Marking Scheme PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2012 Mathematics T Paper 1 (954/1)/ Mathematics S Paper 1 (950/1) 1.(a) log 4 x + log x 4 = 2.5

log 4 x + x4

4

log4log

= 25 M1

log 4 x + x4log

1 = 25

2(log 4 x) 2 - 5 log 4 x + 2 = 0 M1 ( 2log 4 x - 1) (log 4 x - 2) = 0 M1 2log 4 x = 1 , log 4 x = 2 M1 x 2 = 4 x = 4 2 x = 2 , x = 16 A1 [5]

2. y = x

xcos3

xy = 3cos x B1

xdx

dy + y = -3 sin x B1

x 22

dx

yd + dx

dy + dx

dy = -3 cos x B1

x 22

dx

yd + 2dx

dy = - xy B1

x 22

dx

yd + 2dx

dy + xy = 0 A1 [ 5 ]

3. GP , T 3 = ar 2 = 6 ,

T 6 = ar 5 = - 92 either one correct M1

r = - 31 , a = 54 both correct B1

(a) S n =

)31(1

)31(1

54

n

M1

=

n)

31(1

281 A1

(b) S α = )

31(1

54

M1

= 281 A1 [6]

4. 12)1)(2(

4122

2

x

CBx

x

A

xx

xx M1

1 – 4x - x 2 = A( x 2 + 1 ) + ( Bx + C)( x + 2) When x = -2 , 5 = 5A A = 1 Equating coefficient of x 2 , -1 = 1 + B B = -2 Equating constants , 1 = 1 + 2C C = 0

Hence, f(x) = )1)(2(

412

2

xx

xx = 1

22

12

x

x

x A1

1

02

1

0 12

21()( dx

x

x

xdxxf M1

= [ln | x + 2 | - ln | x 2 + 1| ] 10 M1

= 1

02 |1

2|ln

x

x M1

= ln 23 - ln 2

= ln 43 A1 [6]

5. ( A – B ) ∩ ( A ∩ B)’ = A ∩ B’ Use left hand side: = ( A ∩ B’ ) ∩ ( A’ U B’) A1 = [(A ∩ B’) ∩ A’] U [(A ∩ B’) ∩ B’] A1

= [ (B’ ∩ A) ∩ A ’)] U [(A ∩ B’) ∩ B’)] A1 = [B’ ∩ (A ∩ A ’)] U [A ∩ (B’ ∩ B’)] A1 = [B’ ∩ Ø] U [A ∩ B’ ] A1 = Ø U [A ∩ B’ ] A1 = A ∩ B ‘ A1 [7]

6. (a) 0)21(3

212

kk

k

)1)(13(21kk

k 0 B1

Solve the inequality using appropriate method M1

k < -1 or 21

31

k A1

6. (b) 1531

x

x

|53|1 xx 22 )53()1( xx M1 0)53(1 22 xx M1 -8x2 +32x-24 < 0 -8(x – 1)(x-3) < 0 M1 The solution set is { x : x < 1 , x > 3} A1 [7] 7. (a) )ln( 22 yxy yxy lnln 22

dx

dy

yxdx

dyy

122 M1A1

xdx

dy

yy

212

)12(

22

yx

y

dx

dy A1

(b) 01.0y

21

ex B1

xdx

dyy

xe

)12(

)1(201.021 M1A1

008.0x (3 d.p.) A1 [7]

8. 2

121

)21()1( xx

=

...)2(!2

23

21

)2(211...

!221

21

)(211 22 xxxx M1

8. =

...

231...

81

211 22 xxxx A1A1

= ...8

15231 2 xx A1

The expansion is valid when 1x and 12 x

21

x M1

21

21

x

Since

21,

21

41 , therefore

41 can be used to estimate the value of 2 A1

2

41

815

41

231

4121

411

M1

12895

21

4844.12 or 1.3474 A1 [8]

9. (a) P(x) = 2x3 – 3ax2 + ax + b Find two linear equations and solve for a and b M1

2(1)3 – 3a(1)2 +a(1) + b = 0

-2a + b = -2 P(-2) = -54 2(-2)3 – 3a(-2)2 +a(-2) + b = -54 -14a + b = -38

a = 3, b = 4 A1 P(x) = 2x3 – 9x2 + 3x + 4 = (x – 1)(2x2 – 7x – 4) M1 = (x – 1)(2x + 1)(x – 4) A1 2x6 – 9x4 + 3x2+ 4 = (x2 – 1)(2x2 + 1)(x2 – 4) M1 The zeroes are -2 , -1 , 1 and 2 A1

(b) )2)(1(

93

xx

xy

93)2( 2 xxxy yx2 – yx - 2y = 3x-9 yx2 – (y+3)x + (9 – 2y) = 0 M1

(9) For all real values of x is real, 042 acb (-y-3)2 – 4(y)(9 – 2y) 0 M1 3y2 – 10y + 3 0 (3y – 1)(y – 3) 0

331

yory A1

y does not have any real value between 31 and 3 for all real values of x. A1 [10]

10. (a) 12

yt

21

21

2yy

x M1

14

2

y

x

y2 = 4(x + 1) A1

(i) Vertex = (-1, 0) B1 (ii) Focus = (0, 0) B1 (iii) Directrix , x = -2 B1

(b) x = t(t-2) y = 2(t – 1)

4 = 2(t – 1) t = 3 x = 3 B1

22 tdt

dx 2dt

dy

11

222

ttdx

dy M1

21

dx

dy A1

The equation of normal: y – 4 = -2(x – 3) M1 y = -2x + 10

1105

yx A1

(c) 2x + y – 10 = 0 , (-1, 0)

22 12

100)1(2

d M1

= 5

12

=5

512 A1 [12]

11(a) 1812

411012

011018

0

A M1

= 24 A1 Since 0A , therefore matrix A is non-singular A1 (b)(i) k = 23 k = 23M

= 10420

M1

= 8 A1

(ii)

0832044682

2411A M1

=

00

31

34

61

61

41

31

121

A1

(c) 2b – c = 1 4a + 10b – c = 29 8b – c = 16 M1A1

(d)

29161

1104180120

c

b

a

B1

29161

0832044682

241

c

b

a

M1

=

32025

2 A1

a = 2, b = 25 , c =

320 A1 [13]

12.(a) (i)1

)1(2lim)(lim11

x

xxf

xx

= )1()1(2lim

1

x

x

x M1

= -2 A1

12(a) (ii) )(lim)(lim11

xfxfxx

2)4(lim

1

kxx

M1

2)1(4 k k = 6 A1

(b)(i) )1,0(fR or {y : 0 < y < 1} B1 ]0,1[gR or {y : –1 ≤ y ≤ 0} B1

(ii) ]0,1[gR ; )0,(fD

Since )0,(]0,1[ because ]0,1[0 but )0,(0 M1 That is fg DR gf does not exist. A1

(iii) From the graph of function f, function f is one-to-one in the domain given because any horizontal line y = c , c constant, that cuts the graph of f, cuts only at a point. Therefore function f has an inverse function 1f . B1

Let yxf )(1

)0i.e(DyandRybecause1

11

1)(

ff

2

2

1-

yx

xy

x

xy

yx

yfx

x

xxf

1)(1

D1 : curve y = f(x) or y = g(x) D1 : curves y = f(x) and y = g(x) D1 : both graphs correct. x

y

1 – 1

– 1

1 y = f(x)

y = g(x)

M1

Therefore 10,1:1 x

x

xxf A1 [14]

MARKING SCHEME FOR MATHEMATICS S PAPER 2

No. Solution Mark

1(a) ∑

∑ ∑

= 3581 – ( )( )

= 11

∑

(∑ )

= 3625 –

= 250

∑

(∑ )

=

Pearson’s correlation coefficient,

r =

√

√( )(

)

M1 (Calculating any of the Sxx, Syy, Sxy) M1 A1

1 (b) Coefficient of determinant, ( )

3.08% variation in y can be explained by x.

M1 A1 B1

2(a) Z0.025 = 1.96 A 95% confidence interval for the proportion of lecturers who agreed that the standard of higher education is high

= ( √

( )

)

= ( √ ( )

)

= (0.24 ) = (0.203, 0.277)

B1 M1 for 0.24 M1-correct K A1

2(b)

√ ( )

√ ( )

(

)

( )( )

M1 for M1-substitution B1-z value 1.645 M1 A1

3(a)

B1 A1

3(b) f(x) x

1 2 3 4 5

D1-Uniform scale and at least 3 correct D1-all correct

3(c) Mean =E(x) = ∑ ( )

= (

) (

) (

) (

) (

)

=

( ) ∑ f(x)

= (

) (

) (

) (

) (

)

=

Standard deviation = √ ( ) [ ( )]

= √

(

)

= 1.28

B1 B1 M1 A1

4(a) Unbiased estimate for the population mean µ

= ∑

= 80 Unbiased estimate for the population variance

= ∑ (∑ )

( )

( ) ( )

( )

= 16

B1 M1 A1

4(b) Z0.025 = 1.96 A 95% confidence interval for the population proportion of students who completed the examination

= ( √

( )

)

= ( √ ( )

)

= (0.8 ) = (0.745, 0.855)

B1 M1 for 0.8 M1-correct K A1

4(c) 80% lies in the interval of (0.745, 0.855). Therefore, teacher Tan’s claim is true.

B1-true B1-reason

5(a) ( )

( )

( )=

( )

( )

( )

( )

( )

M1

( ) ( )

[ ( )]

( )

[

( )]

( )

( )

M1 M1 A1

5(b)(i) ( )

( )

(

)

M1 A1

5(b)(ii) ( )

( )

( )

M1A1

5(c) Since ( ) ( ) A and B are not independent events.

B1-deduction B1-reason

6(a) ( )

(

)

----I ( )

(

)

I-II: 0.901 = 2.927 (3dp) Substitute into I, ( ) µ = 18.138 (3dp)

M1 M1 B1-both equations correct A1 A1

6(b) ̅ (

)

( ̅ )

(

)

( ) ( ) ( ) = 0.1957 – 0.00252 = 0.193

B1 B1 for ̅ M1 A1

7 1 2 3 4

2005 - - 1.486 0.736

2006 0.568 1.011 1.556 0.860

2007 0.599 1.106 - -

Average S 0.584 1.059 1.521 0.798

Correction factor

1.0096 1.0096 1.0096 1.0096

Adjusted S 0.590 1.069 1.536 0.806

A1-correct

values of

M1-correct average B1-correction factor M1-adjusted S

8(a)

The distribution is skewed to right.

D1-uniform scale and correct values D1-all correct B1

8(b) Mean = ∑

∑

Standard deviation = √∑

∑ ( ̅)

√

(

)

= 4.97 minutes

M1 A1 M1 A1

9(a)

D1 D1

9(b) ∑ ∑

∑

∑ ∑

B1

0

10

20

30

40

50

60

-0.05 2.95 5.95 8.95 11.95 14.95 17.95 20.95

Frequency

Duration (Minutes)

0

10

20

30

40

50

60

70

80

0 5 10 15

Uniform scale

and at least 5

correct points

All correct

∑ ̅ ̅

∑ ̅

( ) ( ) (

)

( ) ( )

=4.49 ̅ ̅

=

(

)

= 15.27 The equation of the regression line: y = 15.27 + 4.49x

( ̅ ̅) ( ) Straight line passing through his ( ̅ ̅)

M1 (his value) A1 B1 A1 D1 D1

9(c) y = 15.27 + 4.49 (8.5) = 53.44 (nearest integer)

M1 A1

10 (a) Type of handphone

Sales quantity

2009 (q0) 2010 (qn)

A 26 40 153.85

B 18 38 211.11

C 19 21 110.53

D 30 25 83.33

Average relative quantity index = ∑

=

= 139.71

B1 M1 A1

10(b) Paasche price index = ∑

∑

= ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

= 123.23 The price of handphones in 2010 has increased by 23.23% compared to that of 2009

M1 A1 B1

11(a)

D1-correct sequence and flow D1-correct activities and duration

1

000

0 0

2

000

2 5

3

000

3 14

4

000

6 6

A

C

2

6

B

3

5

000

11 14

7

000

19 19

6

000

15 15

D

E

5

F

G

H

5

9

4

9

x

11(b) Activity Total float

A 3

B 11

C 0

D 3

E 11

F 0

G 3

H 0

Since the total float for activities C, F and H are zeroes, therefore the critical path is C-F-H and the minimum completion time is 19 weeks.

B1-correct values of EST B1-correct values of EFT B1- correct values of LST B1-correct values of LFT B1-correct total float B1-critical path B1-minimum completion time

11(c) There is no effect on the completion time of the project. Reason: Total float for activity E represents the maximum time that it can be delayed ( 8 weeks < total float for activity E)

B1-no effect B1-reason

12(a) Maximize profit: Subject to:

B1 B1 B1 B1

12(b) Tableau 2

Tableau 3

Basic P X1 X2 X3 S1 S2 S3 Solution

P 1 -20 0 -2 6 0 0 3000

X2 0 1

0 0 50

S2 0 -

0

1 0 50

S3 0 2 0 1 0 0 1 180

Basic P X1 X2 X3 S1 S2 S3 Solution

P 1 0 0 8 6 0 10 4800

X2 0 0 1

0

5

S2 0 0 0

1

95

X1 0 1 0 0 0

90

M1-attempt to do row operation A1-all values in solution column correct M1-attempt to do row operation A1-all values in tableau correct

12(c) 90 units of tables, 5 units of sofas and none of chairs should be produced per week in order to maximise the total profit. The maximum profit is RM4800.

A1 A1

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