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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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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


RESTRICTED* PERCUBAAN STPM 2011

954/2 1 RESTRICTED

This document is specially for the use of authorized examiner only.

954/2 PERCUBAAN

STPM 2011

MARKING SCHEME

MATHEMATICS T ( MATEMATIK T)

954/2

PAPER 2 ( KERTAS 2)



954/2 2 RESTRICTED

1. The variables x and y are connected by

ye

x

dx

dy

where x ≥ 0. Find y in terms of x given y = 0 when x = 0. [5 marks]

Solution:

√ dy = √ dx

∫

dy = ∫

dx B1

2

=

+ c M1

x = 0, y = 0 c = 2 M1

2

=

+ 2

= ln (

+ 1 ) M1

y = 2 ln (

+ 1 ) A1 (5 marks)



954/2 3 RESTRICTED

2. Show that tan x + cot x = 2 cosec 2x [ 3 marks] Hence, solve the equation 2 cosec 2x = 3 tan x + 1 for 0 ≤ x ≤ 2π giving your answers in terms of π. [ 4 marks]

Solution: LHS = tan x + cot x

= x

x

cossin +

x

x

sincos

= xx

xx

cossincossin 22 M1

= xx cossin

1

= xxcossin2

2 M1

= 2 cosec 2x A1 (3 marks) 2 cosec 2x = 3 tan x + 1

tan x + cot x = 3 tan x + 1 B1

tan 2 x + 1 = 3 tan 2 x + tan x

2 tan 2 x+ tan x – 1 = 0 M1

( 2 tan x – 1) ( tan x +1)= 0

tan x = 21 or tan x = –1 M1

x =0.148 π , 1.148 π or x = 4

3 , 4

7

Solution : x = 0.148 π , 43 ,1.148 π ,

47 A1 ( 4 marks)



954/2 4 RESTRICTED

3. The diagram below shows ABCD is a cyclic quadrilateral. The tangent to the circle at A is parallel to the line BC. Prove that triangle ABC is an

isosceles triangle. [3 marks]

The lines AD and BC extended meet at the point E. Prove that AB is a tangent to the

circle passing through the points B, D and E. [4 marks]

Solution :

ACB = XAC (alternate angles, AX // BC) B1 ABC = XAC (alternate segment theorem) B1 ACB = ABC triangle ABC is an isosceles triangle. (Base angles equal) B1 (3 marks) DEB = XAD (alternate angles, AX // BC) B1 ABD = XAD (alternate segment theorem) B1 ABD = DEB Since any three points can determine a unique circle passing through them there exist a circle passing through the points B, D and E B1 The line AD is tangent to the circle passing through the points B, D and E. ( because angle between the tangent and the chord equals to the angle opposite the chord) B1 (4 marks)

E

A

C

D

B

E

A

C

D

B



954/2 5 RESTRICTED

4. Using the substitution v = x + y, show that the equation dy

dx =

x + y + 2x + y + 1 can be reduced to

dv

dx =

2v + 3v + 1 . [3 marks]

Hence, show that 2y–2x – ln(2x + 2y + 3) + 10 = 0 is the particular solution that satisfies

x = 2 and y = – 3. [5 marks]

Solution: v = x + y

dvdx = 1 +

dydx

dydx =

x + y + 2x + y + 1 reduced to

dvdx – 1 =

v + 2v + 1

dvdx =

2v + 3v + 1 A1

v + 12v + 3 dv =

dx

12

1 – 1

2v + 3 dv =

dx

v – 12 ln(2v + 3) = 2x + c

when x = 2, y = –3, c = – 5

the particular solution is:

2y –2x – ln(2x + 2y + 3) + 10 = 0.

B1

M1

B1

M1

A1

M1

M1

(3 marks)

(5 marks)



954/2 6 RESTRICTED

5. The diagram below shows that ABT is a straight line parallel to DC. CT is a tangent to the circle, centre O. If AOD = 2BTC, prove that

(a) AC = CT, [5 marks]

(b) BT = BC, [2 marks]

(c) BCT and DAC are congruent, [2 marks]

(d) DB parallel to CT. [2 marks]

Solution:

(a) Let BTC = AOD = 2 (Given)

Let ACD= AOD = 2 ( at centre = 2× at circumference)

=

Let CAT= = (alt. s , AB // DC)

=

ACT is isosceles (Base s are equal)

AC = CT

B1

B1

B1

B1

B1 (5 marks)

C

D

T B

A

O



954/2 7 RESTRICTED

(b) Let BCT= = ( between chord and tangent = in alt segment)

= since = from (a)

BC = BT (BCT is isosceles)

(c) Let TBC = b and ADC= d.

then b = d (ext of cyclic quadrilateral = int. opp. )

= and AC = CT (from (a))

BCT and DAC are congruent (ASA)

(d) AOD = 2×ABD ( at centre = 2× at circumference)

ABD = = BTC

DB // CT (corresponding s are equal)

B1

B1

B1

B1

B1

B1

(2 marks)

(2 marks)

(2 marks)



954/2 8 RESTRICTED

6. An aircraft leaves P to fly to Q which is 200 km due North of P. The pilot sets a course due North but after 20 minutes he realizes that, owing to a wind blowing from East, the plane is at a point R, where R is 80 km from P and 26 km west of the line joining P and Q. Find

(a) the speed, in km h-1, of the wind. [2 marks] (b) the speed, in km h-1, of the aircraft in still air. [2 marks] (c) the course the pilot should have set from P in order to have arrived directly at Q.

[2 marks]

Find also the course the pilot should set from R in order to fly directly to Q and, in this case, the time, to the nearest minute, taken for the journey from P to Q via R. [7 marks]

Solution :

(a) sin =

=

M1

u = 78 kmj-1 A1 (2 marks)

(b) V = √ M1 = 226.97 kmj-1 A1 (2 marks)

(c)

sin =

M1

= 020.1 A1 (2 marks)

tan =

= 78.19

v 240

u

75.66

124.34 26

80

d

α



954/2 9 RESTRICTED

=

M1

= 19.660

course = 19.660 + 11.810 M1

= 031.5o A1

= 58.53o

=

M1

w = 197.77 km/h

d = √ = 127.03 km/h M1

t =

= 38.5 min M1

total time = 20 + 38.5 59 minutes A1 (7 marks)

226.97

101.81

11.81

78

W



954/2 10 RESTRICTED

7. 10 students are seated at random in a straight line. Find the probability that

(a) three specific students sit next to each other. [ 2 marks]

(b) two brothers in the group do not sit next to each other. [ 2 marks]

Solution:

(a) P( 3 specific students sit next to each other) = ！

！！

1038 M1

= 151 A1 ( 2 marks)

(b) P ( 2 brothers do not sit next to each other)

= 1 –P ( the 2 brothers sit next to each other)

= 1– !10!2!9 M1

= 54 A1 (2 marks)

8. The random variable X is normally distributed with mean and standard deviation . It

is known that P( X > 30 ) = 0.6 and P( X > 38 ) = 0.2. Find the value of and .

[5 marks]

Solution: XN(, 2)

P(X > 30) = 0.6 – 0.253 = 30 –

P(X > 38) = 0.2 0.842 = 38 –

= 7.306

= 38 – 0.842(7.306)

= 7.31, = 31.8 (3 s. f.)

A1

B1M1

(5 marks)

M1

A1



954/2 11 RESTRICTED

9. A discrete random variable X takes the integer value x with probability P(x) defined by

otherwise,06,5,4),9(3,2,1),2(

)( xxk

xxk

xP

(a) Find the value of k, [2 marks]

(b) Determine the mean and variance of the probability distribution and hence state the

mean and variance of the variable Y where Y = 4X – 2. [6 marks]

Solution: (a) ( ) 1x

P x

24 1

124

k

k

(b) E(X) = ( )x

xP x

= 84k = 3.5

Var(X) = E(X2) – [E(X)]2

= 352k – 3.52

= 2.42.

Y = 4X – 2

E(Y) = 4E(X) – 2 = 12

Var(Y) = 42Var(X) = 38.72

M1

M1A1

M1

B1

B1

A1 ( 2 marks)

( 6 marks)

A1



954/2 12 RESTRICTED

10. The continuous random variable X has the cumulative distribution function

F(x) =

1110

0,0

x

xxk

x

，

，

(a) Find the value of k if the median of X is 91 . [2 marks]

(b) Write down an expression for the probability density function of X, f(x). [1 mark]

Hence,

(c) show that E(X) = 21 . [3 marks]

(d) find Var (X). [3 marks]

Solution :

(a) k√ =

M1

k = A1 (2 marks)

(b) f(x) =

√ , 0 < x < 1 B1 (1 mark)

(c) E(X) = ∫ (

√ )

M1

= [

]

M1

= A1 (3 marks)

(d) E(X2) = ∫ (

√ )

= [

]

=

M1

Var(X) =

– ( )2 M1

=

A1 (3 marks)



954/2 13 RESTRICTED

11. During each working day in a certain factory a number of accidents occur independently

according to a Poisson distribution with mean 0.5.

Calculate the probability that

(a) during any one day , there are two or more accidents, [3 marks]

(b) during two consecutive days there are exactly three accidents altogether. [3 marks]

By using a suitable approximation, calculate the probability that there are 3 weeks that

are accident –free, out of 50 consecutive five –day weeks . [5 marks]

Solution:

(a) X Po( 0.5) : P ( X ≥2) = 1 – 5.0e – 5.0

21 e

M1

= 1 – 5.0

23 e

M1

= 0.0902 A1 (3 marks)

(b) X Po() where = 0.5 x 2 = 1 B1

P ( X = 3) = ！31 31e = 0.0613 M1A1 (3 marks)

X Po() where = 5 x 0.5 = 2.5

P ( no accident in a specific 5- day working week)

= P ( X = 0 )

= 5.2e or 0.0821 B1

Y = number of 5-day working weeks that are accident– free , in 50 consecutive weeks.

Y B ( 50 , 5.2e ) B1( can be implied)

Approximately , Y Po( 4.11) B1

P ( Y = 3 ) =

!311.4 3

11.4e M1

= 0.190 A1 ( 5 marks)



954/2 14 RESTRICTED

12. The following table shows the height ( in cm) of 400 students chosen at random.

Draw a cumulative frequency curve for this distribution. [ 3 marks]

Hence, use your cumulative frequency curve to estimate the median and semi interquartile range for the distribution. [3 marks]

Estimate the mean and standard deviation for the data in this grouped frequency distribution. [6 marks]

Solution:

median = 129 cm, Q1 = 121 cm, Q3 = 138 cm B1(any one)

Semi interquartile range = 5.812113821

cm M1A1 (3 marks)

Height ( cm) Cumulative frequency

< 100 0 < 110 27 < 120 85 < 130 215 < 140 320 < 150 370 < 160 395 < 170 400

R1 ( points correct and

uniform scale)

Cumulative frequency

( No. of students)

Height/cm

100

300

200

400

121 100 129 138 170

R1 ( smooth curve passes

through all points)

R1 ( all correct and

label on axes)



954/2 15 RESTRICTED

B1( midpoints) B1(frequency)

7.129

40051880

mean M1A1

1.13400

51880400

6797600..2

devstd M1A1 ( 6 marks)

End of Marking Scheme

Midpoint,x / cm 105 115 125 135 145 155 165 frequency 27 58 130 105 50 25 5


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