JEE Mathematics Previous Year Questions MCQs Set 02

Answer: (c) 2

Explanation:
1. On (−3, −1], \( f'(x)=3(2+x)^2 \) vanishes only at x = −2, where f is increasing on both sides (an inflection), so no extremum.
2. At x = −1, f(−1) = 1, values to the left are below 1 and to the right (\( x^{2/3} \) near 1) are also below 1: a local maximum.
3. At x = 0, \( x^{2/3} \) has a cusp minimum. Total: 2.

Teacher's Note:
1. C
2. h
3. e
4. c
5. k
6.
7. e
8. a
9. c
10. h
11.
12. c
13. r
14. i
15. t
16. i
17. c
18. a
19. l
20.
21. p
22. o
23. i
24. n
25. t
26.
27. a
28. n
29. d
30.
31. e
32. a
33. c
34. h
35.
36. b
37. r
38. e
39. a
40. k
41.
42. p
43. o
44. i
45. n
46. t
47. .
48.
49. A
50.
51. p
52. o
53. i
54. n
55. t
56.
57. w
58. h
59. e
60. r
61. e
62.
63. t
64. h
65. e
66.
67. d
68. e
69. r
70. i
71. v
72. a
73. t
74. i
75. v
76. e
77.
78. v
79. a
80. n
81. i
82. s
83. h
84. e
85. s
86.
87. b
88. u
89. t
90.
91. t
92. h
93. e
94.
95. s
96. i
97. g
98. n
99.
100. d
101. o
102. e
103. s
104.
105. n
106. o
107. t
108.
109. c
110. h
111. a
112. n
113. g
114. e
115.
116. i
117. s
118.
119. n
120. o
121. t
122.
123. a
124. n
125.
126. e
127. x
128. t
129. r
130. e
131. m
132. u
133. m
134. .

Answer: (c) \( \frac12\log\left(\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+c \)

Explanation:
1. Multiply numerator and denominator of J by \( e^{4x} \): \( J=\int\frac{e^{3x}}{e^{4x}+e^{2x}+1}dx \).
2. So \( J-I=\int\frac{e^{3x}-e^x}{e^{4x}+e^{2x}+1}dx \).
3. Differentiating the option (c) expression, \( \frac12\frac{d}{dx}\log\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1} \), gives exactly \( \frac{e^{3x}-e^x}{e^{4x}+e^{2x}+1} \) (checked symbolically).

Teacher's Note:
1. R
2. e
3. w
4. r
5. i
6. t
7. e
8.
9. J
10.
11. w
12. i
13. t
14. h
15.
16. t
17. h
18. e
19.
20. s
21. a
22. m
23. e
24.
25. d
26. e
27. n
28. o
29. m
30. i
31. n
32. a
33. t
34. o
35. r
36.
37. a
38. s
39.
40. I
41. .
42.
43. V
44. e
45. r
46. i
47. f
48. y
49.
50. t
51. h
52. e
53.
54. a
55. n
56. t
57. i
58. d
59. e
60. r
61. i
62. v
63. a
64. t
65. i
66. v
67. e
68.
69. b
70. y
71.
72. d
73. i
74. f
75. f
76. e
77. r
78. e
79. n
80. t
81. i
82. a
83. t
84. i
85. n
86. g
87. .

Answer: (c) odd and strictly increasing in (−∞, ∞)

Explanation:
1. \( g(-u)=2\tan^{-1}(e^{-u})-\frac\pi2=2\left(\frac\pi2-\tan^{-1}e^u\right)-\frac\pi2=\frac\pi2-2\tan^{-1}e^u=-g(u) \), so g is odd.
2. Also \( g'(u)=\frac{2e^u}{1+e^{2u}}>0 \), so g is strictly increasing everywhere.

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. a
7. n
8. ⁻
9. ¹
10. (
11. 1
12. /
13. t
14. )
15.
16. =
17.
18. π
19. /
20. 2
21.
22. −
23.
24. t
25. a
26. n
27. ⁻
28. ¹
29.
30. t
31.
32. f
33. o
34. r
35.
36. t
37.
38. >
39.
40. 0
41.
42. t
43. o
44.
45. t
46. e
47. s
48. t
49.
50. o
51. d
52. d
53. n
54. e
55. s
56. s
57. .
58.
59. T
60. h
61. e
62.
63. d
64. e
65. r
66. i
67. v
68. a
69. t
70. i
71. v
72. e
73.
74. i
75. s
76.
77. a
78. l
79. w
80. a
81. y
82. s
83.
84. p
85. o
86. s
87. i
88. t
89. i
90. v
91. e
92. .

Answer: (b) C₁ and C₂ touch each other exactly at two points

Explanation:
1. Substitute \( y^2=4x \) in the circle: \( x^2+4x-6x+1=0 \), so \( (x-1)^2=0 \) and x = 1.
2. Then \( y=\pm2 \): the curves meet at (1, 2) and (1, −2).
3. The repeated root means they touch (tangent) at each of these two points.

Teacher's Note:
1. A
2.
3. d
4. o
5. u
6. b
7. l
8. e
9.
10. r
11. o
12. o
13. t
14.
15. i
16. n
17.
18. t
19. h
20. e
21.
22. e
23. l
24. i
25. m
26. i
27. n
28. a
29. t
30. i
31. o
32. n
33.
34. e
35. q
36. u
37. a
38. t
39. i
40. o
41. n
42.
43. s
44. i
45. g
46. n
47. a
48. l
49. s
50.
51. t
52. a
53. n
54. g
55. e
56. n
57. c
58. y
59. .
60.
61. C
62. o
63. u
64. n
65. t
66.
67. t
68. h
69. e
70.
71. d
72. i
73. s
74. t
75. i
76. n
77. c
78. t
79.
80. y
81. -
82. v
83. a
84. l
85. u
86. e
87. s
88.
89. t
90. o
91.
92. g
93. e
94. t
95.
96. t
97. h
98. e
99.
100. n
101. u
102. m
103. b
104. e
105. r
106.
107. o
108. f
109.
110. p
111. o
112. i
113. n
114. t
115. s
116. .

Answer: (a) \( \frac{1}{\sqrt2} \)

Explanation:
1. Volume squared equals the Gram determinant: \( V^2=\begin{vmatrix}1&\frac12&\frac12\\\frac12&1&\frac12\\\frac12&\frac12&1\end{vmatrix} \).
2. This is \( 1+\frac14-\frac34=\frac12 \) (using 1 + 2·(1/8) − 3·(1/4)).
3. So \( V=\frac{1}{\sqrt2} \).

Teacher's Note:
1. V
2. o
3. l
4. u
5. m
6. e
7.
8. o
9. f
10.
11. a
12.
13. p
14. a
15. r
16. a
17. l
18. l
19. e
20. l
21. e
22. p
23. i
24. p
25. e
26. d
27.
28. i
29. s
30.
31. t
32. h
33. e
34.
35. s
36. q
37. u
38. a
39. r
40. e
41.
42. r
43. o
44. o
45. t
46.
47. o
48. f
49.
50. t
51. h
52. e
53.
54. G
55. r
56. a
57. m
58.
59. d
60. e
61. t
62. e
63. r
64. m
65. i
66. n
67. a
68. n
69. t
70.
71. o
72. f
73.
74. i
75. t
76. s
77.
78. e
79. d
80. g
81. e
82.
83. v
84. e
85. c
86. t
87. o
88. r
89. s
90. .

Answer: (d) 5 or 10

Explanation:
1. Independence: \( P(A\cap B)=P(A)P(B) \), so \( \frac{|A\cap B|}{10}=\frac{4}{10}\cdot\frac{|B|}{10} \), i.e. \( |A\cap B|=\frac{2|B|}{5} \).
2. So |B| must be a multiple of 5: |B| = 5 gives |A∩B| = 2 (possible), |B| = 10 gives |A∩B| = 4 (possible).
3. Hence B has 5 or 10 outcomes.

Teacher's Note:
1. W
2. r
3. i
4. t
5. e
6.
7. t
8. h
9. e
10.
11. i
12. n
13. d
14. e
15. p
16. e
17. n
18. d
19. e
20. n
21. c
22. e
23.
24. c
25. o
26. n
27. d
28. i
29. t
30. i
31. o
32. n
33.
34. i
35. n
36.
37. t
38. e
39. r
40. m
41. s
42.
43. o
44. f
45.
46. c
47. o
48. u
49. n
50. t
51. s
52. .
53.
54. C
55. h
56. e
57. c
58. k
59.
60. t
61. h
62. a
63. t
64.
65. t
66. h
67. e
68.
69. i
70. n
71. t
72. e
73. r
74. s
75. e
76. c
77. t
78. i
79. o
80. n
81.
82. s
83. i
84. z
85. e
86.
87. i
88. s
89.
90. f
91. e
92. a
93. s
94. i
95. b
96. l
97. e
98. .

Answer: (c) \( x\sqrt{1+x^2} \)

Explanation:
1. Let \( \theta=\cot^{-1}x \), so \( \cos\theta=\frac{x}{\sqrt{1+x^2}} \) and \( \sin\theta=\frac1{\sqrt{1+x^2}} \).
2. Then \( x\cos\theta+\sin\theta=\frac{x^2+1}{\sqrt{1+x^2}}=\sqrt{1+x^2} \).
3. The bracket is \( (1+x^2)-1=x^2 \) with square root x, so the value is \( x\sqrt{1+x^2} \).

Teacher's Note:
1. D
2. r
3. a
4. w
5.
6. a
7.
8. r
9. i
10. g
11. h
12. t
13.
14. t
15. r
16. i
17. a
18. n
19. g
20. l
21. e
22.
23. f
24. o
25. r
26.
27. c
28. o
29. t
30. ⁻
31. ¹
32.
33. x
34.
35. t
36. o
37.
38. g
39. e
40. t
41.
42. s
43. i
44. n
45. e
46.
47. a
48. n
49. d
50.
51. c
52. o
53. s
54. i
55. n
56. e
57.
58. e
59. x
60. a
61. c
62. t
63. l
64. y
65. .

Answer: (a) \( \hat u=\frac{\hat a+\hat b}{|\hat a+\hat b|} \) and \( M=(1+\hat a\cdot\hat b)^{1/2} \)

Explanation:
1. \( |\vec{OP}|^2=\cos^2t+\sin^2t+2\hat a\cdot\hat b\sin t\cos t=1+(\hat a\cdot\hat b)\sin2t \).
2. Since a·b > 0 (acute), this is largest when sin 2t = 1, i.e. t = π/4.
3. Then \( \vec{OP}=\frac{\hat a+\hat b}{\sqrt2} \), \( M^2=1+\hat a\cdot\hat b \) and \( \hat u=\frac{\hat a+\hat b}{|\hat a+\hat b|} \).

Teacher's Note:
1. E
2. x
3. p
4. a
5. n
6. d
7.
8. t
9. h
10. e
11.
12. s
13. q
14. u
15. a
16. r
17. e
18. d
19.
20. l
21. e
22. n
23. g
24. t
25. h
26. .
27.
28. T
29. h
30. e
31.
32. m
33. a
34. x
35. i
36. m
37. u
38. m
39.
40. o
41. c
42. c
43. u
44. r
45. s
46.
47. a
48. t
49.
50. t
51.
52. =
53.
54. π
55. /
56. 4
57.
58. b
59. e
60. c
61. a
62. u
63. s
64. e
65.
66. t
67. h
68. e
69.
70. a
71. n
72. g
73. l
74. e
75.
76. i
77. s
78.
79. a
80. c
81. u
82. t
83. e
84. .

Answer: (d) −6 + 7i

Explanation:
1. Horizontally away from the origin by 5: \( 1+2i\to6+2i \). Vertically away by 3: \( 6+2i\to6+5i=z_1 \).
2. Moving √2 units along \( \hat i+\hat j \) adds (1, 1): \( 7+6i \).
3. Rotating through π/2 anticlockwise multiplies by i: \( (7+6i)i=-6+7i \).

Teacher's Note:
1. T
2. r
3. a
4. c
5. k
6.
7. t
8. h
9. e
10.
11. p
12. o
13. i
14. n
15. t
16.
17. s
18. t
19. e
20. p
21.
22. b
23. y
24.
25. s
26. t
27. e
28. p
29. .
30.
31. A
32.
33. r
34. o
35. t
36. a
37. t
38. i
39. o
40. n
41.
42. o
43. f
44.
45. 9
46. 0
47.
48. d
49. e
50. g
51. r
52. e
53. e
54. s
55.
56. a
57. b
58. o
59. u
60. t
61.
62. t
63. h
64. e
65.
66. o
67. r
68. i
69. g
70. i
71. n
72.
73. i
74. s
75.
76. m
77. u
78. l
79. t
80. i
81. p
82. l
83. i
84. c
85. a
86. t
87. i
88. o
89. n
90.
91. b
92. y
93.
94. i
95. .

Answer: (d) All of: f″ vanishes at least twice on [0, 1]; f′(1/2) = 0; \( \int_{-1/2}^{1/2}f(x+\tfrac12)\sin x\,dx=0 \); and \( \int_0^{1/2}f(t)e^{\sin\pi t}dt=\int_{1/2}^1f(1-t)e^{\sin\pi t}dt \)

Explanation:
1. Differentiating \( f(x)=f(1-x) \): \( f'(x)=-f'(1-x) \), so \( f'(1/2)=0 \) and \( f'(3/4)=-f'(1/4)=0 \).
2. f′ vanishes at 1/4, 1/2, 3/4, so by Rolle's theorem f″ vanishes at least twice in [0, 1].
3. The function \( x\mapsto f(x+\tfrac12) \) is even (symmetry about 1/2) and sin x is odd, so the integral is 0.
4. Substituting t → 1 − t shows the two integrals in the last statement are equal. All four statements are true.

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. s
10. y
11. m
12. m
13. e
14. t
15. r
16. y
17.
18. f
19. (
20. x
21. )
22.
23. =
24.
25. f
26. (
27. 1
28.
29. −
30.
31. x
32. )
33.
34. f
35. o
36. r
37.
38. d
39. e
40. r
41. i
42. v
43. a
44. t
45. i
46. v
47. e
48. s
49.
50. a
51. n
52. d
53.
54. i
55. n
56. t
57. e
58. g
59. r
60. a
61. l
62. s
63. .
64.
65. R
66. o
67. l
68. l
69. e
70. '
71. s
72.
73. t
74. h
75. e
76. o
77. r
78. e
79. m
80.
81. g
82. i
83. v
84. e
85. s
86.
87. t
88. h
89. e
90.
91. z
92. e
93. r
94. o
95. s
96.
97. o
98. f
99.
100. f
101. ″
102. .

Answer: (b) 1

Explanation:
1. \( \operatorname{Re}((1-i)(x+iy))=x+y \), so C is the line \( x+y=\sqrt2 \). B is the circle with centre (2, 1) and radius 3.
2. The line meets the circle at \( (3.17,-1.76) \) and \( (-0.76,2.17) \) (approximately).
3. Only the second point has y ≥ 1, so A ∩ B ∩ C has exactly 1 element.

Teacher's Note:
1. C
2. o
3. n
4. v
5. e
6. r
7. t
8.
9. e
10. a
11. c
12. h
13.
14. s
15. e
16. t
17.
18. t
19. o
20.
21. a
22.
23. g
24. e
25. o
26. m
27. e
28. t
29. r
30. i
31. c
32.
33. o
34. b
35. j
36. e
37. c
38. t
39.
40. a
41. n
42. d
43.
44. i
45. n
46. t
47. e
48. r
49. s
50. e
51. c
52. t
53. .
54.
55. T
56. h
57. e
58.
59. h
60. a
61. l
62. f
63. -
64. p
65. l
66. a
67. n
68. e
69.
70. c
71. o
72. n
73. d
74. i
75. t
76. i
77. o
78. n
79.
80. r
81. e
82. m
83. o
84. v
85. e
86. s
87.
88. o
89. n
90. e
91.
92. o
93. f
94.
95. t
96. h
97. e
98.
99. t
100. w
101. o
102.
103. i
104. n
105. t
106. e
107. r
108. s
109. e
110. c
111. t
112. i
113. o
114. n
115.
116. p
117. o
118. i
119. n
120. t
121. s
122. .

Answer: (c) 35 and 39

Explanation:
1. The points \( -1+i \) and \( 5+i \) are the ends of a diameter of B (centre 2 + i, radius 3).
2. For a point z on a circle, the sum of squares of distances from the ends of a diameter is constant: \( (2r)^2=36 \) (Pythagoras for the right angle in a semicircle).
3. So the value is 36, which lies between 35 and 39.

Teacher's Note:
1. T
2. h
3. e
4.
5. t
6. w
7. o
8.
9. p
10. o
11. i
12. n
13. t
14. s
15.
16. a
17. r
18. e
19.
20. d
21. i
22. a
23. m
24. e
25. t
26. r
27. i
28. c
29. a
30. l
31. l
32. y
33.
34. o
35. p
36. p
37. o
38. s
39. i
40. t
41. e
42.
43. o
44. n
45.
46. t
47. h
48. e
49.
50. c
51. i
52. r
53. c
54. l
55. e
56. ,
57.
58. s
59. o
60.
61. t
62. h
63. e
64.
65. a
66. n
67. g
68. l
69. e
70.
71. a
72. t
73.
74. z
75.
76. i
77. s
78.
79. a
80.
81. r
82. i
83. g
84. h
85. t
86.
87. a
88. n
89. g
90. l
91. e
92. .