trial perlis 2020

SULIT

PROGRAM GEMPUR KECEMERLANGAN
SIJIL PELAJARAN MALAYSIA 2020
NEGERI PERLIS

SIJIL PELAJARAN MALAYSIA 2020 3472/2(PP)

MATEMATIK TAMBAHAN

Kertas 2
Peraturan Pemarkahan
Oktober

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 19 halaman bercetak

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SULIT

SULIT 2 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

1(a) f ( x) = sin 2x sin x P1 2
2x P1
2
(b) 2 cot x sin2 x = *sin 2x
K1
LHS = 2 cot x sin2 x N1

= 2  cos x  sin 2 x cot x = cos x K1
 sin x  sin x N1

= 2sin x cos x LHS = sin 2x

= sin 2x

= RHS

(c) 2 cot x sin2 x = f ( x) ...

6 cot x sin 2 x = 2 − 3x f ( x) = 2 − x N1
2π 3 2π

( )3 3x
2 cot x sin2 x = 2 − 2π

2 cot x sin2 x = 2 − x ... Sketch the straight line involving
3 2π x and y with *gradient or
*y-intercept correct.
Substitute into
Number of solutions = 5
f ( x) = 2 − x
3 2π

x 0 2π
y 2 −1

33

Number of solution = 5 37
SULIT
3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis

SULIT 3 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

2(a) Let,

O = centre of the circle Find the radius of the circle P1
r = radius of the circle
Then, OA = OB = OC = OD = r r=  x 2 + 42
 2 
 x 2
r=  2  + 42

r = x2 +16 Area of circle − Area of rectangle K1
4
2
The area of the shaded region, = π  * x2 +16  − 8x
 4
L = πr2 − 8x L

 x2 2 − 8x
L = π  4 +16 

 L = πx2 − 8x +16π (Proved) L = πx2 − 8x +16π (Proved) N1 3

4 4

(b) dL = 2πx − 8 = πx − 8 Find dL and equate to 0 K1
dx 4 2 dx

When L is minimum, dL = 0 πx − 8 = 0
dx 2

Then, πx − 8 = 0
2

x = 16 cm d2L K1
π Find dx2

d2L = π  0 π 0
dx2 2 2
 The area of the shaded region is

minimum when x = 16 cm The area of the shaded region is
π
16 N1
minimum when x = π cm 3 6

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SULIT

SULIT 4 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

3 π y − 2  π x  = 15π P1
Arc PU = Arc ST = x  2 
2

Then, y − 2  π x  = 15π xy + π x2 = 3437.5π P1
 2  2

y − πx = 15π ...

Area of the field, y =15π + πx P1

xy + π x2 = 3437.5π ...
2

From : y = 15π + πx ...

Substitute into Eliminate x or y (involving one

x (15π + πx) + π (15π + πx)2 = 3437.5 linear and one non-linear K1
equations in term of x and y)
2
3x2 + 30x − 6875 = 0 x (15π + πx) + π (15π + πx)2 = 3437.5

−(30)  (30)2 − 4(3)(6875) 2
x = 2(3)

x1 = 43.13 x2 = −53.13 (Rejected) Solve the *quadratic equation K1
Formulae
y1 = 58.13π
 x = 43.13 ; y = 58.13π // 182.64

−(30)  (30)2 − 4(3)(6875)
x = 2(3)

a, b, c must be correct

x1 = 43.13 N1

x2 = −53.13 (Rejected)

y = 58.13π // 182.64 N1 77

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 5 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

4(a) LHS = log2 P + log2 Q Use law loga b + loga c = loga bc K1 4
= log2 PQ log2 PQ
= log4 P 37
log4 2 Change to base 4 K1
= log4 P  log2 4
log4 P
= (log4 P) 2 log4 2
= 2(log4 P)

= RHS (Proved)

Use law a logc b = logc ba K1
2log2 2

LHS = 2(log4 P) (Proved) N1

(b) 2x+1 3x−2 = 8 Use law ab  ac = ab+c K1
2x  2 or 3x  3−2
2x  2  3x  3−2 = 8

2x  2 3x  1 = 8 Use law ac  bc = (ab)c K1
9 ( 2  3)x

2x  3x = 36

(2 3)x = 36

6x = 62

x = 2

x = 2 N1

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SULIT

SULIT 6 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

5(a) s1 = ( j1 + j2 )θ s1 = 31.5θ or j2 = 3.5θ (seen) P1

j1 + j2 = 28 + 3.5 = 31.5

Perimeter of the glued fabric 2(28) + (31.5)θ + 3.5θ = 105π + 56 K1

=  105π + 56  cm 4
 4  θ = 2.3565rad N1

2( j1 ) + ( j1 + j2 )θ + j2θ = 105π + 56
4

2(28) + (31.5)θ + 3.5θ = 105π + 56 A1 = 1  31.52  2.3565
2
4

56 + 35θ = 105π + 56 or K1
4
A2 = 1  3.52  2.3565
35θ = 105π 2
4

θ = 105 ( 3.142 ) 2*A1 − *A2  K1
4 ( 35)

θ = 2.3565 rad 2309.37cm2 N1

Area of fabric needed

= 2  1 ( j1 + j2 )2 θ − 1 j22θ 
 2 2 

 1  31.52  2.3565  − 
 2  

= 2  
 1  
2  3.52  2.3565  

= 2(1169.1186 −14.4336)

 2309.37 cm2 66

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 7 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

6(a) Syuhada's annual salaries form a 3

geometric progression with, r = 1.05 P1 2

a = 18000 (Year 2002) T6 =18000(*1.05)6−1 K1 27

r = 1.05 RM22973 N1
Annual salary in 2007 = T6

T6 = 18000(1.05)6−1

= 22973.07
 Annual salary in 2007 = RM22973

(b) Annual salary in nth year exceed

RM36000,

Tn  36000 18000(*1.05)n−1  36000 K1

18000(1.05)n−1  36000
(1.05)n−1  2

(n −1) log10 1.05  log10 2 n = 16 N1

n −1  log10 2
log10 1.05

n  15.21
Thus the minimum value of n is 16

(c) The total salary for the years

2002 to 2007 = S6 ( )18000 *1.056 −1 K1

( )18000 1.056 −1 S6 = 1.05 −1

S6 = 1.05 −1

= 122434.43 RM122434 N1
Thus The total salary for the years

2002 to 2007 = RM122434

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SULIT

SULIT 8 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

7(a)(i) At (h, 2),
2 = (h −1)2 +1
h = 2 N1
h2 − 2h = 0
1
h(h − 2) = 0
h = 0 (Rejected)

h = 2

(ii) y = ( x −1)2 +1

y2 = ( x −1)4 + 2 ( x −1)2 +1 Integrate π y2 dx

Solid volume generated,

 = π 2 22 dx − π 2 y2 dx ( x −1)5 2( x −1)3 K1
00 K1
A1 = 5 + 3 + x
 = π 2 4 dx − π 2 ( x −1)4 + 2 ( x −1)2 +1dx
00

 ( x −1)5 2( x −1)3 2  *2
 2 x 
= π   4 x 0 −  + 3 +  Use limit  into
 5 0 
 0

  1 2    *( x −1)5 2( x −1)3 
  5 3    + + x
  + + 2 −    5 3 
  −  
= π  4 (2) −  1  2   
5  3  
+ − + 0 ( )π
 4 x 2 − *A1
0 K1

= π  8 −  43 −  − 13  
 15  15  
  64 π unit3 N1
15
= 64 π unit3
15 4

(b)(i) Gradient of tangent of 1st curve,
dy = 4x − 5
dx 3(2 p − 3) = −1 K1
Gradient of tangent of 2nd curve,
dy = px − 3 p=4 N1
dx 3
When two curves intersect at the
right angle, 2

(4x − 5)( px − 3) = −1

At x = 2,

3(2 p − 3) = −1

6 p − 9 = −1
p= 4

3

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 9 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

7(b)(ii) y1 =  4x − 5 dx Substitute (2, 3) into  4x − 5 dx K1 3 10

y1 = 2x2 − 5x + c 3 = 2(2)2 − 5(2) + c N1
N1
At point (2, 3), or
3 = 2(2)2 − 5(2) + c
Substitute ( 2, 3) into  *4 x −3 dx
c=5 3

 y1 = 2x2 − 5x + 5

y2 =  4 x−3 dx 3 = 2 (2)2 − 3(2) + c
3
3
2
y1 = 3 x2 − 3x + c

At point (2, 3), y = 2x2 − 5x + 5 or y = 2 x2 − 3x + 19
33
3 = 2 (2)2 − 3(2) + c

3

c = 19 y = 2x2 − 5x + 5 and y = 2 x2 − 3x + 19
3 33

 y1 = 2 x2 − 3x + 19
3 3

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SULIT

SULIT 10 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

8(a) X = Events to choose a good Harumanis

X ~ B (8, 0.85)

X = 0, 1, 2, 3, 4, 5, 7, 8 nCr (0.85)r (0.15)n−r K1

Probability that at least 6 Harumanis

are good, P ( X  6) P( X = 6) + P( X = 7) + P( X = 8) P1
= P( X = 6) + P( X = 7) + P( X = 8)

= 8C6 (0.85)6 (0.15)2 + 8C7 (0.85)7 (0.15)1 0.8948 N1
+ 8C8 (0.85)8 (0.15)0
3
= 0.8948

(b)(i) X = Mark obtained by a student 35 − 48 or 66 − 48
66
X ~ N (48, 62 )

X = 0  X  100 −2.167 3 K1
K1
Probability that a student obtained P(−2.167  Z  3)

between 35 and 66 marks, P (35  X  66) or
equivalent
= P  35 − 48  Z  66 − 48 
 6 6  0.9835 N1

= P (−2.167  Z  3) 177 N1

= 0.9835

Number of students,

= 0.9835180

 177 4

(b)(ii) Let k as passing marks, 1.645 P1

P ( X  k ) = 0.05 k − 48 = −1.645 K1
6
k − 48 = −1.645
6 k = 38.13 N1 3 10
k = 38.13

 k = 38

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 11 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

9(a) 5

x2 1.00 4.00 12.25 16.00 25.00 36.00 N1 5 10
xy 3.00 6.50 14.25 18.04 27.00 37.50 N1

Refer graph 9(a) on Page 18. Plot xy against x2 K1

6 *points plotted correctly K1

Line of best fit N1
(At least *5 points plotted)

(b)(i)  = h+ px   x xy = px2 + h P1
y px  p
 

xy = h + px2
p

 xy = px2 + h p = *m K1
p p =1 N1

p = gradient of the line h = *c K1
p N1
= 27.00 − 3.00 h=2
25.00 −1.00

p =1

h = intersect of xy-axis
p
h =2
p
h = 2p

h = 2(1)

h = 2

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SULIT

SULIT 12 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

10(a)(i) a = p i + 8 j ; b = −3i + 4 j 2

If a and b are parallel, then Use a = λb or b = λa K1 2
a = λb p = −6 N1
2
( )p i + 8 j = λ −3i + 4 j 1

4λ = 8 p = −3λ
λ=2  p = −6

( )(ii) a + b = p i + 8 j + −3i + 4 j ( p − 3)2 +122 = 13 K1

= ( p − 3) i +12 j p = −2 , p = 8 N1

a + b = ( p − 3)2 +122

13 = p2 − 6 p + 9 +144
p2 − 6 p −16 = 0

( p + 2)( p −8) = 0

 p = −2 or p = 8

(b)(i) AP = 6 i + 8 j ; AQ = 4 i + 3 j ;

PR = 1 PQ Use AR = AP + 1 PQ K1
2 2

AR = AP + PR AR = 5 i + 11 j N1
2
= AP + 1 PQ
2

( )= AP + 1 − AP + AQ
2

( )= 1 AP + AQ
2

( )= 1 6 i + 8 j + 4 i + 3 j
2

= 1 (10 i +11 j )
2
 AR = 5 i + 11 j

2

(ii) BR = − AR

BR = −  5 i + 11 j  BR = − * 5 i + 11 j  or − 5i − 11 j N1
 2  2  2

 BR = −5 i − 11 j
2

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 13 3472/2
Sub Total
No. Solution and Mark Scheme Marks Marks

10(c) BA = BQ + QA Find BA = BQ + QA K1 3 10
−10i −11 j
= − AP − AQ
= −6 i − 8 j − 4 i − 3 j ( )Use 1 BA = 1 −10 i −11 j K1
22 N1
= −10 i −11 j
BR = −5 i − 11 j (Proved)
( )1 BA = 1 −10 i −11 j
2
22
= −5 i − 11 j
2

= BR (Proved)

11(a) A(2, 2) ; B (6, 2) C ( xC , yC ) Use 1  (6 − 2)  h = 10 or equivalent K1

Area of isosceles ABC, 2
h=5
1  (6 − 2)  h = 10

2
h=5

xC = 2+6 = 4 xC = 2+ 6 = 4 or yC = 2 − 5 = −3 K1
2 2

yC = 2 − 5 = −3

C (4, − 3) C (4, − 3) N1 3
2
(b) B (6, 2) ; C (4, − 3) ; D ( xD , yD ) xD + 4 = 6 or yD + (−3) = 2 2

xD + 4 = 6 yD + (−3) = 2 22 K1
2
2

xD = 8 yD = 7 D(8, 7) N1

 D (8, 7)

(c)(i) A(2, 2) ; C (4, − 3) ; D (8, 7) Use mDE = mAC K1
k−7 =−5
mAC = 2 − (−3) = −5 11 − 8 2
2
2−4

mDE = mAC

k−7 =−5 k = − 1 N1
11− 8 2 2

k = −1
2

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SULIT

SULIT 14 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

11(c)(ii) C (4, − 3) ; E 11, − 1  ; P ( x, y) Use ( x1 − x2 )2 + ( y1 − y2 )2 K1
2 

PC = 1 ; 4PC = PE ( x − 4)2 + ( y + 3)2
PE 4

4 ( x − 4)2 + ( y + 3)2 = or

( x −11)2 +  y + 1 2 ( x − 11)2 +  y + 1 2
 2   2 

( ) ( )16  x2 − 8x +16 + y2 + 6 y + 9  = 4PC = PE (Can be implied) P1
60x2 − 60y2 − 424x + 380y +1115 = 0
( )x2  2 1 
− 22x +121 +  y + y + 4 

15x2 +15y2 −106x + 95y + 1115 = 0 N1
4
3 10
60x2 − 60 y2 − 424x + 380 y +1115 = 0

12(a) 7500x + 4500y  225000 7500x + 4500y  225000 N1
5x + 3y 150
6000x + 7500y  300000 6000x + 7500y  300000 N1
4x + 5y  200
1500x + 3000y  90000 1500x + 3000y  90000 N1
x + 2y  60 or
3
equivalent
3
(b) Refer graph 12(b) Draw correctly at least one straight line from the K1 2
2
on Page 19. *inequalities involve x and y.

Draw correctly all the three *straight lines N1

Region shaded correctly N1

(c)(i) 2000x +1000y = k

Maximum point = (60, 0) H -Ziez = 60 days N1

 H -Ziez = 60 days S-Ziez = 0 day N1
 S -Ziez = 0 day

(ii) = 2000(60) +1000(0) 2000(60) +1000(0) K1

= 120000

Maximum average profit = RM120000 RM120000 N1 10

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 15 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks
13(a) UW = TU 2 + TW 2
= 122 + 52 2
UW = 13 N1 3
UW = 13
UR = 125 // 11.1803 N1 5 10
UR = QU 2 + QR2
= 102 + 52

UR = 125 // 11.1803

(b) Area of plane RUW = 69.2 m2 1 13 125  sin θ = 69.2 K1

1 13 125  sin WUR = 69.2 2

2

WUR = 72.2172 θ = 72.2172 N1

Since WUR is obtuse angle, then WUR =107.7828 N1
 WUR = 107.7828

( ) ( )(c) RW 2 = 132 + 125 2 − 2(13) 125 cos107.7828

RW = 19.5647 m 2
sin UWR = sin107.7828
125 −
125 19.5647 ( )RW 2 = 132 + K1
UWR = 32.9667
( )2(13) 125 cos107.7828

URW =180 −32.9667 −107.7828 19.5647 m N1

URW = 39.2505 sin UWR = sin107.7828
125 19.5647
K1

UWR = 32.9667 N1

URW = 39.2505 N1

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SULIT

SULIT 16 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks
14(a) v=0 Use v = 0 K1
4t − 8 = 0 4t − 8 = 0 2

t=2 0  t  2 N1 3
0  t  2
3
(b) s =  v dt = 2t2 − 8t + c Integrate  v dt K1

t = 0, s = 0, c = 0 s = 2t2 − 8t
s = 2t2 − 8t

When t = 2,

2 2
0
s = v dt = 2t 2 − 8t  Substitute t = *2 and t = 0 into  v dt K1

0 8 m (No) N1

= (8 −16) − 0 = −8 = 8 m

Particle P didn't reach R

(c) 2 5 Substitute t = 4 or t = 5 into K1

s =  v dt +  v dt  v dt

02

( ) ( )=  
−8 +  2(5)2 −8(5) − 2(2)2 −8(2) 

= 8 + 10 − (−8) 25

s =  v dt +  v dt K1

= 26 02

Total distance travelled = 26m 26m N1

(d)

 shape N1

All correct N1

2 10

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SULIT 17 3472/2

No. Solution and Mark Scheme Sub Total
Marks Marks

15(a) I 20 ( A) = 135 x 100 = 135 K1
50
18

x 100 = 135 RM67.50 N1
50

x = RM67.50 2

(b) I 20 ( C ) = 125

18

P20(C) = 18 + P18(C) 18 + y 100 =120 K1
y
z = 18 + y

18 + y 100 = 120 y = RM90.00 N1
y

 y = RM90.00 z = RM108.00 N1

z = 90 +18 3
z = RM108.00

(c)(i) I 20 ( A) = 135 ; I 20 ( B) = 160 160 or 110 P1

18 18 135(1) + *160(1) +120(1) + *110(1)

I 20 (C ) = 120 ; I 20 ( D ) = 110 K1
4
18 18 131.25 N1

I 20 = 135(1) +160(1) +120(1) +110(1)

18 4

 I 20 = 131.25 3

18

(ii) 1716 100 = 131.25 1716 100 K1
P18
P18
 P18 = RM1307.43 131.25 N1 2 10

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3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 18 3472/2
Graph for Question 9(a)

40

×
(36.00, 37.50)

35

30

×
(25.00, 27.00)

25

20
×
(16.00, 18.04)

15
×
(12.25, 14.25)

10

×
5 (4.00, 6.50)

×
(1.00, 3.00)

0 5 10 15 20 25 30 35 40

3472/2 © 2020 Program Gempur Kecemerlangan SPM Negeri Perlis SULIT

SULIT 19 3472/2
Graph for Question 12(b)
y
(S-Ziez)

80
70

R60

50
40
30
20
10

0 10 20 30 40 50 60 70 x (H-Ziez)

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SULIT