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MENJAWAB SOALAN SPM ADDITIONAL MATHEMATICS Kertas 1 dan 2 Skema Jawapan & Pemarkahan Siri Ceramah Di: (i) SMK Dato’ Klana Putra, Lenggeng, Negeri Sembilan, (ii) SMK Engku Husain, Semenyih, Selangor, (iii) SMK Dengkil, Kuala Langat, Selangor, (iv) SMK Taman Jasmin 2, Kajang, Selangor, (v) SMK Jalan 4, Bandar Baru Bangi, Selangor, (vi) SMK Dato’ Ahmad Razali, Amapang, Selangor.

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Page 1: Ceramah AddMath

MENJAWAB SOALAN SPM

ADDITIONAL MATHEMATICS

Kertas 1 dan 2

Skema Jawapan & PemarkahanSiri Ceramah Di:

(i) SMK Dato’ Klana Putra, Lenggeng, Negeri Sembilan,

(ii) SMK Engku Husain, Semenyih, Selangor,

(iii) SMK Dengkil, Kuala Langat, Selangor,

(iv) SMK Taman Jasmin 2, Kajang, Selangor,

(v) SMK Jalan 4, Bandar Baru Bangi, Selangor,

(vi) SMK Dato’ Ahmad Razali, Amapang, Selangor.

Page 2: Ceramah AddMath

Q.1:

(a) (a) --11, , 1 1 / / --1 1 and and 1 1 / {/ {--11, , 11} }

….√(1)

reject: 1 or -1 / (-1, 1) / [-1, 1]

(b) many(b) many--toto--one / one / ….√(1)

many with one /many with one /

many many → one→ one

Page 3: Ceramah AddMath

Q.2(a):

(a)(a) f(x) = 4x + 2f(x) = 4x + 2

f f --11(x) = y (x) = y

f(y) = xf(y) = x

= 4y + 2 = 4y + 2 √ -----------(1)

x x = = 4y + 24y + 2

x x -- 22y = y = —————— √------------(2)

44

Page 4: Ceramah AddMath

Q.2(b)Q.2(b)::

(b) (b) gfgf(2) = g[f(2)](2) = g[f(2)]

= g[4(2) + 2] = g[4(2) + 2] √√………..………..(1)………..………..(1)

= g(10)= g(10)

= 10= 102 2 –– 3(10) 3(10) –– 11

= 69 = 69 √√…………………..(2)…………………..(2)

f(2) = 4(2) + 2 f(2) = 4(2) + 2

= 10 = 10 √√ ………………….(1)………………….(1)

Page 5: Ceramah AddMath

Q.3Q.3::

gh(x) = g[h(x)]

= mx2 + n – 3 ……..(P1)

= 3x2 + 7

Bandingkan:

m = 3 m = 3 ……...(1)……...(1)

n n –– 3 = 7 3 = 7 ………(P1)………(P1)

n = 10 n = 10 ………(2) ………(2) (Kedua-duanya)

Page 6: Ceramah AddMath

Q.Q.44::

2x2x2 +2 + p + 2 = 2px + xp + 2 = 2px + x22

xx2 2 –– 2px + p + 2 = 0 2px + p + 2 = 0 √ ………..(P1)………..(P1)

aa = 1, b = = 1, b = -- 2p, c = p +22p, c = p +2

bb22 -- 4ac > 04ac > 0

((--2p)2p)2 2 –– 4(1)(p + 2) > 04(1)(p + 2) > 0

4p4p2 2 –– 4p 4p –– 8 > 0 8 > 0 √ ………..(P2)………..(P2)

pp2 2 –– p p –– 2 > 0 2 > 0

Let: Let: pp2 2 –– p p –– 2 = 02 = 0

(p (p –– 2)(p+ 1) = 02)(p+ 1) = 0

p = 2 atau p = p = 2 atau p = -- 11

Therefore: p > 2 atau p < Therefore: p > 2 atau p < --1 1 √ ..…….(3)

-1 2x

Page 7: Ceramah AddMath

Q.5Q.5::

(x (x –– ⅔⅔)(x + 4) = 0 )(x + 4) = 0 ….…….(P1) ….…….(P1)

xx22 –– ⅔ x ⅔ x –– 4x 4x –– 8/3 = 08/3 = 0

3x3x22 + 10x + 10x –– 8 = 0 8 = 0 …………(2)…………(2)

OR:

x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(P1)

3x2 + 10x – 8 = 0 …………(2)

Page 8: Ceramah AddMath

Q.6Q.6::

(a) p = 3 (a) p = 3 (axis of symmetry) √...(1)

(b) x = 3 √...(1)

(c) (3, -1) √...(1)

Note: Minimum value = -1

Page 9: Ceramah AddMath

Q.7:

32x + 1 = 4x

(2x + 1)log10 3 = xlog10 4 √ …..…….(P1)

2xlog10 3 + log10 3 = xlog10 4

2x(0.4771) – x(0.6021) = - 0.4771 √.………..(P2)

0.9542x – 0.6021x = - 0.4771

0.3521x = - 0.4771

- 0.4771 x = ————

0.3521

= - 1.355 √ ………...(3)

Page 10: Ceramah AddMath

Q.8:

2log2log22M = 2 + 4logM = 2 + 4log44NN

4log4log22NN

2log2log22M = 2 + M = 2 + —————— √……….(P1)

loglog2244

loglog22MM22 = = loglog224 + 2log4 + 2log22N N √..……..(P2)

loglog22MM22 = log= log224N4N2 2 √..….…(P3)

MM22 = 4N= 4N2 2

M = M = √4N√4N22

M = 2N M = 2N √..…....(4)

Page 11: Ceramah AddMath

Q.9:

(y – 1) – x = (5 + 2x) – (y – 1) √....(P1)

y – 1 – x = 5 + 2x – y + 1

2y = 3x + 7 √….(P2)

3x + 7 y = ——— √….(3)

2

Page 12: Ceramah AddMath

Q.10(a):

a = 108

ar2 = 12

ar2= 12 √ ……….(P1)

a 108

r2 = ¹/9

r = ¹/3 √..……….(2)

Page 13: Ceramah AddMath

Q.10(b):

108

S∞ = ——— √…….….(P1)1 – ¹/3

108= ———

= 162 √…..…….(2)

Page 14: Ceramah AddMath

Q.11:

x

y = ————

a + bx

a + bx

¹/y = ————x

¹/y = a/x + b √ ……….(P1)

a = gradient

= - 6/4 = - 3/2 √ ……….(2)

b = y-intercept

= 6 √……….(1)

Page 15: Ceramah AddMath

Q.12:

xx/y /y -- yy/4 = 1/4 = 1

4x 4x -- 2y = 8 2y = 8 √..……(P1)

2y = 4x 2y = 4x -- 88

y = y = 44//22x x –– 44

gradient, m = 2 m = 2 √..……(P2)

y y –– 7 = 2(x 7 = 2(x –– 2) 2)

y = 2x + 3 y = 2x + 3 √.……(3)

Page 16: Ceramah AddMath

Q.13:

4m + 9s4m + 9s 2m + 6t2m + 6t √…. (P1)

P(2m, m) P(2m, m) Q(s, t)Q(s, t) R(3s, 2t)R(3s, 2t)

3 23 2

5s 5s –– 9s = 4m, 9s = 4m, or 5t 5t –– 6t = 2m6t = 2m

-- 4s = 4m, 4s = 4m, -- t = 2m t = 2m √……(P2)

-- s = m, s = m,

- 2s = - t

s = t/2 √…...…………………...(3)

S = or t =5 5

Page 17: Ceramah AddMath

Q.Q.1414::

PA : PB = 2 : 1 PA : PB = 2 : 1 √..…(P1)

reject: PA/PB = 2/1

PA = 2 √(x – 5)2 + (y – 0) 2 √…..(P1) or

PB = √(x + 2)2 + (y + 3)2 √….(P1)

PA2 = 4[(x – 5)2 + y 2 ] =

PB2 = (x + 2)2 + (y + 3)2 √....(P2) or

4[(x – 5)2 + y 2 ] = (x + 2)2 + (y + 3)2 √…..(P2)

3x2 + 3y2 - 44x + 6y + 87 = 0 √....(3)

Page 18: Ceramah AddMath

Q.15

A B

C

AC = AB + BC

= AB - CB .........(P1)

= (i + 2j) – (-5i – 6j)

= 6i + 8j ….…(2)

6i + 8jUnit vector AC = ———— ……..….(P1)

√ 62 + 82

6i + 8j 3i + 4j

= ———— or ———— …..……(2)

10 5

D

Page 19: Ceramah AddMath

Q.16

P(1, 2)

O

Q(8, -4)

PQ = PO + OQ

= - OP + OQ

= - (i + j) + (8i + 4j)

= 7i + 3j ….(1)

7i + 3j = ma + nb

= m(i + j) + n(2i – 3j)

= (m + 2n)i + (m – 3n)j ……………….…..(P1)

Compare: m + 2n = 7

m – 3n = 3

5n = 4 .….(P1)

n = 4/5 )..(2 either)

m = 27/5 )..(3)(both)

Page 20: Ceramah AddMath

Q.Q.1717::

sec sec θθ = = 11//cos cos θθ

= = 11//t t ………(1) ………(1)

cos (90 cos (90 –– θ) = sin θ θ) = sin θ ……….(1)……….(1)

= √ 1 = √ 1 –– tt22 ……….(2)……….(2)

1 – t21

t

Page 21: Ceramah AddMath

Q.18Q.18:: 2 sec2 sec22θθ –– 3tan3tanθθ = 4= 4

2(1 + tan2(1 + tan22θθ) ) –– 3tan3tanθθ = 4 = 4 …..…..(1)…..…..(1)

2 + 2 tan2 + 2 tan22θθ –– 3tan3tanθθ = 4= 4

2tan2tan22θθ -- 3tan3tanθθ –– 2 = 02 = 0

(2tan(2tanθθ + 1)(tan+ 1)(tanθθ –– 2) = 0 2) = 0 ………..(2)………..(2)

tantanθθ = = --½ ½ atau atau tantanθθ = 2= 2

θθ = (360= (360oo –– 2626o o 34’), (18034’), (180o o -- 2626o o 34’)34’)

θθ = 333= 333oo 26’, 15326’, 153o o 26’ 26’ ………...(3)………...(3)

θθ = 63= 63o o 26’, (18026’, (180o o + 63+ 63o o 26’)26’)

θθ = 63= 630 0 26’, 24326’, 2430 0 26’ 26’ ….....….(3)….....….(3)

θθ = 63= 630 0 26’, 15326’, 153o o 26’, 24326’, 2430 0 26’, 33326’, 333oo 26’.26’. …….….…….….(4)(4)

Page 22: Ceramah AddMath

Q.19:

BOC = BOC = ππ –– 1 1

= = 33..142 142 –– 1 1

= = 22..142 142 ………..(P1)

s = jθ

BC = 15 x 2.142 ………..(P2)

= 32.13 ………..(3)

Page 23: Ceramah AddMath

Q.Q.2020::

dy (x dy (x –– 4)(4x) 4)(4x) –– (2x(2x22

+ 3)(1)+ 3)(1)—— = = —————————————————————— ……(P2)dx (x dx (x -- 4)4)22

4x …….(P1)

2x2x22

–– 16x 16x -- 33= = —————————————— ..……(3)

(x (x -- 4)4)22

Page 24: Ceramah AddMath

Q.21Q.21::

dy/dx = 4x dy/dx = 4x –– 3 3 ………..(1)

δx = 0.01 δx = 0.01 ………..(1)

x = 2x = 2

δyδy∕∕δxδx ≈ ≈ dydy∕∕dxdx

δyδy ≈ ≈ dydy∕∕dxdx (δx)(δx)

= [4(2) = [4(2) –– 3](0.01) 3](0.01) …………(2)

= 5(0.01)= 5(0.01)

= 0.05 = 0.05 …………(3)

Page 25: Ceramah AddMath

Q.22:

3 3 33

∫ f(x) + ∫ (kx)dx = 20 ∫ f(x) + ∫ (kx)dx = 20 1 11 1

33

8 + [ kx8 + [ kx2 2 ∕ 2∕ 2 ] = 20 ] = 20 …..(P1)11

33

[ kx[ kx2 2 ∕ 2∕ 2 ] = 12 ] = 12 …..(P2) 11

9k/2 – k/2 = 12, 8k/2 = 12 …..(P3)

k = 3 …..(4)

Page 26: Ceramah AddMath

Q.Q.2323::

6 46 4

P or P P or P ….(1)4 34 3

6 46 4

P x P = P x P = ((6 6 x x 5 5 x x 4 4 x x 33) x () x (4 4 x x 3 3 x x 22)) ..(P2)4 34 3

= = 360 360 x x 2424

= = 86400 86400 ……...(……...(33))

Page 27: Ceramah AddMath

Q.Q.2424::99 55

(a) C x C (a) C x C ……….(P1)66 44

atauatau 9 x 8 x 79 x 8 x 7

3 x 23 x 2

= 84 = 84 …………(2)

55 9 5 9 5 99 5 9 5 9 (b) C x C + C x C + C x C (b) C x C + C x C + C x C …...….(P1)

5 5 4 6 3 75 5 4 6 3 7

= 948 = 948 …………(2)

Page 28: Ceramah AddMath

Q.25:

0.5 0.5 –– 0.225 0.225 ……….(P1)……….(P1)

= 0.275 = 0.275 ……….(2) ……….(2)

Page 29: Ceramah AddMath

PAPER 2Q1:

x + 4

x = 3y – 4 or y = ——— ………………. √(P1)3

Eliminate x or y:

(3y – 4)2 + y(3y – 4) – 40 = 0 …………………. √(M1)

3y2 -7y – 6 = 0

(3y + 2) (y – 3)= 0 or using quadratic equation …. √(M1)

or completing the square method.

y = - 2/3 , y = 3. ………………………………√(A1)

x = - 6, x = 5 ………………………………√(A1)

Page 30: Ceramah AddMath

Q3:

(a) (a) r = r = 11..05 05 √….(P1)

Use Use TT6 6 = ar= ar 55

TT6 6 = = 18000 18000 x (x (11..0505))5 5 √….(M1)

= RM 22 973.00√….(A1)

(b) Use (b) Use arar n-1 ≥ ≥ 36 000 36 000 √….(M1)

18000 18000 x (x (11..0505))n-1 ≥ ≥ 36 000 36 000

n ≥ n ≥ 1616 √….(A1)

No working minus No working minus 1 1 mark if answer is mark if answer is

correctcorrect

Page 31: Ceramah AddMath

a(ra(rnn –– 11))

(c) (c) Use Use S6 = ———

r - 1

Use Use S6=18000{(1.05)6 – 1}..√(M1)

1.05 – 1

= RM 122 434 √..(A1)

Page 32: Ceramah AddMath

Q8:

(a)(a)

xx 1.5 3.0 4.5 6.0 7.5 9.01.5 3.0 4.5 6.0 7.5 9.0

loglog10 yy 0.40 0.51 0.64 0.76 0.89 1.000.40 0.51 0.64 0.76 0.89 1.00

√.... (P1)

Page 33: Ceramah AddMath

Plot graph log y against x2

log y

x0 1 2 3 4 5 6 7 8

0.2

0.4

0.6

0.8

1.0

0.0

+

+

+

+

+

√ P1

√ P1

√ G1

9

+

Page 34: Ceramah AddMath

(b)(b) Plot Plot log10 y against against xx

(correct axes and uniform scales)(correct axes and uniform scales)…..√(P1)

6 points plotted correctly …………..√(P1)

Lines of best fit …………..√(G1)

(c) log10 y = = log10 h + 2x log10 k …………….√(P1)

c = log10 h .................... √(M1)

h = 1.78 ………….√(A1)

m = 2 log10 k ………….√(M1)

k = 1.09 – 1.12 ………….√(A1)

Page 35: Ceramah AddMath

Q13:PP11

(a) (a) Use Use I = —— x 100 P0

11..8080

h = —— x 100 ………………..√(M1)1.50

= 120 ………………….√(A1)

00..9090

112112..5 5 = —— x 100 ……………… √(M1)*k

k = 0.8 or 80 sen ……...……………√(A1)

Page 36: Ceramah AddMath

I W IWI W IW

150 30 4500150 30 4500

120 120 45 540045 5400

112112..5 15 16875 15 1687..55

105 10 1050105 10 1050

100 12637100 12637..55 …………………………√(P1)

1263712637..55

I = —— x 100 …………√(M1)100

= 126.38 …………√(A1)

Page 37: Ceramah AddMath

(c) (i) (c) (i) 150150

126.38 = —— x 100 ……√(M1)

100

= = 189189..6 6 ……√(A1)

PP11

(c) (ii)(c) (ii) —— x 100 = 189.6 ……√(M1)

25

= = 4747..39 39 ……√(A1)

Page 38: Ceramah AddMath

MENJAWAB SOALAN

ADDITIONAL MATHEMATICS

SPM

PAPER 1 & 2

THE ENDTHE END