msa 1 revision package 2012 solutions
TRANSCRIPT
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MSA 1 Revision Package 2012 Solutions
Chapter 1: Partial Fractions, Surds, Indices, Trigonometry Solutions
1(a) 2
3 1 3 1
2 1 (2 1)( 1)
x x
x x x x
+ += +
=2 1 1
A B
x x+
+
Using Cover-Up Rule,
13( ) 1
121 3
12
A
+= =
,
3(1) 1 4
2(1) 1 3B
+= =
+
.
2
3 1 1 1 4
2 1 3 2 1 1
x
x x x x
+ = + +
(b)2
2 2
2 5 15
(2 )(1 2 ) 2 1 2
x x A Bx C
x x x x
+ + += +
+ +
Using Cover-Up Rule,2
2
2 5(2) 15(2)
1 2(2)A
+ +=
+= 8.
2 22 5 15 (1 2 ) ( )(2 )x x A x Bx C x+ + = + + +
0 : 2 2 3x A C C= = + =
: 5 2 1x B C B= =
2
2 2
2 5 15 8 3
(2 )(1 2 ) 2 1 2
x x x
x x x x
+ + = + + +
(c)2
2 2
2 1
(1 ) 1
x x A B C
x x x x x
+= + +
Using Cover-Up Rule,2
2
2(1) 1 12
1C
+= =
2 22 1 (1 )x x Ax B x Cx + = + +
1
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0 :1
: 1 0
x B
x A B A
= = =
2
2 2
2 1 1 2
(1 ) 1
x x
x x x x
+ = +
(d)
23 23 45 14 453
( 3) ( 3)
14 45
( 3) 3
x x x
x x x x
x A B
x x x x
+ + += +
+ ++
= ++ +
Using Cover-Up Rule,45 14( 3) 45
15, 13 3
A B +
= = = =
23 23 45 15 13
( 3) 3
x x
x x x x
+ + = +
+ +.
(e)
3
3 2
2 2
2 16
28 ( 2)( 2 4)
( 2)( 2 4) 2 2 4
x x x
x x x x
x A Bx C
x x x x x x
= + +
+= +
+ + + +
Using Cover-Up Rule,
2
2
2
3
3 2
2 1
2 2(2) 4 6
( 2 4) ( )( 2)
10 : 0 4 2 2
3
1: 0
6
2 16 1 1 22
8 6 2 2 4
A
x A x x Bx C x
x A C C A
x A B B A
x x x
x x x x
= =+ +
= + + + +
= = = =
= + = =
= + + +
(f)
2
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3
2 2 2 2 2
3 2
3
2
0
3
2 2 2 2 2
3 1
( 1) 1 ( 1)
3 1 ( )( 1)
: 3
: 0
: 0 3
:1 1
3 1 3 1 3
( 1) 1 ( 1)
x Ax B Cx D
x x x
x Ax B x Cx D
x A
x B
x A C C
x B D D
x x x
x x x
+ + += +
+ + +
+ = + + + +
=
== + =
= + =
+ = +
+ + +
2 is obtuse, so cos and tan are both negative.
(a) 1sincos 22 =+ 1cos 22 =+ x
22 1cos x=
21cos x=
(b)21cos
sintan
x
x
==
(c)
212
cossin22sin
xx ==
(d)
2
22
22
21
1
sincos2cos
x
xx
=
=
=
3(a)
( ) ( )
( )4
2 2 2 2 2 2
2 2 2 2 2 2
2 2 2
2
+ = +
=
=
3
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3(b)( ) ( )
( ) ( )
( ) ( )
2
10 51 10 51
10 51 2 10 51 10 51 10 51
20 2 10 51 10 51
20 2 (100 51)
6
+
= + + +
= +
=
=
3(c) 1125 175 28
20
15 5 5 7 2 7
2 5
15 5 3 7 510
515 3 7
10
+ +
= + +
= + +
= +
4 2 5 2 2 5 2 5 2
5 2 5 2 5 2
10 10 2 10 2
5 2
12 3 103
4 10
+ + +=
+
+ + +=
+=
= +
g
4, 10a b = =
4
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Chapter 2: Binomial Expansion Solutions
1 [NJC/2006/CT/1]
( )
( ) ( )
( ) ( )
2
22
2
2
2 2
2
2
22
5 1f
1 21 2
( 2) 5( 2) 1By cover-up rule, 1.
( 2) 1
5 1 ( )( 2) ( 1)
Compare coeff. of
:1 1 2.
:5 2 5 4 1.
5 1 2 1 1.
1 21 2
x x Ax B Cx
x xx x
C
x x Ax B x x
x A A
x A B B
x x x
x xx x
+ + += = +
+ ++ + + +
= = +
+ + = + + +
= == + = =
+ + + =
+ ++ +
2 1 1 1
2
2 3
2
2 33 2
2 3
2 1 1f ( ) (2 1)(1 ) 2 (1 )
1 2 2
1 ( 1)( 2) ( 1)( 2)( 3)(2 1)(1 ...) (1 ...)
2 2 2! 2 3! 2
1( 2 2 1) 1
2 2 4 8
1 9 9 31 .2 4 8 16
x xx x x
x x
x x xx x
x x xx x x
x x x
+= = + + ++ +
= + + + + +
+ + +
= +
2Validity : 1 and 12
1 and 2
1 1 1
xx
x x
x x