chapter 2 answer maths t focus
DESCRIPTION
focus on examTRANSCRIPT
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
cHAPTER 2 sEquEncEs And sERiEs
Focus on Exam 2
1 u13 = 3, S13 = 234
a + 12d = 3 132
(a + u13) = 234
a = 3 12d ...(1) 132
(a + 3) = 234 ...(2)
From (2), a = 33
When a = 33, 33 = 3 12d d = 52
S25 = 252
3(2)(33) + (24)1 5224 =
252
(66 60) = 75
2 (a) a = 1, r = 54
Sn = a(rn 1)
r 1 =
1542n
1
154 12 = 431542
n
14 (b) Sn > 20
431542n
14 > 201542
n
1 > 5
1542n
> 6
n lg 54
> lg 6
n > 8.03 The least number of terms is 9.
3 12 22 + 32 42 + + (2n 1)2 (2n)2
= 3 7 11 a = 3, d = 4
Sn = n2
36 + (n 1)( 4)4
= n2
(6 4n + 4) = n2
( 4n 2)
= n(2n + 1) [Shown]
(a) 12 22 + 32 42 + + (2n 1)2 (2n)2
+ (2n + 1)2
= n(2n + 1) + (2n + 1)2
= (2n + 1)(n + 2n + 1) = (2n + 1)(n + 1) (b) 212 222 + 232 242 + + 392 402
= (12 22 + 32 42 + + 392 402) (12 22 + 32 42 + + 192 202)
= S20 S10= 20(2(20) + 1) [10(2(10) + 1)]= 820 (210) = 610
4 d = 2 un = a + (n 1)(2) = a + 2n 2 When n = 20, u20 = a + 2(20) 2 = a + 38
S20 = 1120202
(a + u20) = 1120
10(a + a + 38) = 11202a = 74
a = 37u20 = a + 38 = 37 + 38 = 75
5 (a) a = 3, d = 4, Sn = 820
Sn = n2
[2a + (n 1)d ]
820 = n2
(6 + 4n 4)
1640 = n(2 + 4n)n(1 + 2n) = 820
n + 2n2 820 = 0(2n + 41)(n 20) = 0
n = 20 since n = 412
is not a solution.
T20 = 3 + 19(4) = 79
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 3ACE AHEAD Mathematics (T) First Term Updated Edition2
(b) Since they form an A.P., (p2 + q2)2 (p2 2pq q2)2
= (p2 + 2pq q2)2 (p2 + q2)2
(p2 + q2 + p2 2pq q2)(p2 + q2 p2
+ 2pq + q2) = (p2 + 2pq + q2 + p2 q2)(p2 + 2pq
q2 p2 q2) (2p2 2pq)(2q2 + 2pq) = (2p2 + 2pq)
(2pq 2q2) 4pq(p q)(q + p) = 4pq(p + q)(p q)
LHS = RHS [Shown]
6 (a) Sn = pn + qn2, S8 = 20, S13 = 39
(i) S8 = 8p + 64q8p = 20 64q
p = 52
8q jS13 = 39
13p + 169q = 39 kSubstitute j into k,
13152 8q2 + 169q = 39652
104q + 169q = 39
65q = 132
q = 110
p = 52
81 1102 = 1710
(ii) un = Sn Sn 1= pn + qn2 [p(n 1) + q(n 1)2]= p(n n + 1) + q[n2 (n 1)2]= p + q[(n + n 1)(n n + 1)]= p + q(2n 1)(1)
= 1710
+ 110
(2n 1)
= 1710
+ n5
110
= 8 + n
5
(iii) To show that it is an A.P.,un un 1 = un + 1 un = d
8 + n5
8 + (n 1)5
= 8 + (n + 1)
5
8 + n5
2(8 + n)5
= 8 + (n + 1) + 8 + (n 1)
52(8 + n) = 16 + 2n2(8 + n) = 2(8 + n)
LHS = RHS [Shown]
d = (8 + n)
5 8 + (n 1)
5
= 8 + n 8 n + 1
5 =
15
= 0.2
7 If J, K, M is a G.P., thenKJ
= MK
K 2 = MJ j(ar k 1)2 = (ar m 1) (ar j 1)
r 2k 2 = rm 1 + j 1
2k 2 = m + j 2k + k = m + jk m = j k k
and2k = m + j l
(k m)lg J + (m j)lg K + ( j k)lg M
= lg J k m + lg Km j + lg M j k
= lg J j k + lg Km j + lg M j k [from k]= ( j k)(lg J + lg M) + lg K m j
= ( j k)lg JM + lg K m j
= ( j k)lg K2 + (m j)lg K
= 2( j k)lg K + (m j)lg K
= (2j 2k + m j)lg K
= ( j + m 2k)lg K
= (2k 2k)lg K [from l]= 0 [Proven]
8 (a) A.P., a = 200, d = 400 Tn = a + (n 1)d = 2000 + (n 1)(400) = 400 n + 1600 = 400 (n + 4)
(b) Sn = n2
(a + l) = n2
[2000 + 400n + 1600]
= n2
(400n + 3600)
= 200n(n + 9)
(c) Sn > 200 000 200n (n + 9) > 200 000
n2 + 9n 1000 > 0 n < 36.44135, n > 27.44135 [ n = 28
9 S5 = 44, a(1 r5)
1 r = 44 j
S10 S5 = 118
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 3ACE AHEAD Mathematics (T) First Term Updated Edition2
a(1 r10)1 r
a(1 r5)
1 r =
118
a(1 + r5)(1 r5)1 r
a(1 r5)
1 r =
118
k
a(1 r5)
1 r (1 + r5 1) =
118
Substitute j into k,44(r5) =
118
r5 = 132
r = 12
[Shown]
a31 1 12254
1 1 122 = 44
a11 + 1322 = 66a = 64
S = a
1 r
= 64
1 1 122= 42
23
10 (a) r = 3 13 + 1
3 13 1
= 3 2 3 + 1
3 1 = 2 3
u3 = u2r
= ( 3 1)(2 3)= 2 3 3 2 + 3= 3 3 5
u4 = u3r
= (3 3 5)(2 3)= 6 3 3(3) 10 + 5 3= 11 3 19
(b) r = 2 3 [< 1]
S = a
1 r =
3 + 11 (2 3 )
= 3 + 13 1
3 + 13 + 1
= 3 + 2 3 + 1
2= 2 + 3
11 (a) S6S3
= 78
, u2 = 4
8S6 = 7S3 j
ar = 4
a = 4r k
83a(1 r6)
1 r 4 = 73a(1 r3)
1 r 48(1 r6) = 7(1 r3)
8 8r6 = 7 7r3
8r6 7r3 1 = 0(r3 1)(8r3 + 1) = 0
r3 = 1 or 18
r = 12
since r = 1 is
not a solution.a =
4
1 122 = 8
(b) r = 85
12
= 45
, a = 2
Sn > 9.9a(1 rn)
1 r > 9.9
231 1452
n41152
> 9.9
1 1452n
> 0.99
1452n
< 0.01
n lg 45
< lg (0.01)
n > 20.64 The least number of terms is 21.
12 u3 = S2ar2 =
a(1 r2)1 r
ar2(1 r) = a(1 + r)(1 r)ar2 = a(1 + r)
r2 r 1 = 0
r = (1) (1)2 4(1)(1)
2
= 1 5
2a = 2
When r = 1 + 5
2 [|r| > 1], S doesnt exist
When r = 1 + 5
2 [|r| < 1],
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 5ACE AHEAD Mathematics (T) First Term Updated Edition4
S = a
1 r
= 2
1 1 5
2
= 2
12 1 + 52 2 =
41 + 5
1 51 5
= 4 4 5
1 5 = 5 1
13 a = 3, r = 0.4 (a) un < 0.02
ar n 1 < 0.023(0.4)n 1 < 0.02
(0.4)n 1 < 1
150
(n 1)lg 0.4 < lg 1
150
n 1 > 5.47n > 6.47
The least number of terms is 7. (b) S Sn < 0.01
a1 r
a(1 rn)
1 r < 0.01
a
1 r (1 1 + r n) < 0.01
3
0.6 (r n) < 0.01
0.4n < 2 103
n lg 0.4 < lg (2 103)n > 6.78
The least number of terms is 7.
14 Sn = a + ar + ar 2 + + ar n 2 + ar n 1 j
rSn = ar + ar 2 + + ar n 2 +
ar n 1 + ar n k j k,
Sn rSn = a ar n
Sn(1 r) = a(1 r n)
Sn = a(1 r n)
1 r [Shown]
For S to exist, |r| < 1,
lim Sn = S = a(1 r )
1 r =
a1 rn
S SnS
=
a1 r
a(1 r n)
1 ra
1 r
=
a1 r
[1 (1 r n)]
a1 r
= rn [Shown]
(a) u4 = 18, u7 = 163
ar3 = 18 j ar6 = 163
k
kj
, r3 = 1163 2 118
r3 = 1 8272r = 1232
a12323
= 18
a = 2434
S = 12434 21 2
3
= 7294
(b) S Sn
S < 0.001
rn < 0.001
1232n
< 0.001
n lg 23
< lg 0.001
n > 17.04 The least value of n is 18.
15 2r 1
r(r 1)
2r + 1r(r + 1)
= (r + 1)(2r 1) (2r + 1)(r 1)
r(r 1)(r + 1)
= 2r2 r + 2r 1 2r2 + 2r r + 1
r(r 1)(r + 1)
= 2r
r(r 1)(r + 1) =
2(r 1)(r + 1)
[Verified]
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 5ACE AHEAD Mathematics (T) First Term Updated Edition4
or = 2
n 2(r 1)(r + 1)
= or = 2
n 3 2r 1r(r 1)
2r + 1r(r + 1)4
= or = 2
2n [ f (r) f (r + 1)] 3f (r) = 2r 1r(r 1)4
= f (2) f (n + 1)
= 32
2n + 1n(n + 1)
[Proven]
12
or = 2
2(r 1)(r + 1)
= 12
lim 332 2n + 1n(n + 1)4n
= 12
lim 332 2n
+ 1n2
1 + 1n4n
= 12
132 02 = 34 16
1r(r + 1)
1
(r + 1)(r + 2) =
r + 2 rr(r + 1)(r + 2)
= 2
r(r + 1)(r + 2)[Shown]
or = 1
n 1r(r + 1)(r + 2)
= 12
or = 1
n 2r(r + 1)(r + 2)
= 12
or = 1
n 3 1r(r + 1)
1(r + 1)(r + 2)4
3f (r) = 1r(r + 1)4 =
12
or = 1
n [ f (r) f (r + 1)]
= 12
[ f (1) f (n + 1)]
= 12
312 1(n + 1)(n + 2)4 =
12
3 n2 + 3n + 2 2
2(n + 1)(n + 2)4 =
n2 + 3n4(n + 1)(n + 2)
17 (a) 1
(2r 1)(2r + 1)
A2r 1
+ B
2r + 11 A(2r + 1) + B(2r 1)
Let r = 12
, Let r = 12
,
l = B(2) 1 = A(2)
B = 12
A = 12
1
(2r 1)(2r + 1) =
12(2r 1)
1
2(2r + 1)
= 121
12r 1
1
2r + 12 (b) o
r = n
2n
1(2r 1)(2r + 1)
= or = n
2n 3 12(2r 1)
12(2r + 1)4
= 12
or = n
2n [ f (r) f (r + 1)] 3f (r) = 12r 14,
= 12
[ f (n) f (2n + 1)]
= 12
3 12n 1 1
4n + 14 =
12
1 4n + 1 2n + 1(2n 1)(4n + 1)2 =
n + 1(2n 1)(4n + 1)
= an + b
(2n 1)(4n + 1) [Shown] a = 1, b = 1
(c) lim 3 n + 1(2n 1)(4n + 1)4n
= lim 31n
+ 1n2
8 2n
1n24 = 0n
18 Let 1(2r + 1)(2r + 3)
A2r + 1
+ B
2r + 3 l A(2r + 3) + B(2r + 1)
Let r = 32
Let r = 12
1 = B(2) 1 = A(2)
B = 12
A = 12
1(2r + 1)(2r + 3)
= 1
2(2r + 1) 1
2(2r + 3)
or = 1
n 1(2r + 1)(2r + 3)
= 12
or = 1
n 3 12r + 1
12r + 34
= 12
or = 1
n [ f (r) f (r + 1)] 3f (r) = 12r + 14
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 7ACE AHEAD Mathematics (T) First Term Updated Edition6
= 12
[ f (1) f (n + 1)]
= 12
313 12n + 34 =
16
12(2n + 3)
To test for the convergence,
lim 316 1
2(2n + 3)4 = 16n
The series converges to 16
.
S` =
16
19 on = 25
N 1 12n 1
12n + 12
= on = 25
N[ f (n) f (n + 1)] 3f (n) = 12n 1 4
= f (25) f (N + 1)
= 1
50 1
12N + 1
= 17
1
2N + 1
on = 25
un = lim 117
12n + 1 2 =
17n
20 1r!
1(r + 1)!
= (r + 1)! r!r!(r + 1)!
= (r + 1)r! r!
r!(r + 1)!
= r
(r + 1)! [Shown]
or = 1
n
r(r + 1)!
= or = 1
n
31r! 1
(r + 1)!4= o
r = 1
n
[ f (r) f (r + 1)],
3f (r) = 1r!4= f (1) f (n + 1)
= 1 1
(n + 1)!
21 r + 1r + 2
rr + 1
= (r + 1)2 r(r + 2)
(r + 1)(r + 2)
= r2 + 2r + 1 r2 2r
(r + 1)(r + 2)
= 1
(r + 1)(r + 2) [Shown]
or = 1
n
1(r + 1)(r + 2)
= or = 1
n
3r + 1r + 2 r
r + 14= o
r = 1
n
[ f (r) f (r 1)],
3f (r) = r + 1r + 24= f (n) f (0)
= n + 1n + 2
12
= 2(n + 1) (n + 2)
2(n + 2)
= 2n + 2 n 2
2(n + 2)
= n
2(n + 2)
22 f (r) f (r 1)
= 1
(2r + 1)(2r + 3)
1(2r 1)(2r + 1)
= 2r 1 2r 3
(2r 1)(2r + 1)(2r + 3)
= 4
(2r 1)(2r + 1)(2r + 3) [Shown]
or = 1
n
1(2r 1)(2r + 1)(2r + 3)
= 14
or = 1
n
4(2r 1)(2r + 1)(2r + 3)
= 14
or = 1
n
[ f (r) f (r 1)]
3f (r) = 1(2r + 1)(2r + 3)4 =
14
[ f (n) f (0)]
= 143
1(2n + 1)(2n + 3)
134
= 112
14(2n + 1)(2n + 3)
= 4n2 + 6n + 2n + 3 3
12(2n + 1)(2n + 3)
= 4n2 + 8n
12(2n + 1)(2n + 3)
= n2 + 2n
3(2n + 1)(2n + 3)
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 7ACE AHEAD Mathematics (T) First Term Updated Edition6
23 Let 1(4n 1)(4n + 3)
A4n 1
+ B
4n + 3 l A(4n + 3) + B(4n 1)
Let n = 34
Let n = 14
1 = B( 4) 1 = A(4)
B = 14
A = 14
1(4n 1)(4n + 3)
= 141
14n 1
14n + 32
Let f (r) = 1
4r + 3
14
or = 1
n
1 14r 1 1
4r + 32 =
14
or = 1
n [ f (r 1) f (r)]
= 14
[ f (0) f (n)]
= 14
313 1
4n + 34 =
14
34n + 3 33(4n + 3) 4 = n
3(4n + 3)
24 f (x) = x + x + 1
1f (x)
= 1
x + x + 1
x x + 1x x + 1
= x x + 1
x (x + 1)= x + 1 x
ox = 1
24 1f (x)
= ox = 1
24 1 x x + 12
= ( 1 2 + 2 3 + + 24 25)= (1 5) = 4
25 Using long division,1
x2 + 6x + 8 2x2 + 6x + 0x2 + 6x + 8
8
x(x + 6)(x + 2)(x + 4)
1 8(x + 2)(x + 4)
Let 8(x + 2)(x + 4)
A(x + 2)
+ B
(x + 4),
8 A(x + 4) + B(x + 2)
Let x = 4, Let x = 2, 8 = B(2) 8 = A(2)
B = 4 A = 4
x(x + 6)(x + 2)(x + 4)
= 1 4
x + 2 +
4x + 4
ox = 1
n 31 4x + 2 +
4x + 44
= ox = 1
n 1 4 o
x = 1
n 3 1x + 2
1x + 44
= n 4313 15
+ 14
16
+ 15
17
+
+ 1
n + 1 1
n + 3 +
1n + 2
1n + 44
= n 4313 + 14
1n + 3
1n + 44
= n 43 712 2n + 7
(n + 3)(n + 4)4 = n 437(n
2 + 7n + 12) 12(2n + 7)12(n + 3)(n + 4) 4
= n 7n2 + 49n + 84 24n 84
3(n + 3)(n + 4)
= n 7n2 + 25n
3(n + 3)(n + 4)
= 3n(n + 3)(n + 4) n(7n + 25)
3(n + 3)(n + 4)
= n[3(n2 + 7n + 12) 7n 25]
3(n + 3)(n + 4)
= n[3n2 + 21n + 36 7n 25]
3(n + 3)(n + 4)
= n(3n2 + 14n + 11)3(n + 3)(n + 4)
= n(n + 1)(3n + 11)3(n + 3)(n + 4)
[Shown]
26 (2 + 3x)4 = 24 + 4(23)(3x) + 6(22)(3x)2 + 4(2) (3x)3 +(3x)4
= 16 + 96x +216x2 + 216x3 + 81x4
[2 + (3x)]4 = 24 + 4(23)(3x) + 6(22) (3x)2 + 4(2)(3x)3 + (3x)4
= 16 96x + 216x2 216x3 + 81x4 (2 + 3x)4 (2 + 3x)4 = 192x + 432x3
Let x = 2 , (2 + 3 2 )4 (2 +3 2 )4 = 192 2 + 432 ( 2 )3
= 1056 2
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 9ACE AHEAD Mathematics (T) First Term Updated Edition8
27 (a) 1x + 1x25
= x5 + 5x4 11x2+ 10x3 11x2
2
+
10x2 11x23
+ 5x 11x24
+ 11x25
= x5 + 5x3 + 10x + 1011x2 + 51 1x32+ 1
1x52
1x 1x23
= x3 + 3x2 1 1x2+ 3x 11x2
2
+
1 1x23
= x3 3x + 3 11x2 1
x3
1x + 1x25
1x 1x23
= 1x5 + 5x3 + 10x + 10 11x2+ 51
1x32+
1x52 31x3 3x + 311x2
1x32
Terms containing x4
= x513x2 + 5x3(3x) + 10x(x3)= 3x4 15x4 + 10x4
= 2x4
Coefficient of x4 = 2
(b) Tr + 1 = 1nr2an r br = 16r2 (x2) 6 r (2x1)r = 16r2 (2)r x12 2r xr = 16r2 (2)r x12 3r The term independent of x is the term
where 12 3r = 0 r = 4
28 (a) 11 32x25
(2 + 3x)6
= 31 + 5C11 32x2 + 5C21 32
x22
+ 4 [26 + 6C1(2)
5(3x) + 6C2(2)4(3x)2 + ]
= 11 152 x + 452
x22(64 + 576x + 2160x2) = 64 + 576x +2160x2 480x 4320x2
+ 1440x2
= 720x2 + 96x + 64 [Shown]
(b) 11 32x x225
= 3 1x 122 (x + 2)45
= 1x 1225
(x + 2)5
= 11225
(1 2x)5(2)5 11 + x225
= [1 + 5(2x) + 10(2x)2 + 10(2x)3
+ 5(2x)4 ] 31 + 51x22 + 101x22
2
+ 101x223
+ 51x22
4
4 = (1 10x + 40x2 80x3 + 80x4)
11 + 52x + 52
x2 + 54
x3 + 516
x42 = 1 +
52
x + 52
x2 + 54
x3 + 516
x4 10x
25x2 25x3 252
x4 + 40x2 + 100x3
+ 100x4 80x3 200x4 + 80x4
= 1 152
x + 352
x2 154
x3 51516
x4
29 (a) 1 + 10(3x + 2x2) + 10(9)
2 (3x + 2x2)2
+ 10(9)(8)
3! (3x + 2x2)3
= 1 + 30x + 20x2 + 45(9x2 + 12x3) + 120(27x3) = 1 + 30x + 425x2 + 3780x3
Coefficient of x3 = 3780
(b) (1 + x)1(4 + x2)
12
= 31 + 11! (x) + 1(2)
2! (x)2 +
1(2)(3)3!
(x)3 + 414 122 31 + x
2
4 4
12
= 12
(1 x + x2 x3 + )11 + 1
21! 1
x2
4 2
+ 1
21 322
2! 1x2
4 22
+ 2 =
12
(1 x + x2 x3 + ) 11 x2
8 + 2
= 12
11 x2
8 x +
x3
8 + x2 x32
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 9ACE AHEAD Mathematics (T) First Term Updated Edition8
= 12
11 x + 78x2 78
x3 + 2where |x| < 1
30 Let f (x) A1 x
+ Bx + C1 + x2
1 + 2x + 3x2 A(1 + x2) + (Bx + C)(1 x) Let x = 1, 1 + 2 + 3 = A(2)
A = 3 Let x = 0,
1 = 3(1) + (0 + C)(1)C = 2
Let x = 1, 1 2 + 3 = 3(2) + [B + (2)](2)
2 = 6 4 2BB = 0
Hence, f (x) = 3
1 x 2
1 + x2
f (x) = 3(1 x)1 2(1 + x2)1
= 331 + 11! (x) + 1(2)2! (x)2
+ 1(2)(3)
3! (x)3 + 4
231 + 11! (x)2 + 4 = 3(1 + x + x2 + x3 + ) 2(1 x2 + )= 3 + 3x + 3x2 + 3x3 2 + 2x2
= 1 + 3x + 5x2 + 3x3 where |x| < 1[Shown]
31 (a) ( 5 + 2)6 ( 5 2)6
8 5
=
[( 5 + 2)3 + ( 5 2)3][( 5 + 2)3
( 5 2)3]8 5
=
( 5 + 2 + 5 2)[( 5 + 2)2 ( 5 + 2)( 5 2) + ( 5 2)2]( 5 + 2 5 + 2)[( 5 + 2)2
+ ( 5 + 2)( 5 2) + ( 5 2)2]8 5
=
[2 5(5 + 4 5 + 4 1 + 5 4 5 + 4)] [(4)(5 + 4 5 + 4 + 1 + 5 4 5 + 4)]
8 5 = (17)(19)= 323
(b) (1 3x)13
= 31 + 13
1! (3x) +
131
232
2! (3x)2
+
131
2321
532
3!(3x)3 + 4
= 1 x x2 53
x3 where |x| < 13
When x = 18
,
(1 3x)13 = 158 2
13 =
52
= 1 18
11822
531
182
3
3 5 = 13151536
(2)
= 1315768
= 1.71 [2 decimal places]
32 Term independent of x is
1642 (2)4 = 240 (1 y)2n = 1 + 2n(y) +
(2n)(2n 1)2!
(y)2 + ...
= 1 2ny + n(2n 1)y2 + ...
(1 + y)2n = 1 + (2n)(y) + (2n)(2n 1)
2! (y)2
+ ... = 1 2ny + n(2n + 1)y2 + ...
11 y1 + y22n
= (1 y)2n(1 + y)2n
= (1 2ny + n(2n 1)y2 + ...) (1 2ny + n(2n + 1)y2 + ...) = 1 2ny + n(2n + 1)y2 2ny + 4n2y2 + n(2n 1)y2
= 1 4ny + 8n2y2
[ 11 y1 + y22n
= 1 4ny + 8n2y2 + ...
Let y = 150
, n = 116
1 1 150
1 + 150
218
1 41 116211502+ 81
1102
2
1 15022
14951218
79 60180 000
-
Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 11ACE AHEAD Mathematics (T) First Term Updated Edition10
33 (1 x)10 = 1 10x + 45x2 + (1 + 2x2)3 = 1 + 6x2 + (1 + ax)5 = 1 + 5ax + 10a2x2 + (1 + bx2)4 = 1 + 4bx2 + (1 x)10 (1 + 2x2)3 (1 + ax)5 (1 + bx2)4
= (1 10x + 45x2 + )(1 + 6x2 + ) (1 + 5ax + 10a2x2 + )(1 + 4bx2 + )
= 1 + 6x2 10x + 45x2 1 4bx2 5ax 10a2x2 +
10 5a = 0 6 + 45 4b 10a2 = 0a = 2 6 + 45 4b 40 = 0
b = 114
34 (a) 1x + 1x25
1x 1x23
= 3x5 + 5(x)411x2 + 10x311x22 + 10x21
1x32
+ 5x1 1x42 + 1x54 3x3 + 3(x)21
1x 2
+ 3x1 1x22
+ 1 1x23
4 = 1x5 + 5x3 + 10x + 10x +
5x3
+ 1x52
1x3 3x + 3x 1x32
To obtain x4 term,
= + x513x2 + 5x3(3x) + 10x(x3) + = 3x4 15x4 + 10x4 = 2x4
The coefficient of x4 term is 2.
(b) (1 + x)15 5 + 3x
5 + 2x
= 31 + 151!
(x) +
151
452
2! (x)2
+
151
4521
952
3!(x)3 + 4
(5 + 3x)(5 + 2x)1
= 11 + 15x 225
x2 + 6
125x32
(5 + 3x)(5)111 + 25x21
= 11 + 15x 225
x2 + 6
125x32
(5 + 3x)115231 + 11! 1
25
x2+
1(2)2!
125x22
+ 1(2)(3)
3! 125
x23
4
= 1 + 15
x 225
x2 + 6
125x3 1
5 (5 + 3x)
11 25x + 425
x2 8125
x32= 1 +
15
x 225
x2 + 6
125x3 1
5 15 2x
+ 45
x2 825
x3 + 3x 65
x2 + 1225
x32= 1 +
15
x 225
x2 + 6
125 x3 1 1
5 x +
225
x2
4125
x3
= 2
125x3 where |x| < 2
5 [Shown]
Hence, p = 2
125By using x = 0.02,
(1.02)15
125350 2125250 2
= 2
125 (0.02)3
= 1.28 107 [Shown]
35 (1 + ax + bx2)7
= [(1 + ax) + (bx2)]7
= 7Cr (1 + ax)7 r (bx2)r
x term: = 7C0(1 + ax)
7
= 7C0[7C1 (ax)]
= 7ax x2 term: = 7C0(1 + ax)
7 + 7C1(1 + ax)6 (bx2)
= 7C0[7C2(ax)
2] + 7C1[6C0(ax)
0 (bx2)] = 21a2x2 + 7bx2
= (21a2 + 7b)x2
7a = 1 21a2 + 7b = 0
a = 17
2111722
+ 7b = 0
7b = 37
b = 349
Let (1 + ax + bx2)7 = 1 + x
(1 + x)17 = 1 +
17
x 349
x2 + where |x| < 1
By using x = 0.014, 7 1.014 = 1 +
17
(0.014) 349
(0.014)2
= 1.001988 [6 decimal places]
-
Oxford Fajar Sdn. Bhd. (008974-T) 2013
Fully Worked Solution 11ACE AHEAD Mathematics (T) First Term Updated Edition10
36 (1 + x)7
1 2x = (1 + x)7 (1 2x)1
= [1 + 7x + 21x2 + 35x3 + ] 31 + 11! (2x) +
1(2)2!
(2x)2 + 1(2)(3)
3! (2x)3 + 4
= (1 + 7x + 21x2 + 35x3 + ) (1 + 2x + 4x2
+ 8x3 + ) = 1 + 2x + 4x2 + 8x3 + 7x + 14x2 + 28x3
+ 21x2 + 42x3 + 35x3 + = 1 + 9x + 39x2 + 113x3 + where |x| <
12
[Shown] Using x = 0.01,
(0.99)7
(1.02) = 1 + 9(0.01) + 39(0.01)2
+ 113(0.01)3 +
= 0.914 [3 decimal places]
37 1 + x = (1 + x)
12
= 1 +
121!
(x) +
121
122
2!(x)2
+
121
1221
322
3! (x)3
+
121
1221
3221
522
4! (x)4 +
= 1 + x2
x2
8 +
x3
16 5
128 x4 + j
14
(6 + x) (2 + x)1
= 6 + x
4 2131 + x24
1
= 6 + x
4 1
231 + 11!
1x2 2 + 1(2)
2! 1x22
2
+ 1(2)(3)
3! 1x22
3
+ 1(2)(3)( 4)
4! 1x22
4
4 =
6 + x4
1211
x2
+ x2
4 x
3
8 +
x4
16 2
= 6 + x
4 1
2 +
x4
x2
8 +
x3
16 x
4
32 +
= 1 + x2
x2
8 +
x3
16 x
4
32 k
To obtain the error,
j k, 11 + x2 x2
8 +
x3
16 5
128x42
11 + x2 x2
8 +
x3
16 x
4
322 =
5128
x4 + x4
32
= x4
128 The error is approximately x
4
128. [Shown]
38 (a + b)12 = 3a11 + ba24
12 = a
12 11 + ba2
12
a12 11 + ba2
12 = a
12 31 + 12 1ba2+
121
12
122!
1ba22
+
121
12
12112 223!
1ba23
+ ...4 = a
12 11 + b2a
b2
8a2 +
b3
16a3 + ...2j
(a b)12 = a
12 11 ba2
12
= a211 b2a b2
8a2 b
3
16a3 + ...2k
j k
(a + b)12 (a b)
12 = a
12 12b2a +
2b3
16a32 = a
12 1ba +
b3
8a32 Let a = 4, b = 1
5 3 = 412 114 +
18(43)2
= 129256
( 5 3) ( 5 + 3) = 5 3 = 2 129
256 ( 5 + 3) = 2
( 5 + 3) = 2 3 256129
= 512129
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Oxford Fajar Sdn. Bhd. (008974-T) 2013
ACE AHEAD Mathematics (T) First Term Updated Edition12
39 y = 1
1 + 3x + 1 + x
1 + 3x 1 + x
1 + 3x 1 + x
= 1 + 3x 1 + x
1 + 3x (1 + x)
= 1
2x 1 1 + 3x 1 + x 2 [Shown]
12x3(1 + 3x)
12 (1 + x)
124
= 1
2x531 + 121!
(3x) +
121
122
2!(3x)2
+
121
1221
322
3!(3x)34 31 +
121!
(x)
+
12
1 1222!
(x)2 +
121
1221
322
3!(x)346
= 1
2x311 + 32
x 98
x2 + 2716
x32 11 + x2 x
2
8 +
x3
1624 =
12x1
x x2 + 138
x32
= 12
x2
+ 1316
x2
Using x = 1
100,
y = 1
1 + 3x + 1 + x
= 1
103100 +
101100
= 10
103 + 101 =
12
1 11002
2 +
13161
11002
2
= 79 213
160 000 [Shown]
40 (a) 3 5x + 3x2
A(1 + x2) + (B + Cx)(1 2x)
Let x = 12
,
54
= 54
(A)
A = 1 Let x = 0,
3 = 1(1 + 0) + (B + 0)(1)B = 2
Let x = 1,1 = 1(2) + (2 + C )(1)
C + 2 = 1C = 1
(b) (1 2x)1 = 31 + 11! (2x) + 1(2)
2! (2x)2
+ 1(2)(3)
3! (2x)34
= 1 + 2x + 4x2 + 8x3
(1 + x2)1 = 31 + 11! (x2) + 4= 1 x2
(c) (3 5x + 3x2)(1 2x)1(1 + x2)1
= 1
1 2x +
2 x1 + x2
= 1(1 2x)1 + (2 x)(1 + x2)1
= 1(1 + 2x + 4x2 + 8x3) + (2 x)(1 x2) = 1 + 2x + 4x2 + 8x3 + 2 2x2 x + x3
= 3 + x + 2x2 + 9x3 a = 1, b = 2, c = 9