1920 SM015_1

1 SESSION 2019/2020 SM015/1

PSPM 1

SM015/1
PSPM 1

2019/2020

14 OCTOBER 2019

KOLEJ MATRIKULASI KEDAH
Authored by: KANG KOOI WEI

1

2 SESSION 2019/2020 SM015/1

PSPM 1 [2 marks]
[7 marks]
Questions [5 marks]
[5 marks]
SECTION A [45 marks] [3 marks]
[4 marks]
This section consists of 5 questions. Answer all questions. [2 marks]

1. Given the complex numbers 1 = − and 2 = 2 + √3. [2 marks]
[5 marks]
a. Express 12 and ̅ 2 in the form + , where , ∈ ℛ.
[2 marks]
b. From part 1(a), find = 12+ ̅ ̅2̅. Hence, find | | and argument . [4 marks]
[4 marks]
1

2. Solve the following:

a. 3(52 ) + 2512 +1 = 200
b. + 4 ≤ 2 + < 12

3. The sum of the first terms of a sequence is given by = 2 + 3−4 .

a. Find the value of constant c such that the -th term is 3−4 .
b. Show that the3 sequence is a geometric series.
c. Find the sum of the infinite series, ∞.

2 30
4. Given matrix = [−5 0 4].

0 21
a. Find the determinant of matrix by expanding the first row.
b. Calculate the adjoin of matrix . Hence, find −1.

1
c. Solve the equation = , where = [2], by using the answer obtained

2
in part 4(b).

5. Given ( ) = 1+11+1 1 .
a. Simplify ( ) and evaluate (12).
b. The domain of ( ) is a set of real number except three numbers.
Determine the numbers.

2

3 SESSION 2019/2020 SM015/1

PSPM 1 [6 marks]
[7 marks]
SECTION B [25 marks] [2 marks]
[7 marks]
1. Solve the following: [3 marks]
a. 22 = 2 4( + 4)

b. 2 | −3 | ≥ 1

2 −1

2. Given a function ( ) = ln(2 + 1)

a. State the domain and range of ( ).

b. Find the inverse function of ( ) and state its domain and range. Hence, find

the value of for which −1( ) = 0.

c. Sketch the graph of ( ) and −1( ) on the same coordinate axes.

END OF QUESTION PAPER

3

4 SESSION 2019/2020 SM015/1

PSPM 1

Question A1

1. Given the complex numbers 1 = − and 2 = 2 + √3.

a. Express 12 and ̅ 2 in the form + , where , ∈ ℛ.

b. From part 1(a), find = 12+̅ ̅2̅. Hence, find | | and argument .

1

SOLUTION

a) = −

= + √
= (− )

= − +
̅ ̅ ̅ = − √

b) = +̅ ̅ ̅



= (− )+( − √ )

(− )

= − √

(− )

= ( − √ )( )
(− )( )

= − √
−

= +√



= √ +

| | = √√ +

4

5 SESSION 2019/2020 SM015/1

PSPM 1

= √ +
=
= − ( )

√

=



5

6 SESSION 2019/2020 SM015/1

PSPM 1

Question A2

2. Solve the following:
a. 3(52 ) + 2512 +1 = 200
b. + 4 ≤ 2 + < 12

SOLUTION

a) 3(52 ) + 2512 +1 = 200

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



= − (ignored)



=
=

b) + 4 ≤ 2 + < 12 And + <
+ − <
+ ≥ + ( + )( − ) <
+ − − ≥ = − =
− ≥
( + )( − ) ≥ 6
= − =

7 SESSION 2019/2020 SM015/1

PSPM 1

(−∞, − ) (− , ) ( , ∞) (−∞, − ) (− , ) ( , ∞)
+
+ - + + + - + +
+
− - - ○+ − - - +

○+ - + ○-

(−∞, − ] ∪ [ , ∞) And (− , )

− −

(− , − ] ∪ [ , )

7

8 SESSION 2019/2020 SM015/1

PSPM 1

Question A3

3. The sum of the first terms of a sequence is given by = 2 + 3−4 .
a. Find the value of constant c such that the -th term is 3−4 .
b. Show that the sequence is a geometric series.
c. Find the sum of the infinite series, ∞.

SOLUTION

a) = 2 + 3−4

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

b) = − ( − )

= − ( − )
− [ − ( − )]
−

= ( − )
[ − + ]
−

= − + −

−

= −

−

8

9 SESSION 2019/2020 SM015/1

PSPM 1

=

−

,

−

c) = − ( − )

= = − ( − ( )) = −


=



∞ =
−

(− )

=

−( )

(− )

=

( )

= −

9

10 SESSION 2019/2020 SM015/1

PSPM 1

Question A4

2 30
4. Given matrix = [−5 0 4].

0 21

a. Find the determinant of matrix by expanding the first row.

b. Calculate the adjoin of matrix . Hence, find −1.

1
c. Solve the equation = , where = [2], by using the answer obtained in

2

part 4(b).

SOLUTION

2 30
a) = [−5 0 4]

0 21

| | = ( ) | | − ( ) |− | + ( ) |− |

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

b) Adjoin of matrix

+ | | − |− | + |− |
, = − | | + | | − | |

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

10

11 SESSION 2019/2020 SM015/1

PSPM 1

− −
= (− − )

−

=

− −
= ( − )

− −

− =
| |

− −
−
= ( − )
− −

−
= (− − )

−

c) =

= −


[− ] [ ] = [ ]

−

[ ] = (− − ) [ ]
−

+ −
= (− − + )

+ −

11

12 SESSION 2019/2020 SM015/1

PSPM 1

−
= ( )

−

∴ = − , = , = −

12

13 SESSION 2019/2020 SM015/1

PSPM 1

Question A5

5. Given ( ) = 1+11+1 1 .
a. Simplify ( ) and evaluate (21).
b. The domain of ( ) is a set of real number except three numbers. Determine

the numbers.

SOLUTION

a) ( ) = 1
1+1+1 1

=
+ +



=

+ +

=
+ +
+

= +

+

( ) = ( )+
( )+


= ( )



=



b) ( ) = 1
1+1+1 1

13

14 SESSION 2019/2020 SM015/1

PSPM 1

1 ≠ 0  ≠ 0



1 + 1 ≠ 0  ≠ −1



1 + 1 ≠ 0  2 + 1 ≠ 0  ≠ −1
1+ 1
2

∴ ≠ 0; ≠ −1; ≠ − 1

2

14

15 SESSION 2019/2020 SM015/1

PSPM 1

Question B1

1. Solve the following:
a. 22 = 2 4( + 4)

b. 2 |2 −−31| ≥ 1

SOLUTION

a) 22 = 2 4( + 4)

= ( + )


= ( + )


= ( + )


= ( + )
( )

= ( + )
= +

=

b) 2 | −3 | ≥ 1

2 −1

| − | ≥

−

− ≥ OR − ≤ −
−
−

− − ≥ − + ≤

− −

( − )−( − ) ≥ ( − )+( − ) ≤
( − ) ( − )

− − + ≥ − + − ≤

− −

15

16 SESSION 2019/2020 SM015/1

PSPM 1 − ≤

− ≥ −

− = ; =

− <
<

(−∞, ) ( , ) ( , ∞)

< +
- +

− +
-

+
-

−

+ ○-

OR ( , ]
(−∞, )






(−∞, ) ∪ ( , ]



16

17 SESSION 2019/2020 SM015/1

PSPM 1

Question B2

2. Given a function ( ) = ln(2 + 1)
a. State the domain and range of ( ).
b. Find the inverse function of ( ) and state its domain and range. Hence, find
the value of for which −1( ) = 0.
c. Sketch the graph of ( ) and −1( ) on the same coordinate axes.

SOLUTION

a) ( ) = ln(2 + 1)

:

: + >

> −



: (− , ∞)

:

= (−∞, ∞)

b) ( ) = ln(2 + 1)

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



:
− = = (−∞, ∞)

:

− = = (− , ∞)



17

18 SESSION 2019/2020 SM015/1

PSPM 1

− ( ) =
− =



− =
=
=
=

c)

18