PSPM 1 Q&A

3 SESSION 2019/2020 SM015/2
[5 marks]
PSPM 1

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



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

the coordinates correct to 3 decimal places.

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]
[3 marks]
→2
[5 marks]
b. Find the values of such that ( ) is discontinuous at = 8. [9 marks]
[3 marks]
2. Given a right circular cylinder with height 2ℎ and radius = √ 2 − ℎ2,
which is inscribed in a sphere of fixed radius a.
a. Show that the volume of the cylinder is = 2 ( 2 − ℎ2)ℎ.
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.
c. Show that the ratio of the volume of the sphere to the maximum
volume of the cylinder is √ : .

END OF QUESTION PAPER

3

351

4 SESSION 2019/2020 SM015/2
[8 marks]
PSPM 1

Question A1 CHOW CHOON WOOI

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

SOLUTION

( ) = − − +

( ) = − + = ( − )( − )

( ) =

( ) =
( ) = ( ) − ( ) − ( ) + =

+ =

= − …………… (1)

( ) = ( ) − ( ) − ( ) + =

− − + =


− − + =

+ = …………… (2)

Substitute (1) into (2)

( − ) + =

− + =

− =

= −

=

( ) = − + +

4

352

5 SESSION 2019/2020 SM015/2

PSPM 1

− + + CHOW CHOON WOOI
− + +

− +
− +
− +

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

Alternative Method:

( ) = − − +
( ) = − + = ( − )( − )
( ) = ➔ ( − ) ( )
( ) = ➔ ( − ) ( )
( ) = ( ) ( )
− − + = ( − )( − )( + )
− − + = ( − − + )( + )
− − + = ( − + )( + )
− − + = + − − + +
− − + = + ( − ) + ( − ) +
Compare the coefficient:
: =

=

5

353

6 SESSION 2019/2020 SM015/2

PSPM 1 = CHOW CHOON WOOI
=
: − = −
: − = ( ) − ( )
=
: − = −
− = ( ) − ( )
= −

= ; = −
( ) = ( − )( − ) ( + )

= ( − )( − ) ( + )

6

354

7 SESSION 2019/2020 SM015/2

PSPM 1 [5 marks]
[5 marks]
Question A2 CHOW CHOON WOOI

2. a) Show that − cot can be simplified as tan 2 .
b) Hence, find the values of ° and °. Give your answer in the form

of + √ , where a, b and c are integers.

SOLUTION

a) − = −



= −



= −( − )


=


=


=



b) = −


° = ( °)

= ° − °

= − °

° °

= − √





= − √

7

355

8 SESSION 2019/2020 SM015/2

PSPM 1

= + CHOW CHOON WOOI

° = ° +
= ( °) +
= ( − √ ) +
= ( + − √ ) +

= − √
= ( − √ )

8

356

9 SESSION 2019/2020 SM015/2

PSPM 1 [7 mark]
[3 mark]
Question A3

3. a) A function ( ) is defined as

− < CHOW CHOON WOOI
− , =
( ) = − ,
− >
{ − ,

Determine whether ( ) is continuous at = .

b) Find −+ − .

→∞

SOLUTION

9− 2 , < 3
= 3
−3 > 3

a) ( ) = −6,

162 −54 2 ,

{ 3−27

At =

( ) = −

( ) = ( − )
→ − → − −

= [( + )( − )]
→ − −

= [−( + )( − )]
→ − −

= [−( + )]

→ −

= −( + )

= −

9

357

10 SESSION 2019/2020 SM015/2

PSPM 1

( ) = ( ) −
−
→ + → +

= [ ]− ( − ) + +

→ + ( − )( + + ) − − CHOW CHOON WOOI
−
= [ ]−
−
→ + ( + + ) −

= − ( ) −
−
+ ( )+

= −


= −

( ) = −

→

( ) = ( ) = − ,

→

( ) =

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

5 3 + 4 − 9

= 3
7 − 3 3
→∞

3

= 5 + 4 − 9
2 3
→∞ 7
3 − 3

5+0−0
= 0−3

5
= −3

10

358

11 SESSION 2019/2020 SM015/2
[5 mark]
PSPM 1 [7 marks]

Question A4

4. a) Given = 2 −3 , find 2 2 . Give your answer in the simplest form. CHOW CHOON WOOI
b) Given = ( + ).

Show that ( + ) + + = ( + ).


SOLUTION

a) = 2 −3

= = −

′ = ′ = − −

= ′ + ′



= ( )(− − ) + ( − )( )

= − [− + ]

= − + = −

′ = − + ′ = − −

= (− + )(− − ) + ( − )(− + )


= − [(− )(− + ) + (− + )]

= − [ − − + ]

= − [ − + ]

b) = ( + )
+ = ( + )



11

359

12 SESSION 2019/2020 SM015/2

PSPM 1

[ + ] + [ + ] = [− ( ) ( + ) + ( + )] Given:
= ( + )


+ + + = − ( + ) + ( + )


+ + = − ( ) + ( + ) CHOW CHOON WOOI


+ + + ( ) = ( + )


+ + ( + ) = ( + )


( + ) + + = ( + )


12

360

13 SESSION 2019/2020 SM015/2
[5 marks]
PSPM 1

Question A5 CHOW CHOON WOOI

5. Given ( ) = 2 − 6, 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.

SOLUTION

( ) = −



= − −

′( ) = + −

= +


′( ) =


+ =
+ =

= −

= − .

= − .

= − . ,

( ) = (− . ) −

− .

= .

= (− . , . )

= − −


= −


= − .

= − = . > ( )
(− . )

∴ (− . , . ) .

13

361

14 SESSION 2019/2020 SM015/2

PSPM 1

Graph: (Only for illustration purpose)

CHOW CHOON WOOI

14

362

15 SESSION 2019/2020 SM015/2

PSPM 1 [5 marks]
[3 marks]
Question B1

1. The function ( ) is defined by

− , ≤ CHOW CHOON WOOI

−
, < ≤
( ) = √ −

| − | >
{ − + ,

Where and is constant.

a. If lim ( ) exists, find the values of .

→2

b. Find the values of such that ( ) is discontinuous at = 8.

SOLUTION

− , ≤

− < ≤
,
( ) = √ −

| − | >
{ − + ,

a) lim ( )

→2

( ) = ( )
→ − → +

− = −
→ − → + √ −

− = − √ +

→ + √ − √ +

− = ( − )(√ + )
→ + −

− = ( − )(√ + )
( − )
→ +

− = √ +

→ +

15

363

16 SESSION 2019/2020 SM015/2

PSPM 1 CHOW CHOON WOOI

− = √ ( )+


− =
=
= ±√

b) ( ) is discontinuous at = 8

( ) = −

√ ( )−

( ) = = | − | = {−( −− ) ,, − ≥
− <


| − | = − + = {− −+ , , ≤
+ >
+
→ + − → + −

= +

→ +

= +

( ) =

+ ≠

≠

∈ ℝ\{ }

16

364

17 SESSION 2019/2020 SM015/2

PSPM 1 [5 marks]
[9 marks]
Question B2 [3 marks] CHOW CHOON WOOI

2. Given a right circular cylinder with height 2ℎ and radius = √ 2 − ℎ2,
which is inscribed in a sphere of fixed radius a.
a. Show that the volume of the cylinder is = 2 ( 2 − ℎ2)ℎ.
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.
c. Show that the ratio of the volume of the sphere to the maximum
volume of the cylinder is √ : .

SOLUTION






a) 2 + ℎ2 = 2

= −

= √ −

= ( )

= (√ −

)

= ( − )

17

365

18 SESSION 2019/2020 SM015/2

PSPM 1

b) = 2 ( 2 − ℎ2)ℎ CHOW CHOON WOOI

= −

= −



= ;



− =

( − ) =

− =

=

=


= −



= − < ( > )


=
= =



= √
= −

= −
=

= ( − )
= ( − )
= ( )
=
= √

= √

18

366

19 SESSION 2019/2020 SM015/2

PSPM 1

c) Volume of sphere = 4 3

3

= CHOW CHOON WOOI

= (√ = ( √ ) = √

)

= √



= √


: = √ :

19

367

2020/2021

368

CHOW CHOON WOOI

SM015 PSPM I 2020/2021

Section A

1. (a) ℎ < 1 CHOW CHOON WOOI
≥ 1, ≠
+ 1 ,
( ) = { 2 − −

− ,

= 1. ℎ . , lim ( ). [ 6 marks]
[3 marks]
→

(b) 1

2 + 5 3
lim (3 4)
−
→+∞

2. (a) ℎ

= 1 + 3 2, = 2 − 4√ , > 0


2 2 [6 marks]
2 . , 2 = 1

(b) 2 [5 marks]
= (1 − 2 ) . ℎ ℎ = −2 2 . [5 marks]

3. A farmer has a 100 meter of fencing wants to fence off a rectangular piece of
land that borders a straight river. The part of the land which borders the river
needs no fencing. Find the dimensions of the enclosed region that has the
maximum area.

Section B

1. (a) ℎ [4 marks]
ln 24 −1 = ln 8 +5 + 2161−2 [5 marks]

2.

(b) 1 = 2 − 3 2 = 4 + 2 .

ℎ , 3 = ̅ 1 ℎ , | 3|.
2

369

SM015 PSPM I 2020/2021
2 ℎ ℎ .

4 [5 marks] CHOW CHOON WOOI
( ) 4 − ≥ 6. [6 marks]
( ) | − 6| < 3| − 2|
3 ℎ ( ) ℎ − 2 ≤ ≤ 5.

(5,11)

(−2,9)



(3, −1)

[1 marks]
(a) ℎ ( ). [4 marks]
[2 marks]
(b) ℎ ( ).
[4 marks]
(c) ( )(0). [5 marks]
[4 marks]
4 Given the function ( ) = 3 , ≠ −2, 2 ( )( ) = 3 , ≠ 5.
2 −4 −5 [8 marks]

(a) ( ) ≥ 0 .

(b) Show that −1( ) exist. Hence, find −1( ) and its domain.

(c) Find the values of and if ℎ( ) = + , such that
(ℎ −1)( ) = 5 2 + 12.

5 Given
2 2 + 2

( ) = 2 − 4
(a) ℎ ℎ ( )

370

SM015 PSPM I 2020/2021

(b) , ℎ ℎ ℎ ( ). [3 marks]
[5 marks]
6 2 [6 marks]
(a) = ( + ) , ℎ ℎ − 2 = 0. [6 marks]
4 2

CHOW CHOON WOOI
(b) ( ) = − 2 , ≠ ln 2 ℎ .

, ℎ ′( ) = −2.

(c) Find the value of tangent to the curve 2 = 3 and 2 2 + 3 2 = , where is
a constant, at the points (1, −1) (1, 1).
Hence, state the relationship between the tangent values at these two points.

7 Consider a cone as shown in FIGURE 2.

°

FIGURE 2

Water flows out from a cone through a hole located at the bottom tip at the [8 marks]
rate of 5 3/ . Find the rate of change of the height of the water when the
volume of water in the cone is 3000 3.

END OF QUESTION PAPER

371

SM015 PSPM I 2020/2021

Question A1 < 1
≥ 1, ≠
(a) ℎ

+ 1 ,
( ) = { 2 − −

,
−

= 1. ℎ . , lim ( ). [ 6 marks] CHOW CHOON WOOI
[3 marks]
→

1
2 + 5 3
(b) lim ( )
→+∞ 3 − 4

SOLUTION

a) + 1 , < 1
( ) = { 2 − − ≥ 1, ≠
− ,

lim ( ) = lim ( )
→1− →1+

lim + 1 = lim 2 − −

→1− →1+ −

12 − 1 −
1 + 1 = 1 −

−
2 = 1 −

2 − 2 = −

= 2

= 2

+ 1 , < 1

( ) = { 2 − − 2 ≥ 1, ≠ 2
− 2 ,

lim ( ) = lim 2 − − 2

→2 →2 − 2

= lim ( − 2)( + 1)

→2 − 2

= lim + 1

→2

= 2+1

=3

372

SM015 PSPM I 2020/2021

b) 1 3√2 + 5

2 + 5 3 lim
lim (3 4) =
− →+∞ 3√3 − 4
→+∞

3√2 + 5

= lim
→+∞ 3√3 4
− CHOW CHOON WOOI

3√2 + 5

= lim
→+∞ 4
3√3 −

3√2 + 0
= 3√3 − 0

1

23
= (3)

373

SM015 PSPM I 2020/2021

Question A2 [6 marks]
[5 marks]
(a) ℎ

= 1 + 3 2, = 2 − 4√ , > 0


2 2 CHOW CHOON WOOI
2 . , 2 = 1

(b) 2
= (1 − 2 ) . ℎ ℎ = −2 2 .

SOLUTION

(a) = 1 + 3 2, = 2 − 4√ , > 0


= −1 + 3 2 1

= 2 − 4 2

= − −2 + 6 = 2 − 2 −12


1 2
= − 2 + 6 = 2 − 1

6 3 − 1 2
= 2
3

2 2 − 2
=1

2


= .

3
2 2 − 2 2
= . 6 3 − 1
1

2

33

2 2 ( 2 − 1)
= 6 3 − 1

3

2 3 − 2 2
= 6 3 − 1


= ( ) .

3
2 3 − 2 2
( ) = ( )
6 3 −1

374

SM015 PSPM I 2020/2021

3 = 6 3 − 1
′ = 18 2
= 2 3 − 2 2

1

′ = 6 − 3 2

2 ′ − ′
2 = ( 2 ) . ( )

13
(6 3 − 1) (6 − 3 2) − (2 3 − 2 2) (18 2)
2 CHOW CHOON WOOI
=[ ] . ( )
(6 3 − 1)2 6 3 − 1

13

2(6 3 − 1) (6 − 3 2) − 18 4 (2 3 − 2 2)
= (6 3 − 1)3

=

13
2 (1)2[6(1)3 − 1] [6(1) − 3(1)2] − 18(1)4 [2(1)3 − 2(1)2]
2 =
[6(1)3 − 1]3

5[3] − 18[2 − 2]
= [6 − 1]3

15
= 125

3
= 25

375

SM015 PSPM I 2020/2021

(b) 2
= (1 − 2 )

1 − cos 2 2
= sin 2 (1 − 2 )

1 − cos 2 (1 − cos 2 )(2 cos 2 ) − (sin 2 )(2 sin 2 )] CHOW CHOON WOOI
= sin 2 [
(1 − cos 2 )2

2cos 2 (1 − cos 2 ) − 2 sin2 2
= sin 2 (1 − cos 2 )

2cos 2 − 2 cos2 2 − 2 sin2 2
= sin 2 (1 − cos 2 )

2cos 2 − 2(cos2 2 + sin2 2 )
= sin 2 (1 − cos 2 )

2cos 2 − 2(1)
= sin 2 (1 − cos 2 )

−2(1 − cos 2 )
= sin 2 (1 − cos 2 )

−2
= sin 2

= −2 2

376

SM015 PSPM I 2020/2021

Question A3

A farmer has a 100 meter of fencing wants to fence off a rectangular piece of land
that borders a straight river. The part of the land which borders the river needs no
fencing. Find the dimensions of the enclosed region that has the maximum area.

[5 marks] CHOW CHOON WOOI

SOLUTION





+ + = 100
= 100 − 2
To maximize the area:

=
= (100 − 2 )
= 100 − 2 2

= 100 − 4


= 0

100 − 4 = 0
= 25
= 100 − 2(25)

= 50

2
2 = −4 ( )

∴ = 50 = 25 ℎ .

377

SM015 PSPM I 2020/2021

Question B1 [4 marks]

(a) ℎ
ln 24 −1 = ln 8 +5 + 2161−2

2.

(b) 1 = 2 − 3 2 = 4 + 2 . CHOW CHOON WOOI

ℎ , 3 = ̅ 1 ℎ , | 3|. [5 marks]
2

SOLUTION

(a) ln 24 −1 = ln 8 +5 + 2161−2 log = log
ln 24 −1 = ln 23( +5) + 224(1−2 )
ln 24 −1 − ln 23 +15 = (4 − 8 ) 22 = 1
24 −1
ln 23 +15 = 4 − 8
ln 24 −1−(3 +15) = 4 − 8 log − log = log
ln 2 −16 = 4 − 8
( − 16) ln 2 = 4 − 8

ln 2 − 16 ln 2 = 4 − 8

ln 2 + 8 = 4 + 16 ln 2
(ln 2 + 8) = 4 + 16 ln 2

4 + 16 ln 2
= ln 2 + 8

378

SM015 PSPM I 2020/2021

(b) 1 = 2 − 3 , 2 = 4 + 2 = +
| | = √ 2 + 2
3 = ̅ 1
2

2 + 3 CHOW CHOON WOOI
= 4 + 2

2 + 3 4 − 2
= 4 + 2 . 4 − 2

8 − 4 + 12 − 6 2
= 16 − 4 2

14 + 8
= 20

14 8
= 20 + 20

72
= 10 + 5

| 3| = √( 7 2 + 22
()
10 )
5

= √ 49 + 4
100 25

= √13
20

379

SM015 PSPM I 2020/2021

Question B2 [5 marks]
[6 marks]
ℎ ℎ .
(4, ∞)
( ) 4 ≥ 6. +
4 − -

( ) | − 6| < 3| − 2| - CHOW CHOON WOOI

SOLUTION

(a) 4
4 − ≥ 6

4
4 − − 6 ≥ 0

4 − 6(4 − )
4 − ≥ 0

4 − 24 + 6
4 − ≥ 0

10 − 24
4 − ≥ 0

:

10 − 24 = 0 ➔ = 12

4 − = 0 5

➔ = 4

10 − 24 12 12
(−∞, 5 ) ( 5 , 4)
4 −
10 − 24 - +
+ +
4 −
- ○+

12
{ : 5 ≤ < 4}

(b) | − 6| < 3| − 2|
( − 6)2 < [3( − 2)]2
2 − 12 + 36 < 9( 2 − 4 + 4)

380

SM015 PSPM I 2020/2021

2 − 12 + 36 < 9 2 − 36 + 36 (3, ∞)
+
8 2 − 24 > 0 +

8 ( − 3) > 0 ○+

: ➔ = 0 CHOW CHOON WOOI
8 = 0 ➔ = 3
− 3 = 0

8 (−∞, 0) (0,3)
− 3 - +
- -
8 ( − 3)
○+ -

{ : < 0 > 3}

381

SM015 PSPM I 2020/2021

Question B3

ℎ ( ) ℎ − 2 ≤ ≤ 5.

(5,11)

(−2,9) CHOW CHOON WOOI



(3, −1) [1 marks]
[4 marks]
(a) ℎ ( ). [2 marks]

(b) ℎ ( ).
(c) ( )(0).

SOLUTION

(a) : [−1, 11]

(b) −2 ≤ ≤ 3: (−2, 9), (3, −1)

− 1 = 2 − 1
− 1 2 − 1

− 9 −1 − 9
+ 2 = 3 + 2

− 9
+ 2 = −2

− 9 = −2 − 4

= −2 + 5

3 ≤ ≤ 5: (5,11), (3, −1)

− 1 = 2 − 1
− 1 2 − 1

382

SM015 PSPM I 2020/2021

− 11 −1 − 11
− 5 = 3 − 5
− 11
− 5 = 6
− 11 = 6 − 30
= 6 − 19

( ) = {−62 −+159 , −2 ≤ ≤ 3 CHOW CHOON WOOI
, 3 ≤ ≤ 5

(c) (0) = [ (0)]
= [−2(0) + 5]
= (5)
= 6(5) − 19
= 11

383

SM015 PSPM I 2020/2021

Question B4

Given the function ( ) = 3 , ≠ −2, 2 ( )( ) = 3 , ≠ 5.
2 −4 −5

(a) ( ) ≥ 0 . [4 marks]
[5 marks]
(b) Show that −1( ) exist. Hence, find −1( ) and its domain. CHOW CHOON WOOI
[4 marks]
(c) Find the values of and if ℎ( ) = + , such that
(ℎ −1)( ) = 5 2 + 12.

SOLUTION

(a) 3
( ) = 2 − 4 , ≠ −2, 2

( )( ) = 3 5 , ≠ 5
−

3
[ ( )] = − 5

33
[ ( )]2 − 4 = − 5

3[ ( )]2 − 12 = 3 − 15

3[ ( )]2 = 3 − 3

[ ( )]2 = − 1

( ) = √ − 1

: − 1 ≥ 0
≥ 1

(b) ( ) = √ − 1
( 1) = ( 2)

√ 1 − 1 = √ 2 − 1

22

(√ 1 − 1) = (√ 2 − 1)
1 − 1 = 2 − 1
1 = 2
1 = 2 ℎ ( 1) = ( 2). ℎ , ( )
−1( ) .

384

SM015 PSPM I 2020/2021

[ −1( ) ] =
√ −1( ) − 1 =
−1( ) − 1 = 2
−1( ) = 2 + 1
−1 = = [0, ∞)

(c) ℎ( ) = + CHOW CHOON WOOI
−1( ) = 2 + 1
(ℎ −1)( ) = 5 2 + 12
ℎ[ −1( )] = 5 2 + 12
[ −1( )] + = 5 2 + 12
[ 2 + 1] + = 5 2 + 12
2 + + = 5 2 + 12
ℎ
2: = 5
: + = 12
5 + = 12
= 7

Alternative Method

ℎ( ) = +

−1( ) = 2 + 1

(ℎ −1)( ) = 5 2 + 12

ℎ[ −1( )] = 5 2 + 12

= 2 + 1 ➔ 2 = − 1

ℎ( ) = 5( − 1) + 12

= 5 − 5 + 12

= 5 + 7

ℎ( ) = 5 + 7

= 5 ; = 7

385

SM015 PSPM I 2020/2021

Question B5 [8 marks] CHOW CHOON WOOI
[3 marks]
Given
2 2 + 2

( ) = 2 − 4
(a) ℎ ℎ ( )

(b) , ℎ ℎ ℎ ( ).

SOLUTION

(a) 2 2 + 2
( ) = 2 − 4

Vertical Asymptotes

2 − 4 = 0

= ±2

2 2 + 2 2(2)2 + 2 lim 2 2 + 2 2(2)2 + 2
lim = =
2 − 4 (2)2 − 4 →2+ 2 − 4 (2)2 − 4
→2−

10 10
=0 =0

= −∞ = +∞

2 2 + 2 2(−2)2 + 2 2 2 + 2 2(−2)2 + 2
→li−m2− 2 − 4 = (−2)2 − 4 lim =
2 − 4 (−2)2 − 4
10 →−2+
=0
= +∞ 10
=0

= −∞

386

SM015 PSPM I 2020/2021

Horizontal Asymptotes

2 2 + 2 2 2 + 2 2 2 + 2 2 2 + 2

lim 2 − 4 = lim 2 lim 2 − 4 = lim 2
2 − 4 2 − 4
→+∞ →+∞ →−∞ →−∞

2 2

= lim 2 + 2 = lim 2 + 2 CHOW CHOON WOOI
1 − 1 −
→+∞ 2 →−∞ 2
4 4

2 2

2+0 2+0
= 1− 0 = 1− 0

=2 =2

∴ : = 2 = −2
: = 2

(b) 2 2 + 2
= 2 − 4

( )

= 2


= −2 = 2

387

SM015 PSPM I 2020/2021

Question B6

2
(a) = ( + ) , ℎ ℎ 2 − 2 = 0.
4
[5 marks]
[6 marks] CHOW CHOON WOOI
(b) ( ) = − 2 , ≠ ln 2 ℎ . [6 marks]

, ℎ ′( ) = −2.

(c) Find the value of tangent to the curve 2 = 3 and 2 2 + 3 2 = , where is
a constant, at the points (1, −1) (1, 1).
Hence, state the relationship between the tangent values at these two points.

SOLUTION

(a)
= (4 + )

= sec2 + )
(

4

2
= 2 sec ( + ) sec ( + )
2 4 4

2 )
= 2 sec ( + ) sec ( + ) tan ( +
2 4 4 4

2 = 2 sec2 + ) tan + )
2 (4 (4

2
2 = 2 ( ) . ( )

2
2 = 2

2
2 − 2 = 0

388

SM015 PSPM I 2020/2021

(b)
( ) = − 2

= = − 2

′ = ′ =

′( ) = ( − 2)( ) − ( )( ) CHOW CHOON WOOI
( − 2)2

( )[( − 2) − ( )]
= ( − 2)2

( )[−2]
= ( − 2)2

−2
= ( − 2)2

′( ) = 2

−2
( − 2)2 = 2

−2 = 2( − 2)2

= 2 − 4 + 4

2 − 5 + 4 = 0
=

2 − 5 + 4 = 0

( − 1)( − 4) = 0

= 1 = 4

= 1 = 4
= 0 ln = ln 4

ln = ln 4

= ln 4

∴ = 0 = ln 4

(c) 2 = 3

2 = 3 2


3 2
= 2

ℎ (1, −1)

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SM015 PSPM I 2020/2021

3(1)2 3
= 2(−1) = − 2

ℎ (1,1)

3(1)2 3
= 2(1) = 2

2 2 + 3 2 = CHOW CHOON WOOI


4 + 6 = 0
4
= − 6
ℎ (1, −1)

4(1)2 2
= − 6(−1) = 3

ℎ (1,1)
4(1)2 2
= − 6(1) = − 3

At these two points, the tangent line of = and + = are perpendicular
to each other.

390

SM015 PSPM I 2020/2021

Question B7

Consider a cone as shown in FIGURE 2.

° CHOW CHOON WOOI

FIGURE 2

Water flows out from a cone through a hole located at the bottom tip at the [8 marks]
rate of 5 3/ . Find the rate of change of the height of the water when the
volume of water in the cone is 3000 3.

SOLUTION Remarks

° = −5 3/


° ℎ
=? ℎ = 3000

ℎ ℎ
= .

ℎ
= . (−5)

391

SM015 PSPM I 2020/2021

= 1 2ℎ
3


CHOW CHOON WOOI
°


tan 30° = ℎ
1
√3 = ℎ

ℎ = √3

ℎ
=

√3

= 1 ℎ2
() ℎ
3 √3

= 1 ℎ3
9

= 1 ℎ2
ℎ 3

ℎ = ℎ . (−5)


3
= ℎ2 . (−5)

15
= ℎ2

=

3000 = 1 ℎ3
9

ℎ3 = 27000

ℎ = 30

ℎ −15
= (30)2
= − 1 /

60

392