Jpwp Mark Scheme Spm2008 Trial Add Maths P2

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JABATAN PELAJARAN WILAYAH PERSEKUTUAN KUALA LUMPUR

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2008

SKEMA PEMARKAHAN

ADDITIONAL MATHEMATICS

PAPER 2

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SECTION A Solution

Sub Mark

3r2 + rs + 6 = 7 ---(1) 3r + 2s = 7 --------(2) 7  3r s …………………………………………………… 2  7  3r  3r 2  r    6  7 ………………………………………….  2 

1

Question 1.

Full Mark

1

3r 2  7r  2  0  7   (7) 2  43 2

…………………………………….

1

r  0.257,  2.591 ……………………………………………….

1

s  3.115, 7.387 ……………………………………………….

1

r

23

[5]

2. (a)

dy  3(3  2 x ) 2 (2) ………………………………………. dx dy ………………………………………………..  6 dx y  1  6( x  1) ………………………………………..

y  6 x  7

………………………………………………..

(b) l  2 r dl (i)  2 dr dr dr dl   dt dl dt 1   (0.2) ……………………………………………….. 2  0.03183 cms 1……………………………………………..

1 1 1 1

4

1 1

(ii) Initial r 

60 2 60  5(0.03 183) ………………. 2 (3.142 ) r  9.707 …………………………

A fter 5s, r 

1 4 1 [8]

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2 3. (a)

(b)

POQ  10 = 11.2 ……………………………………….  POQ = 1.12 radians …………………………………… OR ………………………………………. 10 OR = 4.357 cm …………………………………… RQ = 5.643 cm ……………………………………

Cos 1.12 

1 1

2

1 1 1

3

(c) Area of sector POQ  area of ΔPOR  area of quadrant RSQ 1 1 1   10 2  1.12   10 sin1.12   4.357    5.643 2 2 2 2 2  56  19.609  25.01 

 11.38 cm

2

……………………………………………….

1, 1 3 1 [8]

4. (a)

The amount of savings at the end of every year forms a G.P with a = 5000 r = 1.035 ..............................................

Tn > 6000 5000 (1.035) n – 1 > 6000 ................................................. (1.035) n – 1 > 1.2 ( n – 1) log 1.035 > log 1.2 ........................................... n – 1 > 5.30 n > 6.3 n = 7 ................................................ (b)

5.

(a) (i)

(ii)

T15 = 5000 (1.035) 14 ................................................................................. = 8093.47 ................................................................. 0

1  2 P ( X  0) 10 C 0     3  3  0.01734

1

1 1

1 1 1

[6]

10

…………………………… ……………………………

1 1

2

P ( X  2)  1  P ( X  0)  P ( X  1) 1

= 1  0.01734   0.8960

10

9

1 2 C1     …………  3   3  …………

………………………………..

1 1

2

(b) 1 ( i) x  (600)  200 …………………………………………. 3  1  2  600    ……………………………………  3  3  ……………………………………  1 1 .5 4 7 ……………………………………………

(ii)  

1 1 1

3 [7]

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SULIT 6.

h2 k Let

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 100 ………………………………………………… u  pv hi  k j  3 pi  4 p j ………………………………………. 3p = h or 4p = k

4 3 k  h, or h  k 3 4

OR

h2 

Then, (3p)² + (4p)² = 100 OR

 34 k  Then, h = 6 and k = 8

 43 h  2

2

 1 0 0 or

1

1 1 1

 k 2  100

………………………………………..

1,1 [6]

SECTION B Question

Sub Mark

Solution

Full Mark

7. Refer to the attachment on page 8. 8. 3 1 1  or m PR  2 ………………………. 1 3 2 Point N ( 1, 2) ………………………………………………...

(a) mQS 

The equation of PR is

y – 2 = 2(x – 1) y = 2x

When x = –1 , then Therefore R (–1, –2)

1 (1)  2 x or 3 x=2 or y=4 1 Area of PQRS = 2 1 = 2 = 15

(b) 1 

(c)

OR

2 y 2 1 (1) y  2

y = 2(–1) = –2 . ………………………………………..

2

1 (2)  2 y 3

1 1 1

1

………………..

1

6  2  1  12  4  3  6  2 ………….

1

2 1 1 3 2 4 3 2 1 4

………………………………….…….

3 1

(1  2) 2  (3  4) 2 ……..

1

x 2  4 x  4  y 2  8 y  16  4 (9  1) ……………………. x 2  y 2  4 x  8 y  20  0 …………………………...

1 1

( x  2) 2  ( y  4) 2  2

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4 9.

(a) Construct lines to get modal marks ………………………….. Mode = 66.17 (accept range 65.50 – 66.50) ………………..

 45   7   10  48 ……………………………. (b) Q1  39.5   4 5        3  45   22   Q3  59.5   4  10  69.292 ………………….. 12       Interquartile range = 69.292 – 48 ……………………… = 21.292 ……………………………… (c)

7(34.5)  5(44.5)  10(54.5)  12(64.5)  11(74.5) 7  5  10  12  11  57.833 ……………………………………………

Mean =

1 1

2

1

1 1 1

4

1 1

Standard deviation 7(34.5) 2  5(44.5) 2  10(54.5) 2  12(64.5) 2  11(74.5) 2  57.8332 7  5  10  12  11

= 13.664 ……………………………………………………… 10.

(a)

2

2x + 3 = 4x + 9 ……………………………………………. (x + 1)(x – 3) = 0 …………………………………………… x = -1, h = 5 ……………………………………… x = 3, k = 21

1 4 1

10

1 1 1

3

(b)

1  5  21 4  or 2

3

 2 x3   3  3 x  …………………………….   1

 2  3 3   2   1 3   3 3      3   1   ………………………   3    3     Area of trapezium – area under a curve = 52 – 30 2 ……………………………………………… 3 1 = 2 1 u nit 2 ……………………………………………. 3

1 1

1 1

4

21

(c)

 1  y2    3 y  or 1  (3 2 )( 2 1  9 ) ………………… 2 2 3 3 = 1    2 1 2  3 2 1    3 2  3 3   ……………….        2   2   2

1 1

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Volume of revolution = Volume generated by the curve – Volume of cone = 81 - 36  = 45  unit3 ………………………………………………..

1

3 10

11.

(a)

L.H .S .  sin x cos x (cos 2 x  2 cos 2 x  3)  sin x cos x (2 cos 2 x  1  2 cos 2 x  3) …………… …………..  sin x cos x (2)  2 sin x cos x

………………………………

 sin 2 x

………………………………

1

1 1

 R.H .S

3

(b) y

y = sin 2x

1

y

1 2

O 1

2

-1

 2



3  2

x 1  2 2 x

2

Sketch the graph of y = sin 2x : ……………… sin curve with 2 cycles ……………… ……… max = 1, min = -1 , x- intercepts correct ……… x 1 π x 1 sin x cos x (cos 2 x  2 cos 2 x  3) =  2π 2 x 1 From (a), sin 2 x   2π 2

1, 1 1

2 sin x cos x (cos 2 x  2 cos 2 x  3) =

Recognised the required straight line is y 

Draw the straight line y 

x 1  2 2

………… ………… x 1  2 2

… …

1

1

1 7

 From the graph, no. of solutions = 3

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SECTION C Question

Solution

Sub Mark

Full Mark

1 1

2

12. (a) 15 – 3 t > 0 ………………………………………………. 0 < t < 5 ……………………………………………… 3 (b) s  15t  t 2 2 3 = 15(5) - (5)² ………………………………………….. 2 = 37.5 ……………………………………………………. Particle P reaches B. …………………………………………. (c) When t = 8, s = 15 (8) –

3 (8)² …………………………….. 2

1 1 1

3

1

s = 24 Total distance travelled = 2 (37.5) – 24 or 37.5+(37.5-24) = 51 m …………………………….. (d) sp

1

3

1

37.5 24 t 0 5 8 Shape of the curve ……………………………………………….. Critical points (0,0), (5, 37.5), (8, 24)…………………………….

13.

11 9  sin ABC sin 320 ………………………………………  ABC  40.37o ………………………………………… ABC is an obtuse angle  ABC  139.63o or 139o38’ ………………… 1 (ii) 32.86   11 6  sin ACD …………………………….. 2  ACD  84.72 O or 84o43’……………………… (iii) AD 2  112  6 2  2116 cos 84.72 o ……………………..

(a)(i)

AD  12.04 …………………………………………….

1 1

2 10

1 1 1 1 1 1 1

7

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b) ( i)

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D C 11 cm 9 cm 9 cm

A

32O

B’

B

1

AB’C drawn ………………………………………………… 1 (ii) Area of  AB ' C   11 9  sin107.630 ……………………. 2 = 47.18 cm² ………………………………..

1 1

3 10

14.

Refer to the attachment on page 9.

15. (a) (i) (ii) (iii)

x = RM1.20 ………………………………………. y = 160 ………………………………………… z = RM5.40 ……………………………………….

1 1 1

3

(b) I =

175 (14 )  125 (18 )  160 (16 )  135 (19 )  110 (13 ) 14  18  16  19  13

I = 140.7 ...................................................................................

(c)

(d)

Expenditure 2004

1

 100  140.7 ...................................................

1

Expenditure 2004 = RM962.39 ……………………

1

684



1,1

115  140.7 …………………………………. 100 = 161.8 …………………………………………….

I 2007 / 2000 

3

2

1 1

2 10

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8

QUESTION 7 x+1 log10 y

1.5

2

3

4

5

6

7

0.362

0.415

0.550

0.672

0.799

0.919

1.049

At least 2 decimal places  1

log10 y

1.1 

1.0 

0.9 

0.8



1 2 1

(5)

y = h kx+1

(b) 0.7

  

 axes and scales  plotted points  line of best fit

(a) graph

 log10 y = log10 h + (x+1)log10 k

 1

0.6 the two points



i) Gradient = 0.799 – 0.175  selected must lie 5–0 on the graph 1

0.5 0.48

log10 k = 0.1248 ± 0.01



0.4

k = 1.333 ± 0.1 



1

0.3 ii) y-intercept, log10 h = 0.175

h = 1.496  ± 0.1

0.2 0.175

iii)

0.1

O

 0.01

log 10 y = 0.48 ± 0.01 y = 3.02  ± 0.1

1 1 (5) [10] [10]

1

2 2.4 3

4

5

6

7

8

x+1

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y

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Question 14

110 (a) 100 7x + 4y = 420

70x + 40y  4200 20x + 80y  1600 y  2x

1 1 1 (3)

(b) One correct line All three lines correct Shaded region

90 y = 2x 80

1 1 1

(3)

(c) i) Profit =20x + 10y Max point = (56, 6) 1 Max profit = 20(56) + 10(6) = RM1180 1

(2)

ii) Draw line x = 2y Max number of Powermax x = 46 ± 1

70

1 1

(2) 10

60

50 Profit function : P = 20x + 10y

40

R

30

x = 2y

20

10 X

Maximum point (56, 6) x + 4y = 80

O

10

20

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40

46 50

60

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x