2014 2 SELANGOR SMK Shahbandaraya KLANG_MATHS QA

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2014-2-SELANGOR-SMK Shahbandaraya Klang_MATHS QA

SMK SHAHBANDARAYA KLANG

TRIAL STPM TERM 2 2014

MATHEMATICS (T)

Section A [ 45 marks ]

Answer al l questions in thi s section.

1a. Function fis defined by

33

33)(

3

xifx

xifexf

x

Show that fis continuous at 3x .

b. Evaluate 52

53lim

2

x

x

x

[ 4 marks ]

[ 2 marks ]

2. Given that sin y = x, show that = .

Hence, show that = 1 . [ 5 marks ]3. A curve is defined by the parametric equations = and = 2 ,

where 0.(a)Show that

= 2

+

,

(b) Find the coordinates of points when = .

[ 4 marks ]

[3 marks ]

4. By using the substitution y = vx, where v is a function of x, reduce the

differential equation = to a differential equation that contains v and x only.

Hence, solve the differential equation above given that y = 0 when x = 3.

[ 7 marks ]

5. Use the trapezium rule with five ordinates, evaluate , giving youranswer correct to three decimal places.

By evaluating the integral exactly, show that the error of the approximation is

about 0.28%.

[ 9 marks ]


2/8

6. Given that = t a n, express in terms of tan . Hence show that = 2 8 6 .By using the Maclaurins Theorem, show that if xis small such that the terms

higher than can be neglected, then tan =

.Hence, find an approximation for the value of tan dx. , giving youranswer correct to 2 significant figures.

[ 4 marks ]

[ 4 marks ]

[ 3 marks ]

Section B [ 15 marks ]

Answer any one question in th is section.

7. Two iterations suggested to estimate a root of the equation 5 1 = 0are+ = 1, + = 5 1 .( a ) Show that the equation 5 1 = 0has a root between 0 and 1.( b ) Using =0.5, show that one of the iterations converges to the rootwhereas the other does not.

Use the iteration which converges to the root to determine the root correct to

three decimal places.

[15 marks]

8. The function f is defined by f= , where x> 0( a ) State all the asymptotes of f.

( b ) Find the stationary point of f, and determine its nature.

( c ) Obtain the intervals, where ( i) f is concave upwards, and(ii) f is concave downwards.

Hence, determine the coordinates of the point of inflexion.

( d ) Sketch the graph y= f(x).

[15 marks]

PREPARED BY : TEO JOO AN


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= 2 1 2 = 2 5 2 A1 4 marks

3(b) = 13

2 5

2= 1

3

= 1 = 1 M1A1When = 1 . = 1and = 3; when = 1 , = 1and = 3 coordinates are 1, 3, 1, 3 A1 3 marks

4 By using the substitution y = vx, where v is a function of x, reduce the

differential equation = to a differential equation that contains v and x only.

Hence, solve the differential equation above given that y = 0 when 3x .

[ 7 marks ]

4. y = vx = = =

=

1

4

M1

M1

=

+= | | | | =

+ =

+ = |+| =

M1

M1

M1

M1

x = 3, y = 0,

ln 1 = 3 + c

c = -3

|+| = 3A1


5/8

5 Use the trapezium rule with five ordinates, evaluate , giving youranswer correct to three decimal places.

By evaluating the integral exactly, show that the error of the approximation is

about 0.28%. [ 9 marks ]

h =

=0.5

x 2 2.5 3 3.5 4

y =

0.3466 0.3665 0.3662 0.3579 0.3466

By using trapezium rule,

. {0.3466 + 0.3466 + 2 (0.3665 + 0.3662 +0.3579) } 0.719 (to 3 d.p.)

=

[

]

= [ln4 ln2]= 0.721 (to 3 d.p.)

Error of the approximation =.7.79

.7 x 100%

= 0.28%

B1

M1

M1

A1

M1

M1A1

M1

A1

6 Given that = t a n, express in terms of tan .Hence show that

= 2 8 6 . [ 4 marks]

By using the Maclaurins Theorem, show that if xis small such that the terms higher

than can be neglected, then tan = . [ 4 marks ]Hence, find an approximation for the value of tan dx. , giving your answercorrect to 2 significant figures. [3 marks]

6 = t a n = 2tan sec= 2tan 1 tan=2tan2tan M1A1

= 2sec223tan2sec2=21tan6tan1tan= 2 1 6 1 = 2 8 6 M1

A1

4 marks

= 8 12 B1


6/8

= 8 12

12 ()

= 8 12 1 2 ()

B1

Let = tan20 = 0, 0=0, 0=2,0 = 0, 0 = 16

Maclaurin Series:

= 0 0 02! 2 03! 04! = tan

= 22! 2 164! = 2 2

3

M1

A1

Must

have 4 marks

tan dx.

= (2 23 4) dx

.

= 3 2

15 .

= 0.053 20.05

15 [0]0.000042

M1

A1

A1

Musthave3 marks

7 Two iterations suggested to estimate a root of the equation 5 1 = 0are+ = 1, + = 5 1 .( a ) Show that the equation 5 1 = 0has a root between 0 and 1.( b ) Using =0.5, show that one of the iterations converges to the rootwhereas the other does not.

Use the iteration which converges to the root to determine the root correct to

three decimal places. [ 15 marks ]

(a) Let

=

5 1

0 = 1 > 01 = 1 5 1 = 3Since f(0) and f(1) have different signs, therefore there is a

root between 0 and 1.

B1B1

M1A1

Using =0.5,For + = 1 = 0.5 1

= 0.225

Since

is between 0 and 1, therefore the iteration converges

to the root.

M1

A1

M1A1

For + = 5 1


7/8

= 50.51 = 1.1447

Since > 1, therefore the iteration does not converge to theroot.

M1

A1

M1 A1

(c) Using + = 1= 0.2022= 0.202 (to 3 d.p)= 0.2017= 0.202 (to 3 d.p.)

Therefore, the root of the equation is 0.202

B1

M1A1

8 The function f is defined by

f= , where x> 0(a) State all the asymptotes of f. [ 2 marks ]

(b) Find the stationary point of f, and determine its nature. [ 6 marks ]

(c) Obtain the intervals, where

(i) f is concave upwards, and

(ii) f is concave downwards.

Hence, determine the coordinates of the point of inflexion. [ 5 marks ]

(d) Sketch the graph y= f(x). [ 2 marks ]

8(a) = 0 ; = 0 B1, B1 2 marks(b) f= ln 2

= 2 ln 2

= 12ln2

M1 QuotientRule

Stationary point,

= 0

12ln2 = 0 2 = 122 =

= 12 ; = 2

M1

A1 Both

=

32122

= 6ln25 12 = 4.34 < 0

M1

A1

f(x)


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, is a maximum point. A1 6 marks(c) Concave upwards: > 06ln25 > 0

>12

Interval: ,

M1

A1

Concave downwards: < 06ln25 < 0 < 12

Interval: 0,

M1

A1

Point of inflexion is

,

B1

5 marks

(d)

D1

D1

2 marks

x

y

0.5

(0.824,

0

2014 2 SELANGOR SMK Shahbandaraya KLANG_MATHS QA

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