[edu.joshuatly.com] kedah spm trial 2011 maths ts paper 1 (w ans)

CONFIDENTIAL*/SULIT*

954/1 STPM TRIAL 2011

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MARKING SCHEME FOR

MATHEMATICS T PAPER 1 (954/1)

STPM TRIAL EXAMINATION 2011

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CONFIDENTIAL*/SULIT* 2

954/1 STPM TRIAL 2011 [Please turn over

1. Determine the value of m if i

miz524

5

is a real number and find this real number

[ 4 marks ] Solution:

iix

imiz

524524

5245

M1

=2016

5241054

mmii

= 36

)104(36

5254 imm

M1

Z is a real number, 036

104

m

4m + 10 = 0

m = -25 A1

z = 45 A1

[4]

2. Given that 53x . Express x

x 2 in the form rqp .

State the values of randqp , . [4 marks] Solution:

53

253

=535612

M1

=5353x

535612

M1

= 523

23 M1

5 ,23 ,

23

rqp , All correct A1

[4]




3. A cylinder open at one end is constructed from thin metal. The total surface area of the cylinder is 192 cm 2 . The cylinder has radius of r cm and height of

h cm. Show that the volume V cm 3 of the cylinder is, V = 319221 rr .

Given that the radius of the cylinder can vary, find the maximum value of V.

[6 marks] Solution: 19222 rhr

r

rh2

192 2 B1

rrrhrV

2192 2

22 M1

319221 rrV A1

2

2396 r

drdV

M1

02396,0 2 r

drdV

8,642 rr . B1

.512)8(2

)8(96 33max cmV

A1

[6]

4. In a geometric progression , the third term is 6 and the sixth term is 92

.

Find the first term and the common ratio of the series. [2 marks] Hence, find (a) nS , the sum of the first n terms of the series. [2 marks]

(b) S , the sum to infinity of the series. [2 marks]

Solution: ar2 = 6 , a = 54 either one correct M1

r = -31 , a = 54 both correct A1




(a) nS =

n

311

311

54 M1

=

n

311

281 A1

(b)

311

54S M1

= 281 A1

[6] 5. The function f is defined by

3, 53, 53

)( 2 xxxxx

xf

(a) Without sketching the graph, determine whether f is continuous at x = 3. [4 marks] (b) Sketch the graph of f in the domain [0,5] and state the range of f. [3 marks] Solution: (a) 6)(lim

3

xxf B1

6)(lim

3

xxf B1

6)( xf B1

3

)(limx

xf = 3

)(limx

xf = )3(f =6, f is continuous at x = 3. A1




(b)

Straight line and curve meet at (3,6), all points correct. D2 Range = [0,15] B1 [7]

6. Given that .431)4)(31(

135822

2

xCBx

xA

xxxx

Show that C = 0 and determine the values of A and B. [3 marks]

Hence, evaluate

0

1 2

2

)4)(31(1358 dx

xxxx correct to 3 decimal places. [3 marks]

Solution:.

)31)(()4(1358 22 xCBxxAxx

Let x = 2491

9113

358,

31

AA B1

Let x = 0, 8 = 4A + C 0C B1 Let x = 1, 8 + 5 – 13 = 5A – 2B – 2C 5 B B1

dxx

xx

dxxx

xx

0

1 2

0

1 2

2

45

312

)4)(31(1358

= 0

1

2 4ln2531ln

32

xx M1

=

5ln

254ln

324ln

251ln

32 M1

= 0.366 A1 [6]

6

15

3 5

0 x

y




7. Find the equation of the tangent at the point P

tt 2,2 to the rectangular hyperbola xy = 4.

[3 marks]

The tangent meets the y-axis at point A. The straight line that passes through point A and parallel

to the x-axis meets the hyperbola at point Q. Find the coordinates of point Q. [3 marks] If M is the midpoint of chord PQ , find the equation of the locus of point M. [3 marks]

Solution:

y = x4

24xdx

dy

At point P, 244tdx

dy

= 21t

M1

Equation of tangent at point P is

)2(12

2 txtt

y M1

t2y -2t =-x+2t x + t2y = 4t A1 on y –axis , x=0 t2y = 4t

y = t4 M1

The coordinates of point A are ( 0, t4 )

When y = t4 , xy = 4

x(t4 ) = 4 M1

x = t

the coordinates of point Q are (t, t4 ) A1

the coordinates of point M =

2

42

,2

2 tttt

B1




=

tt 3,

23

Let x = 23t and y =

t3

xy =

tt 3

23 M1

xy = 29

A1 [9] 8 (a) Find the coefficient of 5x in the binomial expansion of 8)3( x . [2 marks]

(b) Expand 21

)21(

x as a series in ascending powers of x up to and including the term in 3x expressing the coefficients in their simplest form. State the range of values of x for which the expansion

is valid. By substituting 100

1x , estimate the value of 2 correct to four decimal places.

[7 marks] Solution:

(a) 6T = 53358

x

M1

Coefficient of 5x = 1512 A1

(b) ...26

)25)(

23)(

21(

22

)23)(

21(

)2(211)21( 322

1

xxxx M1

= ...25

231 32

xxx A1

Range of x for which the expansion is valid : .21

21:

xx B1

Substituting 100

1x

LHS = 21

)100

1(21

= 7

25 M1

RHS = ...100

125

1001

23)

1001(1

32




1.01015 M1

7

25 1.01015 M1

4142.12 A1 [9]

9. Given that f(x) = x4 – 3x3 + ax2 + 15x + 50, where a is a constant, and x + 2 is a factor of f(x), find (a) the value of a, [2 marks] (b) f(5) and hence factorise f(x) completely into linear factors, [3 marks] (c) the set of values of x for which f(x) > 0, [2 marks] (d) the set of values of x for which f(|x|) > 0. [3 marks] Solution:

(a) (x + 2 ) is a factor f(-2) = 0 (-2)4-3(-2)3 + a(-2)2 + 50 = 0 M1 a = -15 A1 (b) f(5) = 54 – 3 (5)3 + (-15)(5)2 +15 (5) +50 = 625 – 375 – 375 + 75 +50 = 0 B1 f(x) = ( x + 2 )( x - 5 )( x2 – 5 ) M1 = ( x + 2 )( x - 5 )( x - 5 )( x + 5 ) A1 (c) f(x) > 0, use graphical or table method, M1 (d) Solution set = 5525,: orxxorxxx A1

For f(|x|) > 0 , |x| < - 5 ( reject) or -2<|x|< 5 or |x| > 5 M1 0 |x|< 5 or |x| > 5 M1 Solution set = 5555,: xororxxxx A1 [10]




10. Given matrices A and B where

682121440

123122121

BandA . Show that matrices A and B abide by the

commutative law of multiplication. [3 marks] Find adjoint A. [3 marks] A hawker offers three type of noodles, Hokkien Noodle, Hailam Noodle and Tomyam Noodle. The price for each bowl of noodle is fixed and the price of 2 bowls of Hailam Noodle is equal to a bowl Hokkien Noodle and a bowl of Tomyam Noodle. A family pays RM23.00 for 3 bowls of Hokkien Noodle, 2 bowls of Hailam Noodle and a bowl of Tomyam Noodle. Another family pays RM19.50 for 2 bowls of Hokkien Noodle, 2 bowls of Hailam Noodleand a bowl of Tomyam Noodle. By using x, y and z to represent the price of a bowl of Hokkein Noodle, Hailam Noodle and Tomyam Noodle respectively, obtain a matrix equation to represent the information. Hence, find the price of a bowl of each kind of noodle. [6 marks]

10. AB =

400040004

682121440

123122121

B1

BA =

400040004

682121121

682121440

B1

As AB = BA, matrices a and B obey the communication law of multiplication. A1

Minors of A =

614824210

B1

Cofactors of A =

614824210

B1

Adjoint A =

682121440

A1




AB = - 4I

682121440

411A B1

x – 2y + z = 0, 2x + 2y + z = 19.50, 3x + 2y + z = 23 B1

2350.19

0

123122121

zyx

B1

235.19

0

682121440

41

zyx

M1

=

50.400.450.3

A1

x = RM3.50, y = RM4.00, z = RM4.50 A1 [12]




11. Sketch the graphs of x

xy 3 and y = 4 on the same coordinate axes. Mark the region

bounded by the two graphs as A. [4 marks] Find the area of A. [4 marks] Find, in terms of , the volume of the solid formed when region A is rotated 360o about the x- axis. [4 marks] Solution:

Sketch graph: Curve G1

Line y = 4, x = 1, x = 9 G2

Area, dxx

xA 39

132

=9

1

23

63232

xx M1

=

16)1(

3296)9(

3232 2

323

M2

= 2units 312932 A1

= 2units 322

y

xxy 3

y=4 4

1 0 9 x

A




Volume,

9

1

22 dx 3)8()4(

xxV M1

=

9

1

dx 69128x

x M1

= 9

1

2

6ln92

xxx M1

=

)9(69ln9

292

)1(6)1ln(9

212

M1

= 78.107 unit3 A1 [12]

12. Given a curve 22

x

x

eey .

(a) Show that .12 2ydxdy

[5 marks]

(b) Show that 0dxdy for all values of x. [3 marks]

(c) Find y

x lim and y

x lim . [2 marks]

(d) Sketch the graph of 22

x

x

eey . [3 marks]

Explain how the number of roots of the equation 22)12(

x

x

eexk ,depends on k.

[2 marks]

Solution:

(a) 2)2( xx eye B1

xxx edxdyeye )2( M1

121

dxdy

ey x




1

122

21

dxdy

yyy M1

1111

dxdy

yyy

11

2

dxdy

yy M1

ydxdy

y

1

12

212 ydxdy

A1

(b) 22 2

22

eeeee

dxdy xxxx

M1

= 22

4

x

x

ee A1

0dxdy for all values of x. A1

(c) 122lim 2

ee x

x A1

122lim 2

ee x

x A1




(d)

curve G1 x = ln 2 G1 point (0,-3) and label G1 Sketch y = k(2x – 1) on the same axes. M1 2 roots when k > 0 and no root when k 0. A1 [15]

y

x x=ln2

y =1

y=-1 O


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