1920 SM015_2 Solution

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PSPM 1 SESSION 2019/2020 SM015/2



SM015/2





PSPM 1




2019/2020












































14 OCTOBER 2019



KOLEJ MATRIKULASI KEDAH
Authored by: CHOW CHOON WOOI






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PSPM 1 SESSION 2019/2020 SM015/2

Questions





SECTION A [45 marks]

This section consists of 5 questions. Answer all questions.



2
2
3
1. Polynomial ( ) = 12 − − + 8 is divisible by 3 − 7 + 4. Find the
values of and . Hence, factorise ( ) completely.
[8 marks]



2. a) Show that − cot can be simplified as tan . [5 marks]
2

b) Hence, find the values of ° and °. Give your answer in the form
+ √ , where a, b and c are integers.
[5 marks]



3. a) A function ( ) is defined as


−
, <
−
( ) = − , =

− , >
{ −


Determine whether ( ) is continuous at = .


[7 mark]


+ −
b) Find . [3 mark]
→∞ −


2
2 −3
4. a) Given = , find . Give your answer in the simplest form. [5 mark]
2


b) Given = ( + ).



Show that ( + ) + + = ( + ).
[7 marks]
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PSPM 1 SESSION 2019/2020 SM015/2



6
2
5. Given ( ) = − , where ≠ 0. Find the coordinate(s) of stationary point


and determine whether it is a relative maximum or a relative minimum. Give

the coordinates correct to 3 decimal places.
[5 marks]





SECTION B [25 marks]

This section consists of 2 question. Answer all questions.

1. The function ( ) is defined by

− , ≤

− , < ≤
( ) = √ −

| − | + , >
{ −
Where and is constant.
a. If lim ( ) exists, find the values of . [5 marks]
→2
b. Find the values of such that ( ) is discontinuous at = 8. [3 marks]



2
2
2. Given a right circular cylinder with height 2ℎ and radius = √ − ℎ ,
which is inscribed in a sphere of fixed radius a.
2
2
a. Show that the volume of the cylinder is = 2 ( − ℎ )ℎ. [5 marks]
b. From part 2(a), find the maximum value of the volume, in terms
of as the height, ℎ varies. Hence, state the value of the volume
[9 marks]
if = 3.
c. Show that the ratio of the volume of the sphere to the maximum

volume of the cylinder is √ : .
[3 marks]





END OF QUESTION PAPER



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PSPM 1 SESSION 2019/2020 SM015/2

Question A1



2
3
2
1. Polynomial ( ) = 12 − − + 8 is divisible by 3 − 7 + 4. Find the
values of and . Hence, factorise ( ) completely.
[8 marks]

SOLUTION





( ) = − − +

( ) = − + = ( − )( − )
( ) =


( ) =



( ) = ( ) − ( ) − ( ) + =
+ =
= − …………… (1)


( ) = ( ) − ( ) − ( ) + =


− − + =

− − + =

+ = …………… (2)
Substitute (1) into (2)

( − ) + =

− + =

− =
= −

=




( ) = − + +






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PSPM 1 SESSION 2019/2020 SM015/2





+



− + − + +




− +

− +

− +




( ) = − + +

= ( − + )( + )
= ( − )( − ) ( + )

= ( − )( − ) ( + )





Alternative Method:



( ) = − − +

( ) = − + = ( − )( − )
( ) =  ( − ) ( )


( ) =  ( − ) ( )

( ) = ( ) ( )



− − + = ( − )( − )( + )



− − + = ( − − + )( + )



− − + = ( − + )( + )





− − + = + − − + +




− − + = + ( − ) + ( − ) +
Compare the coefficient:
3
: 3 = 12
= 4
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PSPM 1 SESSION 2019/2020 SM015/2


0
: 8 = 4
= 2

2
: − = 3 − 7
− = 3(2) − 7(4)
= 22

: − = 4 − 7

− = 4(4) − 7(2)
= −2



= ; = −
( ) = ( − )( − ) ( + )

= ( − )( − ) ( + )
















































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PSPM 1 SESSION 2019/2020 SM015/2

Question A2




2. a) Show that − cot can be simplified as tan . [5 marks]
2

b) Hence, find the values of ° and °. Give your answer in the form
of + √ , where a, b and c are integers.
[5 marks]

SOLUTION


a) − = −

−
=

−( − )
=



=




=



=






b) = −

°
° = ( )

= ° − °
°
= −
° °
√

= −

= − √











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PSPM 1 SESSION 2019/2020 SM015/2




= +


° = ° +

= ( °) +

= ( − √ ) +
= ( + − √ ) +

= − √

= ( − √ )






























































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PSPM 1 SESSION 2019/2020 SM015/2

Question A3



3. a) A function ( ) is defined as


−
, <
−
( ) = − , =
−
, >
{ −


Determine whether ( ) is continuous at = .

[7 mark]

+ − [3 mark]
b) Find .
→∞ −


SOLUTION



9− 2
−3 , < 3
a) ( ) = −6, = 3
162 −54 2
{ −27 , > 3
3


At =

( ) = −

−
( ) = ( )
→ − → − −
( + )( − )
= [ ]
→ − −
−( + )( − )
= [ ]
→ − −
= [−( + )]
→ −

= −( + )

= −



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PSPM 1 SESSION 2019/2020 SM015/2


( ) = ( − )

→ + → + −
− ( − )
= [ ] + +


→ + ( − )( + + )

− −
−
= [ ] −



→ + ( + + )

−
− ( )
= −

+ ( )+
−
− −
=

= −
( ) = −
→

( ) = ( ) = − ,
→

( ) =






3
3
5 +4 −9 5 +4 −9
b) =
→∞ 7−3 3 →∞ 7−3 3
3
5 + 4 − 9
3
=
→∞ 7 − 3 3
3

4 9
5 + 2 − 3
=
→∞ 7 − 3
3

5 + 0 − 0
=
0 − 3

5
= −
3






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PSPM 1 SESSION 2019/2020 SM015/2

Question A4



2
2 −3
4. a) Given = , find . Give your answer in the simplest form. [5 mark]
2


b) Given = ( + ).



Show that ( + ) + + = ( + ).
[7 marks]



SOLUTION



2 −3
a) =

= = −
′
′
= = − −

′
= + ′


= ( )(− − ) + ( − )( )

= − [− + ]

= − + = −
′
′
= − + = − −

= (− + )(− − ) + ( − )(− + )



= − [(− )(− + ) + (− + )]

= − [ − − + ]

= − [ − + ]








b) = ( + )


+ = ( + )



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PSPM 1 SESSION 2019/2020 SM015/2





[ + ] + [ + ] = [− ( ) ( + ) + ( + )]

Given:






+ + + = − ( + ) + ( + ) = ( + )





+ + = − ( ) + ( + )






+ + + ( ) = ( + )




+ + ( + ) = ( + )




( + ) + + = ( + )












































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PSPM 1 SESSION 2019/2020 SM015/2

Question A5



6
2
5. Given ( ) = − , where ≠ 0. Find the coordinate(s) of stationary point


and determine whether it is a relative maximum or a relative minimum. Give the

coordinates correct to 3 decimal places.
[5 marks]


SOLUTION



( ) = −

−

= −
−
′
( ) = +

= +

′
( ) =

+ =


+ =

= −

= − .
= − .
= − . ,


( ) = (− . ) −
− .
= .

= (− . , . )

= −
−


= −

= − .

= − = . > ( )
(− . )


∴ (− . , . ) .


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PSPM 1 SESSION 2019/2020 SM015/2



Graph: (Only for illustration purpose)

































































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PSPM 1 SESSION 2019/2020 SM015/2

Question B1



1. The function ( ) is defined by

− , ≤

− , < ≤
( ) = √ −
| − |
+ , >
{ −
Where and is constant.

a. If lim ( ) exists, find the values of . [5 marks]
→2
b. Find the values of such that ( ) is discontinuous at = 8. [3 marks]



SOLUTION




− , ≤

− , < ≤
( ) = √ −

| − | + , >
{ −


a) lim ( )
→2

( ) = ( )
→ − → +


− = −
+
→ − → √ −

− = − √ +
+
→ √ − √ +
( − )(√ + )

− =
→ + −

− = ( − )(√ + )
→ + ( − )


− = √ +
→ +





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PSPM 1 SESSION 2019/2020 SM015/2



− = √ ( )+


− =

=

= ±√





b) ( ) is discontinuous at = 8


−
( ) =
√ ( )−


( ) = =
− , − ≥
| − | = { −( − ), − <
| − | − + − , ≤
+ = + = { − + , >
→ + − → + −
= +
→ +
= +


( ) =

+ ≠

≠

∈ ℝ\{ }




























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PSPM 1 SESSION 2019/2020 SM015/2

Question B2



2
2
2. Given a right circular cylinder with height 2ℎ and radius = √ − ℎ ,
which is inscribed in a sphere of fixed radius a.
2
2
a. Show that the volume of the cylinder is = 2 ( − ℎ )ℎ. [5 marks]
b. From part 2(a), find the maximum value of the volume, in terms
of as the height, ℎ varies. Hence, state the value of the volume

if = 3. [9 marks]

c. Show that the ratio of the volume of the sphere to the maximum
volume of the cylinder is √ : . [3 marks]




SOLUTION



















2
2
2
a) + ℎ =
= −



= √ −



= ( )


= (√ − )


= ( − )


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PSPM 1 SESSION 2019/2020 SM015/2






2
2
b) = 2 ( − ℎ )ℎ
= −



= −


= ;



− =


( − ) =


− =


=


=


= −


= − < ( > )


=


= =

= √



= −

= −
=


= ( − )


= ( − )

= ( )

=

= √

= √






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PSPM 1 SESSION 2019/2020 SM015/2



4 3
c) Volume of sphere =
3


=


= (√ ) = ( √ ) = √



√
=

√
=

: = √ :





















































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