Jawapan Mt Kertas 2 Pahang 2008

SULIT 3472/2 Additional Mathematics Paper 2 September 2008

SEKTOR PENGURUSAN AKADEMIK JABATAN PELAJARAN PAHANG

PEPERIKSAAN PERCUBAAN SPM TAHUN 2008

ADDITIONAL MATHEMATICS Paper 2

MARKING SCHEME This marking scheme consists of 12 printed pages

WORKING / SOLUTION

QUESTION

1

2 (a)

(b) (c)

3(a)

(b)

3x + 2y +1 =8 x2 + 9 x - y = 8 7 − 3x 7 − 2y y= or x= 2 3 7 − 3x 7 − 2y Substitute y = or x = 2 3 into non linear equation 7 − 3x x2 + 9 x − ( ) = 8 or 2 7 − 2y 2 7 − 2y ( ) + 9( )− y =8 3 3 (2x+23)(x-1)=0 or (4y-83)(y-2) =0 or solve quadratic equation using formula or completing the squares

MARKS

P1

P1

K1

K1

x = -11.5, x =1

N1

y = 20.75, y =2

N1

uuur PR = 2 i − 2 j % r uuu uuur uuu uuur uuur uuur % r QS = QR + RS or QS = QR + QP uuur QS = 4 i + 4 j uuuur %uuur % P ' R = k RS uuuur P ' R = −3 i + 4 j + 5 i + 2 j % uuuur uuuur uuur % % % P ' R = 2 i + 6 j = 2( i + 3 j ) => P ' R = 2 RS =>collinear % % % % P ' R : RS = 2 :1 sin A cos A + , 2sin A cos A 1 + (2 cos 2 A − 1) use of sin2A = 2 sinA cos A or cos 2A = 2 cos 2 A-1 1 1 1 + = 2 cos A 2 cos A cos A = secA 2 ( sec2x -1) = sec x +1, use of 1 + tan2 x = sec2 x ( 2 sec x – 3) ( sec x +1 ) =0 3 or sec x = -1 sec x = 2 2 cos x = or cos x = -1 3 X = 48.19o ,180o ,311.81o or

= 48o11' ,180o ,311o 49'

6

N1

P1 N1 K1

N1 N1

6

P1 K1 N1 P1 K1 K1 K1

N1

2

TOTAL

8

SULIT

Question 4(a)(i)

(ii)

4(b)

Working / Solution 22(18) + 27(22) + 32(29) + 37( p) + 42(25) = 32.75 18 + 22 + 29 + p + 25 p = 26 60 − 40 median = 29.5 + ( )(34.5 − 29.5) 29 = 32.95 Height of the bars proportional to the frequency or Label the lower and upper boundaries/mid points/class interval correctly. Correct way of finding the value of mode.

Marks K1

Total

N1 K1 N1 K1 K1 7

5(a)

Gradient of normal = −

5(b)

N1 K1

Modal mass = 33 dy = 3x 2 + 4 x dx dy (1,-1) , = 7 dx 1 7

K1

Equation of normal: 1 y-(-1) = − ( x-1) 7 7y+x + 6=0 dy = −24 x −4 dx δ y −24 (1.98 − 2) ≈ δ x 24

K1

K1

K1 N1

≈ 0.03

6

6(a)

A =8000 8000, 8000(0.9), 8000(0.9)2,… r =0.9 8000 sα = 1 − 0.9 = 80,000

P1 K1 N1

3

SULIT

Question 6(b)

Working / Solution 8000(0.9) > 1000 ⎛1⎞ (n-1) log10 (0.9) > log10 ⎜ ⎟ , taking log both sides ⎝8⎠ n < 20.74 n = 20 n−1

Marks

Total

P1 K1 K1 N1

7

7(a) x 15 20 25 lg y -0.82 -0.42 -0.022 Plot log10 y against x 6 points plotted correctly

7(b) i.

ii iii. 8(a)

30 0.37

35 0.77

N1

K1

Line of best fit, ( passes through as many points as possible and balance in terms of numbers point appear above and below the line, if any .)

N1

log10 y = − A log10 10 + x log10 b y-intercept, c = -A log10 10 -2.05 = -A log10 10 A = 2.05 m = log10 b = 0.08 b =1.2 37.5 1 + (−3) 2 + 10 , ) Mid point of AC ( 2 2 (-1 , 6 ) 10 − 2 Gradient of AC = 1 − (−3) =2

P1

Gradient of perpendicular line to AC = −

(b)

40 1.17

P1 N1 K1 N1 N1

10

N1

1 2

Use of m1m2 = −1 Equation of perpendicular bisector AC 1 (y-6) = − ( x + 1) 2 2y + x =11 or equivalent y =0, x =11 => B(11, 0) 1 −3 11 1 −3 Area of rhombus ABCD = 2× 2 2 0 10 2 = 120

4

N1

K1 K1 N1 K1 K1 N1

SULIT

Question 8(c)

Working / Solution Use of distance formula for PA or PC

Marks

Total

K1

PA = ( x − 1) 2 + ( y − 10) 2 or PC = ( x + 3) 2 + ( y − 2) 2 Use 2PC = PA,

K1

2 ( x + 3) 2 + ( y − 2) 2 = ( x − 1) 2 + ( y − 10) 2 9(a)

3x 2 + 3 y 2 + 26 x + 4 y − 49 = 0 y =3x , y =4-x2 (x+4)(x-1)=0 solve simultaneous equation x =1, x = 4 4-x2 =0, x= ±2 or find the limits of integration 1 3 Use area of triangle = (1)(3) = or 2 2

(b)

K1

2 ∫ (3x)dx or ∫ (4 − x )dx 1

0

1

2

⎡ 3x ⎤ ⎡ x ⎤ ⎢ 2 ⎥ or ⎢ 4 x − 3 ⎥ ⎣ ⎦0 ⎣ ⎦1 ⎡ 8⎤ ⎡ 1⎤ Substitution, ⎢8 − ⎥ − ⎢ 4 − ⎥ ⎣ 3⎦ ⎣ 3⎦ 2 = 1 unit2 3 3 2 Add up 2 area, + 1 2 3 1 3 unit 2 6 1 Volume of cone = π (3) 2 (1) = 3π or 3 2

10

2

1

Integrate

N1

3

K1

K1

K1 N1

K1

1

⎡ 9 x3 ⎤ π⎢ ⎥ ⎣ 3 ⎦0 3π 2

π ∫ (4 − x 2 ) 2 dx 1

2

⎡ 8 x3 x5 ⎤ = π ⎢16 x − + ⎥ 3 5 ⎦1 ⎣ ⎡ 8(2)3 25 ⎤ ⎡ 8(1)3 13 ⎤ + ⎥ − ⎢16(1) − + ⎥} = π {⎢16(2) − 3 5⎦ ⎣ 3 5⎦ ⎣ 5

K1

K1

SULIT

Question

Working / Solution =3

Total

8 π 15

Volume generated = 3 π + 3

N1

8 π 15

8 π 15 p = 0.3 or q =0.7 P(X=5)= 30C5 (0.3)5 (0.7) 25 Use of P(X=r)=

=6 10(a) i.

Marks

Cr ( p ) r ( q ) n − r =0.04644 P(X<2) = P(X=0) + P(X=1) 30 C0 (0.3)0 (0.7)30 + 30C1 (0.3)1 (0.7) 29

N1

10

P1 K1

n

ii.

(b)i

ii

11(a)

11(b)

=0.0009660 or 9.660 × 10−4 P ( z < c ) =0.202 C =-2.05 25 − µ X −µ −2.05 = , use of z = 3 σ µ = 31.15mm 30 − 31.15 32 − 31.15
N1 K1 N1 N1 K1 N1

K1 N1 K1

10

K1

K1

K1 N1 K1

AC 8

=10.50 K1

1 Area of OAC = (8)(10.50) 2 = 42.01 6

SULIT

Question 11(b)

12

Working / Solution 1 1 Area of sector OAB = (8) 2 (0.92) use of A= r 2θ 2 2 =29.44 Area of the shaded region Q, =42.01-29.44 = 12.57 (a) v = 4 (b) v max , a = 0 a = 2t − 5 = 0 5 t= s 2 2 ⎛5⎞ ⎛5⎞ v max = ⎜ ⎟ − 5⎜ ⎟ + 4 ⎝2⎠ ⎝2⎠ 1 v max = −2 ms −1 4 (c) used v<0 t 2 − 5t + 4 < 0 (t − 1)(t − 4) < 0

1< t < 4 (d) s = ∫ t 2 − 5t + 4 dt

(

Total

K1

K1 N1

10

P1 K1

K1 N1

K1

N1

)

1

13

Marks

4

⎡ t 3 5t 2 ⎤ ⎡ t 3 5t 2 ⎤ =⎢ − + 4t ⎥ + ⎢ − + 4t ⎥ 2 2 ⎦1 ⎣3 ⎦0 ⎣ 3 Substitute the values of t 2 = 23 m 3 P (a) use 1 × 100 P0 x = 1.62, y = 152, z = 3.60 ∑ Iw used I = ∑w

120(4 ) + 152(5) + 150(2) + 125(3) 14 = 136.79 I=

7

K1K1 K1 N1

10

K1N1N1N1 K1 K1(used I) N1

SULIT

Question

14

15

Working / Solution (c) I P = 144, I R = 180 194.4(4) + 152(5) + 108(2) + 125(3) I= 14 = 147.93 (a) I x + y ≤ 500 II y ≤ 3 x III y ≥ 200 Cannot have sign ‘=’ (b) One of graph of straight line is correct All the graph of straight line are correct The shaded region of R is correct (c) (i) 200 (ii) maximum point (300,200) – based on the Graph 25 ( 300 ) + 20 ( 200 ) - substitute any number based on the value in shaded region 11500 (a) used cosine rule 2 2 QS 2 = (10.5) + (12.5) − 2(10.5)(12.5) cos 80 0 QS = 14.86 cm (b) used sine rule sin R sin 35 = 14.86 9.5 sin R = 0.89719 ∠QRS = 116.210 Q

Q'

Marks P1 K1 N1 N1 N1 N1

Total

10

K1 K1 N1 N1 N1

K1 N1

10

K1 N1 K1

N1

S

N1 P

(i) Can see anywhere in the diagram (ii) Find ∠PQS , used sine rule , hence find ∠QPQ ' K1

sin ∠PQS sin 80 o = 12.5 14.86 ∠PQS = 55.93o , ∠QPQ ' = 68.14 o

8

SULIT

Question

Working / Solution Find area of ∆PQS or area of ∆PQQ '

Marks K1 N1

Total

1 (10.5)(12.5)sin 80 o 2 = 64.63cm2

area ∆PQS =

1 (10.5)(10.5)sin 68.14 o 2 = 51.16 cm2 Find the area of ∆Q ' PS = 64.63 − 51.16 = 13.47 Or any other methods area ∆PQQ'=

9

K1 N1

10

SULIT

Graph For Question 4(b) Number of luggage

30

28

26

24

22

20

18

16

10 19.5

29.5 24.5 Modal mass= 33

34.5

39.5

SULIT 44.5

Mass(kg)

No.7(a)

log 10 x 1.5 x

1.0 x

0.5 x

x 0 5

10

15

20

x

25

30

35

40

x

-0.5 x

-1.0 K1

-1.5 N1

-2.0 N1

-2.5

11

SULIT

Graph for Question14(b) 500

400

300

R

(300, 200)

200

100

0

50

100

150

200

250

12

300

350

400 SULIT

450

500