2014 2 SABAH SMKBandaraya2KK Maths QA

954/2 (2014) SMK BANDARAYA II KOTA KINABALU

2014-2-SABAH-SMKBandaraya2KK_MATHS QA 954/2

SECTION A [45 marks]

Answer all the questions in this section

1. The function is defined by () = {(3 )2 , 3

1 +

, > 3

Given that () is continuous at x = 3 , find the value of a . [4 marks]

Sketch the graph of = () [3 marks]

2. (i) Evaluate cot

3

6

[4 marks]

(ii) Using an appropriate substitution, evaluate 1

ln

3

[4 marks]

3. By using the substitution = 2 , find the general solution to the differential equation

=

21

42+2. [7 marks]

4. Using Maclaurins Theorem

() = (0) + (0) +1

2!2(0) +

1

3!3(0) +

Show that the power series of cos up to the term in 4are :-

(i) = 1 + +1

22 +

1

63 +

1

244 + [3 marks]

(ii) cos = 1 1

22 +

1

244 [4 marks]

5. Show that

(ln tan ) =

2

sin 2 . [2 marks]

Hence , find the solution of differential equation

sin 2

= 2(1 ).

For which =1

3 when =

1

4. Express y explicitly in terms of x in your answer. [8 marks]

6. Given that =

1 , show that (1 )

= (1 + 2). [6 marks]

954/2

*This question paper is CONFIDENTIAL, until the examination is over CONFIDENTIAL*


CONFIDENTIAL* 954/2

SECTION B [15 marks]

Answer any one question in this section

7. Sketch on the same coordinate axes, the graphs of =1

2 and = 2 + 1. Show that the equation

4 2 = 0 has a root in the interval [1 , 0] [5 marks]

Use the Newton-Raphson method to find the root, correct to three decimal places. [6 marks]

The other root of the equation lies in the interval [n ,n + 1], where n is an integer.

State the value of n. [4 marks]

8. Sketch on the same axes the curve =4

, 2 = 4( 1)

Find the coordinates of the points of intersection of the two curves. [6 marks]

Show that the area of the region bounded by =4

, 2 = 4( 1) and y = 4 is

(20

3 4 ln 2) . [4 marks]

Calculate the valume of the solid of revolution when this region is rotated through 360

About y-axis. [5 marks]

954/2

*This question paper is CONFIDENTIAL, until the examination is over CONFIDENTIAL*

TERM 2 MATHEMATICS T STPM TRAIL EXAM MARKING SCHEME


Question WORKING SOLUTION MARKS

Q1 () = {

(3 )2 , 3

1 +

, > 3

lim3

(3 )2 B1

lim3+

1 +

B1

lim3

(3 )2 = lim3+

1 +

M1

(3 3)2 = 1 +

3 ; = 3 A1

D1 = (3 )2

D1

= 1 3

D1 (All correct: intersection,

asymptote)

7 marks

Q2 (i) cot =

cos

sin

3

6

3

6

M1

(Integrable)

= [ln sin ]

6

3 A1

= [ln sin

3 ln

6] M1

= ln (3

2) ln

1

2 = ln 3 A1

4 marks

Q2 (ii) 1

ln

3

let u = ln x ; dx = x du

when x = 3 = 3; when x = e then u = 1

M1 (Any substitute form)

= 1

3

1(

1

)

= 1

3

1

M1 (New variable, ignore limits)

= [ln ]13 A1

= ln 3 A1

4 marks

Q3

=(2)1

2(2)+2 ; = 2

=1

2+2

B1

(Substitute 2x-y)

12

10

8

6

4

2

-5 5 10 15 20

y=(3-x) 2

y=1--3

x

y = 1

= (3 )2

= 1 3

9

y = 1


= 2

= 2

= 2

M1

=1

2+2

2

=

1

2+2

= 2

1

2+2

=

3+5

2(+1)

M1 A1 (Change variable)

(+1)

3+5 =

1

2

1

3

2

3(3+5) =

1

2 M1

(Integrable form)

13

[ 2

3ln|3 + 5|] =

1

2 + A1

6 4|3 + 5| = 9 + 18

4|3 + 5| = 6 9 18 ; = 2

|6 3 + 5| =3

4[ 2]

9

2

6 3 + 5 = 3

4(2)

9

2

6 3 + 5 = 3

4(2) ; where =

9

2

A1

7 marks

Q4 (i) Let () = ; () = () = () = B1

Then (0) = (0) = (0) = 0 = 1

= 1 + (1) +1

2!2(1) +

1

3!3(1) +

1

4!4(1)

M1

= 1 + +1

22 +

1

63 +

1

244 + A1

3 marks

Q4(ii) Let () = cos

() = sin ; () = cos ; () = ; () = cos

B1

(0) = cos 0 = 1 ; (0) = sin 0 = 0 ;

(0) = cos 0 = 1 ; (0) = sin 0 = 0 ; (0) = 1

B1

cos = (0) + (0) +

1

2!2(0) +

1

3!3(0) +

1

4!4(0)

Then cos = 1 1

2!2 +

1

4!4

M1

cos = 1 1

22 +

1

244 A1

4 marks

Q5

(ln tan ) =2

tan B1

=1

tan 2


=

sin

1

2

=2

2 sin cos

=2

sin 2

M1

A1

sin 2

= 2(1 )

1

(1 ) =

2

sin 2

B1 (Separate variable)

1

+

1

1 = ln tan + M1: LHS (Partial fraction)

A1: RHS (From 1st part)

ln ln|1 | = ln A( tan x)

M1 (LHS)

1

= M1 (Remove log)

When =

1

3 , =

4

1

3

11

3

=

4

=1

2

A1

1 =

1

2tan

2 = tan tan

=tan

2+

A1

10 marks

Q6 =

1

=

1

+ =

(1)1(1)

(1)2 M1 A1

+ =

1

(1)2 A1

multiplying (1 )2 on both sides

(1 )2

+ (1 )2

(1)=

M1 (Multiply correct

expression)

(1 )2

(1)

+ (1 ) = M1

(Eliminate ) (1 )

= (1 + 2) A1

6 marks


Q 7

D1

( =1

2)

D1 ( = 2 + 1)

D1 (All correct)

Let () =1

2 2 1 B1

(0) =

1

20 2(0) 1 =

1

2< 0

(1) =1

21 2(1) 1 = 1.1839 > 0

M1 (evaluate +/-, any

dummy)

Since f (0) < 0 and f(-1) > 0 has opposite sign hence show that ther is a root in the interval [-1 , 0]

A1

() =1

2 2 1 ; () =

1

2 2 B1

Since f(0) = 1

2 and (1) = 1.1839 , the solution is closer to 0 then -1.

Let 0 = 0.3

1 = (0.3) (0.3)

(0.3)

= 0.3 (0.02959)

1.62959 = 0.31816

M1

(Substitute x0 ) A1

(At least 4 d.p.)

2 = (0.31816) (0.31816)

(0.31816)

= 0.31816 (0.0000631906)

1.636257

= 0.31812

3 = (0.31812) (0.0000022594)

(0.3)

= 0.31812 (0.0000022594)

1.636242

= 0.31812

0.318 (3. . )

The root is 0.318

M1A1 (Stopping criteria)

A1

(1) =

1

21 2(1) 1 = 1.64086

(2) =1

22 2(2) 1 = 1.305472

(3) =1

23 2(3) 1 = 3.04277

Since f(2) and f(3) has a different sign then the other root is in the interval

[ 2 , 3 ], Hence n = 2

M1 (evaluate +/-, any

dummy)

M1 A1

15 marks

10

8

6

4

2

-2 -1 1 2 3 4 5 6

y = 1

2e x

y = 2x+1

D

C


8

D1

(2 = 4( 1))

D1

( =4

)

D1 (All correct)

The intersection of the point

2 = 4( 1) equa (1)

=4

equa (2)

16

2= 4( 1)

M1 (Solve with correct

method)

3 2 4 = 0

( 1)(2 + + 2) = 0

= 2 , = 4

2= 2

The point of intersection is (2 , 2)

M1 A1

Area = (2

4+ 1) (

4

)

4

2 M1

= [3

12+ 4 ln ]2

4 A1

= (64

12+ 4 44) (

8

12+ 2 22) M1

=20

3 4 2 A1

Volume = (2

4+ 1)2 (

4

)2

4

2 M1 A1

= (4

16+

2

2+ 1

16

2)

4

2 A1

= [5

80+

3

6+ +

16

]2

4 M1

= [(64

5+

64

6+ 4 + 4) (

2

5+

8

6+ 2 + 8)]

=296

153

A1

15 marks

5

4

3

2

1

2 4 6 8 10

y = 4

xy2=4(x-1)

y = 4

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2014 2 SABAH SMKBandaraya2KK Maths QA