KEAM 2014 Mathematics Question Paper with Solutions
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Transcript of KEAM 2014 Mathematics Question Paper with Solutions
SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION - 2014 – PAPER II
VERSION – B1
[MATHEMATICS] 1. Ans: 185 Sol: 2 ⊕ 3 = 4 + 9 = 13 (2 ⊕ 3) ⊕ 4 = 13 ⊕ 4 = 132 + 42 = 169 + 16 = 185. 2. Ans: 520 Sol:
30 = ( )30x10010 +
⇒ x = 270 and
30 = ( )y3010012 +
⇒ y = 220 x + y + 30 = 270 + 220 + 30 = 520. 3. Ans: f is not a 1 − 1function Sol: f(x) = |x − 2| f(1) = |1 − 2| = 1 f(3) = |3 − 2| = 1 Not one-one. 4. Ans: 1680
Sol: 8P4 = !4!8
5. Ans: No answer Sol: No correct option is given. Options (A) and
(D) are same. If one of them were [1, ∞), that would have been the correct option.
6. Ans:
Sol: (fοg) (x) = 3x2 −
If the question were domain of (fοg) (x) is, then (E) would be the correct option.
.
7. Ans: 2
Sol: ( ) ( )
( )223
i68
4i3i3
+++
( )( )
36i966416i249i33i933
−+++−−−+=
= ( )( )
22 9628
i9628192i56i9628
192i56
+−−=
+−
= 9216784
i1843253765376i1568+
+−+
= i210000
i20000 =
∴ |z| = 222 = . 8. Ans: Re(z) = 0 Sol: Let ω = u + iv
∴ z = 1ivu1ivu
11
++−+=
+ω−ω
= ( )( )
( )[ ] ( )[ ]( )[ ] ( )[ ]iv1uiv1u
iv1uiv1uiv1uiv1u
−+++−++−=
+++−
= ( ) ( )
22
222
v1u2u
v1uiv1uiv1u
++++++−−−
= 1u2vu
iviv1vu22
22
+++++−+
u22iv20
++=
=+ 1vu 22Q
= u1
iv+
.
9. Ans: 3i
e3π
−
Sol: 3i2
eπ
= cos ω=π+π3
2sini
32
∴ 1 + z + 3z2 + 2z3 + 2z4 + 3z5 = 1 + ω + 3ω2 + 2ω3 + 2ω4 + 3ω5 = 1 + ω + 3ω2 + 2 + 2ω + 3ω2 = 3 + 3ω + 6ω2 = 3(−1 + ω + ω2) + 3ω2
= 3ω2 = 3 3i4
eπ
C
y 30 x
M
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=
π+π34
sini34
cos3
=
π+π+
π+π3
sini3
cos3
= 3i
e33
sini3
cos3π
−=
π−π− .
10. Ans: −128i
Sol: z1 = 22 (1 + i), z2 = 1 + 3 i ∴z1
2 = 4 × 2 (1 + i)2 = 8(1 + 2i − 1) = 16i
z23 = ( )3i31+
= ( )3i33333i31 +×−+
= 1 + i 33i933 −− = −8 ∴ z1
2 z23 = 16i × −8 = −128 i.
11. Ans: 0 Sol: (z3 − z1) = (z2 − z1) [cos90 + isin 90] (z3 − z1) = i(z2 − z1) (z3 − z1)
2 = −(z2 − z1)2
⇒ (z3 − z1)2 + (z2 − z1)
2 = 0. 12. Ans: a2 − 4b = 4 Sol: Let α, α + 2 be two consecutive odd root ∴ α + α + 2 = a and α(α + 2) = b ∴ 2α + 2 = a
∴ α = 2
2a − and α2 + 2zα = b
since α2 + 2α = b
⇒ ( )
b2
2a2
42a 2
=
−+−
⇒ b2a4
4a4a2=−++−
⇒ a2 − 4a + 4 + 4a − 8 = 4b
⇒ a2 − 4b = 4. 13. Ans: a2 − 2b2 Sol: α + β = a, αβ = b2 α2 + β2 = (α + β)2 − 2αβ = a2 − 2b2.
14. Ans: 43
Sol: α + β = −3 αβ = −4
43
4311 =
−−=
αββ+α=
β+
α.
15. Ans: 2
Sol: ( ) 081323 2xx2 =+− +
⇒ 0813.3.23 2xx2 =+−
⇒ 0813183 xx2 =+−
⇒ ( ) 0813183 x2x =+−
⇒ (3x − 9)2 = 0 ⇒ 3x − 9 = 0 ⇒ 3x = 32
⇒ x = 2.
16. Ans: ( )
4
2β−α
Sol: α + β = −2b ⇒ b = 2−β+α
αβ = c
∴ b2 − c = αβ−
−β+α 2
2
= ( ) ( )
444 22 β−α=αβ−β+α
17. Ans: 4x2− 5x + 1 = 0 Sol: 2x2 + 3x + 1 = 0
α + β = 23−
, αβ = 21
∴ α2 + β2 = ( ) αβ−β+α 22
= 21
223
2
×−
−
= 45
149 =−
∴ Required quadratic equation is
041
x45
x2 =+− .
18. Ans: 201
Sol: ( )( ) ( )( )∑∑== ++
−=++
17
8n
17
8n3n2n
233n2n
1
= ( ) ( )( )( )∑
= +++−+17
8n3n2n2n3n
= ∑=
+−
+
17
8n3n
12n
1
= ....121
111
111
101 +
−+
−
…..
−+201
191
= 201
2012
201
101 =−=−
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19. Ans: 3 : 1
Sol: a : b = 3 + 2 2 : 3 − 2 2
326
2ba ==+
11ab ==
∴ 13
M.GM.A = ⇒ A.M : G.M = 3 : 1.
20. Ans: 1020 Sol: x1 + x4 + x9 + x11 + x20 + x22 + x27 + x30 = 272 ⇒ a + a + 3d + a + 8d + a + 10d + a + 19d + a + 21d + a + 26d + a + 29d = 272 ⇒ 8a + 116d = 272 ⇒ 4(2a + 29d) = 272
x1 + x2 + ……. + x30 = S30 = [ ]d29a22
30 +
= 15 × 4
272
= 1020.
21. Ans: 2
189
Sol: ar = 24, ar4 = 3
243
arar4
= ⇒ r3 = 81
⇒ r = 21
∴ a + ar + ar2 + ar3 + ar4 + ar5 = a (1 + r + r2 + r3 + r4 + r5)
= 48
++++52 2
1......
2
121
1
= 48 2
189
21
2
1148
21
1
21
16
6
=
−=
−
−
.
22. Ans: 35
Sol: [ ]d74a22
75 + = 2625
75[a + 37d] = 2625
a + 37d = 3575
2625 = .
23. Ans: −3 Sol: 2k = −6 k = −3. 24. Ans: 9 Sol: Tn = nC3 Tn + 1 − Tn = (n + 1)C3 − nC3 = nC2 = 36
⇒ ( )
362.1
1nn =−
⇒ n(n − 1) = 72
⇒ n2 − n − 72 = 0 ⇒ (n − 9) (n + 8) = 0 ⇒ n = 9. 25. Ans: 10C5
Sol: Tr + 1 = 10Cr rr10
10x
x10
−
T5 + 1 = 5510
510
10x
x10
C
−
= 55
510
10x
x10
C
= 10C5. 26. Ans: −1275 Sol: x49(−1 − 2 −3 − ……. − 50) = −x49(1 + 2 + 3 + …… + 50)
= −x49
×2
5150
= −1275 x49. 27. Ans: 729 Sol: Put x = 1
Then ( ) 666
321x2x1 =+=
+
= 729.
28. Ans: 2
2nn2 −+
Sol: 2P1 + 3P1 + …… + nP1 = 2 + 3 + …… + n
= ( )
12
1nn −+
= 2
2nn2 −+.
29. Ans: 400 Sol: a can take the values 1, 3, 5, 7, 9 b can take 0, 3, 6, 9 c can take 0, 2, 4, 6, 8 d can take 2, 3, 5, 7 ∴ Total number of 4 digit numbers = 5C1 × 4C1 × 5C1 × 4C1 = 5 × 4 × 5 × 4 = 400. 30. Ans: 0
Sol:
0aaaa
0aaaa
1aa
1aa
1aa
1aa
3423
2312
21
43
32
21
−−−−=
R2 → R2 − R1
R3 → R3 − R2
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= 0
0dd
0dd
1aa 21
= .
31. Ans: 81
Sol: Area of the triangle
1cy
cx
1by
bx
1by
ax
21
33
22
11
=
cyx
byx
ayx
abc21
33
22
11
c2yx
b2yx
a2yx
abc1
41
33
22
11
= C3 → 2C3
= 81
2abc
abc1
41 =×× .
32. Ans: a ≠ 2 Sol: 3x + y − z = 2 x + 0 y − z = 1 2x + 2y + a = 5
0
a22
101
113
≠−−
⇒ 3(0 + 2) −1(a + 2) − 1(2 − 0) ≠ 0 ⇒ 6 − a − 2 − 2 ≠ 0 ⇒ 2 − a ≠ 0 ⇒ a ≠ 2. 33. Ans: 4
Sol: 0k31
2k2=
−−
(since matrix is singular)
⇒ (2 − k) (3 − k) − 2 = 0 ⇒ 6 − 5k + k2 − 2 = 0 ⇒ k2 − 5k + 4 = 0 ⇒ −k2 + 5k − 4 = 0 ⇒ 5k − k2 = 4. 34. Ans: zero
Sol:
1logclogc1
log
c1
log1logb1
log
clogblog1log
caa
aba
aaa
=
1logclogclog
clog1logblog
clogblog1log
−−−
= ,0
0clogclog
clog0blog
clogblog0
=−
−− because
the determinant of a skew symmetric matrix of odd order is zero.
35. Ans: 1 Sol: 2x + y − 4 = 0 3x + 2y − 2 = 0 x + y + 2 = 0
( ) ( ) ( )23426242
211
223
412
−−+−+=−−
= 12 − 8 − 4 = 0 ∴ The system of equation has a unique
solution. 36. Ans: 0 Sol: 37. Ans: 14x + 5y ≥ 70, y ≤ 14 and x − y ≥ 5 38. Ans: ~[p ∨ (~q)] ≡ (~p) ∧ q 39. Ans: F, F, T 40. Ans: p ∨ (~q) Sol: ~ [(~p) ∧ q] = ~(~p) ∨ ~q = p ∨ ~q.
41. Ans: sinθ + cosθ ≤ 2
Sol: a sinθ + b cosθ ≤ 22 ba +
⇒ sinθ + cosθ ≤ 22 11 +
≤ 2
42. Ans: 2π
Sol:
2 2 3
1 2 2
3 1
−2 2 0
f(x) = −2x + 2 f(x) = 2x
4
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Given sum = ( )
+ −−
22
1tan22tan 11
= ( ) ( )2
22cot22tan 11 π=+ −− .
43. Ans: ab1ba
−+
Sol: Given sum = 2 tan−1a + 2 tan−1b
= 2 xtan2ab1ba
tan 11 −− =
−+
⇒ x = ab1ba
−+
44. Ans: °2sin
2
Sol: Given sum = tan 1° + cot 1°
= °°°+°
1cos.1sin1cos1sin 22
= °° 1cos.1sin2
2
= °2sin
2.
45. Ans: 0
Sol: S5 = 02
cos105
cos =π=π.
46. Ans: 5
3π
Sol:
π−π=
π −−5
32coscos
57
coscos 11
= 5
35
3coscos 1 π=π− .
47. Ans: 15 Sol: Given sum = 1 + tan2 (tan−1 3) +
1 + cot2 ( )2cot 1−
= 1 + 32 + 1 + 22 = 15. 48. Ans: 2 Sol: sinθ + cosec θ = 2 ⇒ sinθ = 1 ∴ sin6θ + cosec6 θ = 16 + 16 = 2. 49. Ans: 2 Sol: ( )[ ]xfLt
0x +→
++
+++=
+→x
x3sin12x
x5sin6x
x7sinx
x2sin12x
x4sin18x
x6sin7x
x8sin
Lt0x
= 2 Only option (D) matches. 50. Ans: –4 Sol:
M is
=
++3,
27
215
,2
52
Y = 2x + K passes through
3,
27
⇒ K = –4 51. Ans: (5, 2) Sol: Circum centre is the mid point of
Hypotenuse
⇒
+−+2
62,
282
⇒ (5, 2) 52. Ans: 1 :1 Sol: The required ratio is
( ) ( )( ) ( )
−−+−+−−=
++++−
7556277542
cbyaxcbyax
22
11
= 1 :1 53. Ans: 4 Sol: a2 − b2 = 512 ⇒ (a + b) (a − b) = 29 ⇒ (a + b, a − b) = (28, 2), (27, 22), (26, 23), (25, 24) (Q a > b, a + b > a − b), the other
combinations like (24, 25) etc cannot be accepted) 29, 1 also cannot be accepted since a and b are positive integers.
M
D
A (2, 5)
B
C (5, 1)
R (8, 6)
Q (8, –2) P (2, –2)
C
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54. Ans: 51
Sol: The line is ( ) ( )41y
31x + = 1
⇒ 3x + 4y − 1 = 0
P = ( ) ( )
22 43
10403
+
−+ =
51
.
55. Ans: 27−
Sol: Centre : (4, −1) Given parabola is y = (x − 2)2 + 6
⇒ (x − 2)2 = 4 . ( )6y41 −
∴ Vertex is (2, 6)
∴ slope = 27−
.
56. Ans: 21
Sol: The required line is x − 2y + k = 0 passes through (1, 1) ⇒ 1 − 2 + k = 0 ⇒ k = 1 ∴ x − 2y + 1 = 0
y intercept = 21
21
bc =
−−=−
.
57. Ans: a2
Sol: p = θ+θ
−22 eccossec
a
θθθ+θ
=
22
22
cossin
cossin
a
= asinθ cosθ = θ2sin2a
q = θ=θ+θ
θ−2cosa
sincos
2cosa22
∴ 4p2 + q2 = a2(sin2 2θ + cos22θ) = a2 . 58. Ans: 2 Sol: C1 : (1, −2), C2 (−2, 2) r1 = 1 r2 = 2
C1 C2 = 21 rr5169 +>=+
∴ Two circles do not intersect ∴ dmin = C1 C2 − (r1 + r2) = 5 − (1 + 2) = 2. 59. Ans: (9, 8)
Sol:
( )5,52
2k,
21h =
++
⇒ h= 9 and k = 8. 60. Ans: 2 Sol: The two circles touch internally ⇒ C1C2 = |r1 − r2|
⇒ 22 kh + = 22 kh4 +−
⇒ 4kh2 22 =+ ⇒ 2kh 22 =+
∴ r = 2. 61. Ans: (x – 2)2 + (y – 2)2 = 8 Sol: Let the required circle be x2 + y2 + 2gx + 2fy + c = 0 ––––(1)
8cfg 22 =−+ ⇒ g2 + f2 = 8 –––(2)
16 + 8g = 0 ⇒ g = –2 ⇒ f2 = 4 ⇒ f = –2
∴ Req uired circle is x2 + y2 – 4x – 4y = 0 or (x – 2)2 + (y – 2)2 = 8 62. Ans: x = cosθ – 1, y = 2sinθ + 1
Sol: ( ) ( )
12
1y11x
2
22
=−++
⇒ x + 1 = 1 cosθ and y – 1 = 2sinθ ⇒ x = cosθ – 1 and y = 2sinθ + 1 63. Ans: 10
Sol: 53
ePMPF ==
⇒ PM = 10635
PF35 =×=×
64. Ans: 16y
3x 22
=−
(0, 1) (h, K)
x2 + y2 = 16
(0, 0)
(h, K) (1, 2) 5 5
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Sol: a32b34ab2 2
2=⇒=
2a = 3a32 =⇒
⇒ b2 = 6332 =×
⇒ 16y
3x 22
=−
65. Ans: 524
Sol: Focus : (ae, 0) =
0,
4a5
2x + 3y − 6 = 0 passes through
0,
4a5
⇒ ( ) 06034a5
2 =−+
⇒ 62a5 = ⇒ a =
512
∴ 2a = 524
.
66. Ans: Sol: a = cosα
Given ellipse is 19y
16x 22
=+
Foci ⇒ ( ) ( )0,70,916 ±=−± ⇒ a2 + b2 = 7 ⇒ b2 = 7 – cos2α
∴ 1cos7
y
cos
x2
2
2
2=
α−−
α
67. Ans: 4
15
Sol: A = θ=× sinb.a21
ba21
= 4
1521
544121 =××++
68. Ans: a3brr
Sol: cos6θ = ( )
aba
abarrr
rrr
•+
•+
= a.ba
a
aba
b.aa22
rrr
r
rrr
rrr
+=
•+
+
= ba
arr
r
+, where
222babarrrr
+=+
∴ a3b
ba
a
21
22
rr
rr
r
==+
=
69. Ans: Sol:
jkCA −=r
ikAB −=
cosθ = 21
2.2
0001 =+−−
⇒ θ = 60°
θ =3π
70. Ans: 7
Sol: 222
bb.a2abarrrrrr
+−=−
⇒ 7 = 14 –2 22
bbrr
+
⇒ 2
br
= 7
⇒ 7b =r
71. Ans: 222
cbarrr
++
Sol: +++=++2222
cbacbarrrrrr
[ ]c.bc.ab.a2rrrrr
+++
( ) 0cb.a.e.icba =++⊥rrrrrr
–––(1)
Similarly ( ) 0ac.b =+rrr
––––(2)
& ( ) 0ba.c =+rrr
–––––(3) (1) + (2) + (3)
( ) 0c.bc.ab.a2 =++rrrrrr
∴ 2222
cbacbarrrrrr
++=++
72. Ans: –25
Sol: 2222
wvuwvurrrrrr
+++++
[ ]w.uw.vv.u2rrrrrr
+++ = 0
∴ 9 + 16 +25 = –2 [ ]w.uw.vv.urrrrrr
++
[ ]w.uw.vv.urrrrrr
++ = 25250 −=−
73. Ans: 71±
Sol: ( ) 1k6j2i3 =−+λ
A B
C ( )ki +
( )kj + ( )ji +
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= ( ) 1k6j2i3 =−+λ
⇒ 13649 =++λ
|λ| 49 =1
⇒ 71
49
1 ±==λ
74. Ans: k2ji2 +− Sol: The required vector
= ( )knjmi3 ++l
where 194
9a
94 2
=++
⇒ a2 = 1 ⇒ a = 1
∴ vector is
+− k32
j31
i32
3
= k2ji2 +−
75. Ans: ( )k11j3i5.r −+r
= 2
135
Sol: Mid point of PQ is
−29
,27
,23
DR of the normal is 5, 3, –11
∴ Plane is
029
z1127
y323
x5 =
+−
−+
−
⇒ 5x + 3y – 11z = 2
135
⇒ ( )k11j3i5.r −+r
= 2
135
76. Ans: 2π
Sol: Ιst line is 2
5z
23
23
y
11x +=
+=−
2nd line is 0
2z21y
32x −=
−+=−
cosθ = 0049
449
1
033 =++++
−−
θ = 90°
θ =2π
77. Ans: 3
1z43y
52x −=
−−=
−−
Sol: Let DR of the line of intersection of the
planes be a, b, c
⇒ a – 2b – c = 0 ––––(1) a + b + 3c = 0 ––––(2)
3c
4b
5a ==
− ⇒ a = –5k, b = –4k, c= 3k
⇒ 3
1z43y
52x −=
−−=
−−
78. Ans: ( )k4ji326
1 −−
Sol: DR of the line : 3, –1, –4
DC’ S : 26
4,
26
1,
26
3 −−
79. Ans: cos–1
32
Sol: DR’s of the normal : 2, −1, 2 DR’s of Z axis : 0, 0, 1
cosθ = ( )( ) ( ) ( )( )( )
32
100414
120102 =++++
+−+
⇒ θ = cos−1
32 .
80. Ans: (2, –3, 4)
Sol: c
zzb
yya
xx 111 −=−=−
( )
222111
cba
czbyax
++++−=
∴ 12929
4z
3y
2x ===
−=
⇒ x = 2, y= –3, z = 4 81. Ans: 13
Sol: 1325144 =+ 82. Ans: 2 Sol: ∑ ∑ =⇒=− 54x945x ii
∑ =×+×− 459255410x 2i
⇒ ∑ = 360x 2i
⇒ σ = 29
549
3602
=
−
83. Ans: 8
13
Sol: Events: (1, 5), (1, 6), (2, 4), (2, 5), (2, 6) (3, 3), (3, 4), (3, 5), (3, 6), (4, 2) (4, 3), (4, 4), (4, 5), (4, 6), (5, 1) (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
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Probability = 8
1366
26 =×
84. Ans: [1 – P(A)] . P(B’) Sol: P (A ∪ B)’ = P(A’ ∩ B’) = P(A’) P(B’) = [1 – P(A)] . P(B’) 85. Ans: 14
Sol: σ = 12
1nd
2 −
= 12
1497
−
= 7 × 2 = 14
86. Ans: 2165
Sol: 2165
38
357
4=
××
87. Ans: 5
Sol: ( ) ( )
( ) ( )1xx1x
1x1xlim
22
35
1x +−−+−
→
= 1x1x
lim55
1x −−
→ = 5 × 14 = 5
88. Ans: 53
Sol:
( )53
x5x5
1ex3
x3x31log
limx5
2
2
0x=
−
×+
→
89. Ans: x
Sol: 1
1x36x3
21x3
2x
1x32x
f−
−+
+−
+
=
−+
= x
90. Ans: –1 and 2 Sol:
2xlim→
ax + 3 = 2x
lim→
ax2 – 1
⇒ 2a + 3 = 2a2 – 1 ⇒ 2a2 = 2a – 4 = 0 ⇒ a = –1, 2 91. Ans: f’(x) = 0 Sol: |f(x) – f(y)|2 ≤ |x – y|3 –––(1) |f(y) – f(x)|2 ≤ |y – x|3 –––(2) If x > y, 2nd inequality will be wrong
If x < y. 1st inequality will be wrong So only possiblity is f(x) = k(a constant
function) f’(x) = 0 92. Ans: 1
Sol: f(x) = ( )dttcos121
x
1∫ −
= [ ]x1tsint21 −
= [ ]1sin1xsinx21 +−−
f’(x) = [ ]xcos121 −
f’(π) = [ ] 11121 =+
93. Ans: 10
Sol: ( ) x22x'fdxdy 2 +=
1xdxdy
= = 5 x 2 × 1
= 10 94. Ans: –4 Sol: f’(x) = 2x + b 10 + b = 2 × ( 7 + b) = 14 + 2b b = –4
95. Ans: tcos
13
Sol: tcos
1tcostsec
dxdy
3
2==
96. Ans: θ2sec
Sol: θθ−
θθ=sincosa3
cossina3dxdy
2
2
= tanθ
θ=
+ 22
secdxdy
1
97. Ans: 2x1
2
−
Sol: x = sinθ ⇒ y = 2sin–1x
⇒ 2x1
2dxdy
−=
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98. Ans: (–1, 1)
Sol: 2y 4dxdy = ⇒
y2
dxdy =
1y2y2 =⇒=
⇒ 1 = 4x + 5 x = –1 (–1, 1) 99. Ans: x∈ (2, 3) Sol: f’(x) = 6x2 – 30x + 36 = 6(x2 – 5x +6) = 6(x – 2) (x – 3) < 0 x∈ (2, 3) 100. Ans: 15
Sol: y2 exy dxdy
y2.eydxdy
x xy+
+ = 9e–3 . 2x
9. e–3 33 e18dxdy
6.e3dxdy −− −=+
+−
–9y1 + 27 + 6y1 = –18 –3y1 + 27 = –18 3y1 = 45 y1 = 15 101. Ans: –6 Sol: V = πr2h
0dtdr
r2.hdtdh
rdtdv 2 =
+π=
⇒ ( ) 05.30dtdh
.25 =+
625150
dtdh −=−=
102. Ans: f(x) is not differentiable at x = 4 Sol: f(x) is not differentiable at x = 4
103. Ans: 41
Sol: y = x2 − 2x
1
3x
2x2
dxdy += = −2 + −2 = −4
41
dxdy
1−−=
−=
41
.
104. Ans: 2−
Sol: 211 22 −=+−
105. Ans: ( )
Cxx1 4
14
++−
Sol: ∫
+4
3
45
x
11x
dx
1 + ∫−⇒=4
34t
dt41
tx
1
= C4
1t
41 4
1
+−
−−
= Cx
11
41
4+
+−
( )
Cxx1 4
14
++−
106. Ans: – cot (x ex) + C Sol: x ex = t ⇒
∫ dtectcos 2 = – cot t + C
= – cot (x ex) + C
107. Ans: C1x
ex+
+
Sol: ( )∫
+−
+dx
x1
1x1
1e
2x
= C1x
ex+
+
108. Ans: ex sin2 x + C
Sol: ( )∫ + dxxcosxsin2xsine 2x
= ex sin2x + C
109. Ans: 2 C2x
sin2 +
Sol: ∫ dx2xcos2 2
= ∫ dx2x
cos2
= C2
12
xsin2 +
= C2x
sin22 +
110. Ans: Cxsec1x 12 +−− −
Sol: x = sec θ ⇒ ∫ θθθθθ
secdtansectan
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= ( )∫ θ−θ d1sec2
= tan θ - θ + C
= Cxsec1x 12 +−− −
111. Ans: Cx
51
151 2
3
2+
+−
Sol: x = 5 tan θ ⇒ ∫+
dxx
x54
2
= ∫ θθθθ
4
2
tan25
dsec5sec5
= ∫ θθθ
dsin
cos51
4
= ∫ 4t
dt51
= Ct
1151
3+
cosec x = 2
2
x
5x +
= Cx
51
151 2
3
2+
+−
112. Ans:
+2
e1log
e1
Sol: ( )∫ ∫−
− +=
+
1
0
e
1xx
x
x eee
dxe
e1
dx
ex = t ex dx = dt
( )∫ ∫
+−=
+=
e
1
e
1et
1t1
e1
ettdt
= e
1ett
loge1
+
=
+−
e11
loge2
elog
e1
=( )
+ e11
21
loge1
=
+2
e1log
e1
113. Ans: e + 2e1 −
Sol: ex = ex ⇒ x = 0
A = ( )∫−−
1
0
xx dxee
= (ex – e–x) dx
= e + 2e1 − .
114. Ans: 21
Sol: ∫+
e
1
dxx3
xlog1
= ∫2
1
dtt31
=
22
2t
31
= ( )21
1461 =−
115. Ans: 16π
Sol: ∫ +⇒=
1
08
34
x1
dxxtx
= ∫ +
1
02t1
dt41
= ( )101 ttan41 −
= 1644
1 π=π×
116. Ans: ( )22log23
Sol: ∫ ∫=4
2
4log
2log
dxxt
dttlog
=
4log2
2log
2
2x
= ( ) ( )[ ]22 2log2log421 −
= ( )22log321 ×
= ( )22log23
117. Ans: x3 + y3 = 3 k xy + C Sol: k [x dy + y dx] = x2 dx + y2 dy
⇒ k [d (x y) = ( )33 yxd31 +
⇒ k x y = k3
yx 33++
⇒ x3 + y3 = 3 k xy + C = x3 + y3 = 3 k xy + C
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