qs 015/1 matriculation programme examination semester i
TRANSCRIPT
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 2
1. Solve the equation 3๐ฅ + 3(3โx) = 12.
2. Solve the inequality 1
6โ๐ฅ<
1
xโ1.
3. Given matrices ๐ด = [1 0 04 1 0๐ ๐ 1
] and ๐ต = [1 0 0๐ง 1 0๐ฅ ๐ฆ 1
] where ๐ต is the inverse of ๐ด. Find ๐ฅ, ๐ฆ and
๐ง in terms of ๐ and ๐.
4. Using algebraic method, find the least value of n for which the sum of the first n terms of a
geometric series
0.88 + (0.88)2 + (0.88)3 + (0.88)4 + โฏ
is greater than half of its sum to infinity.
5. (a) State the interval for x such that the expansion for (4 + 3๐ฅ)3
2 is valid.
(b) Expand (4 + 3๐ฅ)3
2 in ascending power of x up to the term in ๐ฅ3.
(c) Hence, by substituting an appropriate value of x, evaluate (5)3
2 correct to three decimal
places.
6. (a) Given ๐(๐ฅ) = 2๐ฅ + 1 and ๐(๐ฅ) = ๐ฅ2 + 2x โ 1.
(i) Find (๐ โ ๐)(x).
(ii) Evaluate (3๐ โ 2)(1).
(b) Given ๐(๐ฅ) = โ2๐ฅ +1
2. State the domain and range of ๐(๐ฅ). Hence, on the same axes,
sketch the graph of ๐(๐ฅ) and ๐โ1(x).
7. Let ๐ง = ๐ + ๐๐ be a nonzero complex number.
(a) Show that 1
๐ง=
zฬ
|z|2.
(b) Show that if zฬ = โz, then z is a complex number with only an imaginary part.
(c) Find the value of a and b if ๐ง(2 โ ๐) = (zฬ + 1)(1 + i).
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 3
8. (a) Solve the for the following equation |6๐ฅ2 + ๐ฅ โ 11| = 4.
(b) Find the solution set for the inequality
2 โ (๐ฅ + 2
๐ฅ โ 4) < 5
9. Two companies P and Q decided to award prizes to their employees for three work ethical
values, namely punctuality (x), creativity (y) and efficiency (x). Company P decided to award a
total of RM3850 for the three values to 6, 2 and 3 employees respectively, while company Q
decided to award RM3200 for the three values to 4, 1 and 5 employees respectively. The total
amount for all the three prizes is RM1000.
(a) Construct a system of linear equations to represent the above situation.
(b) By forming a matrix equation, solve this equation system using the elimination
method.
(c) With the same total amount of money spent by company P and Q, is it possible for
company P to award 15 employees for their creativity instead of 2 employees? Give
your reason.
10. (a) Determine wheather ๐(๐ฅ) =1
๐ฅโ4 and ๐(๐ฅ) =
4๐ฅ+1
๐ฅ are inverse function of each other
by computing their composite functions.
(b) Given ๐(๐ฅ) = ln(1 โ 3๐ฅ).
(i) Determine the domain and range of ๐(๐ฅ). Then sketch the graph of ๐(๐ฅ).
(ii) Find ๐โ1(x), if it exists. Hence, state the domain and range of ๐โ1(x).
END OF QUESTION PAPER
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 4
1. Solve the equation 3๐ฅ + 3(3โx) = 12.
SOLUTION
3๐ฅ + 3(3โx) = 12
3๐ฅ + 33. 3โx = 12
3๐ฅ + 33.1
3x= 12
3๐ฅ + 27
3x= 12
Let ๐ฆ = 3๐ฅ
๐ฆ + 27
y= 12
๐ฆ2 + 27
y= 12
๐ฆ2 + 27 = 12y
๐ฆ2 โ 12y + 27 = 0
(y โ 9)(๐ฆ โ 3) = 0
(y โ 9) = 0 (๐ฆ โ 3) = 0
๐ฆ = 9 ๐ฆ = 3
3๐ฅ = 9 3๐ฅ = 3
๐ฅ = 2 ๐ฅ = 1
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 5
2. Solve the inequality 1
6โ๐ฅ<
1
xโ1.
SOLUTION 1
6 โ ๐ฅ<
1
x โ 1
1
6 โ ๐ฅโ
1
x โ 1< 0
(x โ 1) โ (6 โ ๐ฅ)
(6 โ ๐ฅ)(x โ 1)< 0
๐ฅ โ 1 โ 6 + ๐ฅ
(6 โ ๐ฅ)(x โ 1)< 0
2๐ฅ โ 7
(6 โ ๐ฅ)(x โ 1)< 0
2๐ฅ โ 7 = 0 6 โ ๐ฅ = 0 ๐ฅ โ 1 = 0
๐ฅ =7
2 ๐ฅ = 6 ๐ฅ = 1
(โโ, 1) (1,7
2) (
7
2, 6) (6, โ)
2๐ฅ โ 7 - - + +
6 โ ๐ฅ + + + -
x โ 1 - + + + 2๐ฅ โ 7
(6 โ ๐ฅ)(x โ 1) + - + -
โด (1,7
2) โช (6, โ)
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 6
3. Given matrices ๐ด = [1 0 04 1 0๐ ๐ 1
] and ๐ต = [1 0 0๐ง 1 0๐ฅ ๐ฆ 1
] where ๐ต is the inverse of ๐ด. Find ๐ฅ, ๐ฆ and
๐ง in terms of ๐ and ๐.
SOLUTION
๐ด = [1 0 04 1 0๐ ๐ 1
] ๐ต = [1 0 0๐ง 1 0๐ฅ ๐ฆ 1
]
๐๐๐๐๐ ๐ต is the inverse of ๐ด ๐ด๐ต = ๐ผ
๐ด๐ต = ๐ผ
[1 0 04 1 0๐ ๐ 1
] [1 0 0๐ง 1 0๐ฅ ๐ฆ 1
] = [1 0 00 1 00 0 1
]
[1 + 0 + 0 0 + 0 + 0 0 + 0 + 04 + ๐ง + 0 0 + 1 + 0 0 + 0 + 0
๐ + ๐๐ง + ๐ฅ 0 + ๐ + ๐ฆ 0 + 0 + 1] = [
1 0 00 1 00 0 1
]
[1 0 0
4 + ๐ง 1 0๐ + ๐๐ง + ๐ฅ ๐ + ๐ฆ 1
] = [1 0 00 1 00 0 1
]
4 + ๐ง = 0 ๐ง = โ4
๐ + ๐๐ง + ๐ฅ = 0 ๐ฅ = โ๐ + 4๐
๐ + ๐ฆ = 0 ๐ฆ = โ๐
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 7
4. Using algebraic method, find the least value of n for which the sum of the first n terms of a
geometric series
0.88 + (0.88)2 + (0.88)3 + (0.88)4 + โฏ
is greater than half of its sum to infinity.
SOLUTION
๐ ๐ >1
2๐ โ
๐ = 0.88, ๐ = 0.88
๐(1 โ ๐๐)
1 โ ๐>
1
2(
๐
1 โ ๐)
(0.88)[1 โ (0.88)๐]
1 โ 0.88>
1
2(
0.88
1 โ 0.88)
[1 โ (0.88)๐] >1
2(
0.88
1 โ 0.88) (
1 โ 0.88
0.88)
1 โ (0.88)๐ >1
2
1 โ1
2> (0.88)๐
1
2> (0.88)๐
(0.88)๐ <1
2
๐๐๐ (0.88)๐ < ๐๐๐ (1
2)
๐ log 0.88 < โ0.3010
๐(โ0.0555) < โ0.3010
โ0.0555๐ < โ0.3010
0.3010 < 0.0555๐
0.0555๐ > 0.3010
๐ฎ๐๐๐๐๐๐๐๐ ๐บ๐๐๐๐๐
๐ ๐ =๐(1 โ ๐๐)
1 โ ๐
๐ โ=(
๐1โ๐
)
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 9
5. (a) State the interval for x such that the expansion for (4 + 3๐ฅ)3
2 is valid.
(b) Expand (4 + 3๐ฅ)3
2 in ascending power of x up to the term in ๐ฅ3.
(c) Hence, by substituting an appropriate value of x, evaluate (5)3
2 correct to three decimal
places.
SOLUTION
a) (4 + 3๐ฅ)3
2 = [4 (1 +3
4๐ฅ)]
3
2
๐โ๐ ๐๐ฅ๐๐๐๐ ๐๐๐ ๐๐ ๐ฃ๐๐๐๐ ๐๐๐
โ1 <3
4๐ฅ < 1
โ4
3< ๐ฅ <
4
3
b) (4 + 3๐ฅ)3
2 = [4 (1 +3
4๐ฅ)]
3
2
= 432 (1 +
3
4๐ฅ)
32
= 8 (1 +3
4๐ฅ)
32
= 8 [1 +(
32)
1!(
3
4๐ฅ) +
(32) (
12)
2!(
3
4๐ฅ)
2
+(
32) (
12) (โ
12)
3!(
3
4๐ฅ)
3
+ โฏ ]
= 8 [1 +9
8๐ฅ +
3
8(
9
16๐ฅ2) โ
3
48(
27
64๐ฅ3) + โฏ ]
= 8 [1 +9
8๐ฅ +
27
128๐ฅ2 โ
81
3072๐ฅ3 + โฏ ]
= 8 [1 +9
8๐ฅ +
27
128๐ฅ2 โ
27
1024๐ฅ3 + โฏ ]
= 8 + 9๐ฅ +27
16๐ฅ2 โ
27
128๐ฅ3 + โฏ
c) (4 + 3๐ฅ)3
2 = 8 + 9๐ฅ +27
16๐ฅ2 โ
27
128๐ฅ3 + โฏ
(4 + 3๐ฅ)32 = (5)
32
4 + 3๐ฅ = 5
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 10
3๐ฅ = 1
๐ฅ =1
3
(4 + 3๐ฅ)32 = 8 + 9๐ฅ +
27
16๐ฅ2 โ
27
128๐ฅ3 + โฏ
[4 + 3 (1
3)]
32
= 8 + 9 (1
3) +
27
16(
1
3)
2
โ27
128(
1
3)
3
+ โฏ
[5]32 = 8 + 9 (
1
3) +
27
16(
1
3)
2
โ27
128(
1
3)
3
+ โฏ
= 11.180
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 11
6. (a) Given ๐(๐ฅ) = 2๐ฅ + 1 and ๐(๐ฅ) = ๐ฅ2 + 2x โ 1.
(i) Find (๐ โ ๐)(x).
(ii) Evaluate (3๐ โ 2๐)(1).
(b) Given ๐(๐ฅ) = โ2๐ฅ +1
2. State the domain and range of ๐(๐ฅ). Hence, on the same axes,
sketch the graph of ๐(๐ฅ) and ๐โ1(x).
SOLUTION
a) ๐(๐ฅ) = 2๐ฅ + 1 and ๐(๐ฅ) = ๐ฅ2 + 2x โ 1
i. (๐ โ ๐)(x) = ๐(๐ฅ) โ ๐(๐ฅ)
= (2๐ฅ + 1 ) โ (๐ฅ2 + 2x โ 1)
= 2๐ฅ + 1 โ ๐ฅ2 โ 2x + 1
= โ๐ฅ2 + 2
ii. (3๐ โ 2๐)(๐ฅ) = 3๐(๐ฅ) โ 2๐(๐ฅ)
= 3(๐ฅ2 + 2x โ 1) โ 2( 2๐ฅ + 1)
= 3๐ฅ2 + 6x โ 3 โ 4๐ฅ โ 2
= 3๐ฅ2 + 2x โ 5
(3๐ โ 2๐)(1) = 3(1)2 + 2(1) โ 5
= 3 + 2 โ 5
= 0
b) ๐(๐ฅ) = โ2๐ฅ +1
2
๐ท๐: 2๐ฅ +1
2โฅ 0
2๐ฅ โฅ โ1
2
๐ฅ โฅ โ1
4
๐ท๐ = {๐ฅ: ๐ฅ โฅ โ1
4}
๐ ๐ = {๐ฆ: ๐ฆ โฅ 0}
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 12
x
y
โ1
4
โ1
4
๐(๐ฅ)
๐โ1(๐ฅ)
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 13
7. Let ๐ง = ๐ + ๐๐ be a nonzero complex number.
a. Show that 1
๐ง=
zฬ
|z|2.
b. Show that if zฬ = โz, then z is a complex number with only an imaginary part.
c. Find the value of a and b if ๐ง(2 โ ๐) = (zฬ + 1)(1 + i).
SOLUTION
a) ๐ง = ๐ + ๐๐
zฬ
|z|2 =๐โ๐๐
|๐+๐๐|2
=๐ โ ๐๐
(โa2 + ๐2)2
=๐ โ ๐๐
a2 + ๐2
=๐ โ ๐๐
a2 + ๐2 (
๐ + ๐๐
๐ + ๐๐)
=๐2 + ๐๐๐ โ ๐๐๐ โ (๐๐)2
(a2 + ๐2)(๐ + ๐๐)
=๐2 โ ๐2๐2
(a2 + ๐2)(๐ + ๐๐)
=๐2 โ ๐2(โ1)
(a2 + ๐2)(๐ + ๐๐)
=๐2 + ๐2
(a2 + ๐2)(๐ + ๐๐)
=1
(๐ + ๐๐)
=1
z
PSPM I QS 015/1 Session
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Chow Choon Wooi Page 14
b) zฬ = โz
๐ โ ๐๐ = โ(๐ + ๐๐)
๐ โ ๐๐ = โ๐ โ ๐๐
๐ = โ๐
2๐ = 0
๐ = 0
๐๐๐๐๐ ๐ง = ๐ + ๐๐
= 0 + ๐๐
= ๐๐
โด z is a complex number with only an imaginary part when zฬ = โz
c) ๐ง(2 โ ๐) = (zฬ + 1)(1 + i)
(๐ + ๐๐)(2 โ ๐) = [(a โ bi) + 1](1 + i)
2๐ + 2๐๐ โ ๐๐ โ ๐๐2 = (a โ bi)(1 + i) + (1 + i)
2๐ + 2๐๐ โ ๐๐ + ๐ = (๐ โ ๐๐ + ๐๐ โ ๐๐2) + (1 + i)
2๐ + ๐ + (2๐ โ ๐)๐ = ๐ โ ๐๐ + ๐๐ + ๐ + 1 + ๐
2๐ + (2๐ โ ๐)๐ = (๐ + 1) + (๐ โ ๐ + 1)๐
By equating real part
2๐ = ๐ + 1
๐ = 1
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 15
By equating imaginary part
(2๐ โ ๐) = (๐ โ ๐ + 1)
3๐ โ 2๐ = 1
3๐ โ 2(1) = 1
3๐ โ 2 = 1
3๐ = 3
๐ = 1
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 16
8. (a) Solve the for the following equation |6๐ฅ2 + ๐ฅ โ 11| = 4.
(b) Find the solution set for the inequality
2 โ (๐ฅ + 2
๐ฅ โ 4) < 5
SOLUTION
a) |6๐ฅ2 + ๐ฅ โ 11| = 4
6๐ฅ2 + ๐ฅ โ 11 = 4 or 6๐ฅ2 + ๐ฅ โ 11 = โ4
6๐ฅ2 + ๐ฅ โ 15 = 0 6๐ฅ2 + ๐ฅ โ 7 = 0
(3๐ฅ + 5)(2๐ฅ โ 3) = 0 (6๐ฅ + 7)(๐ฅ โ 1) = 0
3๐ฅ + 5 = 0 2๐ฅ โ 3 = 0 6๐ฅ + 7 = 0 ๐ฅ โ 1 = 0
๐ฅ = โ5
3, ๐ฅ =
3
2 ๐ฅ = โ
7
6, ๐ฅ = 1
โด= โ5
3, ๐ฅ =
3
2, ๐ฅ = โ
7
6, ๐ฅ = 1
b) 2 โ (๐ฅ+2
๐ฅโ4) < 5
2 โ (๐ฅ + 2
๐ฅ โ 4) โ 5 < 0
2(๐ฅ โ 4) โ (๐ฅ + 2) โ 5(๐ฅ โ 4)
๐ฅ โ 4< 0
2๐ฅ โ 8 โ ๐ฅ โ 2 โ 5๐ฅ + 20
๐ฅ โ 4< 0
โ4๐ฅ + 10
๐ฅ โ 4< 0
โ4๐ฅ + 10 = 0 ๐ฅ โ 4 = 0
๐ฅ =10
4=
5
2 ๐ฅ = 4
|๐| = ๐ โบ ๐ = ๐ ๐๐ ๐ = โ๐
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 17
(โโ,5
2) (
5
2, 4) (4, โ)
โ4๐ฅ + 10 + - -
๐ฅ โ 4 - - +
โ4๐ฅ + 10
๐ฅ โ 4 - + -
๐โ๐ ๐ ๐๐๐ข๐ก๐๐๐ ๐ ๐๐ก: {๐ฅ: ๐ฅ <5
2 โช ๐ฅ > 4}
PSPM I QS 015/1 Session
2014/2015
Chow Choon Wooi Page 18
9. Two companies P and Q decided to award prizes to their employees for three work ethical
values, namely punctuality (x), creativity (y) and efficiency (z). Company P decided to award
a total of RM3850 for the three values to 6, 2 and 3 employees respectively, while company
Q decided to award RM3200 for the three values to 4, 1 and 5 employees respectively. The
total amount for all the three prizes is RM1000.
(a) Construct a system of linear equations to represent the above situation.
(b) By forming a matrix equation, solve this equation system using the elimination
method.
(c) With the same total amount of money spent by company P and Q, is it possible for
company P to award 15 employees for their creativity instead of 2 employees? Give
your reason.
SOLUTION
a) 6๐ฅ + 2๐ฆ + 3๐ง = 3850
4๐ฅ + ๐ฆ + 5๐ง = 3200
๐ฅ + ๐ฆ + ๐ง = 1000
b) ๐ด๐ = ๐ต
(6 2 34 1 51 1 1
) (๐ฅ๐ฆ๐ง
) = (385032001000
)
(
6 2 34 1 51 1 1
|385032001000
)
๐ 3 โ ๐ 3
(1 1 14 1 56 2 3
|100032003850
)
๐ 2 = 4๐ 1 โ ๐ 2
(1 1 10 3 โ16 2 3
|1000800
3850)
๐ 3 = 6๐ 1 โ ๐ 3
(1 1 10 3 โ10 4 3
|1000800
2150)
๐ 3 = 4๐ 2 โ 3๐ 3
(1 1 10 3 โ10 0 โ13
|1000800
โ3250)
PSPM I QS 015/1 Session
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Chow Choon Wooi Page 19
๐ 3 =1
โ13๐ 3 (
1 1 10 3 โ10 0 1
|1000800250
)
๐ 2 = ๐ 2 + ๐ 3
(1 1 10 3 00 0 1
|10001050250
)
๐ 1 = ๐ 1 โ ๐ 3
(1 1 00 3 00 0 1
|750
1050250
)
๐ 2 =1
3๐ 2 (
1 1 00 1 00 0 1
|750350250
)
๐ 1 = ๐ 1 โ ๐ 2
(1 0 00 1 00 0 1
|400350250
)
โด ๐ฅ = 400, ๐ฆ = 350, ๐ง = 250
c) 6๐ฅ + 15๐ฆ + 3๐ง = 3850
4๐ฅ + ๐ฆ + 5๐ง = 3200
๐ฅ + ๐ฆ + ๐ง = 1000
(6 15 34 1 51 1 1
) (๐ฅ๐ฆ๐ง
) = (385032001000
)
๐ฟ๐๐ก ๐ด = (6 15 34 1 51 1 1
)
|๐ด| = |6 15 34 1 51 1 1
|
= (1) |15 31 5
| โ (1) |6 34 5
| + (1) |6 154 1
|
= (75 โ 3) โ (30 โ 12) + (6 โ 60)
= 0
PSPM I QS 015/1 Session
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Chow Choon Wooi Page 20
๐๐๐๐๐ |๐ด| = 0, ๐ด ๐๐ ๐ ๐๐๐๐ข๐๐๐. ๐กโ๐ข๐ ๐กโ๐ ๐ ๐ฆ๐ ๐ก๐๐ ๐๐ ๐๐๐๐๐๐ ๐๐๐ข๐๐ก๐๐๐ โ๐๐ฃ๐ ๐๐ ๐ข๐๐๐๐ข๐ ๐ ๐๐๐ข๐ก๐๐๐.
โด ๐ผ๐ก ๐๐ ๐๐๐ก ๐๐๐ ๐ ๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐ฆ ๐ ๐ก๐ ๐๐ค๐๐๐ 15 ๐๐๐๐๐๐ฆ๐๐๐ ๐๐๐ ๐กโ๐๐๐ ๐๐๐๐๐ก๐๐ฃ๐๐ก๐ฆ
๐๐๐ ๐ก๐๐๐ ๐๐ 2 ๐ค๐๐กโ ๐กโ๐ ๐ ๐๐๐ ๐ก๐๐๐ ๐๐๐๐ข๐๐ก ๐๐ ๐๐๐๐๐ฆ.
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Chow Choon Wooi Page 21
10. (a) Determine wheather ๐(๐ฅ) =1
๐ฅโ4 and ๐(๐ฅ) =
4๐ฅ+1
๐ฅ are inverse function of each other
by computing their composite functions.
(b) Given ๐(๐ฅ) = ln(1 โ 3๐ฅ).
(i) Determine the domain and range of ๐(๐ฅ). Then sketch the graph of ๐(๐ฅ).
(ii) Find ๐โ1(x), if it exists. Hence, state the domain and range of ๐โ1(x).
SOLUTION
a) ๐(๐ฅ) =1
๐ฅโ4 ๐(๐ฅ) =
4๐ฅ+1
๐ฅ
๐[๐(๐ฅ)] = ๐ [4๐ฅ + 1
๐ฅ]
=1
(4๐ฅ + 1
๐ฅ ) โ 4
=1
4๐ฅ + 1 โ 4๐ฅ๐ฅ
= ๐ฅ
๐๐๐๐๐ ๐[๐(๐ฅ)] = ๐ฅ, ๐กโ๐ข๐ ๐ ๐๐๐ ๐ ๐๐๐ ๐๐๐ฃ๐๐๐ ๐ ๐๐ข๐๐๐ก๐๐๐ ๐๐ ๐๐๐โ ๐๐กโ๐๐.
b) ๐(๐ฅ) = ln(1 โ 3๐ฅ)
i. ๐ท๐: 1 โ 3๐ฅ > 0
1 > 3๐ฅ
3๐ฅ < 1
๐ฅ <1
3
๐ท๐ = {๐ฅ: ๐ฅ <1
3}
๐ ๐ = {๐ฆ โ โ}
๐ฅ
๐(๐ฅ)
1
3
๐(๐ฅ) = ln(1 โ 3๐ฅ)