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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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Teknik Menjawab Addmath SPM Kertas 1 &2

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MENJAWAB SOALAN SPM

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.

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

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

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)

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)

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

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)

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

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)

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)

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

Q.10(a):

a = 108

ar2 = 12

ar2= 12 √ ……….(P1)

a 108

r2 = ¹/9

r = ¹/3 √..……….(2)

Q.10(b):

108

S∞ = ——— √…….….(P1)1 – ¹/3

108= ———

= 162 √…..…….(2)

Q.11:

x

y = ————

a + bx

a + bx

¹/y = ————x

¹/y = a/x + b √ ……….(P1)

= - 6/4 = - 3/2 √ ……….(2)

b = y-intercept

= 6 √……….(1)

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)

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

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)

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

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)

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

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)

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)

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

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)

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)

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))

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)

Q.25:

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

= 0.275 = 0.275 ……….(2) ……….(2)

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)

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

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)

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)

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

+

(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)

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)

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)

(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)

MENJAWAB SOALAN