stpm trials 2009 math t paper 1 answer scheme (kelantan)

8/11/2019 STPM Trials 2009 Math T Paper 1 Answer Scheme (Kelantan)

http://slidepdf.com/reader/full/stpm-trials-2009-math-t-paper-1-answer-scheme-kelantan 1/9

Stpm2009 mls: I

Us

in

g algebraic laws

of

set, prove that [

(A

v

B)nCJv[(A

V

B)nC

J- B

=

A -B

. [6

mar.lsl

So lution: [(Av

B)n C

JV[(AvB)

nC

J -

B

=[(A v B)nCJV [(A V

8)

n C ln

B BI

= (A vB) n (CvC) n B BI

v

~

= (A v B) n B BI

= (AnB ) v (BnB ) BI

=(AnB )v ;

= (A n B ) BI

=A-B

BI

2.

:;

Find r

x

dx

i 2: rr:i

4

mar ; .]

So

lution

:

r

3x f 3U

l 2J7:i

<Ix =,

2u du

MI

, 3

=f

du

2

AIM I

A I

3 On the same diagram sketch the graphs

o

y =

and y ' x

I.

1

marks

I

x + 1

A region R is

bou

nded

by the two

gmphs

and

the y axis. Find the area of r

egio

n R.[7 marh]

A solid

is fonned

by r

ota

t

ing the

region Rco

mpl

ete ly

about

y-axis.

Fi

nd

the

vo

lume of the

sol id

formed

.

[5 morh

So lution:

5 -

--- ---+.. , f - ,-=-- --

+

y

1 1 1



Slpm2009 mls:2

~ = x

H I

S= 2x2 X - 1

2xl

+x - 6 = O

2x -

3 x+2

= 0

MIAI

,

re

- X ~

X 3 X ~

MI I

o

x

1 2 2 2

, 9

=5 In [x+ IU

-3

+

4

MIMI

=

~ ~ AI

2 4

n5

)

I (3

o lum

e =

ily-

3 2

(3)

X

5 10 9

,.. f 1 ' - -+ 1

dy+-1t

1

y

Y 4

(

5 ) 9

= f - 101ny+

y

-If

Y ,

4

M IA I

AI

= ,,[( - ¥ - 10In5 + 5) - ( ¥ - 10In2 2H

1

= I O l n ~ AI

4 5

4. The function

/ i s

defined by

I x)

= { ~

x

';C 3

O x = 3

MI

i) Find

lim

I x) .

lim

I x)

and hence detennine iflhe function/i s continuous al x =3

,.

. r

[3 marls)

(ii) Sketch the g

.ph

off

[2 marls]

So lution:

l

im

lex

=

lim

x - 3

=

I

. -.

,.

.-d· x - 3

lim / ( x) = lim -

H 3

=_ 1 MIAI

.

,. . ,- x - 3

Since lim I{x) '$. lim I x)

. _., . - 0 )

: lis

not

continuous at x = 3. A I



Stpm2009 mls:3

DIDI

5.

Prove that, for all va

lu

es

of k,

the line /l x +y =

8k

is a tangent to the hyperbola xy::: 16.

-Icnee, find the coordinates

of

the point

of

contact. [5

marks1

Solution:

y=8k-.I .

.I (8

k

-

6 =

16

k'x' -

8h

16

- 0

- 4Xh - 4) = 0

-

4

0

4

x=

.:::

There is only

one

point of contact

MIAI

MI

: The line

Itx

+

y

' 8A: is a tangent to the hyperbola .ry

16

. A I

When

x = .

x=4

k

k

: The coordinates

of

the point

of

contact =- ~ . 4 . t ) BI

6.

Sketch the graphs

of y

= and

y

12x

-

l

io

n the sa

me diagram.

Hence. so lve the

Ixl

inequality 12x - I

I>

~ [7

mOTu]

Ixl

So

lution:

... ..

=f

2.1 - 11

Ixl

:::) . . =

2x - 1

.I

=>

2.. -

1

- 1

= 0

y Ilx - II

1 1 1



Slpm2009 mls:4

=>«- I)(2x+ ~ O

I

x : J - 2'

MIAI

For 1 2 x I > ~ the so lution

D

{x:x< ~ x I}

MIAI

(

y - 2

x

7. Find the

values of

x, y

and z i

4x -

4

2x

Y

y z

X

I)

X

is

symmetric

- 1

So

lution

: l

=

4%

- 4

-

4x

- l,

x 'y+z.

MIAI

=>< -

4x + 4 - 0,

=>

(x

- 2)

2

a

0,

=>x=2, y = 4(2) - 1

=

7,z

= 2 - 7

5

MtAIAI

[5 marb)

8. Given thall

(x)

-

x + ax + 8x + b. where a and b

an:

constant. I (x)

-

0 when x = -2. When

f(x)

is

divided

by

x

2,

th

e

remaind

er is

2

Determine the values of a and b

(a)

Show

that the equationj x}

0

has only

one

real root

and

find

the

set of values ofx

such

thatl

(x) > 0

(b) Express + 4 in partial fractions.

I(x)

So luti

on:

I(x) -

x

ax

8x + b

[ (x) -

lx 2ax 8

1 (-2) -

12

- 40 + 8 = 0

40

20

a - 5

Whena - 5. I(x) -

x

+ 5x + 8x + b

1(-2) -

-8 20 -

16 b -

2

-4+b

- 2

b - 6

:.I(x)

-

x 5<

8x

6

a) [(x)

ax +0:< 8x+ b

- 3) = - 27 45 - 24 6 = 0

.: c - -3 i a rool o/ the equationj(x) = O.

Le l[(x) =

x +

5<

8x

6

=(x 3)(< 2x 2)

When[(x) -

O

(x

3)(< 2x 2) - 0

.T _ _ ± ~

2

1)

2 ± ~

x - -3.

x

-

The equalion n,x) - 0 has only one real rool, X

-

3.

Whenl(x) > 0

x 3)(..

2x

2 > 0

B

ut

r2

2.t

2:::

(x

+

1)

2

1

>

O

since

(x

1)

2 > 0

for all

real

va

lueso

f x



Stpm2009 mls:5

Hence (x 3)(x

2

2x

2 > 0

Whenx 3 > O:::: x > 3

The

set

of values o x such that}{x»O is/x: x

>-3

).

b x+4 _ x+4

f(x)

(x 3)(x

2.< 2

l..c x+4

A

nx+C

t (x+3)(x

+

2x+2) · (x+3) (x +2x+ 2)

x 4 .

A

(x

2.r

2 (x

3)(

C

Substitutingx - -3: I =A (9 - 6+2)

A=. .

5

Comparing the coefficients ~ 0 = A B

B- -. .

5

Co

mparing

the constan

t

terms

: 4

.,.

2. 1

JC

4 = ~ +3C

5

3C - . .8.

5

~

5

I I 6

.: A=S,B-

S

andC -

S

x+4 = _ 1_ _

(x - 6)

x+3Xx

2

+

2x+2)

S x+) 5(x

l

+ 2. +2)

9.

Given the

matTi

ces A ~ -2

1

and

1 - 3 - I I

Find

the values

of p

and

,..

Hence

, solve

the simultane

ous

equa

ti

ons

x+ y - z 2

2y +z

Sx-9

1x

- 3y - z =

4

Since AB SA

:

satisfy AB c BA

AB - BA nl, where n is a scalar a

nd

I is the 3 x

J

idenfi

ty

molrix.

~

4 ;

]= n

1 - 3 - I 1 5

r

[5 marks

[5 morbI



10.

mls:6

q - 2

pq 22

7q 14

3p 16 2r = nl

7- r J

9 - r

q - 2

pq 22

7q

- 14

3p 16 2r

= n

-r

J

n

OJ

9 - r 0 0 n

p 10 =

0,

q - 2 = o

and 7

- r = O N o t e t h a t n

p=IO,q=2and r=7

A

= ~

~ ~

~

_ : O 4 ~ I J

7 - 3 - I 7 -

3 - 1

Al

so,

AB

::

BA

'

2J

x y -z ' -2

2y z =

5x

- 9. -IOx + 4y 2z = - 8

7 - 3y z

c

14

The 3 si

multaneous

equ

ations can

be

written

as a matrix equation

as

fo llows

HO _ ~ ~ : J ; ) = ~ ~ 8

J

Premultipllying with

matrix

B

BAUJ

=

B ~ 8 l

V =

B

~ 8 )

2 [ ~ }

=

; J U ~ 8 J S l o c e /

~ ) ~ )

2 r = 4,

2y

c - 2 and

z =

6

x 2.y 1andz ;

Given that

=

. show that O x2}d

y

~

dx dx

4 marks]



Stpm2009

m s

7

x

Ans

wer

:

y

=

- - - ,

l + x1)1

,

y

I + X2)2 - x

y 1 +x, )- i (. .X2x) + 1 +x , )i

l.

= 1

2 dx

y + 1 u , )i l. = 1

dx

2y l. + (I +x, )i d'y + l.(. .Xlu , ) i(2x

)=

O

dx dx dx

2

I+xl) i d

2

y

+

2[2Y

+ (I + x

2

)

i

x

= O

dx dx

I +X, )i d

y

+ 3( _ ~

/x

l

fix

I X1 1

(I+z ' / 'y + 3z l. = O

dx dx

II .

Express 1

as partial fract

ion .

(r + IXr +2)

.

_2 1

Hence or otherwise, find the value of L -

,••• , (r+IXr

+2

)

, · 1..

I

Find

al

so

the value of lim

L

-

. ,

...

,

r + IXr

+2)

An

swer:

Let = _ A _ B_

(r+

I

Xr

+ 2)

(r+l )

(r + 2

I -

A(r+2)

+ B(r + I)

S

ub

s

titute r

2

I

= - B, B :: - I

r

I

if

..

I

(r +

IXr

+

2

(r + l)

(r

+

2

'f

__

;;

'f

~ -

~

,.... , r + IXr + 2) , .... , r + 1 r

2

(_ I

__

+

__

_ ) + (_ I__

n + 2 n + 3 n 3 n + 4 n + 2 n 3

(_ I__ .......+ (-,-

__

_ ) +

n + 4 n + 5 2n 2n + 1

_1____ ) .

2n + 1 2n + 2

1

n 2 2n + 2

[4

mar,u]

[6mar,u]

[4 m

ar

,u]



S pm2009 mls:S

2("+1)("+2)

~ _ l

1

IimL

----

.40, ••• ,

r

+

I

) r

+

2)

lim n

· - 2 +

1)("+2)

l im _ I _

I-

..... n 2

2n 2

o

1

2

By using sketching graph show that

th

e approximate value of the root can

be derived by the

following

Newton

-

Raphson

iterative

fonnula

x , =

x.

- f(x . ) .

[4

marks]

f(x.)

S

how that

the

equation tan

=

2x has

a

root in

the

interval ( ~ .

).

[3 marks1

Hence. find

the

root

of

the

equation for ( ~ ,

correct to three decimal places.

[7

marks]

Answer

.

frx)

f{x.)]

Q(x

. , x,.

01

Refer to

the graph

above

,

Q is the point of intersection

of

tangent at P

and

the x-axis.

Gradien ofll>e langen . P

=

f(x.)

:. Equation of langen . P is:y - j x.) = f(x.) [x

- x l

A Q

.y= O.O - j x.) =

f(x.)[x",

-

x.l MIAI

:. x••, = x. - f(x.) ,

f'(x.)

0 A I

f(x.)

l i ~ t a n x

= 1 X

MI

:. The

equation tan

x

=

r

has

a

root

in

the

i

nterval

( ~ . A I

j x) - lanx -

2x

f(x)

= ec

'

x -

2

Letxo I , XI = X x

o

)

f '(x

. )



SIpm2009 ml

s:

9

= 1_ tanl - 2 = 1.3 105 MIAI

s

l l 2

= 1.3105 _ tan(I.3105) - 2 1.31 05) - 1.2239

X

sec (1.3 105) - 2

=

1.2

239 _ tan(1.2239) - 2(1.2239) _ 1.1760

X

sec (

1.

2239) - 2

x = 1.1760 tan(1.I760) - 2(1.I76O) = 1.1659 MI

sec (1.I76O) - 2

x = 1.1659 tan(1.I659) - 2(1.1659) = 1.1656 MI

sec (1.1659) - 2

:. x = l.l66 AI

stpm trials 2009 math t paper 1 answer scheme (kelantan)

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