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ADDITIONAL MATHEMATICS 3472/2 OKTOBER 2020

MAJLIS PENGETUA SEKOLAH MALAYSIA (MPSM) CAWANGAN KELANTAN

TINGKATAN 5

2020

ADDITIONAL MATHEMATICS

KERTAS 2

UNTUK KEGUNAAN PEMERIKSA SAHAJA

SKEMA PEMARKAHAN

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1.

(a) (b)

(c)

f (x) = - (x2 + 3x + (32)

2 - (32)2 – 4)

OR equavalent f(x) = - (x + %

&)2 + &'

(

(− %

&, &'(

)

Shape Maximum point X intercept 0 ≤ f(x) ≤ &'

(

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2.

(a)

(b)

(c)

T7 = 100π (+&)7-1 100π, 50π, 25π.

&'+,

π

100π (+&)n-1 = &'

,(π

n = 9

100𝜋𝜋

1 −12

200 π m / 628.4 m

3

2

2

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2.

(a)

(b)

(c)

T7 = 100π (+&)7-1 100π, 50π, 25π.

&'+,

π

100π (+&)n-1 = &'

,(π

n = 9

100𝜋𝜋

1 −12

200 π m / 628.4 m

3

2

2

7

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3.

(a) (b)

x + 2( 2x + 6) = 20 or 01,&

+ 2y = 20 (

2', (,')

3(20 −85)

& +(0 −465 )

&

20.57 40 = &:.'<=>?@

80 =

A:.'+(%

41.14

2

4

6

p Qt t=

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4 (a)

(b)

Shape of 𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 Max and min 𝑦𝑦 = −2𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 2 cycles for 0≤ 𝑥𝑥 ≤ 2𝜋𝜋 Modulus graph 𝑦𝑦 = I

&J− 1

Graph straight line c=-1 and m positive Number of solutions = 8

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4 (a)

(b)

Shape of 𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 Max and min 𝑦𝑦 = −2𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 2 cycles for 0≤ 𝑥𝑥 ≤ 2𝜋𝜋 Modulus graph 𝑦𝑦 = I

&J− 1

Graph straight line c=-1 and m positive Number of solutions = 8

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5 𝑥𝑥 − 3𝑦𝑦 + 4 = 5 𝑦𝑦(𝑥𝑥 + 2𝑦𝑦) = 5 𝑥𝑥 = 1 + 3𝑦𝑦 or 𝑦𝑦 = I1+

%

𝑦𝑦(1 + 3𝑦𝑦 + 2𝑦𝑦) = 5 or I1+%K𝑥𝑥 + 2 LI1+

%MN = 5

Solve the quadratic equation Formulae

𝑦𝑦 = 1(+)±P(+)Q1((')(1')&(')

or

𝑥𝑥 = 1(1<)±P(1<)Q1((')(1(%)&(')

𝑦𝑦 = 0.905, 𝑦𝑦 = −1.105 Or 𝑥𝑥 = 3.715, x= −2.315 𝑥𝑥 = 3.715, x= −2.315 Or 𝑦𝑦 = 0.905, 𝑦𝑦 = −1.105

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6(a)

(b)

TU

T== :.%+,1:.:'(

&'

+%+

+&':: or 0.01048

TUTV= 4𝜋𝜋𝑟𝑟& or TX

TV= 8𝜋𝜋𝑟𝑟 Substitute 𝑟𝑟 = 1.2

TX

T== TU

T=× TX

TV× TVTU

+%+

+&'::× 8𝜋𝜋𝑟𝑟 × +

(JVQ

+%+

(': or 0.2911

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4

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7 (a)

(b)(i)

(ii)

(c)

Write triangle law 𝐵𝐵𝐵𝐵\\\\\\⃗ − 10𝑥𝑥 + 10𝑦𝑦 𝐵𝐵𝐵𝐵\\\\\⃗ − 2𝑥𝑥 + 5𝑦𝑦 𝐴𝐴𝐵𝐵\\\\\⃗ = 8𝑥𝑥 + 5𝑦𝑦 𝐴𝐴𝐴𝐴\\\\\⃗ = 𝑘𝑘

1+𝑘𝑘(12𝑥𝑥 − 5𝑦𝑦)

𝐸𝐸𝐵𝐵\\\\\⃗ = −4𝑥𝑥 + 10𝑦𝑦 𝐴𝐴𝐴𝐴\\\\\⃗ = (4 − 4ℎ)𝑥𝑥 + 10ℎ𝑦𝑦 Equate and solve coefficient of x or y

2c+dc

= 4 − 4ℎ or 'c+dc

= 10ℎ 𝑘𝑘 = &

%

ℎ = +

'

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8 (a)(i)

(ii)

(b)(i)

(II)

mean, µ = 500

P(Z > ) = 0.2

= 0.842

m = 584.2 Ahmad qualify/ layak 585 > 584.2

or

0.80541

n =

0.2(1554) 310 / 311

1

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2

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10

100500-m

100500-m

÷øö

çèæ -

££-

100500680

100500400 zP ( )8.11 ££- zP

8054.01252 K1

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9 (a)

(b)(i)

(ii)

120: × J

+2:e

&J

%

𝑆𝑆𝑆𝑆𝑆𝑆60: = I

i

Use 2𝑗𝑗𝑠𝑠𝑆𝑆𝑆𝑆 k

&

2 l√%&𝑗𝑗n

√3𝑗𝑗

𝑠𝑠𝑆𝑆𝑆𝑆60: = √%&

or 𝜃𝜃 = J%

or 𝜃𝜃 = &J%

𝐴𝐴+ = 𝜋𝜋𝑗𝑗& or 𝐴𝐴& =

+&𝑗𝑗& L&J

%M

𝐴𝐴% =+&𝑗𝑗& J

%− +

&𝑗𝑗& √%

&

𝐴𝐴+ − 𝐴𝐴& − 2𝐴𝐴%

JiQ

%+ √%i

&

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9 (a)

(b)(i)

(ii)

120: × J

+2:e

&J

%

𝑆𝑆𝑆𝑆𝑆𝑆60: = I

i

Use 2𝑗𝑗𝑠𝑠𝑆𝑆𝑆𝑆 k

&

2 l√%&𝑗𝑗n

√3𝑗𝑗

𝑠𝑠𝑆𝑆𝑆𝑆60: = √%&

or 𝜃𝜃 = J%

or 𝜃𝜃 = &J%

𝐴𝐴+ = 𝜋𝜋𝑗𝑗& or 𝐴𝐴& =

+&𝑗𝑗& L&J

%M

𝐴𝐴% =+&𝑗𝑗& J

%− +

&𝑗𝑗& √%

&

𝐴𝐴+ − 𝐴𝐴& − 2𝐴𝐴%

JiQ

%+ √%i

&

2

3

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10 (a)

(b)

(c)(i)

1𝑢𝑢

0.80 0.50 0.33 0.25 0.21 0.17

1𝑣𝑣

0.35 1.60 2.25 2.55 2.75 2.90

Plot +

r against +

s

(Correct axes and uniform scales) 6 *points plotted correctly Line of best fit (At least *5 points plotted) 𝑢𝑢 = 2.33 ± 0.1 +

r= &c

ts+ u

t

Use 𝑐𝑐 = u

t or 𝑚𝑚 = &c

t

ℎ = 2.54 𝑘𝑘 = −5.14

2

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If table not shown, all the points are correctly plotted award N1

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11 (a)

(b)

(c)

3 = 4 − IQ

(

𝑥𝑥 = 2, 𝑥𝑥 = −2 𝐵𝐵𝐵𝐵 = 4

∫ l4 − IQ

(n&

1& 𝑑𝑑𝑥𝑥

l4𝑥𝑥 − Iz

+&n &1&

l4(2) − &z

+&n − l4(−2) − (1&)z

+&n

14.67 or ((

%

4(1) ((

%+ 4(1)

18.67 or ',

%

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12

(a)

(b)

(c)

r = 6 substitute t=3 in equation V p(3)2 + q (3) + 6 = 42 TrT== 2𝑝𝑝𝑝𝑝 + 𝑞𝑞 9p + 3q = 36 OR

6p +q = 0 p = −4 and q = 24

t =−(−24)±~(24)2−4(−4)(6)

2(−4)

-4t2 + 24t + 6 = 0 t = 3 + +

2 √672 s = (

%𝑝𝑝% +&(

&𝑝𝑝& + 6𝑝𝑝

substitute t = 6 or t = 5 in equation S s = (

%(6)% +&(

&(6)& + 6(6) or

s = (

%(5)% +&(

&(5)& + 6(5)

<<2%

5

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12

(a)

(b)

(c)

r = 6 substitute t=3 in equation V p(3)2 + q (3) + 6 = 42 TrT== 2𝑝𝑝𝑝𝑝 + 𝑞𝑞 9p + 3q = 36 OR

6p +q = 0 p = −4 and q = 24

t =−(−24)±~(24)2−4(−4)(6)

2(−4)

-4t2 + 24t + 6 = 0 t = 3 + +

2 √672 s = (

%𝑝𝑝% +&(

&𝑝𝑝& + 6𝑝𝑝

substitute t = 6 or t = 5 in equation S s = (

%(6)% +&(

&(6)& + 6(6) or

s = (

%(5)% +&(

&(5)& + 6(5)

<<2%

5

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13. a) (b) (c)

(i) x ≥ +

%𝑦𝑦 or

(ii) 12x + 8y ≤ 6000 or 3x + 2y ≤ 1500

(iii) 8x + 4y ≥ 2400 or 2x + y ≤ 600

graph- attachment

- draw correctly at least one straight line from

* inequalities involves x and y

- draw correctly ALL the

* straight line

Note: Accept dotted line

Region shaded correctly.

(i) 120

(ii) (200, 450) seen 450

8(200) + 4 (450)

8x + 4y ( profit equation)

RM 3400.00

Noted: inequality for which no symbol "=" is accepted

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3y x£ N1

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x

y

x = 200

2x + y = 600 3x + 2y = 1 500

x = 𝟏𝟏𝟑𝟑 y

R

100 200 O 300 400 500 600

100

200

300

400

500

600

700

800

450

120

x

y

x = 200

2x + y = 600 3x + 2y = 1 500

x = 𝟏𝟏𝟑𝟑 y

R

100 200 O 300 400 500 600

100

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500

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450

120

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14. (a) (b)

(i) 135 = &+:ÅQeÇÉ

𝑋𝑋100

RM 155.56

(ii) +&:+::X+%:

+::𝑋𝑋100

156

Increase 56 %

(i) p = 15

(+%'Ö':)d(0Ö&')d(+&:Ö+')d(<:Ö+:)':d&'dÜd(Ü1')

= 119.50

108

(ii) 108 = ÅQeQe+%:

𝑋𝑋100

RM 140.40

2 3 3 2

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15. (a) (b) (c)

(i) BD2 = 3.742 + 82 – 2(3.74)(8)(Cos 450)

5.973 km

(ii) á>à∠Xäã

2= á>à('

e

<.2'

46.110

Noted: ∠𝐴𝐴𝐴𝐴𝐴𝐴 = ∠𝐵𝐵𝐴𝐴𝐴𝐴

(iii) 88.890

A = +&x8x7.85xSin88.890

31.39 km2

The farthest hotel from hotel A is hotel C because

the distance between them faces the largest angle

+&x8x(shortestdistance) = 31.39

7.848 km

2 2 3 1 2

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