

x2 = (5x – 12)(x – 4) x2 = 5x2 – 32x + 48 4x2 – 32x + 48 = 0 x2 – 8x + 12 = 0 (x – 2)(x – 6) = 0 x = 2 or x = 6

(b) x – 2——–x + 1

=

1—2

x——–x – 2

x2 – 4x + 4 = 1—2

x2 + 1—2

x

2x2 – 8x + 8 = x2 + x

x2 – 9x + 8 = 0 (x – 1)(x – 8) = 0 x = 1 or x = 8

(c) x + 3——–x + 1

= x + 8——–x + 3

x2 + 6x + 9 = x2 + 9x + 8 3x = 1

x = 1—3

(d) x – 1——–x – 2

= 3x – 5——–x – 1

x2 – 2x + 1 = 3x2 – 11x + 10 2x2 – 9x + 9 = 0 (2x – 3)(x – 3) = 0 x = 3—

2 or x = 3

24 (a) T

1 = 32(1)

= 9 T

2 = 32(2)

= 81

(b) r = 81—–9

= 9

25 –2(–2)n–1 = 1024 (–2)n–1 = –512 (–2)n–1 = (–2)9

n–1 = 9 n = 10

26 3(2)n–1 � 600 (n – 1)log

102 � log

10 200

n – 1 � 7.644 n � 8.644∴ n = 9

27 ar3 = 24 … 1 ar 6 = 192 … 22 ÷ 1 :

r3 = 8 r = 2Substitute r = 2 into 1 : a(2)3 = 24 a = 3∴ T

13 = ar12

= 3(2)12

= 12 288 28 (a) 6h + k———

2h + k = 14h + k———

6h + k 36h2 + 12hk + k2 = 28h2 + 16hk + k2

8h2 = 4hk k = 2h(b) T

1 = 2h + 2h

= 4h T

2 = 6h + 2h

= 8h

r = 8h—–4h

= 2

29 (a) x + 4——–x

= 2x + 2——–x + 4

x2 + 8x + 16 = 2x2 + 2x x2 – 6x – 16 = 0

23– —– 2

10n

Page 101: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(x + 2)(x – 8) = 0 x = –2 or x = 8 ∴ x = 8 (�0)

(b) r = 12—–8

= 3—2

ar2 = 8 a� 3—

2 �2

= 8 a = 8� 4—

9 � = 32—–9

T6 = ar5

= 32—–

9 � 3—

2 �5

= � 32—–

9 ��243—–32 �

= 27

30 –2(–2)n–1 = 1024 (–2)n–1 = –512 (–2)n–1 = (–2)9

n–1 = 9 n = 10

31 ar – a = 4 a(r – 1) = 4 … 1 ar2 – ar = 16 ar(r – 1) = 16 … 2

2 ÷ 1 : r = 4

Substitute r = 4 into 1 : 3a = 4 a = 4—

3 ∴ T

5 = ar4

= 4—

3 (4)4

= 4—

3 (256)

= 341 1—

3

32 a + ar = 17 1—2

a(1 + r) = 35—–2

… 1 ar2 = 14—–

3 … 2

2 ÷ 1 : r 2

——1 + r

= � 14—–3 �� 2—–

35 � 105r2 = 28 + 28r 105r2 – 28r – 28 = 0 15r2 – 4r – 4 = 0 (5r + 2)(3r – 2) = 0

r = – 2—5

or r = 2—3

∴ r = 2—3

(�0)

33 a = 8 and r = 24—–8

= 3The sum from 8th term to 10th term= S

10 – S

7

= 8(310 – 1)———–

3 – 1 –

8(37– 1)———–

3 – 1= 236 192 – 8 744= 227 448

34 (a) T

1 = S

1 = 4 – 4—

21 = 2 T

2 = S

2 – T

1

= �4 – 4—–22 � – 2

= 1 r =

T2—–

T1

= 1—2

(b) T5 = ar4

= 2� 1—

2 �4

= 2� 1—–16 �

= 1—8

35 16�—–�p

q——– = ——–

4

16�—–�p

4q = 256——p2

q = 64—–p2

36 (a) ar4 = 32

2r4 = 32 r4 = 16 r4 = 24

r = 2 (b) S

8 = 2(28 – 1)————

2 – 1 = 510

37 (a) r = 12—–4

= 3(b) S

n = 13 120

4(3n – 1)

————3 – 1

= 13 120 3n – 1 = 6 560 3n = 6 561 3n = 38

n = 8

38 (a) T4 = 24

81r3 = 24

r3 = 8—–27

r3 = � 2—3 �3

r = 2—3

(b) S∞ =

81———1 –

2—3

= 243

39

a(1 – r3)———–

1 – r = 14 … 1

a——–1 – r

= 16 … 2 Substitute 2 into 1 : 16(1 – r3) = 14 1 – r3 = 7—

8

r3 = 1—8

r3 = � 1—2 �3

r = 1—

2 Substitute r = 1—

2 into 2 :

a———

1 – 1—2

= 16 a = 16� 1—

2 � = 8

40 (a) S∞ = 1———1 – 1—

3 = 3—

2

(b) S∞ =

1�—�2————–1 – 1�– —� 2

= 1—

2 ÷ 3—

2

= 1—3

(c) S∞ = 0.2———1 – 0.1

= 2—9

(d) S∞ = 24———1 – 1—

2 = 48

••41 (a) 0.15 = 0.15 + 0.0015 + 0.000015

+ ... S∞ = 0.15———–

1 – 0.01 = 0.15——

0.99

= 5—–33

••(b) 0.06 = 0.06 + 0.0006 + 0.000006

+ ... S∞ = 0.06———–

1 – 0.01

= 0.06——0.99

= 2—–33 •

(c) 2.4 = 2 + 0.4 + 0.04 + 0.004 + ...

= 2 + 0.4———1 – 0.1

= 2 + 4—9

= 2 4—9

Page 102: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

••(d) 1.45 = 1 + 0.45 + 0.0045 + 0.000045 + ...

= 1 + 0.45———–1 – 0.01

= 1 + 5—–11

= 1 5—–11

42 ar3 = 8a r3 = 8 r = 2 S

4 = 360

a(24 – 1)————

2 – 1 = 360

15a = 360 a = 24∴ The area of the largest sector = 24°, 48°, 96°, 192°

SPM Appraisal Zone

1 (a) T6 = 2(6) + 7 = 19

(b) T5 = 2(5) + 7 = 17

d = T6 – T

5

= 19 – 17 = 2

2 T15

= 2T5

–4 + 14d = 2(–4 + 4d) –4 + 14d = –8 + 8d 6d = –4

d = – 2—3

T13

= a + 12d

= –4 + 12�– 2—3 �

= –12

3 (a) For an arithmetic progression a, a + 4, a + 8, ...

d = (a + 4) – a = 4 T

8 = 33

a + 7(4) = 33 a = 5(b) T

10 = a + 9d

= 5 + 9(4) = 41

4 T16

= 3T5

a + 15d = 3(a + 4d) a + 15d = 3a + 12d 2a – 3d = 0 … 1 T

12 – T

7 = 20

a + 11d – (a + 6d) = 20 5d = 20 d = 4Substitute d = 4 into 1 : 2a – 3(4) = 0 2a = 12 a = 6∴ a = 6, d = 4

5 (a) a + 4d = –4 … 1 a + 9d = 16 … 2

2 – 1 : 5d = 20 d = 5 Substitute d = 5 into 1 : a + 4(5) = –4 a = –24 ∴ a = –24, d = 5(b) T

20 = –24 + 19(5)

= 71

6 x + 3 – (x + 2) = 2x2 + 1 – (x + 3) 1 = 2x2 – x – 2 2x2 – x – 3 = 0 (2x – 3)(x + 1) = 0

x = 3—2

or x = –1

7 (a) a + 9d = 21 … 1

S30

– S20

= 675

30—–2

(2a + 29d) – 20—–2

(2a + 19d)

= 675 10a + 245d = 675 2a + 49d = 135 … 2

1 × 2: 2a + 18d = 42 … 3

2 – 3 : 31d = 93 d = 3 Substitute d = 3 into 1 : a + 9(3) = 21 a = –6 ∴ a = –6, d = 3

(b) S10

= 10—–2

[2(–6) + 9(3)]

= 5(–12 + 27) = 75

8 S10

= 80 2a + 9d = 16 … 1 S

22 – S

10 = 624

22—–2

(2a + 21d) – 80 = 624

22a + 231d = 704 2a + 21d = 64 … 22 – 1 : 12d = 48

d = 4

Substitute d = 4 into 1 : 2a + 9(4) = 16 2a = –20 a = –10∴ T

1 = –10

T2 = –10 + 4 = –6

T3 = –6 + 4 = –2

∴ The fi rst three terms is –10, –6 and –2.

9 (a) 2x + 6 – (x + 3) = 8 – (2x + 6) x + 3 = 2 – 2x 3x = –1

x = – 1—3

(b) The fi rst three terms of an arithmetic

progression are 8—3

, 16—–3

and 8 with

a = 8—3

and d = 8—3

∴ S8 = 8—

2 �2� 8—3 � + 7� 8—

3 �� = 4(24) = 96

10 (a) S6 = 6

3[2(–9) + 5d] = 6 –18 + 5d = 2 5d = 20 d = 4(b) S

n = 90

n—2

[2(–9) + (n – 1)(4)] = 90

n—2

(4n – 22) = 90

4n2 – 22n = 180 2n2 – 11n – 90 = 0 (2n + 9)(n – 10) = 0

n = – 9—2

or n = 10 ∴ The number of terms in the

arithmetric progression is 10.(c) T

10 = a + 9d

= –9 + 9(4) = 27

11 S20

= 140

20—–2

(3 + l) = 140

3 + l = 14 l = 11

12 (a) x + 3——–x + 1

= x + 8——–x + 3

x2 + 6x + 9 = x2 + 9x + 8 3x = 1

x = 1—3

(b) r =

1—3

+ 3———1—3

+ 1

= 21—2

13 ar3 = 3 … 1

ar7 = 1—–27

… 2

2 ÷ 1 : r 4 = 1—–81

r 4 = �± 1—3 �

4

r = ±

1—3

Substitute r = 1—3

into 1 :

a� 1—3 �

3

= 3

1—–27

a = 3 a = 81 Substitute r = – 1—

3 into 1 :

a�– 1—3 �3 = 3

– 1—–27

a = 3 a = –81

∴ a = –81, r = – 1—3

and a = 81, r = 1—3

Page 103: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

14 (a) ar2 = 21—4

… 1

ar5 = – 2—3

… 2

2 ÷ 1 : r3 = – 8—–27

r3 = �– 2—3 �

3

r = – 2—3

Substitute r = – 2—3

into 1 :

a�– 2—3 �

2

= 9—4

4—9

a = 9—4

a = 81—–16

∴ a = 81—–16

, r = – 2—3

(b) T2 = ar

= � 81—–16 ��– 2—

3 � = – 27—–

8

15 a = 3—4

and r = 1—2

÷ 3—4

= 2—3

3—4 � 2—

3 �n – 1

= 4—–27

� 2—3 �

n–1

= 16—–81

� 2—3 �

n–1

= � 2—3 �

4

n – 1 = 4 n = 5

16 ar + ar2 = 30 ar (1 + r) = 30 … 1 ar3 = 27 … 2

2 ÷ 1 : r2

——–1 + r

= 9—–10

10r2 = 9 + 9r 10r2 – 9r – 9 = 0 (5r + 3)(2r – 3) = 0

r = – 3—5

or r = 3—2

17 For a geometric progression 2, 3, 9—2

, ...

a = 2 and r = 3—2

Sn � 30

2�� 3—

2 �n

– 1�—————–

3—2

– 1 � 30

� 3—2 �

n

– 1 � 71—2

n log10

� 3—2 � � log

10 �8 1—

2 � n � 5.278 n = 6

18 (a) a(ar6) = ar3

ar3 = 1 … 1 a + ar3 = 9 … 2 Substitute 1 into 2 :

a + 1 = 9 a = 8 Substitute a = 8 into 1 : 8r3 = 1

r3 = 1—8

r = 1—2

∴ a = 8, r = 1—2

(b) S4 =

8�1 – � 1—2 �

4

�—————

1 – 1—2

= 15

19 (a) T2 = S

2 – S

1

= 15�1 – 1—9 � – 15�1 –

1—3 �

= 131—3

– 10

= 31—3

(b) S∞ = 15 �1 – 1—3

∞� = 15

20

� 1—12�

——–1 – r

= 2—3

1—–12

= 2—3

(1 – r)

1—8

= 1 – r

r = 7—8

21 a——1 – r

= 9

a = 9(1 – r) … 1 ar = 2

a = 2—r

… 2

From 1 and 2 : 2—r = 9(1 – r)

2 = 9r – 9r2

9r2 – 9r + 2 = 0 (3r – 1)(3r – 2) = 0

r = 1—3

or r = 2—3

Substitute r = 1—3

into 2 :

a = 2—–

� 1—3 �

= 6

Substitute r = 2—3

into 2 :

a = 2—–

� 2—3 �

= 3

∴ a = 3, r = 2—3

and a = 6, r = 1—3

22 (a) 1 + 3h———1 + h

= 1 + 4h———1 + 3h

1 + 6h + 9h2 = 1 + 5h + 4h2

5h2 + h = 0

h(5h + 1) = 0

h = 0 or h = – 1—5

∴ h = – 1—5

(b) The fi rst 3 terms are 4—5

, 2—5

and 1—5

with a = 4—5

and r = 1—2

S∞ =

� 4—5 �

———1 – 1—

2 = 1

3—5

23 (a) ar2 = 22—3

… 1

ar5 = 8—–81

… 2

2 ÷ 1 r3 = 1—–27

r = 1—3

Substitute r = 1—3

into 1 :

a� 1—3 �

2

= 8—3

a = � 8—3 �(9)

= 24

∴ a = 24, r = 1—3

(b) S∞ = 24———1 – 1—

3 = 36

24 2.555 ... = 2 + 0.5 + 0.05 + 0.005 + ... = 2 +

0.5———1 – 0.1

= 2 +

5—9

= 25—9

25 (a) Let PQ and ∠QPR be x cm and θ respectively.

A

1 =

1—2

x2 sin θ A

2 =

1—2

� x—2 �� x—

2 � sin θ

= 1—8

x2 sin θ A

3 =

1—2

� x—4 �� x—

4 � sin θ

= 1—–32

x2 sin θ

A2—–

A1

= A

3—–A

2

= 1—4

Since A

2—–A

1

= A

3—–A

2

, thus the areas of

triangles form a geometric

progression with common ratio 1—4

.

Page 104: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) S∞ =

1—2

x2 sinθ————

1 – 1—4

= 2—3

x2 sin θ

2—3

x2 sinθ————1—2

x2 sinθ =

2—3

÷ 1—2

= 4—3

26 (a) a = 80 and r =

3—4

T4 = ar3

= 80� 3—

4 �3

= 33.75° (b) S∞ = 80———

1 – 3—4

= 320°

27 (a) h1 = 32� 3—

4 � = 24 cm h

2 = 24� 3—

4 � = 18 cm h

3 = 18� 3—

4 � = 13.5 cm The respective height of the ball

are as follows: 24 cm, 18 cm, 13.5 cm

h

2—h

1

= 18—–24

= 3—4

h

3—h

2

= 13.5——18

= 3—4

Since the ratios are equal, the height of the ball form a geometric progression with the common ratio

of 3—4

.

(b) The height of 6th bounce, T

6 = ar 5

= 24� 3—4 �

5

= 5.695 cm(c) The total distance traveled = 32 + 2(24) + 2(18) + 2(13.5) + ...

= 32 + 48———1 – 3—

4 = 32 + 192 = 224 cm

28 (a) A1 = x2

A2 = � x—

2 �� x—2 �

= 1—

4x2

A3 = � x—

4 �� x—4 �

= 1—–16

x2

A

3—–A

2

= A

2—–A

1

= 1—4

Since A

3—–A

2

= A

2—–A

1

, so the areas of

the squares form a geometric

progression with common ratio 1—4

.

(b) (i) A1 = 162

= 256 cm

T3 = 256� 1—

4 �2

= 16 cm2

∴ The area of the 3rd square is 16 cm2.

(ii) S∞ = 256———

1 – 1—4

= 341

1—3

cm

29 (a) P

1 = 3x

P

2 = 3—

2x

P

3 = 3—

4x

P

3—–P

2

= P

2—–P

1

= 1—2

Since

P3—–

P2

= P

2—–P

1

so the perimeters

of the triangles form a geometric

progression with common ratio 1—2

.

(b) (i) P1 = 3(64)

= 192 cm T

9 = 192 � 1—

2 �8

= 0.75 cm So, the length of the side of

the 9th triangle

= 0.75——

3 = 0.25 cm

(ii) S∞ = 192———1 – 1—

2 = 384 cm

30 (a) x——–

x + 5 =

x – 4——–x

x2 = x2 + x – 20 x = 20 (b) r =

20—–25

=

4—5

(c) S∞ =

25———1 – 4—

5 = 125 S

3 = 25 + 20 + 16

= 61 ∴ The difference is 125 – 61 = 64

31 (a) Sn = 126

n—2

[2(21) + (n – 1)(–1)] = 126

n—2

(43 – n) = 126 n(43 – n) = 252 n2 – 43n + 252 = 0 (n – 7)(n – 36) = 0 n = 7 or n = 36 ∴ n = 7(b) T

7 = a + 6d

= 21 + 6(–1) = 15

32 (a) S10

= 310 5(2a + 9d) = 310 2a + 9d = 62 … 1 S

3 = 114

3—2

(2a + 2d) = 114 2a + 2d = 76 … 2 1 – 2 : 7d = –14 d = –2 Substitute d = –2 into 1 : 2a + 9(–2) = 62 2a = 80 a = 40 ∴ a = 40, d = –2(b) T

n = 30

40 + (n – 1)(–2) = 30 2(n – 1) = 10 n – 1 = 5 n = 6 ∴ The 6th part has a length of

30 cm. (c) S

7 =

7—2

[2(40) + 6(–2)] =

7—2

(68) = 238 ∴ The sum of the last three parts = 310 – 238 = 72 cm

33 (a) S2 = 22 + 3(2)

= 10(b) T

2 = S

2 – S

1

= 10 – (1 + 3) = 6(c) d = T

2 – T

1

= 6 – 4 = 2

34 (a) Sn = 40

n—2

(2 + 14) = 40

n—2

(16) = 40 8n = 40 n = 5

(b) 5—2

[2(2)] + 4d] = 40 4 + 4d = 16 4d = 12 d = 3 ∴ The lengths of the other sides

= 5 cm, 8 cm, 11 cm

Page 105: Analysis Spm Additional Mathematics

13 Linear Law

1

Booster Zone

1 (a) y = ax + b—x

y—x

= a + b—x2

= b � 1—x2 � + a

where Y = y—

x and X = 1—

x2 ,

and the gradient, m = bthe Y-intercept = a

(b) y = a—x + b x

y

—–x

= a——x x + b

= a� 1——x x � + b

where Y = y

—–x

and X = 1——x x

,

and the gradient, m = a the Y-intercept = b

(c) y = a—–x

+ b x

y—–

x = a—x + b

= a� 1—x � + b

where Y = y—x and X = 1—x and the gradient, m = a the Y-intercept = b (d) y = a—x + b—

x2

x2y = ax + b where Y = x2y and X = x, and the gradient, m = a the Y-intercept = b(e) y = abx

log10

y = log10

abx

= log10

a + log10

bx

= x log10

b + log10

a = (log

10 b)x + log

10 a

where Y = log10

y and X = x and the gradient, m = log

10 b

the Y-intercept = log10

a(f) y = axb

log10

y = log10

axb

= log10

a + log10

x b

= b log10

x + log10

a where Y = log

10y and X = log

10x

and the gradient, m = b the Y-intercept = log

10 a

(g) xy = a(x + b) xy = ax + ab where Y = xy and X = x, and the gradient, m = a the Y-intercept = ab

(h) y = ax2 + x + b y – x = ax2 + b where Y = y – x and X = x2, and the gradient, m = a the Y-intercept = b (i) y = x———

ax + b y(ax + b) = x

ax + b = x—y

x—y

= ax + b

where Y = x—y

and X = x,

and the gradient, m = a the Y-intercept = b

(j) y = a——–

x + b y(x + b) = a

x + b———a

= 1—y

1—y

= � 1—a �x + b—

a

where Y = 1—y

and X = x,

and the gradient, m = 1—a

the Y-intercept = b—a

(k) ax + by = xy

ax—–y

+ b = x

ax—–y

= x – b

x—y

= x—a

– b—a

= � 1—a �x – � b—

a � where Y = x—

y and X = x,

and the gradient, m = 1—a

the Y-intercept = – b—a

(l) y = ba–x

log10

y = log10

ba–x

= log10

b + log10

a–x

= –x log10

a + log10

b = (–log

10 a)x + log

10 b

where Y = log10

y and X = x, and the gradient, m = –log

10 a

the Y-intercept = log10

b

2 (a) m = 5 – 1——–3 – 5

= – 4—2

= –2 Y = –2X + c

The line passes through point (5, 1).

1 = –2(5) + c c = 11 The equation of the straight line

is Y = –2X + 11

y

—–x

= –2x2 + 11

y = –2x5—2 + 11 x

(b) m =

1 – (–3)———–

1 – 5 = – 4—

4 = –1 Y = – X + c The line passes through point (1, 1). 1 = –1 + c c = 2 Y = –X + 2

y—x

= –x + 2

y = –x2 + 2x

(c) m = 5 – 2——–2 – 1

= 3 Y = 3X + c The line passes through point (1, 2). 2 = 3(1) + c c = –1 The equation of the straight line

is Y = 3X – 1 log

10y = 3log

10(x + 1) – 1

= 3log10

(x + 1) – log10

10

= log10

(x + 1)3

———10

y = (x + 1)3

———10

(d) m = 3 – 0——–0 – 2

= – 3—2

Y-intercept, c = 3

Y = – 3—2

X + 3

1—y

= – 3—2

� 1—x � + 3

= –3 + 6x———–2x

y = 2x———

6x – 3

3 (a) m = 4 – 1——–4 – 2

= 3—2

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 106: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

The line passes through point (2, 1).

Y = 3—2

X + c

1 = 3—2

(2) + c

c = –2

The equation of the straight line is

Y = 3—2

X – 2 xy = 3—

2x – 2

y = – 2—

x + 3—

2 (b) y = – 2—–

(4) + 3—

2 = 1

4 m = 14 – (–1)————4 – 1

= 15—–3

= 5The line passes through point (1, –1). Y = 5X + c –1 = 5(1) + c c = –6∴ y = 5X – 6

y—x

= 5 x – 6

y—–(1)

= 5 (1) – 6 y = – 1

5 (a) m = 4 – 1——–6 – 0

= 3—6

= 1—2

Y-intercept, c = 1

∴ Y = 1—2

X + 1

log10

y = 1—2

log10

x + 1 2 =

1—2

log10

x + 1

1—2

log10

x = 1 log

10 x = 2

x = 102

= 100

(b) log10

y = 1—2

log10

x + 1

= log10

x1—2 + log

1010

= log10

10x1—2

y = 10 x

6 y = Abx

log10

y = log10

Abx

= log10

A + log10

bx

= (log10

b)x + log10

A

m = 4 – 2——–3 – 2

= 2Y = 2X + log

10A

The line passes through point (2, 2). 2 = 2(2) + log

10A

log10

A = –2 A = 10–2

= 0.01Compare log

10Y = (log

10 b)x + log

10 A

with Y = 2X + log10

A log

10 b = 2

b = 102

= 100

7 m =

1—5——

1– — 2

= – 2—5

Y = – 2—5

X + 1—5

1—y

= – 2—5

� 1—x � + 1—

5

= – 2—–5x

+ x—–5x

5x = –2y + xyxy = 5x + 2yCompare to xy = px + qy∴ p = 5, q = 2

8 m = 2 – 0———

12 – 6

= 2—6

= 1—3

Y = 1—3

X + c

The line passes through point (6, 0). 0 = 1—

3(6) + c

c = –2The equation of the straight line is Y = 1—

3 X – 2

xy = 1—

3x2 – 2

y = 1—

3x – 2—

x 3y = x – 6—x Compare to ky = x + h—

x ∴ k = 3, h = –6

9 2y2 + 6y = x y + 3 = x—–

2y y = 1—

2 � x—y � – 3

where c = –3∴ p = –3

The line passes through point (q, –1).

Y = 1—2

X – 3

–1 = 1—2

q – 3

1—2

q = 2

q = 4∴ p = –3, q = 4

10 y = ax b

log10

y = log10

axb

= log10

a + log10

xb

= b log10

x + log10

a

b = –(–1)——

3 = 1—

3 log

10a = –1

a = 10–1

= 0.1 ∴ a = 0.1, b = 1—

3

11 y (2x + 1) = 3x

2x + 1———

3x = 1—

y

1—y

= 1—3

� 1—x � + 2—

3

The equation of the straight line is

Y = 1—3

X + 2—3

q = 1—3

(7) + 2—3

= 7 + 2——–3

= 3

2 = 1—3

p + 2—3

1—3

p = 2 – 2—3

= 4—3

p = 4∴ p = 4, q = 3

12 xy = – 1—4

x2 + 4 Y = – 1—

4 X + 4

where c = 4 ∴ p = 4The line passes through point (q, 2).

2 = – 1—4

q + 4 1—4

q = 4 – 2 q = 8∴ p = 4, q = 8

13 (a)

x 0.5 1.5 2.5 3.5 4.5 5.5

y—x 20.5 17.5 14.5 11.5 8.5 5.5

(b) y = ax + bx2

y—x = bx + a

Page 107: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(i) a = 22 (intercept on y-axis)

(ii) b = 19 – 7——–1 – 5

= 12—––4

= –3 (iii) When x = 3

y

—3

= –3(3) + 22

y = 13(3) = 39

14 (a) x 1 2 3 4 5

y x 1.50 4.95 8.49 12.20 15.43

(b) y x = ax + b From the graph,

a = 12.2 – 2———–4 – 1.2

= 10.2——2.8

= 3.64 b = –2.2 ∴ a = 3.64, b = –2.2

15 (a) x 1 3 4 7 9

log10 y 1.15 1.41 1.56 1.99 2.24

(b) y = ab x

log10

y = log10

ab x

= log10

a + log10

b x

= (log10

b)x + log10

a log

10 a = 1 (intercept on log

10 y-axis)

a = 10

log10

b = 2.20 – 1.15—————8.4 – 1

= 1.05——7.4

= 0.142 b = 100.142

= 1.3868 ∴ a = 10, b = 1.3868

SPM Appraisal Zone

Paper 1

1 m = 2 – 4——–6 – 0

= – 2—6

= – 1—3

Y = – 1—3

X + 4 xy = – 1—

3x2 + 4

y = – 1—3

x + 4—x

2 (a) m = 11 – 3———5 – 1

= 8—4

= 2 Y = 2X + c 3 = 2(1) + c c = 1 Y = 2X + 1 y = 2� 1—

x � + 1 (b) 5 = 2� 1—

x � + 1 5 – 1 =

2—x

x = 2—4

= 1—2

3 (a) qy2 = –px3 + 1 y2 = –

p—q x3 + 1—q

Y = 2—

3X + c

6 = 2—3

(3) + c c = 6 – 2 = 4 Y = 2—

3 X + 4

y2 = –

p—q x3 + 1—q

1—q = 4

q = 1—4

–p

—–

� 1—4 �

= 2—3

–p = 1—

6 p = – 1—

6 ∴ p = – 1—

6, q = 1—

4 (b) Y = 2—

3X + 4

2 = 2—3

k + 4 2—

3k = –2

k = –3

4 (a) y(2x + k) = h

1—y

= 2x + k———h

1—y

= � 2—h �x +

k—h

2—h

= 6 – 2——–8 – 0

= 4—8

= 1—2

h = 4

k—h

= 2

k—4

= 2

k = 8 ∴ h = 4, k = 8

(b) 4 – 2——–k – 0

= 1—2

2—k

= 1—2

k = 4

5 y = –x2 + 2x

y

—x = –x + 2

Y = –X + 2The line passes through point (1, p). p = –(1) + 2 = 1The line passes through point(q, –3). –3 = –(q) + 2 q = 5∴ p = 1, q = 5

y—x

25

20

15

10

5

01 2 3 4 5 6

x

(5, 7)

(1, 19)22

13

y x

16

14

12

10

8

6

4

2

0

–2

–4

1 2 3 4 5 6x

(1.2, 2)

–2.2

(4, 12.2)

log10

y

2.5

2.0

1.5

1.0

0.5

02 4 6 8 10 12

x

(1, 1.15)

(8.4, 2.20)

Page 108: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

6 4x + 3y = 9xy

4x—y

+ 3 = 9x

4x—y

= 9x – 3

1—y

= – 3—4

� 1—x � + 9—4

Y = – 3—4

X + 9—4

The line passes through point �p, 3—2 �.

3—2

= – 3—4

(p) + 9—4

3—4

p = 9—4

– 3—2

p = 1

The line passes through point (2, q).

q = – 3—4

(2) + 9—4

= 3—4

∴ p = 1, q = 3—4

7 log10

y = log10

axb

= log10

a + log10

xb

= b log10

x + log10

a

b = 1 – 3——–2 – 1

= –2Y = –2X + cThe line passes through point (2, 1). 1 = –2(2) + c c = 5log

10 a = c

= 5 a = 105

= 100 000∴ a = 100 000, b = –2

8 xy = h� x—y � – k

h = 9 – 1——–3 – 1

= 8—2

= 4Y = 4X + cThe line passes through point (1, 1). 1 = 4(1) + c c = –3∴ –k = c = –3 k = 3∴ h = 4, k = 3

9 y x = –px + q

–p = 2 – 8——–4 – 1

= – 6—3

p = 2Y = 2X + cThe line passes through point (1, 8). 8 = 2(1) + c c = 6 q = c = 6∴ p = 2, q = 6

10 xy = –x + h

k – 0——–1 – 3

= –1

k = 2Y = – X + cThe line passes through point (3, 0). 0 = –(3) + c c = 3 h = c = 3∴ h = 3, k = 2

Paper 211 (a) x 1 2 3 4 5

y x 3.1 11.5 19.1 26.4 34.2 (b) y = a x + b—–

x y x = ax + b

(i) a = 26.5 – (– 3.5)——————4 –0

= 7.5 (ii) b = –3.5

12 (a) 1—x 1.00 0.50 0.33 0.25 0.20 0.17

1—y 1.85 1.18 0.89 0.75 0.67 0.61

(b) b—y

= 1 – a—x

1—y

= �– a—b �� 1—

x � + 1—b

1—b

= 0.38 b =

1——0.38

= 2.6316

–a = 1.85 – 0.38—————1 – 0

a = –3.8684

13 (a) x 1.0 1.5 2.0 2.5 3.0

y—x2 2.00 3.56 5.00 6.56 8.11

(b) y

—x2

= ax + b

(i) a = 6.56 – (–1.15)——————2.5 – 0

= 3.084 (ii) b = –1.15 (iii) From the graph, when x = 0.5

y—x2

= 0.4

y = (0.5)2(0.4) = 0.1

14 (a) x x 0.59 1.84 3.04 4.94 7.41

y x 8.62 7.10 5.65 3.41 0.39

y—x2

8

7

6

5

4

3

2

1

0

–1

0.5 1.0 1.5 2.0 2.5 3.0 3.5x

–1.15

y x

10

9

8

7

6

5

4

3

2

1

01 2 3 4 5 6 7 8

x x

y x

40

35

30

25

20

15

10

5

0

–5

1 2 3 4 5 6–3.5

x

(4, 26.5)

1—y

2.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

00.2 0.4 0.6 0.8 1.0 1.2

1—x

0.38

1.1

(1, 1.85)

(2.5, 6.56)

9.35

7.75

Page 109: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) From the graph, y x = ax x + b

(i) a = 0 – 9.35———–

7.75 – 0 = –1.2065 (ii) b = 9.35

15 (a)

x 0.5 1.0 1.5 2.0 2.5 3.0

log10 y 0.792 0.672 0.568 0.462 0.342 0.230

(b) log10

y = log10

ab x

= log10

a + log10

b x

= (log10

b)x + log10

a

From the graph, (i) log

10 a = 0.9

a = 100.9

= 7.9433 (ii) log

10b = 0.9 – 0.23————–

0 – 3 = –0.2233 b = 10–0.2233

= 0.598 (iii) When x = 2.7 log

10 y = 0.295

y = 100.295

= 1.9724

log10

y

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00.5 1.0 1.5 2.0 2.5 3.0 3.5

x

(3, 0.23)

Page 110: Analysis Spm Additional Mathematics

14 Integration

Booster Zone

1 (a) �2x5 dx

= 2x6

—–6

+ c

= 1—3

x6 + c

(b) � 1—3

x dx

= x3—2

——–3� 3—

2 � + c

= 2—9

x3—2 + c

(c) � – 1—x2

dx

= �–x–2 dx

= –x–1

——–1

+ c

= 1—x + c

(d) � 1—–5x6

dx

= � 1—5

x–6 dx

= x–5

——–5(–5)

+ c

= – 1——25x5

+ c

(e) � 4—x

dx

= �4x 1– — 2 dx

= 4x1—2

——1—2

+ c

= 8 x + c

(f) �4x4 dx

= 4x5

——5

+ c

2 (a) �(2 – 1—x

) dx

= �(2 – x 1– — 2 ) dx

= 2x – x1—2

——1—2

+ c

= 2x – 2 x + c

(b) �(3x5 – 4 x ) dx

= (3x5 – 4x1—2 ) dx

= 3x6

——6

– 4x3—2

——3—2

+ c

= 1—2

x6 – 8—3

x3—2 + c

(c) �(6x2 + 4—x2

) dx

= 6x3

—–3

+ 4x–1

—–––1

+ c

= 2x3 – 4—x

+ c

(d) �(1 – 2x + 3x2) dx

= x – 2x2

——2

+ 3x3

——3

+ c

= x – x2 + x3 + c

(e) �(x3 + 1—–x3

) dx

= x4

—–4

+ x–2

—––2

+ c

= 1—4

x4 – 1——2x2

+ c

(f) �(3 – x ) dx

= 3x – x3—2

——3—2

+ c

= 3x – 2—3

x3—2 + c

3 (a) �x2 – 1——–x2

dx

= �(1 – x–2) dx

= x – x–1

——–1

+ c

= x + 1—x + c

(b) � x4 + 7x——–—x3

dx

= �(x + 7x–2) dx

= x2

—–2

+ 7x–1

——–1

+ c = x2

—–2

– 7—x + c

(c) �(2x – 1—2

)2 dx

= �(4x2 – 2x + 1—4

) dx

= 4x3

——3

– 2x2

——2

+ 1—4

x + c

= 4—3

x3 – x2 + 1—4

x + c

(d) �(2 – 3x)2 dx

= �(4 – 12x + 9x2) dx

= 4x – 12x2

——2

+ 9x3

——3

+ c

= 4x – 6x2 + 3x3 + c

(e) �(1 + x)(3 + 2x) dx

= �(3 + 5x + 2x2) dx

= 3x + 5x2

——2

+ 2x3

——3

+ c

(f) �(1 – x)(1 – x ) dx

= �(1 – x – x + x3—2 ) dx

= x – x3—2

——3—2

– x2

—–2

+ x5—2

——5—2

+ c

= x – 2—3

x3—2 – 1—

2x2 + 2—

5x

5—2 + c

(g) � x ( x + 5) dx

= �(x + 5 x ) dx

= x2

—–2

+ 5x3—2

——3—2

+ c

= 1—2

x2 + 10—–3

x3—2 + c

(h) � x (x + 2—x

) dx

= �(x3—2 + 2x

1– — 2 ) dx

= x5—2

——5—2

+ 2x1—2

——1—2

+ c

= 2—5

x5—2 + 4 x + c

(i) �(2 – x)(2 + x)—————–x2

dx

= ��4 – x2

——–x2 � dx

= �(4x–2 – 1) dx

= 4x–1

—–––1

– x + c

= – 4—x – x + c

(j) �� x – 2——–x3 � dx

= �(x –2 – 2x–3) dx

= x –1

—––1

– 2x –2

—–––2

+ c

= – 1—x + 1—–x2 + c

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 111: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(k) �(3x – x )2 dx

= �(9x2 – 6x3—2 + x) dx

= 9x3

—–3

– 6x5—2

——5—2

+ x2

—2

+ c

= 3x3 – 12—–5

x5—2 + 1—

2x2 + c

(l) �(x – 2 x )2 dx

= �(x2 – 4x3—2 + 4x) dx

= x3

—3

– 4x5—2

——5—2

+ 4x2

—–2

+ c

= 1—3

x3 – 8—5

x5—2 + 2x2 + c

4 (a) �(3x – 5)4 dx

= (3x – 5)5

———–5(3)

+ c

= (3x – 5)5

———–15

+ c (b) �(1 – x)5 dx

= (1 – x)6

———–6(–1)

+ c

= – 1—6

(1 – x)6 + c

(c) �(2 – 3x)2 dx

= (2 – 3x)3

———–3(–3)

+ c

= – 1—9

(2 – 3x)3 + c

(d) � 1 – 2x dx

= �(1 – 2x)1—2 dx

= (1 – 2x)3—2

————–3—2

(–2) + c

= – 1—3

(1 – 2x)3—2 + c

(e) � 4x + 5 dx

= (4x + 5)3—2

————–3—2

(4) + c

= 1—6

(4x + 5)3—2 + c

(f) � 4———–(x + 2)3

dx

= �4(x + 2)–3 dx

= 4(x + 2)–2

———–—–2

+ c

= –2———–(x + 2)2

+ c

(g) � 1———–(2x + 1)2

dx

= �(2x + 1)–2 dx

= (2x + 1)–1

———–––1(2)

+ c

= –1———––2(2x + 1)

+ c

(h) �� 3——––2x – 3 �3

dx

= �27(2x – 3)–3 dx

= 27(2x – 3)–2

———––—–2(2)

+ c

= –27———––4(2x – 3)2

+ c

(i) � –6——––(x – 2)5

dx

= �–6(x – 2)–5 dx

= –6(x – 2)–4

———––—–4

+ c

= 3———–—2(x – 2)4

+ c

(j) � 5———2 – 6x

dx

= �5(2 – 6x) 1– — 2 dx

= 5(2 – 6x)1—2

————–1—2

(–6) + c

= – 5—3

2 – 6x + c

5 (a) dy—–dx

= 2x – 5

y = �(2x – 5) dx = x2 – 5x + c At (4, –2) –2 = (4)2 – 5(4) + c –2 = –4 + c c = 2 ∴ y = x2 – 5x + 2

(b) dy—–dx

= 15x2 – 12

y = �(15x2 – 12) dx

= 5x3 – 12x + c At (1, 3), 3 = 5(1)3 – 12(1) + c 3 = –7 + c c = 10 ∴ y = 5x3 – 12x + 10

(c) dy—–dx

= x2(2x + 1)

y = �(2x3 + x2) dx

= 1—2

x 4 + 1—3

x3 + c

At (1, –1),

–1 = 1—2

+ 1—3

+ c

c = – 11—–6

∴ y = 1—2

x 4 + 1—3

x3 – 11—–6

(d) dy—–dx

= x2(x – 3)

y = �(x3 – 3x2) dx

= x4

—–4

– x3 + c

At (2, –6), –6 = 4 – 8 + c c = –2

∴ y = 1—4

x4 – x3 – 2

(e) dy—–dx

= 2x(x – 3)

y = �(2x2 – 6x) dx

= 2—3

x3 – 3x2 + c

At (3, 6), 6 = 18 – 27 + c c = 15

∴ y = 2—3

x3 – 3x2 + 15

(f) dy—–dx

= (3x – 2)2

y = (9x2 – 12x + 4) dx = 3x3 – 6x2 + 4x + c At (1, 2), 2 = 3 – 6 + 4 + c c = 1 ∴ y = 3x3 – 6x2 + 4x + 1

(g) dy—–dx

= 4———–(x + 2)2

y = �4(x + 2)–2 dx

= 4(x + 2)–1

———–—–1(1)

+ c

= –4——–x + 2

+ c

At (2, 7), 7 = –1 + c c = 8 ∴ y = –4——–

x + 2 + 8

(h) dy—–dx

= 2x + 3

y = (2x + 3)1—2 dx

= (2x + 3)3—2

————–3—2

(2) + c

= (2x + 3)3—2

————–3

+ c

At (3, 5),

5 = 93—2

—–3

+ c

5 = 9 + c c = –4

∴ y = (2x + 3)3—2

————–3

– 4

6 dy—–dx

= 3x2 + 2—–x2

y = �(3x2 + 2x–2) dx

= x3 + 2x–1

—–––1

+ c

= x3 – 2—x

+ c

Page 112: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Since (1, 3) lies on the curve,

3 = (1)3 – 2—–(1)

+ c

3 = –1 + cc = 4∴ y = x3 – 2—x + 4

7 dy—–dx

= 1 + 2x

y = � 1 + 2x dx

= (1 + 2x)3—2

————–3—2

(2) + c

= 1—3

(1 + 2x)3—2 + c

When x = 4 and y = 30, we have

30 = 1—3

(9)3—2 + c

30 = 9 + c c = 21

y = 1—3

(1 + 2x)3—2 + 21

Thus, when x = 0,

y = 1—3

(1) + 21

= 64—–3

8 (a) dy—–dx

= kx – 6

At (2, 1), dy—–dx

= 4

2k – 6 = 4 2k = 10 k = 5

(b) dy—–dx

= 5x – 6

y = �(5x – 6) dx

= 5x2

—–2

– 6x + c

At (2, 1),

1 = 5(2)2

—–—2

– 6(2) + c

1 = 10 – 12 + c c = 3 ∴ y = 5—

2x2 – 6x + 3

9 (a) dy—–dx

= p – x

At (2, 3), dy—–dx

= 0

p – 2 = 0 p = 2

(b) dy—–dx

= 2 – x

y = ��(2 – x) dx

= 2x – x2

—2

+ c

At (2, 3),

3 = 2(2) – (2)2

——2

+ c

3 = 4 – 2 + c c = 1

∴ y = 2x – 1—2

x2 + 1

10 (a) dy—–dx

= 3x2 + px + q

At (1, 0), dy—–dx

= 0

3(1)2 + p(1) + q = 0 p + q = –3 ... 1

At (–3, 32), dy—–dx

= 0

32 = 3(–3)2 + p(–3) + q

3p – q = –5 ... 2

1 + 2 : 4p = –8 p = –2 Substitute p = –2 into 1 : –2 + q = –3 q = –1

(b) dy—–dx

= 3x2 – 2x – 1

y = �(3x2 – 2x – 1) dx

= x3 – x2 – x + c At (1, 0) 0 = 1 –1 – 1 + c c = 1 ∴ y = x3 – x2 – x + 1

11 (a) 3

�� 1

�2 – 3—–x2 � dx

3

= �2x + 3—x � 1

= (6 + 1) – (2 + 3) = 2

(b) 4

�� 1

�x2 – 4 + 4—–x2 � dx

4

= � x3

—3

– 4x – 4—x � 1

= � 64—–3

– 16 – 1� – � 1—3

– 4 – 4� = 12

(c) 4

�� 1

(6x – 3 x ) dx

4

= �3x2 – 2x3—2 �

1

= (48 – 16) – (3 – 2) = 31

(d) 4

�� 2

�x3 – 10—–x2 � dx

4

= � 1—4

x4 + 10—–x �

2

= (64 + 5—2

) – (4 + 5)

= 57 1—2

(e)

4—3��

1(3x – 2) dx

= � 3—

2x2 – 2x�

4—3

1

= 3—2 � 16—–

9 � – 2� 4—3 � – � 3—

2 – 2�

= 1—2

(f) 9

�� 4

1—x

dx

9

= �2 x � 4

= 6 – 4 = 2

(g) 27

�� 1

3 x dx

27

= � 3—4

x4—3 �

1

= 243——4

– 3—4

= 60

(h) 2

�� 1

(3 – 2x2 + 4—–x3

) dx

2

= �3x – 2x3

—–3

– 2—–x2 �

1

= �6 – 16—–3

– 1—2 � – �3 – 2—

3 – 2�

= – 1—6

12 (a) 2

�� 1

2x – 1———x3

dx

= 2

�� 1

(2x–2 – x–3) dx

2

= � –2—–x + 1—–2x2 �

1

= �–1 + 1—8 � – �–2 + 1—

2 � = 5—

8

(b) 3

�� 2

(x – 2)(x – 3) dx

= 3

�� 2

(x2 – 5x + 6) dx

3

= � x3

—–3

– 5x2

—–2

+ 6x� 2

= �9 – 45—–2

+ 18� – � 8—3

– 10 + 12� = – 1—

6

(c) 2

�� 1

x2 + 1—–—–x2

dx

= 2

�� 1

(1 + x–2) dx

2

= �x – 1—x � 1

= �2 – 1—2 � – (1 – 1)

= 3—2

Page 113: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(d) 4

�� 1

2 + x—–—–

x2 dx

= 4

�� 1

(2x–2 + x–3—2 ) dx

4

= � –2—–x

– 2—–x �

1

= �– 1—2

– 1� – (–2 – 2)

= 2 1—2

(e) 3

�� 1

1—x2

(4x2 + 9) dx

= 3

�� 1

(4 + 9x–2) dx

3

= �4x – 9—x � 1

= (12 – 3) – (4 – 9) = 14

(f) 4

�� 1

�3x2 + 2 x—–—–—–

x2� dx

= 4

�� 1

(3 + 2x–3—2 ) dx

4

= �3x – 4—–x �

1

= (12 – 2) – (3 – 4) = 11

(g) –1

�� –2

�x2 + 2—x �2

dx

= –1

�� –2

(x4 + 4x + 4x–2) dx

–1

= � x5

—–5

+ 2x2 – 4—x � –2

= �– 1—5

+ 2 + 4� – �– 32—–5

+ 8 + 2� = 2 1—

5

(h) 4

�� 0

x (1 – x ) dx

= 4

�� 0

( x – x) dx

4

= � 2—3

x3—2 – 1—

2x2�

0

= 16—–3

– 8

= – 8—3

13 (a) 1

�� –1

1——–—(x + 3)2

dx

1

= � –1——–x + 3 �

–1

= – 1—4

– �– 1—2 �

= 1—4

(b)

4—3��

1(3x – 2)5 dx

= � (3x – 2)6

————18 �

4—3

1

= 32—–9

– 1—–18

= 3 1—2

(c) 2

�� 1

1———3x – 2

dx

2

= � 2—3

3x – 2 � 1

= 4—3

– 2—3

= 2—3

(d) 1

�� 0

1———–(3x + 2)2

dx

1

= � –1——–—–3(3x + 2)� 0

= – 1—–15

– �– 1—6 �

= 1—–10

(e) 5

�� 1

(3x + 1) 1–— 2 dx

5

= � 2—3

3x + 1� 1

= 8—3

– 4—3

= 4—3

(f) 4

�� 3

(5 – x)5 dx

4

= � –1—–6

(5 – x)6� 3

= – 1—6

– �– 32—–3 �

= 10 1—2

(g) 4

�� 0

2x + 1 dx

4

= � 1—3

(2x + 1)3—2 �

0

= 9 – 1—3

= 26—–3

(h) 3

�� 2

1——––(4 –x)4

dx

3

= � 1————3(4 – x)3 �

2

= 1—3

– 1—–24

= 7—–24

14 (a) 4

�� 0

f(x) dx = 4

�� 1

f(x) dx + 1

�� 0

f(x) dx

1

�� 0

f(x) dx = 5 – 8

= –3

(b) 4

�� 0

[ x + f(x)] dx

= 4

�� 1

x dx + 4

�� 1

f(x) dx

4

= � 2—3

x3—2 � + 8

1

= � 16—–3

– 2—3 � + 8

= 12 2—3

(c) 4

�� 0

[5f(x) – 2x] dx

= 5 4

�� 0

f(x) dx – 4

�� 0

2x dx

4 = 5(5) – �x2� 0

= 25 – 16 = 9

15 (a) 1

�� 3

f(x) dx = – 3

�� 1

f(x) dx

= –2

(b) 4

�� 1

f(x) dx = 4

�� 3

f(x) dx + 3

�� 1

f(x) dx

= 3 + 2 = 5

(c) 3

�� 1

f(x) dx + 3

�� 4

f(x) dx

= 2 – 3 = –1

16 (a) 3

�� 1

1—3

f(x) dx = 1—3

3

�� 1

f(x) dx

= 1—3

(3)

= 1

(b) 3

�� 1

[f(x) + x] dx

= 3

�� 1

f(x) dx + 3

�� 1

x dx

3

= 3 + � x2

—–2 �

1

= 3 + � 9—2

– 1—2 �

= 3 + 4 = 7

(c) 6

�� 1

[2 f(x) + x] dx

= 2 6

�� 1

f(x) dx + 6

�� 1

x dx

= 2 � 6

�� 3

f(x) dx + 3

�� 1

f(x) dx� +

6

� x2

—–2 �

1

= 2(4 + 3) + �18 – 1—2 �

= 14 + 35—–2

= 31 1—2

17 5

�� 1

[f(x) + kx] dx = 28

5

�� 1

f(x) dx + 5

�� 1

kx dx = 28

5

4 + � kx2

—–2 � = 28

1

25—–2

k – 1—2

k = 24

12k = 24 k = 2

Page 114: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

18 (a) 5

�� 1

3—5

f(x) dx

= 3—5

5

�� 1

f(x) dx

= 3—5

(10)

= 6

(b) 1

�� 5

–2 f(x) dx

= –2 1

�� 5

f(x) dx

= –2(–10) = 20

(c) 3

�� 1

f(x) dx + 5

�� 3

[f(x) + 2] dx

= 3

�� 1

f(x) dx + 5

�� 3

f(x) dx + 5

�� 3

2 dx

5 = 10 + �2x� 3

= 10 + (10 – 6) = 14

19 y = x———

1 + 5x

dy—–dx

= 1 + 5x – 5x—————

(1 + 5x)2

= 1————

(1 + 5x)2 (shown)

3

�� 1

� 4———1 + 5x �2

dx = 16 3

�� 1

1————

(1 + 5x)2 dx

3

= 16 � x———1 + 5x �

1

= 16� 3—–16

– 1—6 �

= 1—3

20 y = x 6 + 3x2

dy—–dx

= 6 + 3x2 + 3x2

———–6 + 3x2

= 6 + 3x2 + 3x2

——————6 + 3x2

= 6 + 6x2

———–6 + 3x2

(shown)

6 5

�� 1

1 + x2

———–6 + 3x2

dx = �x 6 + 3x2 �5

1

5

�� 1

1 + x2

———–6 + 3x2

dx = 1—6

(45 – 3)

= 7

21 (a) Area = 3

�� 0

(x2 – 2x + 2) dx

3

= � x3

—–3

– x2 + 2x� 0

= 9 – 9 + 6 = 6 unit2

(b) Area = 3

�� 1

(x2 + 1) dx

3

= � x3

—–3

+ x� 1

= 9 + 3 – � 1—3

+ 1� = 10 2—

3 unit2

(c) Area = 3

�� 2

4—x2

dx

3

= � –4—–x �

2

= – 4—3

– (–2)

= 2 – 4—3

= 2—3

unit2

(d) Area = 3

�� 0

(x – 3)2 dx

3

= �(x – 3)3

———3 �

0

= –(–9) = 9 unit2

22 (a) Area

= 4

�� 1

(y – 3)2 + 2 dy

= 4

�� 1

(y2 – 6y + 11) dy

4

= � y3

—–3

– 3y2 + 11y� 1

= 64—–3

– 48 + 44 – � 1—3

– 3 + 11� = 9 unit2

(b) Area = 2

�� 0

y(y – 2) dy

2

= � y3

—–3

– y2� 0

= 8—3

– 4

= – 4—3

= 4—3

unit2

(c) Area = 27

�� 8

y1—3 dy

27

= � 3—4

y4—3 �

8

= 243——4

– 12

= 48 3—4

unit2

(d) Area = 3

�� 0

y2

—–16

dy

3

= � y3

—–48 �

0

= 9—–16

unit2

23 (a) y = x ... 1 y = 8x – x2 ... 2

Substitute 1 into 2 : x = 8x – x2

x2 – 7x = 0 x(x – 7) = 0 x = 0 or x = 7 Substitute x = 7 into 1 : y = 7 Area of A = 1—

2(7)(7)

= 24 1—2

unit2

At x-axis, y = 0 8x – x2 = 0 x2 – 8x = 0 x(x – 8) = 0 x = 0 or x = 8 Area of B

= 8

�� 7

(8x – x2) dx

8

= �4x2 – x3

—–3 �

7

= �256 – 512——3 � – �196 – 343——

3 � = 3 2—

3 unit2

∴ Area of combined region

= 24 1—2

+ 3 2—3

= 28 1—6

unit2

(b) y = 8x ... 1 y = 9 – x2 ... 2

Substitute 1 into 2 : 8x = 9 – x2

x2 + 8x – 9 = 0 (x + 9)(x – 1) = 0 x = –9 or x = 1 Substitute x = 1 into 1 , y = 8(1) = 8

Area of A = 1—2

(1)(8)

= 4 unit2

At x-axis, y = 0 9 – x2 = 0 x = ±3 Area of B

= 3

�� 1

(9 – x2) dx

3

= �9x – x3

—–3 �

1

= (27 – 9) – �9 – 1—3 �

= 9 1—3

unit2

∴ Area of combined region

= 4 + 9 1—3

= 13 1—3

unit2

Page 115: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(c) Area of A = 2 × 2 = 4 unit2

Area of B = 4

�� 2

8—x2

dx

4

= � –8—–x �

2

= –2 – (–4) = 2 unit2

∴ Area of combined region = 4 + 2 = 6 unit2

(d) At x-axis, y = 0 x(x – 1)(x – 4) = 0 x = 0 or x = 1 or x = 4 Area of A

= 1

�� 0

(x3 – 5x2 + 4x) dx

= � x4

—–4

– 5x3

—–3

+ 2x2�1

0

= 1—4

– 5—3

+ 2

= 7—–12

unit2

Area of B

= 4

�� 1

(x3 – 5x2 + 4x) dx

4

= � x4

—–4

– 5x3

—–3

+ 2x2� 1

= 64 – 320——3

+ 32 – � 1—4

– 5—3

+ 2� = –11 1—

4

= 11 1—4

unit2

Area of combined region

= 7—–12

+ 11 1—4

= 11 5—6

unit2

24 (a) y = 9 – x ... 1 y = x2 – 4x + 5 ... 2 Substitute 1 into 2 : 9 – x = x2 – 4x + 5 x2 – 3x – 4 = 0 (x + 1)(x – 4) = 0 x = –1 or x = 4 Substitute x = –1 into 1 : y = 9 – (–1) = 10 Substitute x = 4 into 1 : y = 9 – 4 = 5 Area of trapezium

= 1—2

(5)(10 + 5)

= 37 1—2

unit2

Area under the curve

= 4

�� –1

(x2 – 4x + 5) dx

4

= � x3

—–3

– 2x2 + 5x� –1

= 64—–3

– 32 + 20 – �– 1—3

– 2 – 5� = 16 2—

3 unit2

∴ Area of the shaded region

= 37 1—2

– 16 2—3

= 20 5—6

unit2

(b) y = 7 ... 1 y = 8x – x2 ... 2 Substitute 1 into 2 : 7 = 8x – x2

x2 – 8x + 7 = 0 (x – 1)(x – 7) = 0 x = 1 or x = 7 Area of square = 6 × 7 = 42 unit2

Area under the curve

= 7

�� 1

(8x – x2) dx

7

= �4x2 – x3

—–3 �

1

= 196 – 343——3

– �4 – 1—3 �

= 78 unit2

∴ Area of the shaded region = 78 – 42 = 36 unit2

(c) Area of trapezium

= 1—2

(1)(1 + 2)

= 3—2

unit2

Area under the curve

= 2

�� 1

1—x2

dx

2

= �– 1—x �

1

= – 1—2

– (–1)

= 1—2

unit2

∴ Area of the shaded region

= 3—2

– 1—2

= 1 unit2

(d) y = 8 ... 1 y = x2 – 8x + 20 ... 2

Substitute 1 into 2 : 8 = x2 – 8x + 20 x2 – 8x + 12 = 0 (x – 2)(x – 6) = 0 x = 2 or x = 6 Area of square = 4 × 8 = 32 unit2

Area under the curve

= 6

�� 2

(x2 – 8x + 20) dx

6

= � x3

—–3

– 4x2 + 20x� 2

= 72 – 144 + 120 – � 8—3

– 16 + 40� = 21 1—

3 unit2

∴ Area of the shaded region

= 32 – 21 1—3

= 10 2—3

unit2

25 (a) k

�� 1

4—x2

dx = 2

k

�– 4—x � = 2 1

– 4—k

– (–4) = 2

4—k

= 2

k = 2

(b) k

� 2

3x(x – 2) dx = 4 k �x3 – 3x2� = 4 2

k3 – 3k2 – (8 – 12) = 4 k3 – 3k2 = 0 k2(k – 3) = 0 ∴ k = 3

(c) 3

�� k

3x2 dx = 26 3 �x3� = 26 k 27 – k3 = 26 k3 = 1 k = 1

(d) k

�� 1

[(y – 2)2 + 1] dy = 6

k

� y3

—–3

– 2y2 + 5y� = 6 1

k3

—3

– 2k2 + 5k – ( 1—3

– 2 + 5) = 6

k3

—3

– 2k2 + 5k – 28—–3

= 0

k3 – 6k2 + 15k – 28 = 0 (k2 – 2k + 7)(k – 4) = 0 ∴ k = 4

26 (a) Volume = π 4

�� 2

144———(x + 2)2

dx

4

= π� –144———x + 2 �

2

= π [–24 – (–36)] = 12π unit3

(b) Volume = π 1

�� –1

1———(x + 2)2

dx

1

= π� 1– ——– x + 2 �

–1

= π �– 1—3

– (–1)� = 2—

3 π unit2

Page 116: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(c) Volume = π 4

�� 1

(5 + 4x – x2) dx

4

= π�5x + 2x2 – x3

—–3 �

1

= π �20 + 32 – 64—–3

�5 + 2 – 1—3 ��

= 24π unit3

(d) Volume = π 4

�� 0

16x dx 4 = π�8x2� 0

= 128π unit3

27 (a) Volume

= π 4

� 3

(4 – y) dy

4

= π�4y – y2

—–2 �

3

= π �16 – 8 – �12 – 9—2 ��

= 1—2

π unit3

(b) Volume = π 4

�� 0

y—2

dy

4

= π � y2

—–4 �

0

= 4π unit3

(c) Volume = π 4

�� 2

(y – 2) dy

4

= π� y2

—–2

– 2y� 2

= π [8 – 8 – (2 – 4)] = 2π unit3

(d) Volume

= π 4

�� 0

(4y – y2)2 dy = π

4

�� 0

(16y2 – 8y3 + y4)dy

4

= π� 16y3

—––3

– 2y4 + y5

—5 �

0

= π �1024——

3 – 512 + 1024——

5 � = 34 2—–

15 π unit3

28 (a) π k

�� 1

8x dx = 32π k �4x2� = 32 1

4k2 – 4 = 32 4k2 = 36 k2 = 9 k = ±3 ∴ k = 3

(b) π 2

�� k

16———(3 – x)2

= 8π

2

� 16——–3 – x � = 8

k

16 – 16——–3 – k

= 8

16——–3 – k

= 8

2 = 3 – k k = 1

(c) π k

�� 1

4—–y2

dy = 3π

k

�– 4—y � = 3 1

– 4—k

– (–4) = 3

4—k

= 1

k = 4

(d) π 6

�� k

(6 – y) dy = 8π

6

�6y – y2

—–2 � = 8

k

36 – 18 – �6k – k2

—–2 � = 8

k2

—–2

– 6k + 10 = 0

k2 – 12k + 20 = 0 (k – 2)(k – 10) = 0 k = 2 or k = 10 ∴ k = 2

SPM Appraisal Zone

1 (a) �(2 – 3x)2 dx

= �(4 – 12x + 9x2) dx

= 4x – 6x2 + 3x3 + c

(b) 9

�� 4

1—x

dx

9 = �2 x �

4

= 6 – 4 = 2

2 (a) �x4 + 7x—–—–x3

dx

= �(x + 7x–2) dx

= x2

—2

+ 7x–1

——–1

+ c

= 1—2

x2 – 7—x

+ c

(b) 3

�� 0

1———1 + 5x

dx

3

= � 2—5

1 + 5x � 0

= 8—5

– 2—5

= 6—5

3 dy—–dx

= 4——–—(x + 2)2

y = �4(x + 2)–2 dx

= 4(x + 2)–1

——–——–1

+ c

= –4——–x + 2

+ c

Since (2, 7) lies on the curve

7 = –4—–4

+ c

c = 8

∴ y = –4——–x + 2

+ 8

4 y = x———1 + 5x

dy—–dx

= 1 + 5x – 5x—————(1 + 5x)2

= 1———–(1 + 5x)2

(shown)

3

�� 1

� 4———1 + 5x �

2

dx = 16 3

�� 1

1————(1 + 5x)2

dx

3

= 16 � x———1 + 5x �

1

= 16 � 3—–16

– 1—6 �

= 1—3

5 (a) (i) 1

�� 5

f(x) dx = – 5

�� 1

f(x) dx

= –4

(ii) 5

�� 1

2 f(x) dx = 2 5

�� 1

f(x) dx

= 2(4) = 8

(b) 5

�� 1

[f(x) + kx] dx = 28

5

�� 1

f(x) dx + 5

�� 1

kx dx = 28

5

4 + � kx2

—–2 � = 28

1

25—–2

k – 1—2

k = 24

12k = 24 k = 2

6 (a) y = x2

———2x – 1

dy—–dx

= 2x(2x – 1) – 2x2

——————–(2x – 1)2

= 2x2 – 2x————–(2x – 1)2

= 2x(x – 1)————–(2x – 1)2

(b) � 2x(x – 1)————–(2x – 1)2

dx = x2

———2x – 1

2

2 2

�� 1

x(x – 1)———––(2x – 1)2

= � x2

———2x – 1 �

1

2

�� 1

x(x – 1)———––(2x – 1)2

= 1—2 � 4—

3 – 1�

= 1—2 � 1—

3 �

= 1—6

Page 117: Analysis Spm Additional Mathematics

8 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

7 (a) (i) 3

�� 0

1—6

f(x) dx = 1—6

3

�� 0

f(x) dx

= 1—6

(12)

= 2

(ii) 0

�� 3

[f(x) – x] dx

= 0

�� 3

f(x) dx – 0

�� 3

x dx

0

= –12 – � x2

—–2 �

3

= –12 – � –9—–2 �

= –7 1—2

(b) 3

�� 0

[f(x) + mx] dx = 39

3

�� 0

f(x) dx + 3

�� 0

mx dx = 39

3

12 + � mx2

—––2 � = 39

0

9—2

m = 27

m = 27� 2—9 �

= 6

8 dy—–dx

= x2 – 4—––—x2

y = �(1 – 4x–2) dx

= x + 4—x

+ c

Since (2, 7) lies on the curve,7 = 2 + 2 + cc = 3

y = x + 4—x

+ 3

When y = 8

8 = x + 4—x

+ 3

8x = x2 + 4 + 3x x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0x = 1 or x = 4

∴ The coordinates are (1, 8) and (4, 8).

9 (a) At point P, dy—–dx

= 5

4x – 3 = 5 4x = 8 x = 2 When x = 2, y = 2(2) = 4 ∴ P(2, 4)

(b) dy—–dx

= 4x – 3

y = �(4x – 3) dx

= 2x2 – 3x + c When x = 2 and y = 4, we have 4 = 2(2)2 – 3(2) + c c = 2 ∴ y = 2x2 – 3x + 2

10 dy—–dx

= x(2 + 3x)

y = �(2x + 3x2) dx

= x2 + x3 + cAt point (1, –3),–3 = 1 + 1 + cc = –5y = x2 + x3 – 5At point (2, k),k = (2)2 + (2)3 – 5 = 4 + 8 – 5 = 7

11 (a) 4

�� 2

�x3 – 8—x2 � dx

4

= � x4

—4

+ 8—x �

2

= 64 + 2 – (4 + 4) = 58

(b) 4

�� 1

�3 x + 2—x � dx

4

= �2x3—2 + 4 x �

1

= 16 + 8 – (2 + 4) = 18

12 ds—–dt

= 3(t – 1)2 + 2

s = �[3(t – 1)2 + 2] dt

= �(3t2 – 6t + 5) dt

= t3 – 3t2 + 5t + cWhen t = 1 and s = 5, we have5 = 1 – 3 + 5 + cc = 2∴ s = t3 – 3t2 + 5t + 2

13 2

�� 1

(2x + k) dx = 5

2 �x2 + kx� = 5 1

4 + 2k – (1 + k) = 5 3 + k = 5 k = 2

14 5

�� 3

[2f(x) – kx] dx = 12

2 5

�� 3

f(x) dx – 5

�� 3

kx dx = 12

5

2(8) – � kx2

—–2 � = 12

3

25—–2

k – 9—2

k = 4

8k = 4

k = 1—2

15 (a) �x2(3 + 1—x4 ) dx

= �(3x2 + x–2) dx

= x3 – 1—x

+ c

(b) �(2x + 5)3 dx

= (2x + 5)4

———––4(2)

+ c

= 1—8

(2x + 5)4 + c

16 4

�� 0

y dx + 8

�� 0

x dy

= 4 × 8= 32

17 Area of the shaded region

= 3

�� 0

(4x – x2) dx – 1—2

(3)(3)

3

= �2x2 – x3

—3 � – 9—

2 0

= 18 – 9 – 9—2

= 4 1—2

unit2

18 (a) x + 1 = (x – 1)2

x + 1 = x2 – 2x + 1 x2 – 3x = 0 x(x – 3) = 0

x = 0 or x = 3 When x = 0, y = 0 + 1 = 1 When x = 3, y = 3 + 1 = 4 ∴ A(0, 1) and B(3, 4)

(b) Area of the shaded region

= 1—2

(3)(1 + 4) – 3

� 0

(x – 1)2 dx

3

= 15—–2

– �(x – 1)3

—–—–3 �

0

= 15—–2

– � 8—3

– �– 1—3 ��

= 4 1—2

unit2

19 x = 8—x2

x3 = 8x = 2When x = 2, y = 2Area of the shaded region

= 1—2

(2)(2) + 4

� 2

� 8—x2 � dx

4

= 2 + � –8—–x �

2

= 2 + [–2 – (–4)]= 4 unit2

20 k

�� 1

8—x2 dx =

2

�� k

8—x2 dx

k 2

�– 8—x � = �– 8—

x � 1 k

– 8—k

– (–8) = –4 – �– 8—k �

8 – 8—k

= 8—k

– 4

16—–k

= 12

k = 11— 3

Page 118: Analysis Spm Additional Mathematics

9© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

21 Area of region M

= 1

�� 0

(x3 – 3x2 + 2x) dx

1

= � x4

—4

– x3 + x2� 0

= 1—4

– 1 + 1

= 1—4

unit2

Area of region N

= 2

�� 1

(x3 – 3x2 + 2x) dx

2

= � x4

—4

– x3 + x2� 1

= 4 – 8 + 4 – � 1—4

– 1 + 1�= – 1—

4

= 1—4

unit2

∴ Areas of regions A and B are equal.

22 Area of region A

= 2

�� 1

� 4—x2 � dx

2

= � –4—–x �

1

= –2 – (–4)= 2 unit2

Area of region B

3

= � –4—–x �

2

= – 4—3

– (–2)

= 2—3

unit2

The ratio of the areas of region A to

region B is 2 : 2—3

= 3 : 1.

23 a

�� 0

2x2 dx = 18

a

� 2x3

—–3 � = 18

0

2a3

—–3

= 18

a3 = 18� 3—2 �

= 27 a = 3

24 k

�� 1

5x4 dx = 31

k

�x5� = 31 1 k5 – 1 = 31 k5 = 32 k5 = 25

k = 2

25 Volume formed = π 4

�� 1

1—x2

dx

4

= π� –1—–x �

1

= π�– 1—4

– (–1)� = 3—

4π unit3

26 Volume generated = π 2

�� 1

y dy

2

= π� y2

—–2 �

1

= π�2 – 1—2 �

= 3—2

π unit3

27 Volume formed

= π 4

�� 2

9 � x2

—4

– 1� dx

4

= π� 3—4

x3 – 9x� 2

= π[48 – 36 – (6 – 18)]= 24π unit3

28 π p

�� 1

4x dx = 6π

p

�2x2� = 6 1

2p2 – 2 = 6 2p2 = 8 p2 = 4 p = ±2∴ p = 2

29 (a) x2 = 8 – x2

2x2 = 8 x2 = 4 x = ±2 When x = 2, y = (2)2

= 4 ∴ P(2, 4)

(b) Area of region A = 2

�� 0

x2 dx

2

= � x3

—–3 �

0

= 8—3

unit2

(c) Volume of region B generated

= π 8

�� 4

(8 – y) dy

8

= π�8y – y2

—–2 �

4

= π[64 – 32 – (32 – 8)] = 8π unit3

30 (a) 8 – x = 12—–x

8x – x2 = 12 x2 – 8x + 12 = 0 (x – 2)(x – 6) = 0 x = 2 or x = 6

When x = 2, y = 12—–2

= 6

When x = 6, y = 12—–6

= 2

∴ A(2, 6) and B(6, 2)

(b) The total area of regions P and Q = Area of trapezium

= 1—2

(6 – 2)(6 + 2)

= 16 unit2

(c) Volume of region Q generated

= π 6

�� 2

144—––x2

dx

6

= π� –144——–x �

2

= π[–24 – (–72)] = 48π unit3

31 (a) 2x = 6x – x2

x2 – 4x = 0 x(x – 4) = 0 x = 0 or x = 4 When x = 4, y = 2(4) = 8 ∴ A(4, 8) At x-axis, y = 0 6x – x2 = 0 x2 – 6x = 0 x(x – 6) = 0 x = 6 ∴ B(6, 0)

Hence, A(4, 8) and B(6, 0)

(b) Area of region P

= 4

�� 0

(6x – x2) dx – 1—2

(4)(8)

4

= �3x2 – x3

—3 � – 16

0

= 48 – 64—–3

– 16

= 10 2—3

unit2

(c) Volume generated

= π 6

�� 4

(6x – x2)2 dx

= π 6

�� 4

(36x2 – 12x3 + x4) dx

6

= π�12x3 – 3x4 + x5

—–5 �

4

= π�2592 – 3888 + 7776——–5

�768 – 768 + 1024——–5 ��

= 54 2—5

π unit3

Page 119: Analysis Spm Additional Mathematics

10 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

32 (a) At y-axis, x = 0 2y = 3(0) + 1

y = 1—2

∴ A�0, 1—2 �

y = 2—x2

When y = 1—2

, 1—2

= 2—x2

x2 = 4 x = ±2

∴ C�2, 1—2 �

Hence, A�0, 1—2 � and C�2, 1—

2 �(b) Area of the shaded region ABC

= 1—2

(1)�2 1—2 � +

2

�� 1

2—x2

dx – 2� 1—2 �

2

= 5—4

+ � –2—–x � – 1

1

= 1—4

+ [– 1 – (–2)]

= 5—4

unit2

(c) Volume generated

= π 2

�� 1

� 4—–x4 � dx

2

= π� –4—–3x3 �

1

= π�– 1—6

– �– 4—3 ��

= 7—6

π unit3

33 (a) y = 4x + 1 At y-axis, x = 0

y = 4(0) + 1 = 1 ∴ A(0, 1)

(b) dy—–dx

= 1—2

(4x + 1) 1– — 2 (4)

= 2———4x + 1

At P(2, 3), dy—–dx

= 2———–4(2) + 1

= 2—3

∴ Equation of the normal PB:

y – 3 = –3—–2

(x – 2)

2y – 6 = –3x + 6 2y + 3x = 12(c) 2y + 3x = 12 At x-axis, y = 0 2(0) + 3x = 12 x = 4 ∴ B(4, 0)

(d) Area of shaded region X

= 2

�� 0

4x + 1 dx

2

= � 1—6

(4x + 1)3—2 �

0

= 9—2

– 1—6

= 4 1—3

unit2

Area of shaded region Y

= 1—2

(4 – 2)(3)

= 3 unit2

(e) Volume generated

= π 2

�� 0

(4x + 1) dx

2

= π�2x2 + x� 0

= 10π unit3

34 (a) 3x = 4 – x2

x2 + 3x – 4 = 0 (x + 4)(x – 1) = 0 x = –4 or x = 1 When x = 1, y = 3(1) = 3 ∴ P(1, 3)

(b) Area of region R

= 1—2

(1)(3) + 2

�� 1

(4 – x2) dx

2

= 3—2

+ �4x – 1—3

x3� 1

= 3—2

+ �8 – 8—3

– �4 – 1—3 ��

= 3—2

+ 5—3

= 3 1—6

unit2

(c) Volume of region S, generated

= 1—3

π(1)2(3) + π 4

�� 3

(4 – y) dy

4

= π + π �4y – y2

—2 �

3

= π + π �16 – 8 – �12 – 9—2 ��

= 1 1—2

π unit3

35 (a) 2x2 + 3x = 27 2x2 + 3x – 27 = 0 (2x + 9)(x – 3) = 0

x = – 9—2

or x = 3

When x = 3, y = 32

= 9 ∴ A(3, 9) At x-axis, y = 0 3x = 27 x = 9 ∴ B(9, 0) Hence, A(3, 9) and B(9, 0).

(b) Area of the shaded region

= 3

�� 0

x2 dx + 1—2

(9 – 3)(9)

3

= � x3

—–3 � + 27

0

= 9 + 27 = 36 unit2

(c)

Volume generated = π 9

� 0

y dy

9

= π� y2

—2 �

0

= 40 1—2

π unit3

36 (a) (i) At x-axis, y = 0 4(0) = x – 1 x = 1 ∴ B(1, 0)(ii) Area of the shaded region

= 1—2

(2 – 1)� 1—4 � +

4

� 2

� 1—x2 � dx

4

= 1—8

+ �– 1—x �

2

= 1—8

+ �– 1—4

– �– 1—2 ��

= 3—8

unit2

(b) π k

�� 2

� 12——–x + 2 �

2

dx = 12π

k

� –144——–x + 2 � = 12

2

–144——–k + 2

– (–36) = 12

144——–k + 2

= 24

k + 2 = 6 k = 4

37 (a) 7 – 3x = 1—4

x2

x2 + 12x – 28 = 0 (x + 14)(x – 2) = 0 x = –14 or x = 2 When x = 2, y + 3(2) = 7 y = 1 ∴ P(2, 1)

(b) Area of region A

= 2

�� 0

1—4

x2 dx

2

= � 1—–12

x3� 0

= 2—3

unit2

y

x0

9 y = x2

Page 120: Analysis Spm Additional Mathematics

11© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(c) Volume of region B, generated

= 1—3

π (2)2(6) + π 1

�� 0

4y dy 1 = 8π + π �2y2�

0

= 8π + 2π = 10π unit3

38 (a) (i) x = x2 – 5x + 8 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x = 2 or x = 4 When x = 2, y = 2 When x = 4, y = 4 ∴ P(2, 2) and Q(4, 4)(ii) Area of the shaded region

= 1—2

(4 – 2)(2 + 4) –

4

�� 2

(x2 – 5x + 8) dx

4

= 6 – � x3

—3

– 5x2

—–2

+ 8x� 2

= 6 – � 64—–3

– 40 + 32 –

� 8—3

– 10 + 16�� = 6 – 4 2—

3

= 1 1—3

unit2

(b) π 4

�� k – 1

4—–x2

dx = 3π

4

� –4—–x � = 3

k – 1

–1 – � –4——–k – 1 � = 3

4——–k – 1

= 4

k – 1 = 1 k = 2

39 (a) y = 6x2 – x3

dy—–dx

= 0

12x – 3x2 = 0 3x2 – 12x = 0 3x(x – 4) = 0 x = 0 or x = 4 When x = 4, y = 6(4)2 – 43

= 32 ∴ P(4, 32) At x-axis, y = 0 0 = 6x2 – x3

x3 – 6x2 = 0 x2(x – 6) = 0 x = 0 or x = 6 ∴ Q(6, 0)

(b) Area of the shaded region

= 6

�� 0

(6x2 – x3) dx – 1—2

(6 – 4)(32)

6

= �2x3 – x4

—4 � – 32

0

= 432 – 324 – 32 = 76 unit2

40 (a) 3 = x2 + 2 x2 = 1 x = ±1 ∴ A(–1, 3) and B(1, 3)(b) The area of the shaded region

= (2 × 3) – 1

�� –1

(x2 + 2) dx

1

= 6 – � x3

—3

+ 2x� –1

= 6 – � 1—3

+ 2 – �– 1—3

– 2�� = 1 1—

3 unit2

(c) Volume of the shaded region generated

= π 3

� 2

(y – 2) dy

3

= π� y2

—2

– 2y� 2

= π� 9—2

– 6 – (2 – 4)� = 1—

2π unit3

Page 121: Analysis Spm Additional Mathematics

15 Vectors

Booster Zone

1 (a) a~ = 2p~

, b~ = –3p~

(b) p~

= 1—2

a~ and p~

= 1–— 3

b~

So, 1—2

a~ = 1–— 3

b~

b~ = 3–— 2

a~

(c) |a~| = |2p~

|

= 2(2) = 4 units |b~| = |–3p

~|

= |–3| × |p~

|

= 3 × 2 = 6 units

2 (a) (i) ⎯→FC = 2a~

(ii) ⎯→EB = –2c~

(iii) ⎯→OF = –a~

(iv) ⎯→AF = c~

(b) (i) |⎯→ED| = 3 units

(ii) ⎯→CF = –2a~

|⎯→CF | = |–2| × |a~|

= 2(3) = 6 units

PQ 3 3 —— = — SR 5

PQ = 3—5

SR

⎯→PQ = 3—

5 (10a~)

= 6a~

4 (a) ⎯→AB = 2u~,

⎯→AC = 2v~

(b) ⎯→AD = 4—

5

⎯→AE

= 4—5

w~

⎯→DE = 1—

5

⎯→AE

= 1—5

w~

⎯→ED = 1–—

5w~

5 u~ = k v~3a~ + 6b~ = k(2a~ + pb~)

3a~ + 6b~ = 2ka~ + kpb~

By comparison,

2k = 3

k = 3—2

and kp = 6

3—2

p = 6

p = 6( 2—3

)

= 4

6 ⎯→AB = p

⎯→PQ

2u~ – 5v~ = p[4u + (k – 6)v~]

2u~ – 5v~ = 4pu~ + p(k – 6)v~

By comparison, 4p = 2

p = 1—2

and p(k – 6) = –5

1—2

(k – 6) = –5

k – 6 = –10 k = –10 + 6 = –4

7 ⎯→PQ = 3a~

a~ = 1—3

⎯→PQ

⎯→QR = 6a~

a = 1—6

⎯→QR

∴ 1—3

⎯→PQ = 1—

6

⎯→QR

⎯→PQ = 1—

2

⎯→QR

Since ⎯→PQ = 1—

2

⎯→QR and Q is the

common point, therefore, P, Q and R are collinear.

8 h – 4 = 0 h = 4

2k – 1 = 0 2k = 1

k = 1—2

9 p + 2q – 5 = 0 p + 2q = 5 1… 2q – p + 3 = 0 p – 2q = 3 2… 1 + 2 : 2p = 8 p = 4Substitute p = 4 into 1 : 4 + 2q = 5 2q = 1

q = 1—2

10 4m – 9n = 5 1… m – 6n = 0 2…

2 × 4 : 4m – 24n = 0 3…1 – 3 : 15n = 5

n = 1—3

Substitute n = 1—3

into 2 :

m – 6( 1—3

) = 0

m – 2 = 0 m = 2

11 (a) ⎯→MD =

⎯→MC +

⎯→CD

= 3b~ – a~

(b) ⎯→ND = 2—

3

⎯→BC

= 2—3

(6b~)

= 4b~

∴ ⎯→DN = –4b~

(c) ⎯→NM =

⎯→ND +

⎯→DM

= 4b~ + a~ – 3b~ = a~ + b~

12 (a) ⎯→PR =

⎯→PS +

⎯→SR

= 2b~ + a~ = a~ + 2b~

⎯→PX = 1—

2

⎯→PR

= 1—2

(a~ + 2b~)

= 1—2

a~ + b~

(b) ⎯→PY = 1—

2

⎯→PX

= 1—2

( 1—2

a~ + b~)

= 1—4

a~ + 1—2

b~

⎯→QY =

⎯→QP +

⎯→PY

= –a~ + 1—4

a~ + 1—2

b~

= – 3—4

a~ + 1—2

b~

(c) ⎯→PZ = 2—

3

⎯→PS

= 2—3

(2b~)

= 4—3

b~

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 122: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

∴ ⎯→XZ =

⎯→XP +

⎯→PZ

= 1–— 2

a~ – b~ + 4—3

b~

= 1–— 2

a~ + 1—3

b~

13 (a) ⎯→QP =

⎯→QR +

⎯→RP

= 4a~ + 2b~

∴ ⎯→ZP =

⎯→ZQ +

⎯→QP

= –a~ – 4a~ – 2b~ = –5a~ – 2b~

(b) ⎯→ZP = 5

⎯→ZY

⎯→ZY = 1—

5

⎯→ZP

= 1—5

(–5a~ – 2b~)

= –a~ – 2—5

b~

∴ ⎯→QY =

⎯→QZ +

⎯→ZY

= a~ + (–a~ – 2—5

b~)

= – 2—5

b~

(c) ⎯→YX =

⎯→YZ +

⎯→ZR +

⎯→RX

= (a~ + 2—5

b~) + 3 a~ + b~

= 4a~ + 7—5

b~

14 (a) (i) ⎯→MQ =

⎯→MP +

⎯→PQ

= –2b~ + 3a~ = 3a~ – 2b~

∴ ⎯→MN = 1—

5

⎯→MQ

= 1—5

(3a~ – 2b~)

= 3—5

a~ – 2—5

b~

(ii) ⎯→SN =

⎯→SP +

⎯→PM +

⎯→MN

= –a~ + 2b~ + ( 3—5

a~ – 2—5

b~) = – 2—

5a~ + 8—

5b~

(iii)

⎯→SR =

⎯→SP +

⎯→PR

= – a~ + 4b~

(b) ⎯→SN = – 2—

5 a~ + 8—

5b~

= 2—5

(–a~ + 4b~)

= 2—5

⎯→SR

∴ The points S, N and R are collinear.

(c) ⎯→SN = 2—

5

⎯→SR

⎯→SN

—— = 2—5

⎯→SR

SN : SR = 2 : 5

∴ SN : NR = 2 : 3

15 (a) ⎯→OM =

⎯→OP +

⎯→PM

= 2a~ + b~

(b) ⎯→ON =

⎯→OR +

⎯→RN

= 2b~ + a~

= a~ + 2b~

(c) ⎯→PN =

⎯→PQ +

⎯→QN

= 2b~ + (–a~)

= –a~ + 2b~

16 (a) ⎯→MC = 3—

2

⎯→ND

= 3—2

(2b~)

= 3b~

(b) ⎯→AC =

⎯→AM +

⎯→MC

= 3—2

(2a~) + 3b~

= 3a~ + 3b~

(c) ⎯→ME =

⎯→MN +

⎯→NE

= –a~ – 2b~

17 (a) ⎯→AB =

⎯→AO +

⎯→OB

= –a~ + b~

(b) (i) ⎯→OR =

⎯→OA +

⎯→AR

= a~ + h(–a~ + b~)

= a~ – ha~ + hb~

= (1 – h)a~ + hb~

(ii) ⎯→OR = k

⎯→OC

= k(a~ + 2b~)

= ka~ + 2kb~ (1 – h)a~ + hb~ = ka~ + 2kb~(c) By comparison, 1 – h = k h + k = 1 1… h = 2k h – 2k = 0 2… 1 – 2 : 3k = 1

k = 1—3

Substitute k = 1—3

into 1 :

h + 1—3

= 1

h = 1 – 1—3

= 2—3

18 (a) ⎯→PS =

⎯→PR +

⎯→RS

= 2q~

+ 2p~

= 2p~

+ 2q~

⎯→QR =

⎯→QP +

⎯→PR

= –p~

+ 2q~

(b) (i) ⎯→PT = λ

⎯→PS

= λ(2p~

+ 2q~

)

= 2λp~

+ 2λq~

(ii) ⎯→QT = �

⎯→QR

= �(–p~

+ 2q~

)

= –�p~

+ 2�q~

(c) ⎯→PT =

⎯→PQ +

⎯→QT

2λp~

+ 2λq~

= p~

– �p~

+ 2�q~

2λp~

+ 2λq~

= (1 – �)p~

+ 2�q~

By comparison, 2λ = 2� λ = �(shown)

19 (a) (i) ⎯→OR = 2—

3

⎯→OQ

= 2—3

(9q~

)

= 6q~

⎯→PR =

⎯→PO +

⎯→OR

= –4p~

+ 6q~

(ii) ⎯→SQ =

⎯→SO +

⎯→OQ

= –2p~

+ 9q~

(b) (i) ⎯→TR = h

⎯→PR

= h(–4p~

+ 6q~

)

= –4hp~

+ 6hq~

(ii) ⎯→TQ = k

⎯→SQ

= k(–2p~

+ 9q~

)

= –2kp~

+ 9kq~

(c) ⎯→RQ =

⎯→RT +

⎯→TQ

3q~

= 4hp~

– 6hq~

– 2kp~

+ 9kq~

3q~

= (4h – 2k)p~

+ (9k – 6h)q~

By comparison, 4h – 2k = 0 1… –6h + 9k = 3 –2h + 3k = 1 2… 2 × 2 : –4h + 6k = 2 3… 1 + 3 : 4k = 2

k = 1—2

1 Substitute k = — into 1 : 2 4h – 2( 1—

2) = 0

4h – 1 = 0

h = 1—4

20 (a) ⎯→AB =

⎯→AO +

⎯→OB

= –a~ + 2b~

(b) ⎯→OD =

⎯→OC +

⎯→CD

= h⎯→OA + k

⎯→CE

= ha~ + k(–ha~ + b~ )

= ha~ – hka~ + kb~

Page 123: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

= (h – hk)a~ + kb~

(c) ⎯→AD =

⎯→AO +

⎯→OD

= –a~ + (h – hk)a~ + kb~

= (h – hk – 1)a~ + kb~ Since A, D and B are collinear,

⎯→AB = p

⎯→AD

–a~ + 2b~ = p[(h – hk – 1)a~ + kb~]

–a~ + 2b~ = (h – hk – 1)pa~ + kpb~

By comparison, kp = 2

p = 2—k

and (h – hk – 1)p = –1

(h – hk – 1)( 2—k

) = –1

h – hk = 1 – 1—2

k

h(1 – k) = 2 – k———2

h = 2 – k————2(1 – k)

21 (a) (i) 3 i~ + 4 j~

(ii) 3� �4

(b) |⎯→O~

A| = 5 units

1(c)

⎯→OA = — (3 i~ + 4 j

~)

5

22 (a) |a~

| = 17 units

82 + k2 = 17 k2 + 64 = 289 k2 = 225

k = 225 = ±15 k = 15, –15

(b) |⎯→PQ | = 13 units

122 + (–k)2 = 13

k2 + 144 = 169 k2 = 25 k = ±5 k = 5, –5

23 (a) 2a~ + b~ = 2

2 � � –6 +

2� �4

= 4� � –12

+ 2� �4

= 6 � � –8

(b) |2a~ + b~| = 62 + (–8)2

= 100 = 10 units(c) The unit vector in the direction of

2a~ + b~

1 = —— (6 i~ – 8 j

~)

10

1 = — (3 i~ – 4 j

~)

5

24 (a) ⎯→AB =

⎯→AO +

⎯→OB

= – i~ + 4 j~

+ 10 i~ + 8 j~

= 9 i~ + 12 j~

|⎯→AB | = 92 + 122

= 225

= 15 units

∴ The unit vector parallel to ⎯→AB

1 = —— (9 i~ + 12 j

~)

15 1 = — (3 i~ + 4 j

~)

5 2(b)

⎯→AP = —

⎯→AB

3

= 2—3

(9 i~ + 12 j~

)

= 6 i~ + 8 j~

⎯→OP =

⎯→OA +

⎯→AP

= i~ – 4 j~

+ 6 i~ + 8 j~

= 7 i~ + 4 j~

39 25 (a)

⎯→OP = ——

–5� � –12 13

= 3 –5� �12

= –15� � 36

25

⎯→OQ = ——

3� �4

5 = 5

3� �4

=

15 � � 20

(b) ⎯→PQ =

⎯→PO +

⎯→OQ

= 15� � –36

+

15 � � 20

= 30� � –16

|⎯→PQ | = 302 + (–16)2

= 1156

= 34 units

26 (a) ⎯→BT =

⎯→BA +

⎯→AT

=

–3 � � –4 +

6� �3

=

3 � � –1

⎯→BC = 2

3 � � –1

=

6 � � –2

⎯→AC =

⎯→AB +

⎯→BC

= 3� �4

+

6 � � –2

= 9� �2

(b) ⎯→CD = –

⎯→AB

=

–3 � � –4

|⎯→CD | = (–3)2 + (–4)2

= 25 = 5 units

27 (a) ⎯→OA =

4� �3

(b) ⎯→AB =

⎯→AO +

⎯→OB

= –4 i~ – 3 j~

+ 5 j~

= –4 i~ + 2 j~

28 ⎯→OB =

⎯→OA +

⎯→OC

= 3 i~ + 4 j~

+ 6 i~ – 2 j~

= 9 i~ + 2 j~

29 a~ – 3b~ =

0 � � 10 – 3

8� �1

= –24� � 7

|a~ – 3b~| = (–24)2 + 72

= 625

= 25 units The unit vector which is parallel to

a~ – 3b~

1= ——

–24� � 7 25

=

24 –—— 25� �

7 —— 25

24 7= –—— i~ + —— j

~

25 25

30 (a) ⎯→PQ =

⎯→PO +

⎯→OQ

=

2 � � –7 +

3 � � –5

= 5� � –12

5� � –12

= m

2 � � –2 + n

1� �6 2m + n = 5 1… –2m + 6n = –12 2… 1 + 2 :7n = –7 n = –1 Substitute n = –1 into 1 : 2m – 1 = 5 2m = 6 m = 3

(b) |⎯→PQ | = 52 + (–12)2

= 169

= 13 units

Page 124: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

The unit vector parallel to ⎯→PQ

1 = ——

5� � –12

13

=

5 —— 13� � –12 —— 13

5 12 = —— i~ – —— j

~

13 13

31 (a) ⎯→AC = p

⎯→AB

⎯→AO +

⎯→OC = p(

⎯→AO +

⎯→OB)

–2 i~ – j~

+ k i~ + 4 j~

= p(–2 i~ – j~

+ 4 i~ + 2 j~

)

(k – 2) i~ + 3 j~

= p(2 i~ + j~

)

By comparison, p = 3 and k – 2 = 2p k – 2 = 6 k = 8

(b) |⎯→AC | = 3|

⎯→AB |

(k – 2)2 + 32 = 3 22 + 12

(k – 2)2 + 9 = 9(5) k2 – 4k + 4 + 9 = 45 k2 – 4k – 32 = 0 (k + 4)(k – 8) = 0 k = –4 or k = 8

32 (a) (i) ⎯→AB =

⎯→AO +

⎯→OB

= –2u~ – v~ + 3u~ – 2v~ = u~ – 3v~

(ii) ⎯→AC =

⎯→AO +

⎯→OC

= –2u~ – v~ + mu~ + 5v~ = (m – 2)u~ + 4v~

(b) ⎯→AC = k

⎯→AB

(m – 2)u~ + 4v~ = k(u~ – 3v~) By comparison 4 = –3k –4 k = —— 3 and m – 2 = k –4 m = —— + 2 3 2 = — 3

SPM Appraisal Zone

1 w~ = 2u~ – 3 v~

ha~ + (h + k + 3)b~

= 2(5a~ + 4b~) – 3(3a~ – b~)

= 10a~ + 8b~ – 9a~ + 3b~

= a~ + 11b~ By comparison, h = 1and h + k + 3 = 11 k + 4 = 11

k = 11 – 4 = 7∴ h = 1, k = 7

2 (a) (i) ⎯→AB =

⎯→AO +

⎯→OB

= –2a~ – b~ + 3a~ – 2b~ = a~ – 3b~

(ii) ⎯→AC =

⎯→AO +

⎯→OC

= –2a~ – b~ + ha~ + 5b~

= (h – 2)a~ + 4b~ (b) Since A, B and C are collinear

⎯→AC = k

⎯→AB

(h – 2)a~ + 4b~ = k(a~ – 3b~) = ka~ – 3kb~ so –3k = 4 –4 k = —— 3 –4 and h – 2 = —— 3 4 h = 2 – — 3 2 = — 3 2 ∴ h = — 3

3 4a~ + p(a~ – b~) = a~ + b~ + q(a~ + b~)4a~ + pa~ – pb~ = a~ + b~ + qa~ + qb~(4 + p)a~ – pb~ = (1 + q)a~ + (1 + q)b~So, 4 + p = 1 + q p – q = –3 1… and –p = 1 + q p + q = –1 2…1 + 2 : 2p = –4 p = –2Substitute p = –2 into 1 : –2 – q = –3 q = 1∴ p = –2, q = 1

4 (a) a~ + 2ub~ = (4 + v)a~ + b~

So, 4 + v = 1 v = –3

and 2u = 1 1 u = — 2 1 ∴ u = —, v = –3 2(b) p

~ = kq

~

a~ + 2ub~

= k[(4 + v)a~ + b~]

= (4 + v)ka~ + kb~ By comparison, k = 2u and (4 + v)k = 1 (4 + v)(2u) = 1 1 4 + v = —— 2u 1 v = —— – 4 2u 1 – 8u = ———— 2u

1 5 (a)

⎯→AP = —

⎯→AC

2 1

= — (–a~ + b~) 2

⎯→OP =

⎯→OA +

⎯→AP

1 = a~ + — (–a~ + b~) 2 1 1 = — a~ + —b~ 2 2 1 = — (a~ + b~) 2 2(b)

⎯→AQ = —

⎯→AB

3 2 = —(–a~ + 2b~) 3

⎯→OQ =

⎯→OA +

⎯→AQ

2 = a~ + — (–a~ + 2b~) 3 1 4 = —a~ + —b~ 3 3 1 = — (a~ + 4b~) 3

6 (a) ⎯→MD =

⎯→MC +

⎯→CD

= 2b~ – a~

3(b)

⎯→DN = —

⎯→DA

4 3 = — (–4b~) 4 = –3b~

(c) ⎯→MN =

⎯→MD +

⎯→DN

= 2b~ – a~ – 3b~ = –a~ – b~

7 (a) ⎯→OP =

⎯→OQ +

⎯→QP

= 2a~ + b~ – a~ + b~ = a~ + 2b~

(b) ⎯→SR = b~ – a~

(c) ⎯→OS =

⎯→OQ +

⎯→QS

= 2a~ + b~ – 2⎯→SR

= 2a~ + b~ – 2b~ + 2a~ = 4a~ – b~

8 (a) ⎯→PR =

⎯→PQ +

⎯→QR

= a~ + b~ 1(b)

⎯→PT = —

⎯→PS

3 1 = —b~ 3

(c) ⎯→QT =

⎯→QP +

⎯→PT

1 = –a~ + —b~ 3

9 (a) ⎯→QP =

⎯→QO +

⎯→OP

= –q~

+ 2p~

= 2p~

– q~

1(b)

⎯→MP = —

⎯→QP

2

Page 125: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

1 = — (2p

~ – q

~)

2 1 = p

~ – — q

~ 2

1

⎯→NP = —

⎯→MP

2 1 1 = —(p

~ – —q

~)

2 2 1 1 = — p

~ – — q

~ 2 4

⎯→ON =

⎯→OP +

⎯→PN

1 1 = 2p

~ – — p

~ + — q

~

2 4 3 1 = — p

~ + — q

~

2 4

10 (a) ⎯→OP = 3a~ + b~

(b)

11 (a) ⎯→OR = 2p

~ + q

~

(b) ⎯→RS = p

~ – 2q

~

12 |a~| = 5 units

k2 + (–3)2 = 5 k2 + 9 = 25 k2 = 16 k = ±4 k = 4 or –4

13 |a~| = |b~|

(–5)2 = 42 + m2

m2 + 16 = 25 m2 = 9 m = ±3 m = 3 or –3

14 (a) ⎯→OA = 6 i~ + 2 j

~

(b) ⎯→OB =

2� �8

15 (a) ⎯→AB =

⎯→AO +

⎯→OB

=

2� � –2 +

2� �5

= 4� �3

1(b)

⎯→AB = —(4 i~ + 3 j

~)

5

16 (a) ⎯→AB =

⎯→AO +

⎯→OB

=

4� � –3 +

8� �6

=

12� � 3

(b) ⎯→AC = k

⎯→OB

4� �m

= k8� �6

So, 8k = 4 1 k = — 2 and m = 6k = 6

1�— � 2 = 3 1 ∴ k = —, m = 3 2

17 (a) ⎯→OQ =

⎯→OP +

⎯→OR

=

–3� � 3 +

6� �3

= 3� �6

(b) ⎯→PR =

⎯→PO +

⎯→OR

=

3� � –3 +

6� �3

= 9� �0

18 (a) ⎯→PQ = k

⎯→AB

2� �p

= k4� �2

so 4k = 2 1 k = — 2 and p = 2k 1 = 2(—) 2 = 1

(b) |⎯→AB | = |

⎯→PQ |

42 + 22 = 22 + p2

20 = p2 + 4 p2 = 16 p = ±4 p = 4 or –4

19 (a) |p~

| = 92 + (–12)2

= 225 = 15 units(b) p

~ = kq

~

9� � –12

= k6� �m

so 6k = 9 3 k = — 2 and km = –12 3 — m = –12 4 2 m = –12 (—) 3 ∴ m = –8

20 (a) ⎯→AB =

⎯→AO +

⎯→OB

8� � –6 =

–4� � –5

+ ⎯→OB

⎯→OB =

8� � –6

–4� � –5

= 12� � –1

∴ B(12, –1)

(b) ⎯→CD = k

⎯→AB

–4� � p = k

8� � –6

so 8k = –4 1 k = –— 2 and p = –6k 1 = –6(–—) 2 = 3 ∴ p = 3

21 ⎯→AB

= k⎯→AC

⎯→AO +

⎯→OB

= k(⎯→AO +

⎯→OC)

– i~ – j~

+ 3 i~ – 2 j~

= k(– i~ – j~

+ 4 i~ + m j~

)

2 i~ – 3 j~

= k[3 i~ + (m – 1) j~

]

So, 3k = 2 2 k = — 3and (m – 1)k = –3 2 (m – 1)(—) = –3 3 9 m – 1 = –— 2 9 m = 1 –— 2 7 = –— 2 7 ∴ m = –— 2

22 m2� �1

+ n1� �3

= 8� �9

2m + n = 8 1… m + 3n = 9 2…1 × 3 : 6m + 3n = 24 3…3 – 2 : 5m = 15 m = 3Substitute m = 3 into 1 : 2(3) + n = 8 n = 8 – 6 = 2 ∴ m = 3, n = 2

23 p1� �3

+ q2� �1

=

1 � � 13 p + 2q = 1 1… 3p + q = 13 2…1 × 3 : 3p + 6q = 3 3…3 – 2 : 5q = –18 q = –2

Q

O

3a

1– —b 2

Page 126: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

Substitute q = –2 into 1 : p + 2(–2) = 1

p = 5∴ p = 5, q = –2

24 2⎯→AB +

⎯→BC = 8 i~ – 2 j

~

2(⎯→AO +

⎯→OB) +

⎯→BO +

⎯→OC = 8 i~ – 2 j

~ 2(3 i~ – 5 j

~ + 4 i~ + 6 j

~) – 4 i~ – 6 j

~ +

p i~ + q j~

= 8 i~ – 2 j~

14 i~ + 2 j~

+ (p – 4) i~ + (q – 6) j~

= 8 i~ – 2 j~

(p + 10) i~ + (q – 4) j~

= 8 i~ – 2 j~

So, p + 10 = 8 p = –2and q – 4 = –2 q = 2∴ p = –2, q = 2

25 (a) ⎯→PQ =

⎯→PO +

⎯→OQ

=

–2� � 5 +

8� �3

= 6� �8

(b) |⎯→PQ | = 62 + 82

= 100 = 10 units The unit vector in the direction of

⎯→PQ

1 = ——

6� �8 10

=

3 — 5� � 4 — 5

3 4 = — i~ + — j

~

5 5

26 (a) (i) ⎯→DB =

⎯→DO +

⎯→OB

= –y~

+ 4 x~ = 4 x~ – y

~

(ii) ⎯→CA =

⎯→CO +

⎯→OA

= –2 x~ + 3⎯→OD

= –2 x~ + 3y~

(b) (i) ⎯→OE =

⎯→OD +

⎯→DE

= y~

+ h(4 x~ – y~

)

= 4h x~ + (1 – h)y~

(ii) ⎯→OE =

⎯→OC +

⎯→CE

= 2 x~ + k(–2 x~ + 3y~

)

= (2 – 2k) x~ + 3ky~

From (i) and (ii), 4h x~ + (1 – h)y

~

= (2 – 2k) x~ + 3ky~

So, 4h = 2 – 2k 4h + 2k = 2 2h + k = 1 1… and 1 – h = 3k h + 3k = 1 2… 1 × 3: 6h + 3k = 3 3… 3 – 2 : 5h = 2 2 h = — 5 2

Substitute h = — into 1 : 5 2 2(—) + k = 1 5 4 k = 1–— 5 1 = — 5

2 1∴ h = —, k = — 5 5

27 (a) ⎯→CD =

⎯→CO +

⎯→OD

1 = –3a~ + —b~ 3

⎯→AB =

⎯→AO +

⎯→OB

= –a~ + b~

(b) (i) ⎯→OP =

⎯→OC +

⎯→CP

1 = 3a~ + p(–3a~ + —b~) 3 1 = (3 – 3p)a~ + —pb~ 3

(ii) ⎯→OP =

⎯→OA +

⎯→AP

= a~ + q(–a~ + b~)

= (1 – q)a~ + qb~(c) From (i) and (ii), 1 (3 – 3p)a~ + — pb~ = (1 – q)a~ + qb~ 3 So, 3 – 3p = 1 – q

3p – q = 2 1… 1 and — p = q 3 p – 3q = 0 2… 1 × 3: 9p – 3q = 6 3… 3 – 2 : 8p = 6 3 p = — 4 3 Substitute p = — into 1 : 4 3 3(—) – q = 2 4 9 q = — – 2 4 1 = — 4 3 1 ∴ p = —, q = — 4 4

28 (a) (i) ⎯→AC =

⎯→AO +

⎯→OC

= – x~ + y~

(ii) ⎯→OE =

⎯→OA +

⎯→AE

1 = x~ + — (– x~ + 2y

~)

3 2 2 = — x~ + — y

~ 3 3

(b) (i) ⎯→OD =

⎯→OA +

⎯→AD

= x~ + m(– x~ + y~

)

= (1 – m) x~ + my~

(ii) ⎯→OD = n

⎯→OE

2 2 = n(— x~ + —y

~)

3 3 2 2 = —n x~ + —ny

~ 3 3(c) From (i) and (ii), 2 (1 – m) x~ + my

~ = — n x~ +

3 2 —ny

~

3 2 So, 1 – m = — n 3 3 – 3m = 2n 3m + 2n = 3 1…

2 and m = —n 3 3m – 2n = 0 2…1 + 2 : 6m = 3 1 m = — 2 1Substitute m = — into 1 : 2 1 3(—) + 2n = 3 2 3 2n = 3 –— 2 3 = — 2 3 n = — 4 1 3∴ m = —, n = — 2 4

29 (a) ⎯→AB =

⎯→AO +

⎯→OB

= –a~ + 2b~

(b) (i) ⎯→OD =

⎯→OA +

⎯→AD

= a~ + h(–a~ + 2b~)

= (1 – h)a~ + 2hb~

(ii) ⎯→OD = k

⎯→OC

5 = —k(a~ + b~) 3 5 5 = — ka~ + — kb~ 3 3(c) From (i) and (ii), 5 (1 – h) a~ + 2h b~ = — k a~ + 3 5 — kb~ 3 5 So, 1 – h = — k 3 3 – 3h = 5k 3h + 5k = 3 1… 5 2h = — k 3

Page 127: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

6h – 5k = 0 2… 1 + 2 : 9h = 3 1 h = — 3 1 Substitute h = — into 1 : 3 1 3(—) + 5k = 3 3 5k = 3 – 1 = 2 2 k = — 5 1 2 ∴ h = —, k = — 3 5

30 (a) ⎯→PR =

⎯→PO +

⎯→OR

= –3u~ + v~

(b) (i) ⎯→OS =

⎯→OR +

⎯→RS

= v~ + m(3u~ – v~)

= 3mu~ + (1 – m) v~

(ii) ⎯→OS = n

⎯→OQ

4n = —— (2u~ + v~) 5 8 4 = — nu~ + — n v~

5 5(c) From (i) and (ii), 8 4 3mu~

+ (1 – m)v~ = — nu~ + — nv~ 5 5 8 So, 3m = — n 5 15m – 8n = 0 1… 4 1 – m = — n 5 and 5 – 5m = 4n

5m + 4n = 5 2…

2 × 2: 10m + 8n = 10 3… 1 + 3 : 25m = 10 2 m = — 5 2 Substitute m = — into 1 : 5 2 15(—) – 8n = 0 5 8n = 6 3 n = — 4 2 3 ∴ m = —, n = — 5 4

31 (a) (i) ⎯→PR =

⎯→PQ +

⎯→QR

= 4a~ + 2b~

(ii) ⎯→SM =

⎯→SR +

⎯→RM

= 4a~ – b~

(b) (i) ⎯→PN = λ

⎯→PR

= λ(4a~ + 2b~)

= 4λ a~ + 2λ b~

(ii) ⎯→PN =

⎯→PS +

⎯→SN

= 2b~ + �⎯→SM

= 2b~ + �(4a~ – b~)

= 4ua~ + (2 – u)b~(c) From (i) and (ii), 4λ a~ + 2λ b~ = 4�a~ + (2 – �)b~ So, 4λ = 4� λ = �

32 (a) ⎯→XY = – x~ + y

~

⎯→OP =

⎯→OX +

⎯→XP

1 = x~ + —(– x~ + y

~)

5 4 1 = — x~ + —y

~ 5 5

⎯→YQ =

⎯→YO +

⎯→OQ

4 1 = –y

~ + 3(— x~ + —y

~)

5 5 12 2 = —— x~ – —y

~ 5 5

(b) (i) ⎯→YZ =

⎯→YO +

⎯→OZ

= –y~

+ h x~ = h x~ – y

~

(ii) ⎯→YZ = k

⎯→YQ

12 2 = k(——x~ – —y

~)

5 5 12 2 = ——k x~ – —ky

~

5 5(c) From (i) and (ii), 12 2 h x~ – y

~ = ——k x~ – —ky

~

5 5 2 So, – — k = –1 5 5 k = — 2 12 and h = ——k 5 12 5 = —— (—) 5 2 = 6 5 ∴ h = 6, k = — 2

33 (a) ⎯→ON =

⎯→OA +

⎯→AN

= a~ + h(–a~ + b~) = (1 – h)a~ + hb~

(b) ⎯→OL = k

⎯→OM

1 = k(a~ + —b~) 2 1 = ka~ + —kb~ 2 1(c) (1 – h)a~ + hb~ = ka~ + —kb~ 2 So, 1 – h = k h + k = 1 1… 1 and h = — k 2 2h – k = 0 2…1 + 2 : 3h = 1 1 h = — 3

1Substitute h = — into 1 : 3 1 — + k = 1 3 1 k = 1–— 3 2 = — 3 1 2∴ h = —, k = — 3 3

34 (a) (i) ⎯→OB =

⎯→OA +

⎯→OC

= –2 i~ + 2 j~

+ 5 i~ + 2 j~

= 3 i~ + 4 j

~

(ii) |⎯→OB | = 32 + 42

= 25

= 5 units The vector unit in the direction

of ⎯→OB

1 = — (3 i~ + 4 j

~)

5

(iii)

From the diagram,

52 = ( 8 )2 + ( 29 )2 – 2( 8 )( 29 )

cos ∠O⎯→AB

12 cos ∠O

⎯→AB = ——————

2( 8 )( 29 )

= 0.3939

∠O⎯→AB = 66° 48'

(b) (i) ⎯→AD =

⎯→AC +

⎯→CD

= 2 i~ – 2 j~

+ 5 i~ + 2 j~

– 11 i~ + 4 j~

= –4 i~ + 4 j~

(ii) ⎯→OA = 2(– i~ + j

~)

⎯→AD = 4(– i~ + j

~)

So, ⎯→AD = 2

⎯→OA

∴The points O, A andD lie on the same straight line.

35 (a) ⎯→RQ =

⎯→RP +

⎯→PQ

= –4b~ + 4a~

= 4(a~ – b~)

⎯→ST = 2a~ – 2b~

1

⎯→SV = —

⎯→ST

2 = a~ – b~

B

A

O

29

8

5

Page 128: Analysis Spm Additional Mathematics

8 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

⎯→PV =

⎯→PS +

⎯→SV

= 2b~ + a~ – b~ = a~ + b~

(b) (i) ⎯→RU = m

⎯→RQ

= m(4a~ – 4b~) = 4ma~ – 4mb~

(ii) ⎯→PU = n

⎯→PV

= n(a~ + b~) = na~ + nb~

(c) ⎯→PU =

⎯→PR +

⎯→RU

na~ + nb~ = 4b~ + 4ma~ – 4mb~ na~ + nb~ = 4ma~ + (4 – 4m)b~ So, 4m = n 4m – n = 0 1… and 4 – 4m = n 4m + n = 4 2…

1 + 2 : 8m = 4 1 m = — 2 1Substitute m = — into 1 : 2 1 4(—) – n = 0 2 n = 2 1 ∴ m = —, n = 2 2

Page 129: Analysis Spm Additional Mathematics

16 Trigonometry

Booster Zone

1 (a) (b)

(c) (d)

(e) (f)

(g) (h)

(i)

2 (a) 480° (b) 380° (c) –510° (d) –600° (e) 710° (f) –650 3 (a) sin 260° = –sin (260° – 180°) = –sin 80° (b) cos 130° = –cos (180° – 130°) = –cos 50° (c) tan 330° = –tan (360° – 330°) = –tan 30° (d) tan (–250°) = –tan(250° – 180°) = –tan 70° (e) cos (–580°) = cos 580° = cos (580° – 360°) = cos 220° = – cos (220° – 180°) = –cos 40° (f) sin (–675°) = –sin 675° = –sin (675° – 360°) = –sin 315° = –(–sin 45) = sin 45°

4

tan θ � 0 ⇒ θ is in the 2nd quadrant

∴ sin θ = 5

—–13

and tan θ = –5

—–12

5

180° � θ � 360° ⇒ θ is in the 4th quadrant

∴ sin θ = –3—5

and cos θ = 4—5

6

180° � θ � 270° ⇒ θ in the 3rd quadrant

∴ tan θ = 24—–7

and sin θ = –24—–25

7 Since A and B are angles in the same quadrant where sin A � 0 and tan B � 0 ⇒ A and B are in the 3rd quadrant.

(a) cos A = –3—5

(b) tan A = 4—3

(c) cos B = –15—–17

(d) sin B = –8

—–17

8 Since tan θ and sin θ have opposite signs, tan θ � 0 ⇒ sin θ � 0, θ is in the 2nd quadrant.

∴ cos θ = –3

—–10

and sin θ = 1

—–10

9

(a) cos 155° = –cos (180° – 155°) = –cos 25° = –k

(b) sin 25° = 1 – k2

(c) tan (–155°) = –tan 155° = – [–tan (180° – 155°)] = – (–tan 25°) = tan 25°

= 1 – k2

——–k

(d) cos 65° = sin 25° = 1 – k2

10 (a) cos (–600°) = cos 600° = – cos 60°

= –1—2

(b) cos 315° = cos 45°

= 1—2

(c) sin (–210°) = – sin 210° = – (– sin 30°)

= 1—2

(d) tan 480° = – tan 60° = – 3(e) tan (–135°) = – tan 135° = – (– tan 45°) = 1(f) sin (–120°) = – sin 120° = – sin 60

= – 3—–2

(g) tan 420° = tan 60° = 3(h) cos (–45°) = cos 45°

= 1—2

(i) sin 495° = sin 45°

= 1—2

11

Since θ is an acute angle, so θ is in the 1st quadrant.

x

yP

α = 70° 470°x

y

Pα = 20°

520°

x

y

P

α = 30°

390°x

y

Pα = 20°

560°

x

y

P

α = 80°

620°x

y

Pα = 20°700°

x

y

P

α = 40°–680°

x

y

P

α = 40°

–400°x

P

α = 10°

–530°

y

x

y

O

P

θ

12

13

x

y

O

P

Q

αθ

4

3

x

y

O

P

α7

25

x

y

O

P

Q

α

A

45

x

y

B

8

15

x

y

O

P

θ1

3

x

y

O

1 1 – k2

k25°

x

y

O

P

35

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 130: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(a) sec θ = 1

—–—cos θ

= 5—4

(b) cot θ = 1

—–—tan θ

= 4—3

12 cot θ = –3

1

—–—tan θ

= –3

tan θ = –1—3

Since θ is an obtuse angle, so θ is in the 2nd quadrant.

(a) cosec θ = 1

—–—sin θ

= 10

(b) sec θ = 1

—–—cos θ

= –10—–3

13 (a) sec θ = 1

—–—cos θ

= 5—3

cosec θ = 1

—–—sin θ

= 5—4

cot θ = 1

—–—tan θ

= 3—4

(b) sec θ = 1

—–—cos θ

= 5—3

cosec θ = 1

—–—sin θ

= –5—4

cot θ = 1

—–—tan θ

= –3—4

(c) sec θ = 1

—–—cos θ

= –25—–7

cosec θ = 1

—–—sin θ

= 25—–24

cot θ = 1

—–—tan θ

= –7

—–24

(d) sec θ = 1

—–—cos θ

= –17—–15

cosec θ = 1

—–—sin θ

= –17—–8

cot θ = 1

—–—tan θ

= 15—–8

14 (a) cos θ � 0 ⇒ θ is in the 2nd or 3rd quadrant.

cos α = 0.73 α = 43° 7' ∴ θ = 180° – 43° 7', 180° + 43° 7' = 136° 53', 223° 7' (b) tan θ � 0 ⇒ θ is in the 2nd or 4th

quadrant. tan α = 2 α = 63° 26' ∴ θ = 180° – 63° 26', 360° – 63° 26' = 116° 34', 296° 34' (c) sin θ � 0 ⇒ θ is in the 1st or 2nd

quadrant. sin α = 0.6712 α = 42° 10' θ = 42° 10', 180° – 42° 10' = 42° 10', 137° 50' (d) tan θ � 0 ⇒ θ is in the 1st or 3rd

quadrant. tan α = 0.3346 α = 18° 30' ∴ θ = 18° 30', 180° + 18° 30' = 18° 30', 198° 30' (e) cos θ � 0 ⇒ θ is in the 1st or 4th

quadrant. cos α = 0.46 α = 62° 37' ∴ θ = 62° 37', 360° – 62° 37' = 62° 37', 297° 23' (f) sin θ � 0 ⇒ θ is in the 3rd or 4th

quadrant. sin α = 0.866 α = 60° ∴ θ = 180° + 60°, 360° – 60° = 240°, 300° (g) 2 sin2 θ = 1

sin2 θ = 1—2

sin θ = ± 1—2

So, θ is in all the quadrants.

sin α = 1—2

α = 45° ∴ θ = 45°, 135°, 225°, 315° (h) 4 tan2 θ = 1

tan2 θ = 1—4

tan θ = ± 1—2

So, θ is in all the quadrants.

tan α = 1—2

α = 26° 34' ∴ θ = 26° 34', 153° 26', 206° 34',

333° 26'15 (a) sin 2θ � 0 ⇒ 2θ is in the 1st or

2nd quadrant and 0° � θ � 360° ⇒ 0° � 2θ � 720°

sin α = 0.866 α = 60° For angle 2θ in this interval 2θ = 60°, 120°, 60° + 360°, 120°

+ 360° = 60°, 120°, 420°, 480° ∴ θ = 30°, 60°, 210°, 240° (b) tan 2θ � 0 ⇒ 2θ is in 2nd or 4th

quadrant and 0° � θ �360° ⇒ 0° � 2θ � 720° tan α = 1.264 α = 51° 39' 2θ = 128° 21', 308° 21', 488° 21',

688° 21' ∴ θ = 64° 11', 154° 11', 244° 11',

344° 11' (c) cos 2θ � 0 ⇒ 2θ is in 2nd or 3rd

quadrant and 0° � θ � 360° ⇒ 0° � 2θ � 720°. cos α = 0.74 α = 42° 16' 2θ = 137° 44', 222° 16', 497° 44',

582° 16' ∴ θ = 68° 52', 111° 8', 248° 52',

291° 8'

(d) sin 1—2

θ � 0 ⇒ 1—2

θ is in the 1st or

2nd quadrant and 0° � θ � 360°

⇒ 0° � 1—2

θ � 180°

sin α = 4—7

α = 34° 51'

1—2

θ = 34° 51', 145° 9'

∴ θ = 69° 42', 290° 18'

(e) tan 1—2

θ � 0 ⇒ 1—2

θ is in the 1st or

3rd quadrant and 0° � θ � 360°

⇒ 0° � 1—2

θ � 180°

tan α = 0.3 α = 16° 42'

1—2

θ = 16° 42', 196° 42'

θ = 33° 24', 393° 24' ∴ θ = 33° 24' (f) cos 3θ � 0 ⇒ 3θ is in the 2nd or

3rd quadrant and 0° � θ � 360 ⇒ 0° � 3θ � 1080°

cos α = 1—2

α = 45°

x

y

O

P

Q

θ1

3

10

x

y

O

P

Qθ3

54

x

y

O

P

θ

Q3

45

x

y

O

P

8

15

17

x

y

O

P

Q

θ24

25

7

α

α

Page 131: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

3θ = 135°, 225°, 495°, 585°, 855°, 945°

∴ θ = 45°, 75°, 165°, 195°, 285°, 315°

(g) sin (2θ + 30°) � 0 ⇒ 2θ + 30° is in the 1st or 2nd quadrant and0° � θ � 360°

⇒ 30° � 2θ + 30° � 750° sin α = 0.75 α = 48° 35' 2θ + 30° = 48° 35', 131° 25',

408° 35', 491° 25' 2θ = 18° 35', 101° 25', 378° 35',

461° 25' ∴ θ = 9°18', 50°43', 189°18',

230°43' (h) tan (2θ – 40°) � 0 ⇒ 2θ – 40° is

in 2nd or 4th quadrant and 0° � θ � 360°

⇒ –40° � 2θ – 40° � 680° tan α = 0.7 α = 35° 2θ – 40° = –35°, 145°, 325°, 505° 2θ = 5°, 185°, 365°, 545° ∴ θ = 2° 30', 92° 30', 182° 30',

272° 30'

16 (a) 2 sin θ cos θ = sin θ 2 sin θ cos θ – sin θ = 0 sin θ (2 cos θ – 1) = 0

sin θ = 0 or cos θ = 1—2

sin θ = 0 θ = 0°, 180°, 360°

cos θ = 1—2

θ = α, 360° – α = 60°, 300° ∴ θ = 0°, 60°, 180°, 300°, 360° (b) 2 cosec2 θ + cot θ = 8 2(1 + cot2 θ) + cot θ = 8 2 cot2 θ + cot θ – 6 = 0 (2 cot θ – 3)(cot θ + 2) = 0

cot θ = 3—2

or cot θ = –2

cot θ = 3—2

1

——–tan θ

= 3—2

tan θ = 2—3

θ = 33° 41', 213° 41' cot θ = –2

1

——–tan θ

= –2

tan θ = –1—2

θ = 180° – α, 360° – α = 153° 26', 333° 26' ∴ θ = 33° 41', 153° 26', 213° 41',

333° 26' (c) 4 tan θ = sin θ

4� sin θ——–cos θ � = sin θ

4 sin θ = sin θ cos θ

4 sin θ – sin θ cos θ = 0 sin θ (4 – cos θ) = 0 sin θ = 0 or cos θ = 4 sin θ = 0 θ = 0°, 180°, 360° cos θ = 4 (no solution) ∴ θ = 0°, 180°, 360° (d) 3 cos θ = cot θ

3 cos θ = cos θ——–sin θ

3 sin θ cos θ = cos θ 3 sin θ cos θ – cos θ = 0 cos θ (3 sin θ – 1) = 0

cos θ = 0 or sin θ = 1—3

cos θ = 0 θ = 90°, 270° or 3 sin θ – 1 = 0

sin θ = 1—3

θ = 19° 28', 160° 32' ∴ θ = 19° 28', 90°, 160° 32', 270° (e) 2 cos2 θ = 3 cos θ – 1 2 cos2 θ – 3 cos θ + 1 = 0 (2 cos θ – 1)(cos θ – 1) = 0

cot θ = 1—2

or cot θ = 1

cos θ = 1—2

θ = 60°, 300° or cos θ = 1 θ = 0°, 360°

∴ θ = 0°, 60°, 300°, 360° (f) 2 tan2 θ + 1 = 3 tan θ 2 tan2 θ – 3 tan θ + 1 = 0 (2 tan θ – 1)(tan θ – 1) = 0

tan θ = 1—2

or tan θ = 1

tan θ = 1—2

θ = 26° 34', 206° 34' or tan θ = 1 θ = 45°, 225° ∴ θ = 26° 34', 45°, 206° 34', 225° (g) sec θ = 3 cos θ

1

——–cos θ

= 3 cos θ

3 cos2 θ = 1

cos2 θ = 1—3

cos θ = ± 1—3

∴ θ = 54° 44', 125° 16', 234° 44', 305° 16'

(h) 4 cot2 θ + 6 = 11 cot θ 4 cot2 θ – 11 cot θ + 6 = 0 (4 cot θ – 3)(cot θ – 2) = 0

cot θ = 3—4

or cot θ = 2

cot θ = 3—4

tan θ = 4—3

θ = 53° 8', 233° 8'

cot θ = 2

tan θ = 1—2

θ = 26° 34', 206° 34' ∴ θ = 26° 34', 53° 8', 206° 34',

233° 8'17 (a)

(b)

(c)

(d)

(e)

(f)

x

y

y = sin x + 23

2

1

0 90° 180° 270° 360°

x

y

y = 4 cos x4

0

–4

90° 180° 270° 360°

x

y

y = 3 cos x + 5

8

7

6

5

4

3

2

1

O 90° 180° 270° 360°

x

y

y = 3 tan x 135° 315°

3

O–3

45° 90° 180° 225°

x

y

y = tan x – 1

O

–1

–2

45° 135° 225° 315° 90° 180° 270° 360°

x

y

y = 2 sin x2

O

–2

90° 180° 270° 360°

Page 132: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(g)

18 (a)

(b)

(c)

(d)

(e)

(f)

(g)

19 (a) y = 3 cos 2x

(b) y = –2 cos 1—2

x

(c) y = |tan x| (d) y = 2 sin 3x + 220 (a)

∴ The number of solutions = 4

(b)

∴ The number of solutions = 3

(c)

∴ The number of solutions = 4 (d)

∴ The number of solutions = 3

(e)

∴ The number of solutions = 3 (f)

∴ The number of solutions = 5 21 (a) sec x – cos x =

1——–cos x

– cos x

= 1 – cos2 x————

cos x

= sin2 x

———cos x

= sin x � sin x———cos x �

= sin x tan x(b) (cot A + tan A) cos A

= � 1———tan A

+ tan A� cos A

= � 1 + tan2 A

————–tan A � cos A

= �(sec2 A) cos A

———sin A � cos A

= cos2 A

————–—cos2 A sin A

= 1

——–sin A

= cosec A

x

y

y = 4 tan 2 x

4

0–4

67.5° 157.5°

22.5° 45° 90° 112.5° 135° 180°

x

y

y = |2 sin x – 1|

3

2

1

0

–1

–2

–3

90° 180° 270° 360°

x

y

y = |3 cos 2x + 1|

4

3

2

1

0

–1

–2

45° 90° 135° 180°

x

yy = |5 cos x – 2|

7

6

5

4

3

2

1

01

2

3

4

5

6

7

90° 180° 270° 360°

1y = –3 cos —x 2

x

y

3

0

–3

180° 360° 540° 720°

x

y

y = 4 – 2 sin 2x6

4

2

0 45° 90° 135° 180°

x

yy = 1 – cos 2x

2

1

0 45° 90° 135° 180°

x

y

xy = 1 – —– 2π

y = |sin x|1

O

–1

π—2

π 2π 3—π 2

3xy = —– 2π

x

y

y = |2 cos x|

2

O

–2

π 2π 3—π 2

π—2

xy = —– 4π

x

y

y = cos 2x

1

O

–1

π 2π 3—π 2

π—2

3xy = 4 – —– π

x

y

y = 3 sin x + 14

3

2

1

O–1

–2

π 2π 3—π 2

π—2

x

y

3y = 3 cos —x 2

3xy = — – 3 π

3

O

–3

π—3

π 2π 2—π 3

4—π 3

5—π 3

xy = — π

x

y

y = 1 – sin 2x2

1

O π 2π 3—π 2

π—2

x

y

y = 3 cos x – 21

O

–1

–2

–3

–4

–5

90° 180° 270° 360°

Page 133: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(c) cosec A – sin A = 1

——–sin A

– sin A

= 1 – sin2 A————

sin A

= cos2 A———sin A

= cos A � cos A———sin A �

= cos A cot A

(d) sin2 θ(1 + cot2 θ)——————–—

cos2 θ

= sin2 θ(cosec2 θ)————–——

cos2 θ

= sin2 θ

———cos2 θ

� 1———sin2 θ �

= 1

———cos2 A

= sec2 θ

(e) sin A

————1 + cos A

+ 1 + cos A

————–sin A

= sin2 A + (1 + cos A)2

————–——–——–sin A(1 + cos A)

= sin2 A + 1 + 2 cos A + cos2 A————–——–—————–

sin A(1 + cos A)

= 2 + 2 cos A

————–——–sin A(1 + cos A)

= 2(1 + cos A)

————–——–sin A(1 + cos A)

= 2

——–sin A

= 2 cosec A

(f) tan A

————–1 + tan2 θ

= � sin θ——–cos θ �

——––sec2 θ

= sin θ——–cos θ

÷ 1

——–cos2 θ

= sin θ——–cos θ

(cos2 θ)

= cos θ sin θ22 (a) 2 cos2 θ + sin θ = 1 2(1 – sin2 θ) + sin θ = 1 2 sin2 θ – sin θ – 1 = 0 (2 sin θ + 1)(sin θ – 1) = 0

sin θ = –1—2

or sin θ = 1

sin θ = –1—2

θ = 210°, 330° sin θ = 1 θ = 90° ∴ θ = 90°, 210°, 330° (b) 2 cosec2 θ – 1 = 3 cot θ 2(1 + cot2 θ) – 1 = 3 cot θ 2 + 2 cot2 θ – 1 = 3 cot θ 2 cot2 θ – 3 cot θ + 1 = 0 (2 cot θ – 1)(cot θ – 1) = 0

cot θ = 1—2

or cot θ = 1

cot θ = 1—2

tan θ = 2 θ = 63° 26', 243° 26'

cot θ = 1 tan θ = 1 θ = 45°, 225° ∴ θ = 45°, 63° 26', 225°, 243° 26' (c) 2 cot2 θ + cosec θ = 4 2(cosec2 θ – 1) + cosec θ = 4 2 cosec2 θ – 2 + cosec θ = 4 2 cosec2 θ + cosec θ – 6 = 0 (2 cosec θ – 3)(cosec θ + 2) = 0

cosec θ = 3—2

or cosec θ = –2

cosec θ = 3—2

sin θ = 2—3

θ = 41° 49', 138° 11' cosec θ = –2

sin θ = –1—2

θ = 210°, 330° ∴ θ = 41° 49', 138° 11' 210°, 330°

(d) 3 sec2 θ = 5�1 + 1

——–cot θ �

3(1 + tan2 θ) = 5(1 + tan θ) 3 tan2 θ + 3 = 5 + 5 tan θ 3 tan2 θ – 5 tan θ – 2 = 0 (3 tan θ + 1)(tan θ – 2) = 0

tan θ = –1—3

or tan θ = 2

tan θ = –1—3

θ = 161° 34', 341° 34' tan θ = 2 θ = 63° 26', 243° 26' ∴ θ = 63° 26', 161° 34', 243° 26',

341° 34' (e) 5 tan2 θ = 11 sec θ – 7 5(sec2 θ – 1) = 11 sec θ – 7 5 sec2 θ – 11 sec θ + 2 = 0 (5 sec θ – 1)(sec θ – 2) = 0

sec θ = 1—5

or sec θ = 2

sec θ = 1—5

cos θ = 5 (no solution) sec θ = 2

cos θ = 1—2

θ = 60°, 300° ∴ θ = 60°, 300° (f) 3 cosec2 θ = 2 sec θ

3

——–sin2 θ

= 2

——–cos θ

2 sin2 θ = 3 cos θ 2(1 – cos2 θ) = 3 cos θ 2 cos2 θ + 3 cos θ – 2 = 0 (2 cos θ – 1)(cos θ + 2) = 0

cos θ = 1—2

or cos θ = –2

cos θ = 1—2

θ = 60°, 300° cos θ = –2 (no solution) ∴ θ = 60°, 300°

23

(a) sin (A – B) = sin A cos B – cos A sin B

= � 4—5 �� 5

—–13 � – � 3

—5 �� 12

—–13 �

=

20—–65

– 36—–65

= –16—–65

(b) cos (A + B) = cos A cos B – sin A sin B

= � 3—5 �� 5

—–13 � – � 4

—5 �� 12

—–13 �

=

15—–65

– 48—–65

= –33—–65

(c) tan (A – B)

= tan A – tan B

————–——–1 + tan A tan B

=

4—3

– 12—–5————–——–

1 + � 4—3 �� 12

—–5 �

= �– 16

—–15 �

———–

�– 11—–5 �

=

16—–33

24

(a) sin (α + β)—————cos (α – β)

= sin α cos β + cos α sin β———————————cos α cos β + sin α sin β

h�——––�1 + h2

1�——––�1 + k2 +

1�——––�1 + h2

k�——––�1 + k2

= ——————————

1�——––�1 + h2

1�——––�1 + k2 +

h�——––�1 + h2

k�——––�1 + k2

= h + k

————————( 1 + h2 )( 1 + k2 )

÷

1 + hk

————————( 1 + h2 )( 1 + k )

=

h + k——––1 + hk

(b) tan (α + β) = tan α + tan β

————–——–1 – tan α tan β

= h + k

——––1 – hk

x x

y

B0 5

1213

y

A0 3

45

x

1 + h2

y

O

h

x

1 + k2

y

O

k

Page 134: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

25

(a) sin 2A = 2 sin A cos A

= 2� 4—5 �� 3

—5 �

= 24—–25

(b) cos 2A = 1 – 2 sin2 A

= 1 – 2� 4—5 �2

= 1 – 32—–25

= –7

—–25

(c) tan 2A = 2 tan A

————–1 – tan2 A

=

2� 4—3 �

————––1 – � 4

—3 �

2

= � 8—3 �

——–

� –7—–9 �

= –

24—–7

26

(a) sin 2θ = 2 sin θ cos θ = 2m 1 – m2

(b) cos 4θ = 1 – 2 sin2 2θ = 1 – 2 (2m 1 – m2 )2

= 1 – 2 (4m2 [1 – m2)] = 1 – 8m2 + 8m4

(c) cos θ = 1 – 2 sin2 1—2

θ

m = 1 – 2 sin2 1—2

θ

2 sin2 1—2

θ = 1 – m

sin2 1—2

θ = 1 – m

—–—–2

Since θ is an acute angle,

sin 1—2

θ = 1 – m——–2

27

(a) sin 2B = 2 sin B cos B

= 2� 4—5 ��– 3

—5 �

= –24—–25

sin 4B = 1 – 2 sin2 2B

= 1 – 2�– 24—–25 �

2

= –527—––625

(b) cos B = 2 cos2 1—2

B – 1

–3—5

= 2 cos2 1—2

B – 1

2 cos2 1—2

B = 2—5

cos2 1—2

B = 1—5

(c) tan B = 2 tan B

—2————––

1 – tan2 B—2

–4—3

= 2 tan B

—2————––

1 – tan2 B—2

4 tan2 B—2

– 4 = 6 tan B—2

2 tan2 B—2

– 3 tan B—2

– 2 = 0

(2 tan B—2

+ 1)(tan B—2

– 2) = 0

tan B—2

= –1—2

or tan B—2

= 2

Since B is an obtuse angle,

90° � B � 180° ⇒ 45° � B—2

� 90

∴ tan B—2

= 2

28 (a) tan 2A = 2 tan A

————–1 – tan2 A

4—3

= 2 tan A

————–1 – tan2 A

4 – 4 tan2 A = 6 tan A 2 tan2 A + 3 tan A – 2 = 0 (2 tan A – 1)(tan A + 2) = 0

tan A = 1—2

or tan A = –2

Since A is an acute angle,

tan A = 1—2

(b) tan 3A = tan (2A + A)

= tan 2A + tan 2A

————–———1 – tan 2A tan A

=

4—3

+ 1—2————––—

1 – � 4—3 �� 1

—2 �

= � 11—–6 �

——–

� 1—3 �

=

11—–2

29 (a) sin 3A = sin (2A + A) = sin 2A cos A + cos 2A sin A = (2 sin A cos A)(cos A) + (1 – 2 sin2 A)(sin A) = 2 sin A cos2 A + sin A – 2 sin3 A = 2 sin A (1 – sin2 A) + sin A – 2 sin3 A = 2 sin A – 2 sin3 A + sin A – 2 sin3 A = 3 sin A – 4 sin3 A

(b) sin 2A

————–1 – cos 2A

= 2 sin A cos A

————–———1 – (1 – 2 sin2 A)

= 2 sin A cos A

————–——2 sin2 A

= cos A

———sin A

= cot A

(c) cos (A + B)—————cos A cos B

=

cos A cos B – sin A sin B———————————–

cos A cos B

= cos A cos B—————cos A cos B

– sin A sin B—————cos A cos B

= 1 – tan A tan B

(d) cot (A + B) = 1

————–tan (A + B)

= 1

———————

� tan A + tan B——————–1 – tan A tan B �

=

1 – tan A tan B———————

tan A + tan B

= � 11 – ————— cot A cot B �

——–————— 1 1——– + ——– cot A cot B

= � cot A cot B – 1———————

cot A cot B �——–—————

� cot A + cot B——————

cot A cot B � =

cot A cot B – 1———————

cot A + cot B30 (a) 4 sin 2θ = sin θ 4(2 sin θ cos θ) – sin θ = 0 sin θ (8 cos θ – 1) = 0

sin θ = 0 or cos θ = 1—8

sin θ = 0 θ = 0°, 180°, 360°

cos θ = 1—8

θ = 82° 49', 277° 11' ∴ θ = 0°, 82° 49', 180°, 277° 11', 360° (b) tan 2θ = 2 cot θ

2 tan θ

————–1 – tan2 θ

= 2

—–—–tan θ

2 tan2 θ = 2 – 2 tan2 θ 4 tan2 θ = 2

tan2 θ = 1—2

tan θ = ± 1—2

x

y

m

11 – m2

x

y

O

B

3

45

x

y

AO 3

45

Page 135: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

∴ θ = 35° 16', 144° 44', 215° 16', 324° 44'

(c) cos 2θ = cos θ + 2 3(2 cos2 θ – 1) = cos θ + 2 6 cos2 θ – cos θ – 5 = 0 (6 cos θ + 5)(cos θ – 1) = 0

cos θ = –5—6

or cos θ = 1

cos θ = –5—6

θ = 146° 27', 213° 33' cos θ = 1 θ = 0°, 360° ∴ θ = 0°, 146° 27', 213° 33°, 360° (d) 4 sin θ cos θ = 1 2(2 sin θ cos θ) = 1

sin 2θ = 1—2

sin 2θ � 0 ⇒ 2θ is in the 1st or 2nd quadrant and 0° � θ � 360° ⇒ 0° � 2θ � 720°

For angle 2θ in this interval 2θ = 30°, 150°, 390°, 510° ∴ θ = 15°, 75°, 195°, 255° (e) 3 tan θ = 2 tan (45° – θ)

= 2� tan 45° – tan θ———————–1 + tan 45° tan θ �

= 2� 1 – tan θ————–1 + tan θ �

3 tan θ + 3 tan2 θ = 2 – 2 tan θ 3 tan2 θ + 5 tan θ – 2 = 0 (3 tan θ – 1)(tan θ + 2) = 0

tan θ = 1—3

or tan θ = –2

tan θ = 1—3

θ = 18° 26', 198° 26' tan θ = –2 θ = 116° 34', 296° 34' ∴ θ = 18° 26', 116° 34', 198° 26',

296° 34' (f) sin 2θ + 3 cos2 θ = 0 2 sin θ cos θ + 3 cos2 θ = 0 cos θ (2 sin θ + 3 cos θ) = 0 cos θ = 0 or 2 sin θ = –3 cos θ cos θ = 0 θ = 90°, 270° 2 sin θ = –3 cos θ

sin θ

—–—–cos θ

= –3—2

tan θ = –3—2

θ = 123° 41', 303° 41' ∴ θ = 90°, 123° 41', 270°, 303° 41'

SPM Appraisal Zone

1 (a) 0.87(b) –0.5

(c) tan 120° = 0.87—–––0.5

= –1.74

2 270° � θ � 360° ⇒ θ is in the 4th quadrant.

(a) cos (–θ) = cos θ

= 5

—–41

(b) sin 2θ = 2 sin θ cos θ

= 2�– 4—–41 �� 5

—–41 �

= –40—–41

3

(a) cot A = 1

———tan A

= 1—p

(b) cos (90° – A) = sin A

= p

—–—–1 + p2

4

(a) tan2 θ = � 1– m2

—–—–m �

2

= 1– m2

—–—–m2

(b) cos θ = 2 cos2 1—2

θ – 1

m = 2 cos2 1—2

θ – 1

cos2 1—2

θ = m + 1

—–—–2

5 (a) cos 2A = 2 cos2 A – 1

= 2� 3—5 �

2

– 1

= –7

—–25

(b) cos 4 A = 2 cos2 2A – 1

= 2�– 7—–25 �

2

– 1

= –527—––625

6

(a) cos 65° = sin (90° – 65°) = sin 25° = 1 – h2

(b) tan (–155°) = –tan 155° = –(–tan 25°) = tan 25°

= 1– h2

—–—–h

7

(a) sec A = 1

———cos A

= 1

———

�– 3—5 �

= –

5—3

(b) cosec A = 1

———sin A

= 1

——

� 4—5 �

=

5—4

(c) cot A = 1

———tan A

= 1

——

�– 4—3 �

= –

3—4

8 (a) cos x = 1—2

x = 45°, 315° (b) 2 sin x cos x = cos x 2 sin x cos x – cos x = 0 cos x (2 sin x – 1) = 0

cos x = 0 or sin x = 1—2

cos x = 0 x = 90°, 270°

sin x = 1—2

x = 30°, 150° ∴ x = 30°, 90°, 150°, 270°

9 2 cos2 x = 3 cos x – 2 cos 60°

2 cos2 x – 3 cos x + 2� 1—2 � = 0

2 cos2 x – 3 cos x + 1 = 0 (2 cos x – 1)(cos x – 1) = 0

cos x = 1—2

or cos x = 1

cos x = 1—2

x = 60°, 300° cos x = 1 x = 0°, 360°

∴ x = 0°, 60°, 300°, 360°

x

y

5

441

x

y

OA

p

1

1 + p2

x

y

1

m

1 – m2

x

y

O

1

h

1 – h2

25°

x

y

O3

45

A

Page 136: Analysis Spm Additional Mathematics

8 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

10

From the sketch, a = 5 and b = 3.11 a = 4, b = 4 and c = 1 ∴ y = 4 cos 4x + 1

12 (a) a = 3 and period = 360—––

2 = 180°

(b) (90°, 1)13

∴ Range = 2 � y � 4

14 a = 2, b = 315 a = 3, b = 416 y = –2 cos 3x

17 (a) a = 2, b = 3—2

and c = 1

(b) y = 1 – 2 sin 3—2

x

18 y = |4 cos 2x|

19 cot θ – tan θ

——————cot θ + tan θ

=

1——–tan θ

– tan θ————––—

1——–tan θ

+ tan θ

= � 1 – tan2 θ

—————tan θ �

——–————

� 1 + tan2 θ—————

tan θ �=

1 – tan2 θ—————1 + tan2 θ

= 1 – tan2 θ

—————sec2 θ

= 1

———sec2 θ

– � sin2 θ———cos2 θ �

——–——

� 1———cos2 θ �

= cos2 θ – sin2 θ

20 1 – cot2 θ

—————1 + cot2 θ

= 1 – cot2 θ

—————cosec2 θ

= 1

———–cosec2 θ

– � cos2 θ———sin2 θ �

——–——1

———sin2 θ

= sin2 θ – cos2 θ = 1 – cos2 θ – cos2 θ = 1 – 2 cos2 θ ∴ a = 1 and b = –221 5 + cot y = 2 cosec2 y 5 + cot y = 2(1 + cot2 y) 2 cot2 y – cot y – 3 = 0 (2 cot y – 3)(cot y + 1) = 0

cot y = 3—2

or cot y = –1

cot y = 3—2

1

——–tan y

= 3—2

tan y = 2—3

y = 33° 41', 213° 41' cot y = 1

1

——–tan y

= 1

tan y = 1 y = 45°, 225°∴ y = 33° 41', 45°, 213° 41', 225°

22 3 cos2 θ + 3 cos θ = sin2 θ 3 cos2 θ + 3 cos θ = 1 – cos2 θ 4 cos2 θ + 3 cos θ – 1 = 0 (4 cos θ – 1)(cos θ + 1) = 0

cos θ = 1—4

or cos θ = –1

cos θ = 1—4

θ = 75° 31', 284° 29' cos θ = –1 θ = 180°

∴ θ = 75° 31', 180°, 284° 29'23 sin (A + B)

= sin A cos B + cos A sin B

= � 15—–25 �� 24

—–25 � + � 20

—–25 �� 7

—–25 �

= 360——625

+ 140——625

= 500——625

= 4—5

24 tan (α – β) = tan α – tan β

——————–1 + tan α tan β

=

3—4

– 12—–5————––—

1 + � 3—4 �� 12

—–5 �

= �– 33

—–20 �

——––

� 14—–5 �

= –

33—–56

25 cos 2A = 2 cos2 A – 1

= 2� 1—–——1 + m2 �2

– 1

= 2

———1 + m2

– 1

= 2 – (1 + m2)—————–

1 + m2

= 1 – m2

———1 + m2

26

sin A = 2 sin A—2

cos A—2

= 2� t—–——

1 + t2 �� 1—–——

1 + t2 � =

2t———1 + t2

27

(a) sin (A + B) = sin A cos B + cos A sin B

= �– 15—–17 ��– 3

—5 � + �– 8

—–17 ��– 4

—5 �

=

45—–85

+ 32—–85

= 77—–85

(b) cos B = 2 cos2 B—2

– 1

–3—5

= 2 cos2 B—2

– 1

2 cos2 B—2

= 1 – 3—5

cos2 B—2

= 1—5

cos 1—2

B = ± 1—5

180° � B � 270°

⇒ 90° � B—2

� 135°

∴ cos 1—2

B = –1—5

28 (a) cos 2A = 2 cos2 A – 1

= 2� 3—5 �

2

– 1

= –7

—–25

(b) tan A = 2 tan

A—2————––

1 – tan2 A—2

4—3

= 2 tan

A—2————––

1 – tan2 A—2

x

y

O 1

t

1 + t2

A—2

x x

y

O

A8

1517

y

O

B3

45

x

y

y = a sin x + b

8

3

O

–2π 2π 3—π

2π—2

x

yy = 3 – cos 2x

4

3

2

1

0 π 3—π 2

π—4

π—2

Page 137: Analysis Spm Additional Mathematics

9© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

4�1 – tan2 A—2 � = 6 tan

A—2

4 tan2 A—2

+ 6 tan A—2

– 4 = 0

2 tan2 A—2

+ 3 tan A—2

– 2 = 0

�2 tan A—2

– 1��tan A—2

+ 2� = 0

tan A—2

= 1—2

or tan A—2

= –2

0° � A � 90° ⇒ 0° � A—2

� 45°

∴ tan 1—2

A = 1—2

29

(a) cos (A + B) = cos A cos B – sin A sin B

= � 3—5 ��– 24

—–25 � – � 4

—5 �� 7

—–25 �

= –72

—––125

– 28

——125

= –100—––125

= –4—5

(b) tan (B – A) = tan B – tan A

——————–1 + tan B tan A

= –

7—–24

– 4—3————––—–

1 – �– 7—–24 �� 4

—3 �

=

�– 13—–8 �

——––

� 25—–18 �

= –117—––100

30 1 + cos 2θ

—————sin 2θ

= 1 + 2 cos2 θ – 1————–——–

2 sin θ cos θ

= 2 cos2 θ

————–—2 sin θ cos θ

= cos θ——–sin θ

= cot θ31 2 cos 2x = 2 – 3 sin x

2(1 – 2 sin2 x) = 2 – 3 sin x 4 sin2x – 3 sin x = 0 sin x (4 sin x – 3) = 0

sin x = 0 or sin x = 3—4

sin x = 0 x = 0°, 180°, 360°

sin x = 3—4

x = 48° 35', 131° 25'∴ x = 0°, 48° 35', 131° 25', 180°, 360°

32 sin 2θ + 3 cos2θ = 0 2 sin θ cos θ + 3 cos2 θ = 0 cos θ (2 sin θ + 3 cos θ) = 0

cos θ = 0 or 2 sin θ = –3 cos θ cos θ = 0 θ = 90° 270°

2 sin θ = –3 cos θ

tan θ = –3—2

θ = 123° 41', 303° 41' ∴ θ = 90°, 123° 41', 270°, 303° 41'

33 (a), (b)

x = π |sin 2x|

|sin 2x| = x—π

y = x—π

x 0 πy 0 1

∴ The number of solutions = 4

34 (a), (b)

x = π (1 + cos 2x)

x—π = 1 + cos 2x

2 + 2 cos 2x = 2x—π

y = 2x—π

x 0 πy 0 2

∴ The number of solutions = 2

35

∴ The number of solutions = 2

36 (a) 1 – tan2 x

———–—1 + tan2 x

= � sin2 x1 – ——– cos2 x�——–——–

� sin2 x1 + ——– cos2 x �

= cos2 x – sin2 x

————–——cos2 x + sin2 x

= cos2 x – (1 – cos2 x) ————–———–cos2 x + 1 – cos2 x

= 2 cos2 x – 1 = cos 2x (b) (i), (ii)

1 – tan2 x

———–—1 + tan2 x

+ 2x—π = 1

cos 2x = 1 – 2x—π

–y = 1 – 2x—π

y = 2x—π – 1

x 0 πy –1 1

∴ The number of solutions = 237

∴ The number of solutions = 638 (a), (b)

3x—π + tan 2x = 2

y = 2 – 3x—π

x 0 πy 2 –1

∴ The number of solutions = 3

x

y

O

A 3

45

x

y

O

B

24

725

y

xO

1

π—4

π—2

3—π 4

π

y = |sin 2x| xy = — π

y

xO

3

π—2

3—π 2

π

y = –3 cos x

xy = 1– — π

–3

y

xO

1

π—4

π—2

3—π 4

π

2xy = — – 1 π

–1

y

xO

2

π—3

2—π 3

π

xy = 2 – — π

4—π 3

5—π 3

3y = |2 sin —x| 2

y

π—4

3—π 4

π

y = tan 2x

3xy = 2 – — π

Oxπ—

2

y

xO

4

π—4

π—2

3—π 4

π

y = 2 + 2 cos 2x

2xy = — π2

Page 138: Analysis Spm Additional Mathematics

10 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

39

∴ The number of solutions = 240

(a) 1

——–sin A

+ 1

——–cot B

= 1

——–sin A

+ tan B

= 1

——–

� 12—–13 �

+ �– 3—4 �

=

1—3

(b) sin A cos B

————————–sin A sin B + cos B

= � 12—–13 ��– 4

—5 �

————––—–—–

� 12—–13 �� 3

—5 � + �– 4

—5 �

= �– 48

—–65 �

——––

�– 16—–65 �

= 3

(c) sin (A – B)

—————–cos (A + B)

= sin A cos B – cos A sin B————————–——–cos A cos B – sin A sin B

=

� 12—–13 ��– 4

—5 � – �– 5

—–13 �� 3

—5 �

————––—–—–———

�– 5—–13 ��– 4

—5 � – � 12

—–13 �� 3

—5 �

=

�– 33—–65 �

——––

�– 16—–65 �

=

33—–16

41

(a) cos 2A = 1 – 2 sin2 A

= 1 – 2� 4—5 �

2

= –7

—–25

(b) sin 2A = 2 sin A cos A

= 2� 4—5 ��– 3

—5 �

= –24—–25

(c) cos A = 1 – 2 sin2 A—2

–3—5

= 1 – 2 sin2 A—2

2 sin2 A—2

= 1 + 3—5

= 8—5

sin2 A—2

= 4—5

sin A—2

= ± 2—5

90 � A � 180 ⇒ 45 � A—2

� 90

∴ sin 1—2

A = 2—5

42

(a) cos 2θ = 4—5

(b) cos 2θ = 1 – 2 sin2 θ

4—5

= 1 – 2 sin2 θ

2 sin2 θ = 1—5

sin2 θ = 1

—–10

sin θ = ± 1—10

Since θ is an acute angle,

sin θ = 1—10

(c)

tan θ = 1—3

(d) tan 3θ = tan (θ + 2θ)

= tan θ + tan 2θ

———————–1 – tan θ tan 2θ

=

1—3

+ 3—4————––—

1 – � 1—3 �� 3

—4 �

= � 13—–12 �

——––

� 3—4 �

=

13—–9

x

y

O

A

3

45

x

y

101

3

y

x0

53

4

y

xO

3

π

xy = — – 3 π

1y = 3 cos —x 2

–3

3π 4π

x x

y

O

A

5

12

13

y

O

B3

4

5

Page 139: Analysis Spm Additional Mathematics

17 Permutations and Combinations

1

Booster Zone

1 (a) 9! = 362 880(b) 8! = 40 320(c) 7! = 5040(d) 6! = 720

2 (a) 5P4 = 120

(b) 6P4 = 360

(c) 8P4 = 1680

(d) 7P4 = 840

3 (a) 5! = 120(b) (i) 24 × 2! = 48

(ii) 120 – 48 = 72

4 (a) 6 × 4! = 144(b)

The number of arrangements = 4 × 4 × 3 × 2 × 1 × 2 = 192

5 (a) 8! = 40 320(b) (i)

The number of arrangements = 5 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 25 200

(ii)

The number of arrangements = 7 × 6 × 5 × 4 × 3 × 2 × 1 × 3 = 15 120

(iii) 720 × 3! = 4320

6 (a) 7! = 5040(b) (i)

The number of arrangements = 1 × 6 × 5 × 4 × 3 × 2 × 1 = 720(ii)

or

The number of arrangements = 5! × 2 = 240

7 (a)

The number of arrangements = 60 480

(b)

The number of arrangements = 100 800(c) 120 × 5! = 14 400(d)

= 5! × 4! = 2880

8 6P4 = 360

9

The number of arrangements = 2 × 6 × 5 × 4 × 3= 720

10 (a) 7P5 = 2520

(b) (i)

The number of arrangements = 4 × 6 × 5 × 4 × 3 = 1440(ii)

The number of arrangements = 6 × 5 × 4 × 3 × 4 = 1440

11 30C5 = 142 506

12 20C4 = 4845

13 5C3 × 10C

8 = 450

14 6C3 × 4C

2 = 120

15 (a) 6C4 × 2C

2 = 15

(b) 8C5 × 3C

3 = 56

16 (a) 16C7 = 11 440

(b) 10C4 × 6C

3 = 4200

(c) 11 440 – (10C7 × 6C

0) = 11 320

17 (a) (8C4 × 4C

0) + (8C

0 × 4C

4)

= 70 + 1 = 71(b) 8C

2 × 4C

2 = 168

18 (a) 16C4 = 1820

(b) 1 × 14C2 = 91

(c) (8C3 × 8C

1) + (8C

4 × 8C

0)

= 448 + 70 = 518

19 (a) 6C2 × 3C

2 = 45

(b) 6C4 × 3C

0 = 15

20 (a) 10C3 = 120

(b) (1 × 10C4) + (1 × 10C

4) = 420

SPM Appraisal Zone

1 (a) 8! = 40 320(b) 5040 × 2! = 10 080

2

The number of arrangements = 3 × 4 × 3 × 3= 108

3

The number of arrangements = 7 × 7 × 6 × 5 × 4 × 3 × 2 × 1= 35 280

4 (a) 8P5 = 6720

(b) = 240

or = 240

or = 240

or = 240

∴ The number of ways that 5-digit numbers can be formed

= 240 + 240 + 240 + 240 = 960

5 (a) 6! = 720(b)

The number of permutations = 1 × 4 × 3 × 2 × 1 × 1 = 24

6 = 12

or = 18

The number of arrangements = 12 + 18= 30∴ There are 30 odd 4-digit numbers

less than 3000 which can be formed using the digits 1, 2, 3, 4 and 5 without repetition.

7

The number of arrangements = 6 × 5 × 3= 90

C V

4 4 3 2 1 2

V

5 7 6 5 4 3 2 1

C

7 6 5 4 3 2 1 3

1 6 5 4 3 2 1

RA

1 5 4 3 2 1 1

R A

1 5 4 3 2 1 1

BB

4 3 7 6 5 4 3 2 1

5 4 4 3 3 2BG

2 1 1BG BG BG G

V

2 6 5 4 3

6 5 4 3 4O

4 6 5 4 3

7 7 6 5 4 3 2 1

6

6

6 5 4 2 1

5 2 1 4

2 1 5 4

2 1 6 5 4

1 3 2 2

1 3 2 3

01

02

6 5 3E

G B

5 7 6 5 4 3 2 1 4

3 4 3 3CV

T

1 4 3 2 1 1C

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 140: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

8 (a)

The number of permutations = 3 × 2 × 2 × 1 × 1 = 12(b) 24 × 2! = 48

9 6P4 × 5P

3 = 21 600

10 (a) 9C3 = 84

(b) (5C2 × 4C

1) + (5C

3 × 4C

0)

The number of arrangements = 40 + 10 = 50

11 (a) 8C4 = 70

(b) 1 × 7C3 = 35

12 (a) 7C3 × 8C

2 = 980

(b) 7C5 + 8C

5 = 21 + 56

= 77

13 (a) 12C5 = 792

(b) The number of arrangements = 792 – [(7C

0 × 5C

5) + (7C

1 × 5C

4)

+ (7C2 × 5C

3)]

= 792 – (1 + 35 + 210) = 546

14 (6C3 × 4C

2) + (6C

4 × 4C

1)

= 120 + 60= 180

15 (a) 1 × 12C6 = 924

(b) 4C2 × 9C

5 = 756

(c) (8C2 × 5C

5) + (8C

3 × 5C

4) +

(8C4 × 5C

3) + (8C

5 × 5C

2)

= 28 + 280 + 700 + 560 = 1568

M

3 2 2 1 1M SMS

Page 141: Analysis Spm Additional Mathematics

18 Probability

1

Booster Zone

2 9 1 (a) —– (b) —– 13 26

1 2 2 (a) — (b) — 2 3 1 1 (c) — (d) — 3 2

1 3 3 (a) — (b) — 4 8 7 1 (c) —– (d) —– 24 12

1 5 4 (a) — (b) — 2 6

24 1 5 (a) —– (b) — 35 7 3 (c) — 8

k 2 6 ——–– = — k + 4 3 3k = 2k + 8 k = 8

x 3 7 (a) —– = — 30 5 3 x = — (30) 5 = 18 ∴ The number of yellow beads = 30 – 18 = 12

18 + y 5 (b) ——––– = — 30 + y 8 144 + 8y = 150 + 5 3y = 6 y = 2 ∴ 2 blue beads should be added

to the bag.

1 8 8 (a) — (b) — 3 9

1 5 9 (a) — (b) — 3 6

1 710 (a) — (b) — 2 8

2 3 11 (a) — (b) — 3 5 11 (c) 0 (d) —– 15

12 P(A � B) = P(A) + P(B) 0.8 = 0.55 + x x = 0.25

2 113 (a) — (b) — 5 3 11 (c) —– (d) 0 15

2 113 (a) — (b) — 5 3 11 (c) —– (d) 0 15

2 114 (a) — (b) — 9 9 1 (c) — (d) 1 3

15 (a) P(Zhao Ming or Mukhriz wins) = P(Zhao Ming wins) + P(Mukhriz

wins) 2 1 = — + —– 5 10 1 = — 2 1 (b) P(some one else wins) = 1 – — 2 1 = — 2

16 (a) P(P or Q wins)

= P(P wins) + P(Q wins) 1 1 = — + — 4 6 5 = —– 12 (b) P(Q or R wins) = P(Q wins) + P(Q wins) 1 1 = — + —– 6 12 1 = — 4 P(neither Q nor R wins) = 1 – P(Q or R wins) 1 = 1 – — 4 3 = — 4

17 (a) P(motorcycle or car) = P(motorcycle) + P(car) = 0.15 + 0.55 = 0.7

(b) P(none of these vehicles) = 1 – 0.15 – 0.55 – 0.2 = 0.1

18 (a) 0 (b) P(Zaki or Hong Ping or Raju

wins) = P(Zaki wins) + P(Hong Ping

wins) + P(Raju wins) = 0.4 + 0.3 + 0.2 = 0.9

319 (a) P(white) = —– 12 1 = — 4 (b) P(neither white nor black) = 1 – P(white or black) 3 4 = 1 – (—– + —–) 12 12 5 = —– 12 (c) P(either black or red) = P(black) + P(red) 4 5 = —– + —– 12 12 3 = — 4

220 (a) —– 5 (b) P(either odd or divisible by 4) = P(odd) + P(divisible by 4) 5 2 = —– + —– 10 10 7 = —– 10

21 P(A � B) = P(A) × P(B) 5 2 —– = — × P(B) 12 3 5 3 P(B) = —– × — 12 2 5 = — 8

22 P(C � D) = P(C) × P(D) 0.1 = k(k + 0.3) k2 + 0.3k – 0.1 = 0 10k2 + 3k – 1 = 0 (5k – 1)(2k + 1) = 0 1 1 k = — or k = – — 5 2 1 ∴ k = — 5

2 323 (a) P(RB) = — × — 9 9 2 = — 3

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 142: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) P(different colours) = 1 – P(same colour) 4 9 16 = 1 – (—– + —– + —– 81 81 81 52 = —– 81

24 (a) P(BG) = P(B) × P(G) 1 2 = — × — 5 5 2 = —– 25 (b) P(the same colour) = P(RR or GG) 3 3 = —– + —– 10 25 21 = —– 50

25 (a) P(at least one wins) = P(WW or WL or LW) 6 9 2 = —– + —– + —– 20 20 20 17 = —– 20 (b) P(both win) = P(WW) 3 2 = — × — 4 5 3 = —– 10

26 (a) P(all qualify) = P(QQQ) 1 2 5 = — × — × — 2 3 6 5 = —– 18 (b) P(only one qualifi es) = P(QQ'Q' or Q'QQ' or Q'Q'Q) 1 1 5 = —– + —– + —– 36 18 36 2 = — 9

27 (a) P(R � G) = P(RRR or RRG or RGR or

GRR) 1 1 1 1 = —– + —– + —– + — 15 21 10 6 8 = —– 21 (b) P(at most two red apples) = 1 – P(RRR or GGG) 1 5 = 1 – (—– + —–) 15 28 317 = —–– 420

28 (a) P(Hamdan passes both subjects) 3 2 = — × — 4 3 1 = — 2 (b) P(Xue Ming passes only one

subject)

= P(PF or FP) 4 2 1 1 = (— × —) + (— × —) 5 3 5 3 3 = — 5

29 (a) P(a Mathematics book and a history book)

= P(MH or HM) 6 9 9 6 = (—– × —–) + (—– × —–) 15 15 15 15 12 = —– 25 (b) P(at least one Mathematics book) = 1 – P(HH) 9 9 = 1 – (—– × —–) 15 15 16 = —– 25

30 (a) P(the same colour) = P(WW or BB) 5 1 = — + —– 9 18 11 = —– 18 (b) P(one white) = P(WB or BW) 1 5 = — + —– 9 18 7 = —– 18

SPM Appraisal Zone

4 3 1 (a) P(RR) = —– × —– 12 11 12 = —–– 132 1 = —– 11 (b) P(at least one rotten) = 1 – P(GG) 8 7 = 1 – (—– × —–) 12 11 19 = —– 33

2 (a) P(at least one brown) = P(BrBr or BrB1 or BrR or B1Br

or RBr) 3 2 3 8

= (—– × —–) + (—– × —–) + 12 11 12 11 3 1 8 3 (—– × —–) + (—– × —–) 12 11 12 11 1 3 + (—– × —–) 12 11 5 = —– 11 (b) P(of the same colour) = P(BrBr or B1B1) 3 2 8 7 = (—– × —–) + (—– × —–) 12 11 12 11

31 = —– 66

3 (a) P(selecting a point in the circle) area of the circle = —–—–—–—–—–—– area of the square 22

49π 49(—–)

77 7 = —–– = —–––––––– = —––

256 256 128 (b) P(selecting a point outside the

circle) area of the shaded region = —––––––––—––––––––––– area of the square 256 – 49π = —–––––––– 256 22 256 – 49 (—–) 7 = —–—––—––—––– 256 102 = —–– 256 51 = —–– 128

4 P(of the same colour) = P(RR or BB or GG) 5 4 3 2 2 1 = (—– × —) + (—– × —) + (—– × —) 10 9 10 9 10 9 14 = —– 45

6 2 5 (a) — = — k 3 2k = 18 k = 9 3 2 (b) P(GG) = — × — 9 8 6 = —– 72 1 = —– 12

6 (a) P(begins with a vowel) 240 = —–– 720 1 = — 3 (b) P(ends with a consonant) 480 = —–– 720 2 = — 3

7 (a)

O

DO

0.3

0.7

C

DC

0.6

0.4

C

DC

0.2

0.8

Page 143: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

O : Oversleeps DO : Does not oversleep C : Cycles to school DC : Does not cycle to school

(b) P(does not cycle to school) = (0.3 × 0.4) + (0.7 × 0.8) = 0.12 + 0.56 = 0.68

8 (a) P(choosing a letter I) 2 = —– 14 1 = — 7 (b) P(choosing a consonant) 9 = —– 14

2 2 9 (a) P(EE) = — × — 4 4 4 = —– 16 1 = — 4 (b) 2nd spin

1 2 3 41st spin

4

3

2

1

The ∆ indicates the event where the sum is odd

∴ Probability that the sum is odd 8 = —– 16 1 = — 2

2 110 (a) P(B1B1) = — × — 3 4 1 = — 6

(b) P(choosing a pair of brown shoes)

= P(BrB or BrB1 or BrBr) 1 1 1 1 1 2 = (— × —) + (— × —) + (— × —) 3 4 3 4 3 4 1 = — 3

11 (a) S A B

16 3 9

2

2

6

4

C8

(b) (i) P(reads at least one) = 1 – P(reads none) 8 = 1 – —– 50 21 = —– 25

(ii) P(reads only one) = P(reads only A) + P(reads

only B) + P(reads only C) 16 9 6 = —– + —– + —– 50 50 50 31 = —– 50

12 L: person had bought laptops D: person had bought digital cameras

(a) P(L or D) = 1 – P(neither L nor D) = 1 – 0.15 = 0.85(b) P(L or D) = P(L) + P(D) – P(L and D) 0.85 = 0.7 + 0.5 – P(L and D) P(L and D) = 0.7 + 0.5 – 0.85 = 0.35

13 (a) P(passes the point P)

1 1 = — × — 4 3

1 = —– 12 (b) P(passes the point Q)

3 1 1 1 = (— × —) + (— × —) 4 3 4 3 1 = — 3

14 (a) (i) P(one is dotted and the other is stripped)

= P(DS or SD) 3 4 4 3 = (—– × —–) + (—– × —–) 12 12 12 12 1 = — 6

(ii) P(taking of the same pattern)

P(DD or SS or SiSi) 3 3 4 4

= (—– × —–) + (—– × —–) 12 12 12 12 5 5 + (—– × —–) 12 12 25 = —– 72 (b) P(all three are silk ties)

= P(SiSiSi) 5 5 5 = —– × —– × —– 12 12 12 125 = —–– 144

15 (a) P(all three complete) = P(A) × P(B) × P(C) = 0.8 × 0.7 × 0.5 = 0.28 (b) P(at least two complete) = P(ABC or ABC ' or AB 'C or A'BC) = 0.28 + (0.8 × 0.7 × 0.5) + (0.8 ×

0.3 × 0.5) + (0.2 × 0.7 × 0.5) = 0.28 + 0.28 + 0.12 + 0.07 = 0.75

Page 144: Analysis Spm Additional Mathematics

19 Probability Distributions

1

Booster Zone

1 p = 0.2, q = 1 – 0.2 = 0.8 and n = 6.(a) P(X = 3) = 6C

3 (0.2)3(0.8)3

= 0.0819

(b) P(2 � X � 5) = P(X = 3) + P(X = 4) + P(X = 5) = 0.0819 + 6C

4(0.2)4(0.8)2 +

6C5(0.2)5(0.8)

= 0.0819 + 0.0154 + 0.0015 = 0.0988

(c) P(X � 1) = 1 – P(X � 1) = 1 – P(X = 0) – P(X = 1) = 1 – 6C

0(0.2)0(0.8)6 –

6C1(0.2)(0.8)5

= 1 – 0.2621 – 0.3932 = 0.3447

2 X ~ B(8, 0.4)(a) P(X = 4) = 8C

4(0.4)4(0.6)4

= 0.2322

(b) =(X � 3) = P(X = 0) + P(X = 1) + P(X = 2) = 8C

0(0.4)0(0.6)8 + 8C

1(0.4)(0.6)7

+ 8C2(0.4)2(0.6)6

= 0.0168 + 0.0896 + 0.2090 = 0.3154

3 X ~ B(n, p) with n = 7 and p = 0.25.(a) P(X � 2) = 1 – P(X � 1) = 1 – P(X = 0) – P(X = 1) = 1 – 7C

0(0.25)0(0.75)7

– 7C1(0.25)(0.75)6

= 1 – 0.13348 – 0.31146 = 0.5551

(b) P(X � 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.13348 + 0.31146 +

7C2(0.25)2(0.75)5

= 0.44494 + 0.31146 = 0.7564

4 X ~ B(n, 0.5) with p = 0.5 and q = 0.5. P(X � 1) � 0.95 1 – P(X = 0) � 0.95 1 – nC

0(0.5)0(0.5)n � 0.95

1 – 0.5n � 0.95 0.5n � 0.05 n log

10 0.5 � log

10 0.05

n � log

10 0.05

–––––––––log

10 0.5

n � 4.32

∴ The least number of shots is 5.

5 (a) P(X � 1) = 0.0081 P(X = 0) = 0.0081 nC

0(0.7)0(0.3)n = 0.0081

0.3n = 0.34

∴ n = 4(b) P(X = 2) = 4C

2(0.7)2(0.3)2

= 0.2646

6 X ~ B(4, 0.5)P(X � 3)= P(X = 3) + P(X = 4)= 4C

3(0.5)3(0.5) + 4C

4(0.5)4(0.5)0

= 0.25 + 0.0625= 0.3125

7 X ~ B(n, p) with n = 20 and p = 0.1The mean, E(X ) = np = 20(0.1) = 2Standard deviation, σ = npq

= 2(0.9) = 1.8 = 1.342

8 (a) E(X ) = 3.2 0.4n = 3.2 n = 8

(b) Standard deviation, σ = npq = 3.2(0.6) = 1.92 = 1.386

9 np = 5 … 1 npq = 4 … 2

2 ÷ 1 :

q = 4––5

So, p = 1 – 4––5

= 1––5

Substitute p = 1––5

into 1 :

n� 1—5 � = 5

n = 25

10 (a) np = 2 … 1 npq = 1.6 … 2 2 – 1 : q = 0.8 So p = 1 – q = 1 – 0.8 = 0.2 Substitute p = 0.2 into 1 : n(0.2) = 2 n = 10 ∴ p = 0.2

(b) P(X = 4) = 10C4(0.2)4(0.8)6

= 0.0881

11 (a) P(Z � 1.9) = 0.0287

(b) P(Z � – 0.75) = P(Z � 0.75) = 0.2266

(c) P(Z � –2.45) = 1 – P(Z � –2.45) = 1 – P(Z � 2.45) = 1 – 0.00714 = 0.99286

f(z)

zO–2.45

P(Z > –2.45)

(d) P(Z � 1.34) = 1 – P(Z � 1.34) = 1 – 0.0901 = 0.9099

f(z)

zO 1.34

P(Z < 1.34)

(e) P(0.83 � Z � 1.85) = P(Z � 0.83) – P(Z � 1.85) = 0.2033 – 0.0322 = 0.1711

f(z)

zO 0.83 1.85

P(0.83 < Z < 1.85)

(f) P(–1.764 � Z � – 0.246) = P(Z � 0.246) – P(Z � 1.764) = 0.4029 – 0.0388 = 0.3641

f(z)

zO

P(–1.764 < Z < –0.246)

–0.246–1.764

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 145: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(g) P(–2.57 � Z � 0.132) = 1 – P(Z � – 2.57) – P(Z � 0.132) = 1 – 0.00508 – 0.4475 = 0.5474

f(z)

zO

P(–2.57 < Z < 0.132)

0.132–2.57

(h) P(–1.68 � Z � 1.725) = 1 – P(Z � – 1.68) – P(Z � 1.725) = 1 – 0.0465 – 0.0423 = 0.9112

f(z)

zO

P(–1.68 < Z < 1.725)

1.725–1.68

(i) P(–2.05 � Z � 0) = 0.5 – P(Z � 2.05) = 0.5 – 0.0202 = 0.4798

f(z)

zO

P(–2.05 < Z < 0)

–2.05

(j) P(0 � Z � 1.76) = 0.5 – P(Z � 1.76) = 0.5 – 0.0392 = 0.4608

f(z)

zO

P(0 < Z < 1.76)

1.76

(k) P(|Z | � 1.64) = P(–1.64 � Z � 1.64) = 1 – 2P(Z � 1.64) = 1 – 2(0.0505) = 0.8990

f (z)

zO 1.64

P(|Z| < 1.64)

–1.64

(l) P(|Z| � 2.326) = 2P(Z � 2.326) = 2(0.01) = 0.02

f(z)

zO 2.326–2.326

12 (a) 0.0668(b) 0.00714(c) 1 – 0.0344 = 0.9656(d) 1 – 0.0104 = 0.9896(e) 0.2119 – 0.0082 = 0.2037(f) 1 – 0.1056 – 0.0139 = 0.8805(g) 0.5 – 0.3372 = 0.1628(h) 2(0.1587) = 0.3174

13 (a) f(z)

zO a

0.0778

∴ a = 1.42

(b) f(z)

zOa

0.1515

∴ a = –1.03

(c) f(z)

zOa

0.229 0.771

∴ a = –0.742

(d) f(z)

zO a

0.47720.5228

∴ a = 0.057

(e) f(z)

zO a

0.05 0.05

–a

0.9

∴ a = 1.645

(f) f(z)

zO a–a

0.0485

0.903

0.0485

∴ a = 1.66

14 (a) z = 0.44(b) z = – 1.85(c) z = – 1.2(d) z = 2(e) z = 1.1(f) z = 0.7(g) z = 1.2(h) z = – 1

15 (a) x – 12–––––––3

= 1.2

x – 12 = 3.6 x = 15.6

(b) x – 6––––––2

= – 0.7

x – 6 = – 1.4 x = 4.6

(c) x – 50–––––––5

= – 0.742

x – 50 = – 3.71 x = 46.29

(d) x – 200––––––––6

= 1.4

x – 200 = 8.4 x = 208.4

16 X is the height, in centimeters, of a boy.X ~ N(150, 25)(a) P(X � 158)

= P�Z � 158 – 150–––––––––

5 � = P(Z � 1.6) = 0.0548

f(z)

zO 1.6

Page 146: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) P(X � 146) = P�Z � 146 – 150–––––––––

5 � = P(Z � – 0.8) = P(Z � 0.8) = 0.2119

f(z)

zO–0.8

17 X is the examination markX ~ N(45, 202)

(a) P(X � 40) = P�Z � 40 – 45–––––––

20 � = P(Z � – 0.25) = 1 – P(Z � – 0.25) = 1 – 0.4013 = 0.5987

f(z)

zO–0.25

Since there are 200 candidates, the number of candidates who pass the examination is 200 × 0.5987 = 119.74

∴ 120 candidates passed

(b) Given P(X � x) = 0.05 Standardising

P�Z � x – 45––––––

20 � = 0.05

So, x – 45––––––20

= 1.645

x – 45 = 32.9 x = 77.9 = 78 (1 d.p.)

f(z)

zO z

0.05

∴ A distinction is awarded for a mark of 78 or more.

18 P(X � 106) = 0.8849

P�Z � 106 – 100–––––––––

σ � = 0.8849

So, 106 – 100–––––––––σ

= 1.2

1.2σ = 6 σ = 5

f (z)

zO z

0.8849

0.1151

19 X is the volume, in m�, of a can of drinks.X ~ N(350, σ 2)(a) P(X � 357) = 0.2 Standardising

P�Z � 357 – 350–––––––––

σ � = 0.2

So 357 – 350–––––––––

σ = 0.842

7 = 0.842σ σ = 8.31(2 d.p.)∴ The standard deviation, σ = 8.31 m�

O z

f(z)

z

0.2

(b) P(X � 340) = P�Z � 340 – 350–––––––––

8.31 � = P(Z � – 1.203) = 0.1145

∴ The percentage of cans that contain less than 340 m� is 0.1145 × 100 = 11.45%.

O

f(z)

z–1.203

20 X is life, in hours, of a batteryX ~ N(160, 302)P(150 � X � 175)

= P� 150 – 160–––––––––30

� Z � 175 – 160–––––––––

30 �= P(–0.3 � Z � 0.5)= 1 – P(Z � 0.3) – P(Z � 0.5)= 1 – 0.3821 – 0.3085= 0.3094∴ The percentage of batteries which

have a life between 151 hours and 175 hours is 0.3094 × 100 = 30.94%.

O

f(z)

z–0.3 0.5

SPM Appraisal Zone

1 (a) P(X = 0) = 1 – [P(X = 1) + P(X = 2) +

P(X = 3) + P(X = 4) + P(X = 5)] = 1 – (0.05 + 0.1 + 0.25 + 0.35

+ 0.2) = 1 – 0.95 = 0.05

(b) P(X � 2) = 1 – P(X = 0) – P(X = 1) = 1 – 0.05 – 0.05 = 0.9

2 X ~ B(4, p) P(X = 4) = 0.0256 4C

4(p4)(1 – p)0 = 0.0256

p4 = 0.44

∴ p = 0.4

3 X ~ B(8, 0.4)(a) P(X = 3) = 8C

3(0.4)3(0.6)5

= 0.2787

(b) P(X � 6) = P(X = 7) + P(X = 8) = 8C

7(0.4)7(0.6) + 8C

8(0.4)8(0.6)0

= 0.00852

4 np = 20 … 1 npq = 4 npq = 16 … 2

Substitute 1 into 2 : 20q = 16

q = 4––5

So, p = 1 – q

= 1 – 4––5

p = 1––5

Substitute p = 1––5

into 1 : 1––5

n = 20

n = 100

∴ p = 1––5

, n = 100

5 P(X � 1) � 0.95 1 – P(X = 0) � 0.95 1 – nC

0(0.1)0(0.9)n � 0.95

1 – 0.9n � 0.95 0.9n � 0.05 n log

10 0.9 � log

10 0.05

n � log

10 0.05

–––––––––log

10 0.9

n � 28.43∴ The least value of n is 29.

6 (a) P(Z � b) = 0.3 + 0.6 = 0.9

(b) P(Z � a) = 1 – 0.3 = 0.7

7 (a) P(Z � a) = 0.45 From the table, a = 0.125

(b) P(a � Z � b) = 0.45 – 0.3 = 0.15

8 P(X � a) = 0.242

a – 12–––––––

2 = 0.7

a – 12 = 1.4 a = 13.4

9 (a) P(X � 56) = P�Z � 56 – 50–––––––

5 � = P(Z � 1.2) = 0.1151

Page 147: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) P(X � 42) = P�Z � 42 – 50–––––––

5 � = P(Z � – 1.6) = P(Z � 1.6 = 0.0548

10 X is the mass, in grams, of a cabbageX ~ N(1000, 1502)P(715 � X � 1264)

= P� 715 – 1000––––––––––150

� Z � 1264 – 1000––––––––––

150 �= P(–1.9 � Z � 1.76)= 1 – 0.0287 – 0.0392= 0.9321∴ There are 500 × 0.9321 = 466

cabbages have a mass between 715 g and 1264 g.

11 X ~ N(4.5, 1.21)

(a) z = 6.7 – 4.5––––––––

1.1

= 2

(b) P(3.4 � X � 6.7)

= P� 3.4 – 4.5––––––––1.1

� Z � 6.7 – 4.5––––––––

1.1 � = P(–1 � Z � 2) = 1 – 0.1587 – 0.0228 = 0.8185

12 P(X � 20) = 0.0228

P�Z � 20 – �

–––––––15 � = 0.0228

So, 20 – �

–––––––15

= – 2

20 – � = – 30 � = 50

13 P[|Z | � a] = 0.6 P(0 � Z � a) = 0.3So, P(Z � a) = 0.5 – 0.3 = 0.2From the table, a = 0.842.

14 X ~ N(�, 52) P(X � 42) = 0.0808

P�Z � 42 – �

–––––––5 � = 0.0808

So, 42 – �

–––––––5

= 1.4

42 – � = 7 � = 35

15 X is the consultation time, in minutes, of a patient.X ~ N(10, 52)

(a) P(X � 15) = P�Z � 15 – 10–––––––

5 � = P(Z � 1) = 0.1587

(b) P(8 � X � 14)

= P� 8 – 10––––––5

� Z � 14 – 10–––––––

5 � = P(–0.4 � Z � 0.8) = 1 – 0.3446 – 0.2119 = 0.4435

16 (a) X ~ N(�, σ2) P(X � 6.35) = 0.02

P�Z � 6.35 – �––––––––

σ � = 0.02

6.35 – �––––––––

σ = – 2.054

� – 2.054σ = 6.35 … 1

f(x)

x6.35

0.02

P(X � 7.55) = 0.05

P�Z � 7.55 – �––––––––

σ � = 0.05

7.55 – �––––––––

σ = 1.645

� + 1.645 σ = 7.55 … 2

f(x)

x7.55

0.05

2 – 1 : 3.699σ = 1.2 σ = 0.324Substitute σ = 0.324 into 1 :

� – 2.054(0.324) = 6.35 � = 7.015

∴ � = 7.015, σ = 0.324

(b) (i) P(X = 3) = 10C

3(0.02)3(0.98)7

= 0.00083

(ii) P(X � 2) = 1 – P(X = 0) + P(X = 1) = 1 – 10C

0(0.05)0(0.95)10

– 10C1(0.05)(0.95)9

= 1 – 0.59874 – 0.31512 = 0.0861

17 X ~ N(50, 102)

(a) (i) P(X � 45) = P�Z � 45 – 50–––––––

10 � = P(Z � – 0.5) = P(Z � 0.5) = 0.3085

(ii) P(42 � X � 56)

= P�42 – 50–––––––

10 � Z �

56 – 50–––––––

10 � = P(–0.8 � Z � 0.6) = 1 – P(Z � 0.8) – P(Z � 0.6) = 1 – 0.2119 – 0.2743 = 0.5138

∴ 0.5138� 365––––7 � = 26.79

= 27 weeks

(b) P(X � n) = 0.6

P�Z � n – 50

–––––––10 � = 0.6

So, n – 50

–––––––10

= – 0.253

n – 50 = – 2.53 n = 47.47 = 47

f(z)

0.60.4

nx

18 (a) (i) P(X = 4) = 10C4(0.15)4(0.85)6

= 0.0401

(ii) P(2 � X � 5) = P(X = 3) + P(X = 4) = 10C

3(0.15)3(0.85)7 + 0.0401

= 0.1298 + 0.0401 = 0.1699

(b) Mean, � = np = 10(0.15) = 1.5

Variance, σ 2 = npq = 1.275

(c) P(X � 1) � 0.95 1 – P(X = 0) � 0.95

1 – nC0(0.15)0(0.85)n � 0.95

1 – 0.85n � 0.95 0.85n � 0.05 n log

10 0.85 � log

10 0.05

n � log

10 0.05

–––––––––log

10 0.85

n � 18.43 ∴ The least value of n is 19.

19 X is the Mathematics mark.X ~ N(55, 82)(a) P(45 � X � 60)

= P� 45 – 55–––––––

8 � Z �

60 – 55–––––––

8 � = P(–1.25 � Z � 0.625) = 1 – P(Z � 1.25) – P(Z � 0.625) = 1 – 0.1056 – 0.2660 = 0.6284

Page 148: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) P(X � 70) = P�Z � 70 – 55–––––––

8 � = P(Z � 1.875) = 0.0303

∴ The number of candidates with distinction

= 0.0303 × 200 = 6.06 = 6 students

(c) P(X � x) = 0.97

P�Z � x – 55

–––––––8 � = 0.97

So, x – 55

–––––––8

= – 1.881

x – 55 = – 15.048 x = 39.95 = 40 (2 s.f.) ∴ Lowest passing mark = 40

f (x)

0.03 0.97

55x

x

P(X = 4) = 5C4(0.97)4(0.03)

= 0.132820 X is the length, in centimeters, of a

bottleX ~ N(25, 22)

(a) (i) P(X � 28) = P�Z � 28 – 25–––––––

2 � = P(Z � 1.5) = 0.0668

(ii) P(23 � X � 28)

= P� 23 – 25–––––––2

� Z � 1.5� = P(–1 � Z � 1.5) = 1 – P(Z � 1) – P(Z � 1.5) = 1 – 0.1587 – 0.0668 = 0.7745

(b) P(X � 23) = P(Z � – 1) = 0.1587 ∴ The number of bottles = 0.1587 × 500 = 79.35 = 79(2 s.f.)

(c) P(X � L) = 0.1

P�Z � L – 25–––––––

2 � = 0.1

So, L – 25–––––––2

= – 1.281

L – 25 = – 2.562 L = 22.44 = 22 cm (2 s.f.)

f(z)

25x

L

0.1

21 (a) (i) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 5C0� 1––

4 �0� 3––4 �5

5C1� 1––

4 �� 3––4 �4

= 1 – 0.2373 – 0.3955 = 0.3672

(ii) P(X � 1) � 0.9 1 – P(X = 0) � 0.9

1 – nC

0� 1––4 �0� 3––

4 �n � 0.9

1 – � 3––4 �n

� 0.9

� 3––4 �n

� 0.1

n log10� 3––

4 � � log10

0.1

n � log

10 0.1

–––––––––––log

10 � 3––

4 � n � 8 ∴ n = 9

(b) X is the Chemistry mark. X ~ N(65, 102)

(i) P(X � 40) = P�Z � 40 – 65–––––––

10 � = P(Z � – 2.5) = 0.00621

(ii) P(X � 80) = P�Z � 80 – 65–––––––

10 � = P(Z � 1.5) = 0.0668 ∴ The percentage of students

who obtained Al grade = 0.0668 × 100 = 6.68%

22 (a) (i) P(X = 2) = 3C2� 3––

7 �2� 4––7 �

= 0.3149

(ii) P(X � 2) = 1 – P(X = 3)

= 1 – 3C3� 4––

7 �3� 3––7 �0

= 0.8134

(b) X is the mass, in kilograms, of a student.

X ~ N(55, 42) (i) P(X � 62)

= P�Z � 62 – 55–––––––

4 � = P(Z � 1.75) = 0.0401

(ii) P(X � m) = 0.22

P�Z � m – 55–––––––

4 � = 0.22

So, m – 55–––––––4

= 0.772

m – 55 = 3.088 m = 58.09 (2 d.p.)

f(x)

55x

m

0.22

23 (a) np = 2 … 1 npq = 1.5 … 2

2 ÷ 1 : q = 1.5––––2

= 0.75 So, p = 1 – q = 1 – 0.75 = 0.25 Substitute p = 0.25 into 1 : n(0.25) = 2 n = 8 ∴ p = 0.25, n = 8

(b) (i) P(30 � X � 42) = 1 – 0.1587 – 0.0808 = 0.7605

(ii) P(X � 42) = 0.0808

P�Z � 42 – �

––––––––σ � = 0.0808

So, 42 – �

––––––––σ

= 1.4

42 – � = 1.4σ � + 1.4σ = 42 … 1

and P(X � 30) = 0.1587

P�Z � 30 – �

––––––––σ � = 0.1587

So, 30 – �

––––––––σ

= –1

30 – � = –σ

� – σ = 30 … 2

1 – 2 : 2.4σ = 12 σ = 5

Substitute σ = 5 into 1 : � + 1.4(5) = 42 � = 35 ∴ � = 35, σ = 5

24 (a) X ~ B(6, 0.9) (i) P(X = 3) = 6C

3(0.9)3(0.1)3

= 0.0146

(ii) P(X � 3) = P(X = 3) + P(X = 4) +

P(X = 5) + P(X = 6) = 0.0146 + 6C

4(0.9)4(0.1)2

+ 6C5(0.9)5(0.1) +

6C6(0.9)6(0.1)0

= 0.0146 + 0.09842 + 0.35429 + 0.53144

= 0.9988

Page 149: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) X is the length, in centimeters, of a pencil.

X ~ N(�, 0.52) P(X � 15) = 0.05

P�Z � 15 – �

––––––––0.5 � = 0.05

So, 15 – �

––––––––0.5

= 1.645

15 – � = 0.8225 � = 14.18 (2 d.p.)

f(x)

15xµ

0.05

25 (a) (i) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 10C0� 3––

5 �0� 2––5 �10

10C1� 3––

5 �1� 2––5 �9

= 0.9983

(ii) Mean, � = np = 500 × 3––

5 = 300 Standard deviation, σ = npq

= 300� 2––5 �

= 120 = 10.95 ∴ � = 300, σ = 10.95

(b) (i) P(X � a) = 1 – 0.9772 = 0.0228

(ii) P(X � a) = 0.9772

P�Z � a – 50

–––––––5 � = 0.9772

So, a – 50

–––––––5

= – 2

a – 50 = – 10 a = 50 – 10 = 40

Page 150: Analysis Spm Additional Mathematics

Motion along a Straight Line

1

Booster Zone

1 (a) s2 = 2(2 – 4)2

= 2(–2)2

= 8 m

(b) s4 = 4(4 – 4)2

= 0

(c) s6 = 6(6 – 4)2

= 6(4) = 24 m

(d) s9 = 9(9 – 4)2

= 9(25) = 225 m

2 (a) s1 = 6(1)2 – 13

= 5 m

(b) s4 = 6(4)2 – 43

= 32 m

(c) s5 = 6(5)2 – 53

= 25 m

(d) s8 = 6(8)2 – 83

= –128 m

3 (a) (i) s2.5

= 5(2.5) – (2.5)2

= 12.5 – 6.25 = 6.25 m

(ii) s5 = 5(5) – 52

= 0 m

(b) s = –24 5t – t 2 = –24 t 2 – 5t – 24 = 0 (t – 8)(t + 3) = 0 t = 8 or t = –3 ∴ t = 8 s

(c) s = 0 m 5t – t 2 = 0 t 2 – 5t = 0 t(t – 5) = 0 t = 0 or t = 5 ∴ The particle passes through O

again when t = 5 s.

4 (a) s2 = 2(2)2 – 8(2)

= 8 – 16 = –8 m

(b) (i) s = 0 2t 2 – 8t = 0 2t(t – 4) = 0 t = 0 or t = 4 ∴ t = 4 s

(ii) s = 10 2t 2 – 8t = 10 t 2 – 4t – 5 = 0 (t – 5)(t + 1) = 0 ∴ t = 5 s

(c) s � 0 2t 2 – 8t � 0 2t(t – 4) � 0

40t

∴ t � 4

5 (a)

0

t = 0s = 0

s

t = 6s = 12

t = 2s = –4

(b) (i) s = (2)2 – 4(2) = –4 = 4 m

(ii) s3 = 32 – 4(3)

= 9 – 12 = –3 m ∴ The distance in 4th second = s

4 – s

3

= 0 – (–3) = 3 m

(iii) The distance in the interval 0 � t � 6

= 2(4) + 12 = 20 m

Average velocity during the fi rst 6 seconds

= distance travelled

––––––––––––––––time taken

= 20–––6

= 31––3

m s–1

6 (a)

6520

7

5

9

t (s)

s (m)s = 5 + 4t – t2

(b) (i) s1 = 5 + 4(1) – (1)2

= 8 s

0 = 5 + 4(0) – (0)2

= 5 ∴ s

1 – s

0 = 8 – 5

= 3 m

(ii) s4 = 5 + 4(4) – 42

= 5 m∴ The distance in the 5th

second = |s

5 – s

4|

= |0 – 5| = 5 m

Average speed during the fi rst 6 seconds

= distance travelled

––––––––––––––––time taken

= 4 + 9 + 7

–––––––––6

= 20–––6

= 31––3

m s–1

7 (a) (i) 8 m

(ii) 20 m

(iii) –8 m

(b) Average speed

= (20 – 8) + 20 + 8

––––––––––––––––10

= 40–––10

= 4 m s–1

8 (a) (i) s0 = (0 – 2)2 + 5

= 9 m

s2 = (2 – 2)2 + 5

= 5 m∴ The distance during the

next 2 seconds = �5 – 9� = 4 m

(ii) s4 = (4 – 2)2 + 5

= 9 m∴ The distance during the

next 2 seconds = 9 – 5 = 4 m

(b) Average velocity

= (9 – 5) + (30 – 5)

––––––––––––––––7

= 41––7

m s–1

9 (a) v = ds

–––dt

= 3t 2 – 6t ∴ v

2 = 3(2)2 – 6(2)

= 12 – 12 = 0 m s–1

20

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 151: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) s = 0 t 3 – 3t 2 = 0 t 2(t – 3) = 0 t = 0 or t = 3 ∴ v

3 = 3(3)2 – 6(3)

= 27 – 18 = 9 m s–1

(c) v = 9 3t 2 – 6t = 9 t 2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3 or t = –1 ∴ t = 3 s

10 (a) v = ds

–––dt

= 4t – 5 ∴ v

0 = 4(0) – 5

= –5 m s–1

(b) s = 8 2t 2 – 5t + 5 = 8 2t 2 – 5t – 3 = 0 (2t + 1)(t – 3) = 0

t = – 1––2

or t = 3

∴ v3 = 4(3) – 5

= 12 – 5 = 7 m s–1

(c) v = 3 4t – 5 = 3 4t = 8 t = 2 ∴ s

3 = 2(3)2 – 5(3) + 5

= 18 – 15 + 5 = 8 m

11 (a) s = t 3 – 3t 2 – 9t + 5

v = ds

–––dt

= 3t 2 – 6t – 9 When the particle reverses its

direction, v = 0 3t 2 – 6t – 9 = 0 t 2 – 2t – 3 = 0 (t + 1)(t – 3) = 0 t = –1 or t = 3 ∴ t = 3 s

(b) v � 0 3t 2 – 6t – 9 � 0 t 2 – 2t – 3 � 0 (t + 1)(t – 3) � 0

t3–1

∴ t � 3

12 (a) v = ds

–––dt

= 3t 2 – 18t + 24 ∴ v

3 = 3(3)2 – 18(3) + 24

= –3 m s–1

(b) When the particle is instantaneously at rest,

v = 0 3t 2 – 18t + 24 = 0 t 2 – 6t + 8 = 0 (t – 2)(t – 4) = 0 t = 2 or t = 4

(c) v � 0 t 2 – 6t + 8 � 0 (t – 2)(t – 4) � 0

42t

∴ 2 � t � 4

13 (a) v = 8 – 4t s = �v dt = 8t – 2t 2 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = 8t – 2t 2

When t = 1, s = 8(1) – 2(1)2

= 6 ∴ The displacement of Q from O

is 6 m.

(b) For maximum displacement, v = 0 8 – 4t = 0 t = 2 ∴ Maximum displacement = 8(2) – 2(2)2

= 16 – 8 = 8 m

(c) s4 = 8(4) – 2(4)2

= 32 – 32 = 0

0s

2

t = 0s = 0

t = 4s = 0

t = 2s = 8

∴ The total distance travelled during the fi rst 4 seconds

= 2(8) = 16 m

14 (a) v = 3t 2 – 6t

s = �v dt = t 3 – 3t 2 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = t 3 – 3t 2

When t = 5, s = (5)3 – 3(5)2

= 50 ∴ The displacement of particle

when t = 5 is 50 m.

(b) When the particle is momentarily at rest,

v = 0 3t(t – 2) = 0 t = 0 or t = 2 ∴ s

2 = 23 – 3(2)2

= 8 – 12 = –4 m

15 (a) v = 12t – 3t 2

s = �v dt = 6t 2 – t 3 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = 6t 2 – t 3

s4 = 6(4)2 – 43

= 32 m s

3 = 6(3)2 – 33

= 54 – 27 = 27 m ∴ The distance travelled in the 4th

second = 32 – 27 = 5 m

(b) When t = 2, s = 6(2)2 – 23

= 24 – 8 = 16 m When t = 6, s = 6(6)2 – 63

= 216 – 216 = 0 m ∴ The motion of the particle is as

shown below.

0

t = 0s = 0

s

t = 6s = 0

t = 2s = 16

t = 4s = 32

∴ The distance travelled from t = 2 to t = 6 = (32 – 16) + 32 = 48 m

16 (a) v = 10 – 2t s = �v dt = 10t – t 2 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = 10t – t 2

(i) When the velocity is 8 m s–1, 10 – 2t = 8 2t = 2 t = 1 ∴ s

1 = 10(1) – 12

= 9 m

(ii) When the particle in instantaneously at rest,

v = 0 10 – 2t = 0 t = 5 ∴ s

5 = 10(5) – 52

= 25 m

(b) When the particle passes through O again,

s = 0 10t – t 2 = 0 t (10 – t) = 0 t = 0 or t = 10 ∴ v

10 = 10 – 2(10)

= 10 – 20 = –10 m s–1

Page 152: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

17 (a) When the particle is instantaneously at rest,

v = 0 t 2 – 4t + 3 = 0 (t – 1)(t – 3) = 0 t = 1 or t = 3

(b) s = �(t 2 – 4t + 3)dt

= t 3

–––3

– 2t 2 + 3t + c

When t = 0, s = 0 and so c = 0 Hence at time t,

s = t 3

–––3

– 2t 2 + 3t

When t = 1, s = 1––3

– 2 + 3

= 1 1––3

m

When t = 3, s = 9 – 18 + 9 = 0 m

∴ The distance between the positions when the particle is

at rest = 1 1––3

m

18 s = �v dt

= 3t + 5t 2

–––2

– 2t 3

–––3

+ c

When t = 0, s = 0 and so c = 0

Hence, s = 3t + 5––2

t 2 – 2––3

t 3

(a) s3 = 3(3) + 5––

2(3)2 – 2––

3(3)3

= 131––2

m

s2 = 3(2) + 5––

2(2)2 – 2––

3(2)3

= 102––3

m

∴ The distance travelled in the 3rd second

= 131––2

– 102––3

= 25––6

m

(b) s6 = 3(6) + 5––

2(6)2 – 2––

3(6)3

= –36 m The motion of the particle is as

shown below.

0

t = 0s = 0

s

t = 6s = –36

t = 3s = 13

1––2

∴ The distance travelled during the fi rst 6 seconds

= 2�131––2 � + 36

= 63 m

19 s = t 3 – 4t 2 – 12t

v = ds

–––dt

= 3t 2 – 8t – 12

a = dv

–––dt

= 6t – 8

(a) When t = 0, a = 6(0) – 8 = –8 m s–2

(b) When t = 2, a = 6(2) – 8 = 4 m s–2

(c) When the particle passes O again, s = 0 t 3 – 4t 2 – 12t = 0 t (t 2 – 4t – 12) = 0 t (t + 2)(t – 6) = 0 t = 0 or t = –2 or t = 6 ∴ a

6 = 6(6) – 8

= 28 m s–2

(d) a � 0 6t – 8 � 0 6t � 8

t � 4––3

20 v = t 2 + 2t – 8

a = dv

–––dt

= 2t + 2

(a) When the particle is instantaneously at rest,

v = 0 t 2 + 2t – 8 = 0 (t + 4)(t – 2) = 0 t = –4 or t = 2 ∴ a

2 = 2(2) + 2

= 6 m s–2

(b) t 2 + 2t – 8 = 16 t 2 + 2t – 24 = 0 (t + 6)(t – 4) = 0 t = –6 or t = 4 ∴ a

4 = 2(4) + 2

= 10 m s–2

21 a = 8 – 4tv = �a dt = 8t – 2t 2 + cWhen t = 0, v = 3 and so c = 3∴ At time t, v = 8t – 2t 2 + 3

(a) v3 = 8(3) – 2(3)2 + 3

= 9 m s–1

(b) For maximum velocity, a = 0 8 – 4t = 0 t = 2 ∴ Maximum velocity = 8(2) – 2(2)2 + 3 = 11 m s–1

22 (a) a = 2t – 6 ∴ a

0 = 2(0) – 6

= –6 m s–2

(b) v = �a dt = t 2 – 6t + c When t = 0, v = 2 and so c = 2 Hence at time t, v = t 2 – 6t + 2 For minimum velocity, a = 0 2t – 6 = 0 t = 3 ∴ Minimum velocity = 32 – 6(3) + 2 = –7 m s–1

23 a = 6 – 2tv = �a dt = 6t – t 2 + cWhen t = 0, v = 16 and so c = 16Hence, v = 6t – t2 + 16s = �v dt

= 3t 2 – t 3

––3

+ 16t + c

When t = 0, s = 0 and so c = 0

Hence, s = 16t + 3t 2 – 1––3

t 3

(a) For maximum displacement, v = 0 6t – t 2 + 16 = 0 t 2 – 6t – 16 = 0 (t – 8)(t + 2) = 0 t = 8 or t = –2 ∴ Maximum displacement

= 16(8) + 3(8)2 – 1––3

(8)3

= 128 + 192 – 512––––3

= 149 1––3

m

(b) s3 = 16(3) + 3(3)2 – 1––

3(3)3

= 48 + 27 – 9 = 66 m

24 a = 3 – tv = �a dt

= 3t – 1––2

t 2 + c

When t = 0, v = 8 and so c = 8

Hence, v = 8 + 3t – 1––2

t 2

s = �v dt

= 8t + 3––2

t 2 – 1––6

t 3 + c

When t = 0, s = 0 and so c = 0

Hence, s = 8t + 3––2

t 2 – 1––6

t 3

(a) v4 = 8 + 3(4) – 1––

2(4)2

= 12 m s–1

(b) For maximum velocity, a = 0 3 – t = 0 t = 3 ∴ Maximum velocity

= 8 + 3(3) – 1––2

(3)2

= 12 1––2

m s–1

Page 153: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(c) When the particle is at rest, v = 0

8 + 3t – 1––2

t 2 = 0

t 2 – 6t – 16 = 0 (t + 2)(t – 8) = 0 t = –2 or t = 8

∴ s8 = 8(8) + 3––

2(8)2 – 1––

6(8)3

= 64 + 96 – 256––––3

= 74 2––3

m

(d) s10

= 8(10) + 3––2

(10)2 – 1––6

(10)3

= 80 + 150 – 500––––3

= 63 1––3

m

∴ Average speed during the fi rst 10 seconds

=

74 2––3

+ �74 2––3

– 63 1––3 �

–––––––––––––––––––––10

= 8.6 m s–1

25 a = 12 – 6tv = �a dt = 12t – 3t 2 + cWhen t = 0, v = 0 and so c = 0Hence, v = 12t – 3t 2

s = �v dt = 6t 2 – t 3 + cWhen t = 0, s = 0 and so c = 0Hence, s = 6t 2 – t 3

(a) When the particles is at rest, v = 0 12t – 3t 2 = 0 3t(4 – t) = 0 t = 0 or t = 4 ∴ The time taken to reach A is

4 s.

(b) s4 = 6(4)2 – 43

= 32 m ∴ The distance OA is 32 m.

(c) For maximum speed, a = 0 12 – 6t = 0 t = 2 ∴ Maximum speed = 12(2) – 3(2)2

= 24 – 12 = 12 m s–1

26 a = 6 – 2tv = �a dt = 6t – t 2 + cWhen t = 1, v = 10 and so c = 5Hence, v = 5 + 6t – t 2

(a) v0 = 5 + 6(0) – 02

= 5 m s–1

(b) a = 0 6 – 2t = 0 t = 3 ∴ v

3 = 5 + 6(3) – 32

= 14 m s–1

27 a = 1 – 2tv = �a dt = t – t 2 + cWhen t = 0, v = 6 and so c = 6Hence, v = t – t 2 + 6s = �v dt

= 6t + 1––2

t 2 – 1––3

t 3 + c

When t = 0, s = 0 and so c = 0

Hence, s = 6t + 1––2

t 2 – 1––3

t 3

(a) When the particle reverses its direction,

v = 0 t – t 2 + 6 = 0 t 2 – t – 6 = 0 (t + 2)(t – 3) = 0 t = –2 or t = 3 ∴ The particle reverses its direction

when t = 3 s.

(b) s3 = 6(3) + 1––

2(3)2 – 1––

3(3)3

= 18 + 9––2

– 9

= 13 1––2

m

28 a = 6t – 12v = 3t 2 – 12t + cWhen t = 0, v = 9 and so c = 9Hence, v = 3t 2 – 12t + 9s = t 3 – 6t 2 + 9t + cWhen t = 0, s = 0 and so c = 0Hence s = t 3 – 6t 2 + 9t

(a) When the particle is at rest, v = 0 3t 2 – 12t + 9 = 0 t 2 – 4t + 3 = 0 (t – 1)(t – 3) = 0 t = 1 or t = 3 ∴ The particle is at rest when

t = 1 s and t = 3 s.

(b) s1 = (1)3 – 6(1)2 + 9(1)

= 4 m s

2 = (2)3 – 6(2)2 + 9(2)

= 2 m ∴ The distance of the particle

from O is = 4 + (4 – 2) = 6 m

SPM Appraisal Zone

1 v = 12t – 3t 2

s = �v dt = 6t 2 – t 3 + cWhen t = 0, s = 0 and so c = 0Hence at time t, s = 6t 2 – t 3

a = dv

–––dt

= 12 – 6t

(a) For maximum velocity, a = 0 12 – 6t = 0 t = 2

Furthermore, da–––dt

= –6 � 0

⇒ v is maximum when t = 2 ∴ Maximum velocity = 12(2) – 3(2)2

= 24 – 12 = 12 m s–1

(b) When the particle reverses its direction,

v = 0 12t – 3t 2 = 0 3t (4 – t) = 0 t = 0 or t = 4 ∴ The particle reverses its direction

when t = 4 s.

(c) s4 = 6(4)2 – 43

= 96 – 64 = 32 m s

3 = 6(3)2 – 33

= 54 – 27 = 27 m ∴ The distance travelled in the

4th second = 32 – 27 = 5 m

(d) When the particle passes O again, s = 0 6t 2 – t 3 = 0 t 2(6 – t) = 0 t = 0 or t = 6 ∴ The particle passes O again

when t = 6 s.

2 a = 8 – 2tv = 8t – t 2 + cWhen t = 0, v = 9 and so c = 9Hence, v = 8t – t 2 + 9

(a) (i) For maximum velocity, a = 0 8 – 2t = 0 t = 4 ∴ Maximum velocity = 8(4) – 42 + 9 = 32 – 16 + 9 = 25 m s–1

Page 154: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(ii) When the particle stops, v = 0 8t – t 2 + 9 = 0 t 2 – 8t – 9 = 0 (t + 1)(t – 9) = 0 t = –1 or t = 9 ∴ The particle stops after

9 s, so m = 9.

(b)

Ot (s)

v = 9 + 8t – t 2

v (m s–1)

9

25

4 9

From the graph,

s = �9

0 (9 + 8t – t 2)dt

= �9t + 4t 2 – 1––3

t 3�9

0

= 81 + 324 – 243 = 162 m ∴ The total distance travelled is

162 m.

3 a = 12 – 6tv = 12t – 3t 2 + cWhen t = 0, v = 15 and so c = 15Hence, v = 12t – 3t 2 + 15s = �v dt = 15t + 6t 2 – t 3 + cWhen t = 0, s = 0 and so c = 0Hence, s = 15t + 6t 2 – t 3

(a) (i) s1 = 15(1) + 6(1)2 – (1)3

= 20 m

(ii) v = 0 12t – 3t 2 + 15 = 0 t 2 – 4t – 5 = 0 (t + 1)(t – 5) = 0 t = –1 or t = 5 ∴ s

5 = 15(5) + 6(5)2 – (5)3

= 75 + 150 – 125 = 100 m

(iii) a = 0 12 – 6t = 0 t = 2 ∴ s

2 = 15(2) + 6(2)2 – (2)3

= 30 + 24 – 8 = 46 m

(b) The motion of the particle is shown below.

O

t = 2s = 46

s

t = 5s = 100

t = 0s = 0

t = 6s = 90

∴ The total distance from t = 2 to t = 6 is

= (100 – 46) + (100 – 90) = 64 m

4 (a) v = t 2 + pt + 8

a = dv

–––dt

= 2t + p When t = 4, a = 2 2t + p = 2 8 + p = 2 p = –6

(b) (i) When the particle is at rest, v = 0 t 2 – 6t + 8 = 0 (t – 2)(t – 4) = 0 t = 2 or t = 4 ∴ The par t ic le i s

instantaneously at rest when t = 2 s and t = 4 s.

(ii) s = �v dt

= 1––3

t 3 – 3t 2 + 8t + c

When t = 0, s = 0 and so c = 0

Hence s = 1––3

t 3 – 3t 2 + 8t

s2 = 1––

3(8) – 3(4) + 8(2)

= 6 2––3

m

s4 = 1––

3(4)3 – 3(16) + 8(4)

= 5 1––3

m

∴ The distance of AB

= 6 2––3

– 5 1––3

= 1 1––3

m

(c)

Ot (s)

v = t2 – 6t + 8

v (m s–1)

7432

8

15

From the graph, the acceleration is negative when 0 � t � 3.

5 (a) v � 0 2t – 6 � 0 t � 3

(b) s = �v dt = t 2 – 6t + c When t = 0, s = 10 and so c = 10 Hence, s = t 2 – 6t + 10

(i) When the particle reverses its direction,

v = 0 2t – 6 = 0 t = 3 s

3 = 32 – 6(3) + 10

= 9 – 18 + 10 = 1 m

The motion of the particle is as shown below.

Y Xs

t = 3s = 1

t = 0s = 10

∴ The particle will not reach Y.

(ii) s8 = 82 – 6(8) + 10

= 26 m ∴ The total distance travelled

during the fi rst 8 seconds = (10 – 1) + (26 – 1) = 9 + 25 = 34 m

(iii) s = t 2 – 6t + 10

Ot (s)

s = t2 – 6t + 10

s (m)

3 8

10

1

26

6 (a) s = 5 + 3t 2 – t 3

When t = 0, s = 5 + 3(0)2 – 03

= 5 ∴ The distance of OA = 5 m

(b) v = ds

–––dt

= 6t – 3t 2

When the velocity is negative, v � 0 6t – 3t 2 � 0 3t 2 – 6t � 0 3t (t – 2) � 0

O

t2

∴ t � 2

(c) a = dv

–––dt

= 6 – 6t

Page 155: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

For a maximum velocity, a = 0 6 – 6t = 0 t = 1 ∴ Maximum velocity = 6(1) – 3(1)2

= 3 m s–1

(d) s2 = 5 + 3(2)2 – 23

= 9 m s

4 = 5 + 3(4)2 – 43

= –11 m

The motion of the particle is as shown below.

O

t = 0s = 5

t = 2s = 9

t = 4s = –11

∴ The total distance travelled by the particle in the fi rst 4 seconds

= (9 – 5) + 9 + 11 = 24 m

7 (a) s = 2t 2 – 8 When t = 0, s = 2(0)2 – 8 = –8 ∴ The distance OA = 8 m

(b) When the particle passes O, s = 0 2t 2 – 8 = 0 t 2 = 4 t = ±2 t = 2 s

v = ds

–––dt

= 4t ∴ v = 4(2) = 8 m s–1

(c) s3 = 2(3)2 – 8

= 18 – 8 = 10 m

The motion of the particle is as shown below.

OA

t = 3s = 10

t = 0s = –8

∴ Average speed = 8 + 10––––––3

= 6 m s–1

8 (a) (i) a = dv

–––dt

= 8 – 6t

(ii) s = �v dt = 5t + 4t 2 – t 3 + c When t = 0, s = 0 and so c = 0 Hence, at time t, s = 5t + 4t2 – t3

(b) (i) When the particle passes O again,

s = 0 5t + 4t 2 – t 3 = 0 t 3 – 4t 2 – 5t = 0 t(t 2 – 4t – 5) = 0 t(t + 1)(t – 5) = 0 t = 0 or t = –1 or t = 5 ∴ t = 5 s

(ii) When the velocity is 2 m s–1, v = 2 5 + 8t – 3t 2 = 2 3t 2 – 8t – 3 = 0 (3t + 1)(t – 3) = 0

t = – 1––3

or t = 3

∴ t = 3 s

(iii) When the acceleration is 6 m s–2,

a = 6 8 – 6t = 6 6t = 2

t = 1––3

∴ t = 1––3

s

9 (a) (i) a = dv

–––dt

= 12t – 4 When t = 0, a = 12(0) – 4 = –4 m s–2

(ii) For minimum velocity, a = 0 12t – 4 = 0

t = 1––3

∴ Minimum velocity

= 6� 1––3 �

2

– 4� 1––3 � – 2

= 2––3

– 4––3

– 2

= –2 2––3

m s–1

(iii) When P moves towards the left,

v � 0 6t 2 – 4t – 2 � 0 3t 2 – 2t – 1 � 0 (3t + 1)(t – 1) � 0

t

11––3

∴ 0 � t � 1

(b)

0–2

t (s)1 3

40

v (m s–1)v = 6t2 – 4t – 2

∴ The distance travelled during the fi rst 3 seconds

= ��1

0 (6t 2 – 9t – 2) dt � +

�3

1 (6t 2 – 4t – 2) dt

= [2t 3 – 2t 2 – 2t]1

0 +

[2t 3 – 2t 2 – 2t]3

1

= 2 + [30 – (–2)] = 34 m

10 (a) a = 4t – 16 When t = 0, a = 4(0) – 16 = –16 m s–2

(b) v = �a dt

= � (4t – 16)dt = 2t 2 – 16t + c When t = 0, v = 30 and so c = 30 Hence, at time t, v = 2t 2 – 16t + 30

(c) When the particle reverses its direction of motion,

v = 0 2t 2 – 16t + 30 = 0 t 2 – 8t + 15 = 0 (t – 3)(t – 5) = 0 t = 3 or t = 5 ∴ t = 3 s and t = 5 s

(d) s = �v dt

= 2––3

t 3 – 8t 2 + 30t + c

When t = 0, s = 0 and so c = 0 Hence, at time t,

s = 2––3

t 3 – 8t 2 + 30t

When t = 3,

s = 2––3

(3)3 – 8(3)2 + 30(3)

= 36 m

When t = 5,

s = 2––3

(5)3 – 8(5)2 + 30(5)

= 250––––3

– 200 + 150

= 33 1––3

m

The motion of the particle is as shown below.

Os

t = 3s = 36

t = 0s = 0

t = 5

s = 331––3

∴ The distance travelled by the particle in the fi rst 5 seconds

= 36 + �36 – 33 1––3 �

= 38 2––3

m

Page 156: Analysis Spm Additional Mathematics

21 Linear Programming

1

Booster Zone 1 (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

2 (a)

(b)

(c)

(d)

(e)

(f)

3 (a) y � – 1 (b) x � 5(c) y � 2x (d) x + y � 4(e) y – 2x � 8 (f) 2x – y � 10

4 (a)

(b)

(c)

(d)

5 (a) x � 5, y � x + 5, x + y � 5(b) y � 2, y � 2x, y � x(c) y � 6, y � x, x + y � 6(d) 2y � x, y � 3x, 3y + 2x � 24

6 x = curry puffs, y = doughnutsI : x + y � 120II : x � 2yIII : x – y � 40∴ x + y � 120, x � 2y, x – y � 40

7 I : x � yII : x � 2yIII : 8x + 5y � 5000∴ x � y, x � 2y, 8x + 5y � 5000

8 (a) 2 � x � 4(b) At point (3, 6), 3 + 2(6) = 15

9 (a) Maximum value of x = 10 Maximum value of y = 12(b) 2x + 3y = 6 At point (5, 2), 2(5) + 3(2) = 16 ∴ The minimum value of

2x + 3y is 16.10 (a) I : 60x + 30y � 50 × 60

2x + y � 100 II : 30x + 60y � 30 × 60 x + 2y � 60

III : y

—x � 2—1

y � 2x ∴ 2x + y � 100, x + 2y � 60, y � 2x

O 2

y

x

x = 2

O–1

y

x

x = –1

O3

y

xy = 3

O–3

y

xy = –3

O

5

5

y

xx + y = 5

O

(1, 2)

y

x

y = 2x

O

–6

6

y

xx – y = 6

O–2

1

y

x

2y = x + 2

O

–4

3

y

x

4x – 3y = 12

O

2

y

x

y = 2

O–4

y

x

x = –4

O

6

3

y

xy + 2x = 6

O

–4

8

y

xx – 2y = 8

O

(1, 3)

y

x

y = 3x

O

6

6

y

x

O

2

y

x

y = xy = 2

x + y = 2

O

3

y

x

y = 3

4y = 3x +

12

y = 3x – 4

O

y

xx = 1 y + 2x = 6

2y = x

O

y

xx + 2y = 4

y = 2x

2y + 3x = 12

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 157: Analysis Spm Additional Mathematics

2© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b)

(c) (i) 47 chairs (ii) 20 chairs

(d) 20x + 25y = k At point (25, 50), 20(25) + 25(50) = 1750 ∴ The maximum profit can

be obtained by the factory is RM1750.

SPM Appraisal Zone

1 (a) I : x + y � 150

II : y � 1—2

x

III : 400x + 200y � 80 000 2x + y � 400 ∴ x + y � 150, y �

1—2 x,

2x + y � 400(b)

(c) (i) When x = 100, y = 200 (ii) When y = x, the minimum

number printer A = 75 and printer B = 75.

(d) 120x + 60y = k At point (160, 80), 120(160) + 60(80) = 24 000 ∴ The maximum profi t = RM24 000

2 (a) I : x + y � 30 II : x – y � 20 III : 20x + 40y � 1600 x + 2y � 80 ∴ x + y � 30, x – y � 20, x + 2y � 80(b)

(c) (i) 3x + 2y = 6 (ii) At point (40, 20), 3(40) + 2(20) = 160 ∴ The maximum amount = RM160

3 (a) I : x + y � 140 II : y � 3x III : x – y � 40 ∴ x + y � 140, y � 3x, x – y � 40(b)

(c) (i) When y = 60, 20 � x � 80 (ii) 60x + 40y = 2400, x = 50. At point (50, 90), 60(50) + 40(90) = 6600 ∴ The maximum fees

collected = RM6600

4 (a) I : x � 2y

II : x—y

� 2—7

x � 2—7

y

III : 5x + 10y � 800 x + 2y � 160

∴ x � 2y, x � 2—7

y, x + 2y � 160(b)

(c) (i) Maximum point is (20, 70). Number of slippers = 20 Number of sandals = 70

(ii) 3x + 8y = k At point (20, 70), 3(20) + 8(70) = 620 ∴ The maximum amount

= RM620 5 (a) I : x + y � 80

II : y – 2x � 20 III : 5x + 10y � 400 x + 2y � 80 ∴ x + y � 80, y – 2x � 20, x + 2y � 80(b)

(c) (i) When x = 30, 25 � y � 50 (ii) 4x + 8y = k At point (20, 60), 4(20) + 8(60) = 560 ∴ The maximum profit

obtained = RM560

10

100

20

20

30

30

40

40

50

50

60

60

70

70

80

80

90

100

y

x

2x + y = 100

(25, 50)

x + 2y = 60

y = 2x

R

50

500

100

100

150

150

200

200

250

250

300

300

350

350

400

400

450

y

x

(160, 80)

x + y = 150

1y = — x 2

y = x

R

2x + y = 400

x = 100

10

100

20

20

30

30

40

40

50

50

60

60 70 803x + 2y = 6

x + y = 30

y

x

(40, 20)

x + 2y = 80

x – y = 20

R

200

20

40 60

40

80 100

60

120 140

80

160

100

120

140

160

60x + 40y = 2400

y

x

y = 3x

x = 50

R

x + y = 140

y = 60

x – y = 40

100

10

20 30

20

40 50

30

60 70

40

80

50

60

70

80

y

x

y – 2x = 20

(20, 60)

x = 30

R

x + 2y = 80

x + y = 80

200

20

40 60

40

80 100

60

120 140

80

160

100

120

140

y

x

(20, 70)

2x = — y 7

Rx + 2y = 160

x = 2y

Page 158: Analysis Spm Additional Mathematics

1

SPM-Cloned Questions

Chapter 12: Progressions 1 (a) 2m + 1 – m = 5m – 1 – (2m + 1)

m + 1 = 3m – 2 2m = 3 m = 3—

2 (b) The fi rst three terms are

3—2

, 4, 13—–2

, ...

So, a = 3—2

and d = 4 – 3—2

= 5—2

∴ S

11 =

11—–2

�2� 3—2 � + 10 � 5—

2 �� =

11—–2

(3 + 25) = 154

2 For –9, –5, –1, ... d = –5 – (–9) = 4 S

4 = 100

4—2

[2a + 3(4)] = 100 2(2a + 12) = 100 2a + 12 = 50 2a = 38 a = 19∴ The four consecutive terms which sum up to 100 are 19, 23, 27 and 31.

3 (a) d = 9 – 4 = 5(b) The next four terms are 19, 24, 29

and 34. So, the sum of these terms are

19 + 24 + 29 + 34 = 106.

4 For 18, 6, 2, ... a = 18 and r =

6—–18

= 1—3

∴ S∞ = 18———1 – 1—

3 = 18——

� 2—3 �

= 27

5 x—–16

= 16—––8

x = –2(16) = –32

6 (a) 8, 16, 24(b) d = 16 – 8 = 8

7 (a) T4 = 155

k + 3m = 155 … 1 S

8 = 1340

8—2

(2k +7m) = 1340

2k + 7m = 335 … 2 1 × 2 2k + 6m = 310 … 3 2 – 3 m = 25 Substitute m = 25 into 1 : k + 3(25) = 155 k = 155 – 75 = 80(b) 80 + (n – 1)25 = 120 + (n – 1)15 80 + 25n – 25 = 120 + 15n – 15 25n + 55 = 105 + 15n 10n = 50 n = 5

8 (a) T1 = 6� 2—

3 � = 4m T

2 = 4� 2—

3 � = 8—3

m This is in the form of a geometric

progression with a = 4 and r = 8—

3 ÷ 4

= 2—

3

∴ T10

= 4� 2—3 �

9

= 0.104 m(b) The total distance travelled = 6 + 2(4) + 2� 8—

3 � + ... = 6 + 8———

1 – 2—3

= 6 + 24 = 30 m

Chapter 13: Linear Law

1 y = p x – q

—–x

y

—–x

= –q� 1—x � + p

From the graph, –q = 4 – 1———

1—2

– 2 =

3——–

� 3– — 2 �

–q = –2 q = 2

y

—–x

= –2� 1—x � + p At (2, 1), 1 = –2(2) + p

p = 1 + 4 = 5∴ p = 5, q = 2

2 3y = 8x3 + x 3y

—–x

= 8x2 + 1 y—x

= 8—3

x2 + 1—3

At (h, 3), 3 = 8—3

h + 1—3

8—3

h = 8—3

h = 1 At (4, k), k = 8—

3(4) + 1—

3

= 33—–3

= 11∴ h = 1, k = 11

3 y = 2x2 – 4 y

—x2

= 2 – 4—x2

y

—x2

= 2 – 4� 1—x2 �

4 x + a—

x = by

bxy = x2 + a xy = 1—

bx2 + a—

b From the graph, 1—b

= 2——–12 – 6

= 1—3

b = 3 At (6, 0), 0 = 1—

3(6) + a—

3 a—

3 = –2

a = –6∴ a = –6, b = 3

5 (a) y = k4x

log10

y = log10

k + log10

4x

log10

y = log10

4(x) + log10

k(b) log

10 k = 2

k = 102

= 100

6 (a) x2 xy

1 15.5

4 20.0

9 27.6

16 38.0

25 51.5

36 67.8

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 159: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(b) y = hx + k—x xy = hx2 + k

From the graph,

(i) h = 38 – 14———–16 – 0

= 1.5 (ii) k = 14

7 (a) 1—x1—y

1.00 0.75

0.50 1.00

0.33 1.09

0.25 1.12

0.20 1.15

(b)

(c) a—y = – b—x + 4 1—y = – b—a � 1—x � + 4—a From the graph,

(i) 4—a = 1.26 a =

4——1.26

= 3.175

(ii) – b—a = 0.75 – 1.26—————

1.0 – 0 –

b——–3.175

= –0.51

b = 1.619

Chapter 14: Integration 1 Area = 9 unit2

�k

0x2 dx = 9

�x3

—3 �

k

0

= 9

k3

—3

= 9

k3 = 27 k = 3

2 �

3

1 f(x) dx + �

5

3[f(x) + 2] dx

= �3

1f(x) dx + �

5

3f(x) dx + �

5

32 dx

= �5

1f(x) dx + �2x�

5

3

= 8 + (10 – 6)= 12

3 (a) �

2

3 –2h(x) dx = –2 �

2

3h(x) dx

= –2(–4) = 8

(b) �3

2[6 – h(x)] dx = �

3

26 dx – �

3

2h(x) dx

= �6x�3

2 – 4

= (18 – 12) – 4 = 2

4 (a) �(3x2 – 5) dx

= x3 – 5x + c Compare with x3 – px + c So, p = 5

(b) �(3x2 – 5) dx = 4 x3 – 5x + c = 4 When x = 2, 23 – 5(2) + c = 4 8 – 10 + c = 4 c = 6

5 (a) At turning point (2, 5), k(2) – 4 = 0 2k = 4 k = 2 (b)

dy—–dx

= 2x – 4 y = �(2x – 4) dx

= x2 – 4x + c Since (2, 5) lies on the curve, 5 = 22 – 4(2) + c 5 = 4 – 8 + c c = 9

∴ The equation of the curve is y = x2 – 4x + 9.

6 �5

hf(y) dy = 3(5) – 11

= 15 – 11 = 4 unit2

7 (a) y = k(x + 1)3

dy—–dx

= 3k(x + 1)2

At x = –2,

dy—–dx

= 6

3k(–2 + 1)2 = 6 3k = 6 k = 2(b) (i) Area of the shaded region, P

= �–1

–2 2(x + 1)3 dx

= (x + 1)4�———�

–1

–2 2

= 0 – 1—2

= – 1—2

= 1—2

unit2

(ii) Volume generated = π�

0

–14(x + 1)6 dx

= π 4(x + 1)7�————�

0

–1 7 = 4—

7 π unit3

8 (a) y = x2 + 4 dy

—–dx

= 2x At A(–1, 5), dy

—–dx

= 2(–1) = –2 ∴ Equation of the tangent at A is y – 5 = –2 (x + 1) y – 5 = –2x – 2 y = –2x + 3(b) Area under the curve = �

0

–1(x2 + 4) dx

= �x3

—3

+ 4x�0

–1

= 0 – �– 1—3

– 4� = 13—–

3 unit2

Area of trapezium

= 1—2

(1)(3 + 5)

= 4 unit2

∴ Area of the shaded region

= 13—–3

– 4

= 1—3

unit2

(c) Volume generated

= π�6

4(y – 4) dy

= π�y2

—2

– 4y�6

4 = π[18 – 24 – (8 – 16)] = 2π unit3

Chapter 15: Vector

1 (a) →OP = �2

3 �(b)

→PQ =

→PO +

→OQ

= –2 i~ – 3 j~

+ 4 i~ + 4 j~

= 2 i~ + j~

2 →PQ =

→PO +

→OQ

= 8 i~ – 4 j~

+ 5 i~ + 3 j~

= 13 i~ – j~

3 →OD = 3—

4

→OB

= 3—4

(8x~ + 4y~

)

= 6x~ + 3y~

xy

x2

70

60

50

40

30

20

10

0 5 10 15 20 25 30 35 40

(0, 14)

(16, 38)

1—y(0, 1.26)

1.2

1.0

0.8

0.6

0.4

0.2

00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1—x

(1.0, 0.75)

Page 160: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

4 (a) a~ – b~ = 12 i~ + j~

– (9 i~ – k j~

) = 3 i~ + (1 + k) j

~(b) �a~ – b~� = 5

32 + (1 + k)2 = 5 9 + 1 + 2k + k2 = 25 k2 + 2k – 15 = 0 (k + 5)(k – 3) = 0 k = –5 or k = 3

5 (a) →QR =

→QP +

→PR

= –2a~ + 5b~ (b)

→QS =

1—4

→QR

=

1—4

(–2a~ + 5b~)

= – 1—2

a~ + 5—4

b~

→PS =

→PQ +

→QS

= 2a + �– 1—2

a~ + 5—4

b~� =

3—2

a~ + 5—4

b~

6 (a) (i)

→OP =

1—3

→OA

= 1—3

(6x~)

= 2x~

→BP =

→BO +

→OP

= 2x~ – 4y~

(ii) →AQ =

1—2

→AB

= 1—2

(–6x~ + 4y~

)

= –3x~ + 2y~

→OQ =

→OQ +

→AQ

= 6x~ + (–3x~ + 2y~

) = 3x~ + 2y

~

(b) →OR =

→OB +

→BR

h(3x~ + 2y~

) = 4y~

+ k(2x~ – 4y~

) 3hx~ + 2hy

~ = 2kx~ + (4 – 4k)y

~ Equating the coeffi cients, we have: 3h – 2k = 0 … 1 2h = 4 – 4k 2h + 4k = 4 h + 2k = 2 … 2 1 + 2 : 4h = 2 h = 1—

2 Substitute h = 1—

2 into 1 :

3� 1—2 � – 2k = 0

2k = 3—2

k = 3—4

∴ h = 1—2

, k = 3—4

(c) � →AB � = 122 + 42

= 160

= 12.649

7 (a) (i) →AP =

1—3

→AD

→AD = 3

→AP

= 3y~

→DB =

→DA +

→AB

= –3y~

+ x~ = x~ – 3y

~

(ii) →BR =

1—3

→BD

= 1—3

(–x~ + 3y~

)

= – 1—3

x~ + y~

→AR =

→AB +

→BR

= x~ + �– 1—3

x~ + y~�

= 2—3

x~ + y~

(b) →AR = h

→AC

= h(→AD +

→DC)

= h�3y~

+ kx~ – 3—2

y~�

= hkx~ + 3—2

hy~

2—3

x~ + y~

= hkx~ + 3—2

hy~

Equating the coeffi cients,

3—2

h = 1 and hk = 2—3

h = 2—3

2—3

k = 2—3

k = 1 ∴ h =

2—3

, k = 1

8 (a) (i) →AP =

→AO +

→OP

= 2p – 2q

(ii) →PC =

1—3

→PQ

= 1—3

(–2p + 4q)

= – 2—3

p + 4—3

q

→OC =

→OP +

→PC

= 2p – 2—3

p + 4—3

q

= 4—3

p + 4—3

q

(b) →AB =

→AO +

→OB

h(2p~

– 2q~

) = –2q~

+ k� 4—3

p~

+ 4—3

q~�

2hp~

– 2hq~

= 4—3

kp~

+ � 4—3

k – 2�q~

Equating the coeffi cients,

2h = 4—3

k

6h – 4k = 0 3h – 2k = 0 … 1

–2h = 4—3

k – 2

–6h = 4k – 6 6h + 4k = 6 3h + 2k = 3 … 2

1 + 2 : 6h = 3

h = 1—2

Substitute h = 1—2

into 1 :

3� 1—2 � – 2k = 0

2k = 3—2

= 3—4

∴ h = 1—2

, k = 3—4

Chapter 16: Trigonometry 1

(a) sec θ = 1——–

cos θ = 1————–

1 + m2

m�———–� =

1 + m2

———–m

(b) cos (90° – θ) = sin θ

= 1 + m2

1———–

2 2 cos 2x + 4 sin x + 1 = 0 2(1 – 2 sin2 x) + 4 sin x + 1 = 0 2 – 4 sin2 x + 4 sin x + 1 = 0 4 sin2 x – 4 sin x – 3 = 0 (2 sin x + 1)(2 sin x – 3) = 0 2 sin x + 1 = 0 sin x = – 1—

2 x = 210°, 330° or sin x = 3—

2 (no solution)

∴ x = 210°, 330°

3 3 sin2 x – 5 cos x = 5 3(1 – cos2 x) – 5 cos x = 5 3 – 3 cos3 x – 5 cos x = 5 3 cos2 x + 5 cos x + 2 = 0 (3 cos x + 2)(cos x + 1) = 0 3 cos x + 2 = 0

cos x = – 2—3

x = (180° – 48° 11'), (180° + 48° 11') = 131° 49', 228° 11'or cos x + 1 = 0 cos x = –1 x = 180°∴ x = 131° 49', 180°, 228° 11'

4

m

1

θ

1 + m2

pA

1 – p21

Page 161: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

(a) sin 2A = 2 sin A cos A = 2p 1 – p2

(b) cos A = 2 cos2 A—2

– 1

p = 2 cos2 A—2

– 1

p + 1 = 2 cos2 A—2

cos2 A—2

= p + 1——–

2

cos A—2

= p + 1——–

2

5

(a) tan A = – 3—4

(b) cos (A + B) = cos A cos B – sin A sin B

= �– 4—5 �� 5—–

13 � – � 3—5 �� 12—–

13 � = – 20—–

65 – 36—–

65

= – 56—–65

6 (a),(b)

2 sin2 x = x——180°

1 – 2 sin2 x = 1 – x——180°

cos 2x = 1 – x——180°

y = 1 – x——180°

x 0 180°

y 1 0

Number of solutions = 2

7 (a) cos4 x – sin4 x = (cos2 x + sin2 x)(cos2 x – sin2 x) = cos2 x – sin2 x = 1 – sin2 x – sin2 x = 1 – 2 sin2 x = cos 2x

(b) (i),(ii)

2(cos4 x – sin4 x) =

x—π – 1 2 cos 2x = x—

π – 1

cos 2x = x—–

2π – 1—

2

y = x—–2π

– 1—2

x 0 π

y – 1—2

0

Number of solutions = 2

8 (a),(b)

2x—–3π

– sin 2x = 1

sin 2x = 2x—–3π

– 1

3—2

sin 2x = x—π – 3—2

y = x—π – 3—2

x 0 π

y – 3—2

– 1—2

Number of solutions = 3

Chapter 17: Permutations and Combinations

1 4P3 × 5P

4 = 24 × 120

= 2880

2 (a) 6C4 × 5C

2 = 15 × 10

= 150(b) (6C

2 × 5C

4) + (6C

1 × 5C

5)

= 75 + 6 = 81

3 (a) 9! = 362 880

(b) 5 7 6 5 4 3 2 1 4 V C = 5 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 4 = 100 800

4 (a) 18C11

= 31 824(b) 3C

1 × 5C

3 × 4C

4 × 6C

3

= 600

5 (a) 5 4 3 = 5 × 4 × 3 = 60(b) 4 3 4 = 4 × 3 × 4 C = 48

6 (a) 5C4 × 4C

3 = 5 × 4

= 20(b) 4! × 4! = 576

7 (a) 6 5 4 3 = 6 × 5 × 4 × 3 = 360(b) 5 4 3 4 = 5 × 4 × 3 × 4 O = 240

8 (a) 9! = 362 880(b) 6! × 4! = 17 280

Chapter 18: Probability 1 Area of big circle = π(6)2

= 36π cm2

Area of small circle = π(3)2

= 9π cm2

(a) Probability of hitting the unshaded area

area of small circle = ———————— area of big circle

= 9π——36π

= 1—4

(b) Area of shaded area = (36π – 9π) cm2

= 27π cm2

Probability of hitting the shaded area

Area of the shaded area = —————————— Area of big circle

= 27π——36π

= 3—4

2 The tree diagram is drawn and shown as follows:

1st set 2nd set 3rd set

y

5

x

3

O

A

y

x

1312

OB

x

y

y = cos 2x

45° 90° 135° 180°

1

0

–1

y = 1 – x——180°

yy = cos 2x

π—4

3—4

π

1

0

–1

π—2

xπy = x—–

2π – 1—

2

2—5

3—5

Win

Not win

2—5

3—5

2—5

3—5

Win

Not win

Win

Not win

Win

2—5

Win

2—5

0xπ—

4π—2

3—4

π 5—4

ππ

y = x—π – 3—2

3—2

y

3—2

y = 3—2

sin 2x

3—2

π

Page 162: Analysis Spm Additional Mathematics

5© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

From the tree diagram,(a) P(the match ends in two sets

only)

= � 2—5

× 2—5 � + � 3—

5 × 3—

5 � = 4—–

25 + 9—–

25

= 13—–25

(b) P(Pawaz wins after competing three sets)

= � 2—5

× 3—5

× 2—5 � + � 3—

5 × 2—

5 × 2—

5 � = 12—–

125 + 12—–

125

= 24—–125

3 We defi ne the following events as follows:A : getting the same number on the two

spinners.B : the fi rst spinner shows the larger

numbers.

The possibility diagram is shown below:

From the possibility diagram,(a) The dots enclosed in the loop above

represent the possible outcomes in event A.

n(A) = 4.

∴ P(A) = n(A)——n(S)

= 4—–16

= 1—4

(b) The triangle in the possibility diagram above contains the dots representing the outcomes in event B.

n(B) = 6 ∴ P(B) = n(B)——

n(S)

= 6—–16

= 3—8

4 P(G) = 5—

8 x—–

32 = 5—

8

x = 5—8

× 32

= 20

The number of marbles which are not green = 32 – 20 = 12

5 P(two chips of the same colour)= P(RR) + P(BB)

= � 4—7 �� 3—

5 � + � 3—7 �� 2—

5 �= 12—–

35 + 6—–

35

= 18—–25

6 (a) P(Mukhriz or Haziq wins) = P(Mukhriz wins) + P(Haziq wins)

= 1—3

+ 1—4

= 7—–12

(b) P(neither Mukhriz nor Haziq wins) = 1 – P(Mukhriz or Haziq wins)

= 1 – 7—–12

= 5—–12

7 (a) P(neither of them is chosen) = 2—

5 × 7—–

15

= 7—–25

(b) P(one of them is chosen)

= � 2—5

× 7—–15 � + � 3—

5 × 8—–

15 � = 14—–

75 + 24—–

75

= 38—–75

8 (a) P(all three are prefects) = 0.4 × 0.4 × 0.4 = 0.064(b) P(only one of them is prefect) = (0.4 × 0.6 × 0.6) + (0.6 × 0.4

× 0.6) + (0.6 × 0.6 × 0.4) = 0.432

Chapter 19: Probability Distributions

1 (a) X ~ B(10, 0.6) P(x � 8) = P(X = 8) + P(X = 9) + P(X = 10) = 10C

8(0.6)8(0.4)2 + 10C

9(0.6)9(0.4)

+ 10C10

(0.6)10(0.4)0

= 0.12093 + 0.04031 + 0.00605 = 0.1673(b) X ~ N(55, 52)

(i) z = 62 – 55———–

5 = 1.4 (ii) z = –1.4

x – 55———5

= –1.4

x = 55 – 7 = 48 kg

(iii) P(46 � X � 58)

= P� 46 – 55———–5

� Z � 58 – 55———–5 �

= P(–1.8 � Z � 0.6) = 1 – 0.0359 – 0.2743 = 0.6898

2 (a) X ~ B(8, 0.1) (i) P(X = 3) = 8C

3(0.1)3(0.9)5

= 0.0331 (ii) P(X � 3) = P(X = 0) + P(X = 1) + P(X = 2) = 8C

0(0.1)0(0.9)8 + 8C

1(0.1)

(0.9)7 + 8C2(0.1)2(0.9)6

= 0.43047 + 0.38264 + 0.1488 = 0.9619

(b) X ~ N(600, 202) (i) P(X � 575)

= P�Z � 575 – 600—–———20 �

= P(Z � –1.25) = 0.1056 (ii) P(570 � X � 610)

= P�570 – 600————20

� Z � 610 – 600————20 �

= P(–1.5 � Z � 0.5) = 1 – 0.0668 – 0.3085 = 0.6247

(c) The number of cabbages in the market

= 50———0.6247

= 80

3 P(0 � Z � k) = 0.5 – 0.2842 = 0.2158

4 X ~ B(8, 0.3)P(X = 3) = 8C

3 (0.3)3(0.7)5

= 0.2541

5 (a) z = 1.5

x – 220———–20

= 1.5

x – 220 = 30 x = 250

(b) P(X � 210) = P�Z � 210 – 220————–

20 � = P(Z � –0.5) = 1 – P(Z � –0.5) = 1 – 0.3085 = 0.6915

2nd spinner

1st spinner

4

3

2

1

1 2 3 4

f(z)

z–1.8 0.6O

f(z)

z–1.5 0.5O

Page 163: Analysis Spm Additional Mathematics

6 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

6 (a) (i) P(X � 2) = 1 – P(X � 2) = 1 – P(X = 0) – P(X = 1) = 1 – 10C

0 (0.05)0(0.95)10

– 10C1 (0.05)(0.95)9

= 1 – 0.59874 – 0.3151 = 0.0862 (ii) npq = 5.7 n(0.05)(0.95) = 5.7

n = 5.7———0.0475

= 120(b) X ~ N(62, 82) P(55 � x � 72)

= P� 55 – 62———–8

� Z � 72 – 62———–8 �

= P(–0.875 � Z � 1.25) = 1 – P(Z � –0.875) – P(Z � 1.25) = 1 – 0.1908 – 0.1056 = 0.7036 ∴ The total number of workers

= 76———0.7036

= 108

7 (a) X ~ B�6, 3—5 �

(i) P(X = 4) = 6C4 � 3—

5 �4

� 2—5 �

2

= 0.3110 (ii) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 6C0 � 3—

5 �0

� 2—5 �

6

– 6C1 � 3—

5 �� 2—5 �

5

= 1 – 0.004096 – 0.036864 = 0.9590

(b) X ~ N(70, 252) (i) P(65 � X � 85)

= P�65 – 70———25

� Z � 85 – 70———25 �

= P(–0.2 � Z � 0.6) = 1 – P(Z � –0.2) – P(Z � 0.6) = 1 – 0.4207 – 0.2743 = 0.3050 (ii) P(X � 100)

= P�Z � 100 – 70———–20 �

= P(Z � 1.5) = 0.0668 The total number of workers

= 25———0.0668

= 374

8 (a) (i) E(X) = np

= 20� 1—4 �

= 5 (ii) σ = npq

= 5� 3—4 �

= 3.75 = 1.9365

(b) (i) P(X = 3) = 8C3 � 1—

4 �3

� 3—4 �

5

= 0.2076 (ii) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 8C0�1—

4 �0

�3—4 �

8

– 8C1�1—

4 ��3—4 �

7

= 1 – 0.1001129 – 0.2669678 = 0.6329

Chapter 20: Motion along a Straight Line

1 (a) a = dv—–dt

= 6t – 8 When the particle is at rest, v = 0 3t2 – 8t – 3 = 0 (3t + 1)(t – 3) = 0 Since t � 0, t = 3 ⇒ a = 6(3) – 8 = 18 – 8 = 10 m s–2

(b) s = ∫∫v dt = t3 – 4t2 – 3t + c When t = 0, s = 0 and so c = 0 Hence, s = t3 – 4t2 – 3t When t = 2, s = (2)3 – 4(2)2 – 3(2) = 8 – 16 – 6 = –14 When t = 3, s = (3)3 – 4(3)2 – 3(3) = 27 – 36 – 9 = –18 Thus, the distance of AB is = 18 – 14 = 4 m

2 v = ds—–dt

= 3pt2 + 2qt + 4

a = dv—–dt

= 6pt + 2qWhen t = 2, a = 0 6p(2) + 2q = 0 12p + 2q = 0 6p + q = 0 … 1and v = –8 3p(2)2 + 2q(2) + 4 = –8 12p + 4q = –12 3p + q = –3 … 21 – 2 : 3p = 3 p = 1

Substitute p = 1 into 1 : 6(1) + q = 0 q = –6∴ p = 1, q = –6

3 (a) v = ∫a dt = 12t – 3t2 + c When t = 0, v = 15 and so c = 15 Hence, v = 12t – 3t2 + 15(b) s = ∫v dt = 6t2 – t3 + 15t + c When t = 0, s = 0 and so c = 0 Hence, s = 6t2 – t3 + 15t When t = 5, s = 6(5)2 – 53 + 15(5) = 150 – 125 + 75 = 100 Thus, the distance travelled in the

fi rst 5 seconds is 100 m.

4 (a) (i) a = 6 – 2t v = ∫a dt = ∫(6 – 2t) dt = 6t – t2 + c When t = 0, v = 16 and so c = 16 Hence, v = 6t – t2 + 16 For maximum velocity, a = 0 6 – 2t = 0 t = 3

Moreover, da—–dt

= –2 � 0, v is

maximum when t = 3. Thus, maximum velocity = 6(3) – 32 + 16 = 18 – 9 + 16 = 25 m s–1

(ii) When the particle stops, v = 0 6t – t2 + 16 = 0 t2 – 6t – 16 = 0 (t – 8)(t + 2) = 0 Since t � 0, t = 8 ∴ k = 8

(b)

Total distance travelled

= �8

0(6t – t2 + 16) dt

= �3t2 – 1—3

t3 + 16t�8

0

= 3(64) – 1—3

(512) + 16(8)

= 149 1—3

m

f(z)

z–0.5 O

f(z)

z–0.875 1.25O

v

25

16

0 3 8t

v = 6t – t2 + 16

Page 164: Analysis Spm Additional Mathematics

7© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

5 (a) v = 2t (4 – t) = 8t – 2t2

a = dv—–dt

= 8 – 4t For maximum velocity, a = 0 8 – 4t = 0 t = 2

Moreover, da—–dt

= –4 � 0 ⇒ v is

maximum when t = 2. Thus, maximum velocity = 8(2) – 2(2)2

= 16 – 8 = 8 m s–1

(b) s = ∫v dt

= 4t2 – 2—3

t3 + c

When t = 0, s = 0 and so c = 0

Hence, s = 4t2 – 2—3

t3

s4 = 4(4)2 – 2—

3(4)3

= 64 – 128—–3

= 21 1—3

s3 = 4(3)2 – 2—

3 (3)3

= 36 – 18 = 18 m ∴ The distance travelled in the 4th

second

= 21 1—3

– 18

= 3 1—3

(c) When the particle passes O again, s = 0

4t2 – 2—3

t3 = 0

2t2�2 – 1—3

t� = 0

t = 0 or = 6 ∴ t = 6 s(d) When the particle reverses its

direction of motion v = 0 8t – 2t2 = 0 2t (4 – t) = 0 t = 0 or t = 4 ∴ t = 4 s

6 (a) v = t2 – 8t + 12 When t = 0, v = 02 – 8(0) + 12 = 12 Therefore, the initial velocity of

the particle is 12 m s–1.

(b) a = dv—–dt

= 2t – 8 For minimum velocity, a = 0 2t – 8 = 0 t = 4

Moreover, da—–dt

= 2 � 0, v is

minimum when t = 4. Thus, minimum velocity = 42 – 8(4) + 12 = 16 – 32 + 12 = –4 m s–1

(c) When the particle moves to the left,

v � 0 t2 – 8t + 12 � 0 (t – 2)(t – 6) � 0

∴ 2 � t � 6(d) s = ∫v dt

= ∫(t2 – 8t + 12) dt

= 1—3

t3 – 4t2 + 12t + c When t = 0, s = 0 and so c = 0

Hence, s = 1—3

t3 – 4t2 + 12t

s2 = 1—

3(23) – 4(2)2 + 12(2)

= 8—3

– 16 + 24

= 10 2—3

m

s6 = 1—

3(6)3 – 4(6)2 + 12(6)

= 72 – 144 + 72 = 0 m The motion of the particle is shown

as follows:

∴ The total distance travelled in the fi rst 6 seconds

= 2�10 2—3 �

= 21 1—3

m

7 (a) v = 8 + 2t – t2

a = dv—–dt

= 2 – 2t When the particle is at rest, v = 0 8 + 2t – t2 = 0 t2 – 2t – 8 = 0 (t – 4)(t + 2) = 0 Since t � 0, t = 4 ∴ The acceleration of the particle

at R = 2 – 2(4) = 2 – 8 = –6 m s–2

(b) For maximum velocity, a = 0 2 – 2t = 0 t = 1

Moreover, da—–dt

= –2 � 0 ⇒ v is

maximum when t = 1. Thus, maximum velocity = 8 + 2(1) – (1)2

= 9 m s–1

(c) s = ∫v dt = ∫(8 + 2t – t2) dt

= 8t + t – 1—3

t3 + c When t = 0, s = 0 and so c = 0

Hence, s = 8t + t – 1—3

t3

s4 = 8(4) + 4 – 1—

3 (4)3

= 36 – 64—–3

= 14 2—3

m

s6 = 8(6) + 6 – 1—

3(6)3

= 54 – 72 = –18 m The motion of the particle is shown

as follows:

∴ The total distance travelled in the fi rst 6 seconds

= 2�14 2—3 � + 18

= 47 1—3

m

8 (a) v = 3t – 1—2

t2 + 8

When t = 0, v = 3(0) – 1—2

(0)2 + 8 = 8 Therefore, the initial velocity of

the particle is 8 m s–1.

(b) a = dv—–dt

= 3 – t When t = 0, a = 3 Therefore, the initial acceleration

of the particle is 3 m s–2.(c) For maximum velocity, a = 0 3 – t = 0 t = 3

Moreover, da—–dt

= –1 � 0 ⇒ v is

maximum when t = 3. Thus, the maximum velocity

= 3(3) – 1—2

(3)2 + 8

= 12 1—2

m s–1

(d) When the particle stops instantaneously, v = 0.

3t – 1—2

t2 + 8 = 0

2 6t

t = 0s = 0

t = 6s = 0

t = 2s = 10 2—

3

OS

t = 0s = 0

t = 6s = –18

t = 4s = 14 2—

3

O

Page 165: Analysis Spm Additional Mathematics

8 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

1—2

t2 – 3t – 8 = 0

t2 – 6t – 16 = 0 (t – 8)(t + 2) Since t � 0, t = 8 s = ∫v dt

= ∫∫�3t – 1—2

t2 + 8� dt

= 3—2

t2 – 1—6

t3 + 8t + c

When t = 0, s = 0 and so c = 0

Hence, s = 3—2

t2 – 1—6

t3 + 8t

s8 = 3—

2(8)2 – 1—

6(8)3 + 8(8)

= 96 – 256—–3

+ 64

= 74 2—3

m

Chapter 21: Linear Programming

1 (a) I: (0, 0) and (2, 6)

m = 6—2

= 3 The equation is y – 6 = 3(x – 2) y – 6 = 3x – 6 y = 3x ∴ y � 3x II: (0, 0) and (5, 3)

m = 3—5

The equation is

y – 3 = 3—5

(x – 5)

y – 3 = 3—5

x – 3

y = 3—5

x

∴ y � 3—5

x III: (2, 6) and (5, 3)

m = 3 – 6——–5 – 2

= –1 The equation is y – 6 = –1 (x – 2) y – 6 = –x + 2 y + x = 8 ∴ y + x � 8(b) The maximum value is at point (5, 3). Thus, the maximum value of 2x + y = 2(5) + 3 = 13

2 (a) I: (4, 0) and (6, 4)

m = 4—2

= 2 The equation is y – 4 = 2(x – 6) y – 4 = 2x – 12

y = 2x – 8 ∴ y � 2x – 8

II: (0, 8) and (6, 4)

m = – 4—6

= – 2—3

The equation is

y – 8 = – 2—3

(x – 0)

3y – 24 = –2x 3y + 2x = 24 ∴ 3y + 2x � 24

III: x � 2(b) The number of solutions is 17.

3 (a) I: A(2, 0) and E(5, 12)

m = 12—–3

= 4 The equation is y = 4(x – 2) y = 4x – 8 ∴ y � 4x – 8 y – 4x � –8

II: B(6, 0) and D(0, 8)

m = – 8—6

= – 4—3

The equation is

y = – 4—3

(x – 6)

y = – 4—3

x + 8

∴ y � – 4—3

x + 8

3y + 4x � 24

III: D(0, 8) and E(5, 12)

m = 4—5

The equation is

y – 8 = 4—5

(x – 0)

5y – 40 = 4x 5y – 4x = 40 ∴ 5y – 4x � 40

(b) y – 4x = –8 … 1 3y + 4x = 24 … 2

1 + 2 : 4y = 16 y = 4 Substitute y = 4 into 1 : 4 – 4x = –8 4x = 12 x = 3 ∴ C(3, 4) Thus, the value of z x + 2y = 3 + 2(4) = 11

4 (a) I: x + y � 80 II: y � 5x

III: y � 1—2

x + 5

(b)

(c) (i) 15 � y � 60 (ii) The maximum total fees = 60x + 40y = 60(50) + 40(30) = 3000 + 1200 = RM4200

5 (a) I: 45x + 30y � 600 3x + 2y � 40

II: 50x + 70y � 350 5x + 7y � 35

III: x—y

� 4—5

y � 5—4

x

(b)

(c) (i) 14 shirts (ii) Maximum profi t = 16x + 8y = 16(7) + 8(9) = 112 + 72 = RM184

6 (a) I: x + y � 400 II: x � 2y

y � 1—2

x

III: 8x + 4y � 1200 2x + y � 300

y

80

70

60

50

40

30

2015

10

y = 5x

0 10 20 30 40 50 60 70 80

R

k = 60x + 40y

(50, 30)

y = 1—2

x + 5

x

y

20

18

16

14

12

10

8

6

4

2

0 1 2 3 4 5 6 7 8x

R

3x + 2y = 40

(7.25, 9.1)

y = 5—4

x

k = 16x + 8y 5x + 7y = 35

Page 166: Analysis Spm Additional Mathematics

9© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b)

(c) (i) 250 tubes (ii) Maximum profi t = 8x + 4y = 8(267) + 4(134) = RM2672

7 (a) I: x + y � 7 II: x � 3y

y � 1—3

x

III: 400x + 200y � 2000 2x + y � 10

(b)

(c) (i) 1 van (ii) Maximum number of students

that can be accommodated = 40x + 8y = 40(4) + 8(1) = 168

8 (a) I: x + y � 160 II: x � 3y

y � 1—3

x

III: y – x � 80(b)

(c) (i) 120 (ii) Maximum cost = 0.5x + 0.8y = 0.5(40) + 0.8(120) = 20 + 96 = RM116

y

R

x

2x + y = 10

x + y = 7

(4.3, 1.4)

10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7

y = 1—3

x

y = 1—3

x

y

R

x

k = 0.5x + 0.8y

160

140

120

100

80

60

40

20

0 20 40 60 80 100 120 140 160

(40, 120)

y – x = 80x + y = 160

y

400

350

300

250

200

150

100

50

0 50 100 150 200 250 300 350 400

x + y = 400

(267, 134)R

x

2x + y = 300

y = 1—2

x

Page 167: Analysis Spm Additional Mathematics

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Paper 1 1 (a) (i) k = 2

(ii) 1(b) One to one function

2 (a) h–1(3) = 6 – 3——–3

= 1(b) h–1(y) = x

6 – y——–3

= x

6 – y = 3x y = 6 – 3x h(x) = 6 – 3x ∴ h� 1—

3 � = 6 – 3� 1—3 �

= 5

3 f(x) = 5 – 6—x

y = 5 – 6—x

6—x = 5 – y x = 6——–

5 – y f –1(x) = 6——–

5 – x f(x) = f –1(x)

5 – 6—x = 6——–

5 – x

5x – 6———x

= 6——–5 – x

(5x – 6)(5 – x) = 6x 31x – 5x2 – 30 = 6x 5x2 – 25x + 30 = 0 x2 – 5x + 6 = 0 (x – 2)(x – 3) = 0 x = 2 or x = 3

4 4x2 – (m + 2)x + m – 1 = 0 For equal roots, b2 – 4ac = 0 (–m – 2)2 – 4(4)(m – 1) = 0 m2 + 4m + 4 – 16 m + 16 = 0 m2 – 12m + 20 = 0 (m – 2)(m – 10) = 0 m = 2 or m = 10

5 (a) (i) p = –2 (ii) q = –9 (iii) f(x) = a(x – 2)2 – 9 At (0, –5), –5 = 4a – 9 4a = 4 a = 1

(b) f(x) = (x + 2)2 – 9

6 2(x2 – 2) � 7x 2x2 – 7x – 4 � 0 (2x + 1)(x – 4) � 0

∴ x � – 1—

2 or x � 4

7

�4k4�3—2

———–�27k –3�–

2—3

= 8k6

——1—9

k2

= (8)(9)(k4) = 2332k4

Compare with 2x3yk2, we havex = 3, y = 2, z = 4.

8 log2x – 2 = log

4(x – 4)

log2x –

log2(x – 4)

—————log

24

= 2

log2x – 1—

2 log

2(x – 4) = 2

2 log2x – log

2(x – 4) = 4

log2 x2�——–�x – 4

= 4 x2

——–x – 4

= 16

x2 = 16(x – 4) x2 – 16x + 64 = 0 (x – 8)2 = 0 x = 8

9 (a) 7 – (m + 1) = 2m + 1 – 7 6 – m = 2m – 6 3m = 12 m = 4(b) The fi rst three terms are 5, 7 and

9 with a = 5 and d = 2. ∴ The sum of the next three terms = 11 + 13 + 15 = 39

10 (a) T3 = S

3 – S

2

= 3[2(3) – 3] – 2 [2(2) – 3] = 9 – 2 = 7(b) T

1 = 1[2(1) – 3]

= –1 T

2 = S

2 – S

1

= 2 – (–1) = 3 ∴ d = T

2 – T

1

= 3 – (–1) = 4

11 (a) T1 = 64(3)

= 192 T

2 = 32(3)

= 96

r = T

2—–T

1 = 96—–

192

= 1—2

(b) S∞ = a——1 – r

= 192———1 – 1—

2

= 384

12 xy = p + qx y = p� 1—x � + q From the diagram, p = 8 – 2———–

2 – (–2) = 3—

2

At the point (–2, 2), 2 = 3—

2 (–2) + q

2 = –3 + qq = 5

13 Area of ∆ABC –2 2 4 –21—2 | | –1 –3 3 –1

= 1—2

[6 + 6 – 4 – (–2 – 12 – 6)]

= 1—2

(28)

= 14 unit2

14 2y = x + 6

y = 1—2

x + 3

mPQ

= –2

The equation of PQ: y – 2 = –2 (x – 2) y – 2 = –2x + 4 y + 2x = 6At x-axis, y = 0 So, 2x = 6 x = 3∴ P(3, 0)At y-axis, x = 0 So, y = 6∴ Q(0, 6)

1– — 2

4x

1© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

Page 168: Analysis Spm Additional Mathematics

2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

m(0) + 3n 6m + n(0)�————– , ————–� m + n m + n = (2, 2)

So 3n——–m + n

= 2

3n = 2m + 2n 2m = n

m—n

= 1—2

∴ PR : RQ = 1 : 2

15

16 a = kb ~ ~

�–86 � = k� m

3 � So, we have, 3k = 6 and km = –8 k = 2 2m = –8 m = –4∴ m = –4

17 a = 4, b = 3, c = 2

18 �k

l � 4—

x2 � dx = 2 � 4–— x �k

1 = 2

4–— k

– (–4) = 2

4—k

= 2 k = 2

19 (a) ~x =

35—–7

= 5 σ = 287—–

7 – 52

= 16 = 4(b) New σ = 4k

20 (a) 1—2

(10)2θ = 60 50θ = 60 θ = 60—–

50 = 1.2 radians(b) ∠AOB = 3.142 – 1.2 = 1.942 radians s

AB = 10(1.942)

= 19.42 cm Perimeter of the sector AOB = 2(10) + 19.42 = 39.42 cm

21 y = 2x – 3———x + 5

dy—dx

= 2(x + 5) – (2x – 3)————————(x + 5)2

= 13———(x + 5)2

Compare with h———(x + 5)2

, we have

h = 13

22 v = 4—3

πr3

dv

—–dr

= 4πr2

dv

—–dt

= dv

—–dr

× dr

—–dt

12 = 4πr2 dr

—–dt

When surface area is 25π cm2

4πr2 = 25π

r2 = 25—–4

r = 5—2

(r � 0)

So, 12 = 4π � 5—2 �

2

dr

—–dt

12 = 4π � 25—–4 � dr

—–dt

dr

—–dt

= 12——25π

cm s–1

23 (a) 8 7 6 5 4 = 8 × 7 × 6 × 5 × 4 = 6720(b) 2! 6P

3 × 4 = 960

24 P(blue marble) = 1—

5 x———

2x + 15 = 1—

5 5x = 2x + 15 3x = 15 x = 5Total number of marbles= 8 + x + x + 7= 15 + 2x= 15 + 2(5)= 25

25 (a)

From the table, a = 0.7(b) z = 0.7 x – 55———

10 = 0.7

x – 55 = 7 x = 62

Paper 2Section A 1 y = x – 4 … 1

y2 = 2x2 – 17 … 2Substitute 1 into 2 : (x – 4)2 = 2x2 – 17 x2 – 8x + 16 = 2x2 – 17

x2 + 8x – 33 = 0 (x + 11)(x – 3) = 0x = –11 or x = 3

Substitute x = –11 into 1 : y = –11 – 4 = –15

Substitute x = 3 into 1 : y = 3 – 4 = –1∴ x = –11, y = –15; x = 3, y = –1

2 (a)

dy—–dx

= 3 4x – 5 = 3 4x = 8 x = 2 When x = 2, y = 3(2) – 1 = 5 ∴ R(2, 5) (b) dy

—–dx

= 4x – 5 y = ∫(4x – 5)dx = 2x2 – 5x + c When x = 2 and y = 5, 5 = 2(2)2 – 5(2) + c 5 = 8 – 10 + c c = 7 ∴ y = 2x2 – 5x + 7

3 (a), (b)

4π = 3x———

|cos 2x| |cos 2x| = 3x—–

4π y =

3x—–4π

x 0 π

y 03—4

∴ Number of solutions is 4.

4 (a) 4x – 12 = 2x – 8 2x = 4 x = 2 When x = 2, y = 4(2) – 12 = 8 – 12 = –4 ∴ P(2, –4) Equation of AB: y – (–4) = 1(x – 2) y + 4 = x – 2 y = x – 6 At x-axis, y = 0 0 = x – 6 x = 6 ∴ A(6, 0)

b –2a

f(x)

a zO

0.242

0.258

yy = |cos 2x|

π—4

y = 3x—–4π

3—π 4

–1

1

π—2

x

Page 169: Analysis Spm Additional Mathematics

3© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

At y-axis, x = 0 y = 0 – 6 = –6 ∴ B(0, –6)(b)

m(0) + 6n————–

m + n = 2

6n = 2m + 2n 2m = 4n

m—n

= 2—1

m : n = 2 : 1 ∴ AP : PB = 2 : 1

5 (a) Median = 47.5 39.5 +

20 + k�———�2 – 7�—————�

k10 = 47.5

20 + k———2

– 7 = 0.8k 20 + k———

2 = 7 + 0.8k

20 + k = 14 + 1.6k 0.6k = 6 k = 10(b) (i) Range = 74.5 – 24.5

= 50

(ii) x = 1425——30

= 47.5

σ = 72 917.5———–30

– (47.5)2

= 174.333 = 13.204

6 (a) Let BC and AB be x cm and y cm respectively.

A = 1—2

xy

A1 = 1—

2 � x—2 �� y—

2 � = 1—

8xy

A2 = 1—

2 � x—4 �� y—

4 � = 1—–

32xy

A3 = 1—

2 � x—8 �� y—

8 � = 1—–

128xy

Since A

1—–A

= A

2—–A

1

= A

3—–A

2

= 1—4

,

the areas of triangles form a geometric

progression with common ratio 1—4

.

(b) (i) A1 = 1—

2(12)(6)

= 36 cm2

36� 1—4 �

n–1

= 9—–64

� 1—4 �

n–1

= 1——256

= � 1—4 �

4

∴ n – 1 = 4 n = 5

(ii) S∞ = 36———1 – 1—

4 = 36——

3�—�4 = 48 cm2

Section B 7 (a) x 1 2 3 4 5 6

y—x 5 9 13 15 19.5 23

(b) y = ax2 + bx

y—x

= ax + b

(i) a = 19.5 – 1.5————5 – 0

= 18—–5

= 3.6 (ii) b = 1.5 (iii) When x = 3.5,

y—x

= 14

y—–3.5

= 14

y = 49

8 (a) 2x2 + 3x = 27 2x2 + 3x – 27 = 0 (2x + 9)(x – 3) = 0 x = – 9—

2 or x = 3

When x = 3, y = 32

= 9 ∴ A(3, 9) At x-axis, y = 0 2(0) + 3x = 27 x = 9 ∴ B(9, 0) (b) Area of the shaded region = �

3

0 x2 dx + 1—

2 (9 – 3)(9)

= x3

�—�3

0 3 + 27

= (9 – 0) + 27 = 36 unit2

(c)

Volume = π ��

9

0

y dy

= π y2

�—�9

0 2

= 40 1—2

π unit3

→ → → 9 (a) (i) BA = BO + OA

= –b + a ~ ~ = a – b ~ ~ → → → OQ = OB + BQ → = b + 1—

3BA

~

= b + 1—3

(a – b) ~ ~ ~

= 1—3

a + 2—3

b ~ ~ → → → (ii) BP = BO + OP → = –b + 1—

2OA ~

= –b + 1—2

a ~

= 1—2

a – b ~ ~ → → → (iii) PQ = PA + AQ → =

1—2

a + 2—3

AB

~

= 1—2

a + 2—3

(–a + b) ~ ~ ~

= 1– — 6

a + 2—3

b ~ ~ → →(b) (i) OR = m OQ

= m � 1—3

a + 2—3

b� ~ ~

= 1—3

ma + 2—3

mb ~ ~ → → (ii) BR = n BP

= n� 1—2

a – b� ~ ~ = 1—

2na – nb ~ ~

→ → →(c) OR = OB + BR

1—3

ma + 2—3

mb = b + 1—2

na – nb

~ ~ ~ ~ ~

1—3

ma + 2—3

mb = 1—2

na + (1 – n)b ~ ~ ~ ~

So 1—3

m = 1—2

n

2m = 3n 2m – 3n = 0 … 1

and

2—3

m = 1 – n

2m = 3 – 3n

A(6, 0)

P(2, –4)

B(0, –6)

n

m

30

25

20

15

10

5

0

y—x

(0, 1.5)

14

1 2 3 4 5 6x

(5, 19.5)

y

xO

9

y = x2

Page 170: Analysis Spm Additional Mathematics

4 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

2m + 3n = 3 … 2

1 + 2 : 4m = 3

m = 3—4

Substitute m = 3—4

into 1 :

2� 3—4 � – 3n = 0

3n = 3—2

n = 1—2

∴ m = 3—4

, n = 1—2

10 (a)

tan θ =

7—7

θ = 45° ∴ ∠TSV = 45—–

180 × π

= 0.786 rad(b) Area of the shaded region

= � 1—2

(72)(1.571) – 1—2

(7)(7)� +

1—2

(3.5)2(0.786)

= 13.99 – 4.814 = 9.176 cm2

(c) SSQ

= 7(1.571) = 10.997 cm S

TV = 3.5(0.786)

= 2.751 cm QS = 72 + 72

= 9.899 cm ∴ Perimeter of the shaded region = 10.997 cm + 3.5 cm + 2.751 cm + (9.899 – 3.5) cm = 23.647 cm

11 (a) (i) P(X � 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2)

= 1 – 10C0 � 3—

5 �0

� 2—5 �

10

– 10C1 � 3—

5 �1

� 2—5 �

9

– 10C2 � 3—

5 �2

� 2—5 �

8

= 0.9877 (ii) The number of students who

do not stay in hostel

= 800 × 2—5

= 320

(b) (i) P(X � 73) = P�Z � 73 – 55———

15 � = P(Z � 1.2) = 0.1151

(ii) Given P(X � m) = 0.2 P�Z �

m – 55———15 � = 0.2

So m – 55———15

= 0.842

m – 55 = 12.63 m = 67.63

Section C 12 (a)

P10—–

3 × 100 = 130

P

10 = 130 × 3———–

100 = RM3.90 (b) 130(2n) + 375 + 110 + 120n

————————————3n + 4

= 124.5 380n + 485 = 373.5n + 498 6.5n = 13 n = 2

(c) 24.90——–P

08

× 100 = 124.5

P08

= 24.90 × 100—————

124.5 = RM20 (d) I

12/08 =

130(4) + 125(3) + 110 + 144(2)——————————–

10 = 1293——–

10 = 129.3

13 (a) (i) a = dv—–dt

= 2t + 2 (ii) s = ∫ v dt = t3

—3

+ t2 – 15t + c

When t = 0, s = 8 and so c = 8. Hence at time t,

s = t3

—3

+ t2 – 15t + 8

(b) (i) When Q is instantaneously at rest,

v = 0 t2 + 2t – 15 = 0 (t + 5)(t – 3) = 0 t = –5 or t = 3 When t = 3,

s = 33

—3

+ 32 – 15(3) + 8

= 9 + 9 – 45 + 8 = –19 ∴ The distance of Q from Y = 19 + 8 = 27 m (ii) When t = 9,

s = 93

—3

+ 92 – 15(9) + 8

= 197

∴ The total distance travelled by Q in the time interval from t = 0 to t = 9.

= 8 + 2(19) + 197 = 243 m

14 (a) I: x + y � 180 II: y � 2x III: 2y � x(b)

(c) (i) The minimum number of students for course A is 40.

(ii) k = 500x + 600y ∴ The maximum total fees

per month obtained = 500(56) + 600(110) = RM94 000

15 (a) (i) 8———

sin 40° =

12————sin ∠ABC

sin ∠ABC = 12 × sin 40°—————8

= 0.9642 ∠ABC = 74°37' (ii) 122 = 52 + 92 – 2(5)(9) cos ∠ADC 90 cos ∠ADC = –38 cos ∠ADC = –0.4222 ∠ADC = 114° 58' (iii) The area of ABCD

= 1—2

(8)(12) sin 65°23' +

1—2

(5)(9) sin 114° 58'

= 43.638 + 20.395 = 64.03 cm2

(b) (i)

(ii) AB'————–

sin 34° 37' =

12————–sin 105° 23'

AB' = 12 × sin 34° 37'——————–

sin 105° 23' = 7.07 cm

Q

P S7

7

θ

f(z)

0.2

Oz

m – 55———15

y

180

160

140

120

100

80

60

40

20

0

y = 2x

(56, 110)

y = 802y = x

x + y = 180

k = 500x + 600y

20 40 60 80 100 120 140 160 180x

R

A

105° 23'

40°B'

12 cm

34° 37'

C

B

8 cm

74° 37'