JEE Mathematics Previous Year Questions MCQs Set 07

Answer: (c) \( \frac{n(4n^2-1)c^2}3 \)

Explanation:
1. The kth term is \( S_k-S_{k-1}=c(2k-1) \).
2. Sum of squares \( =c^2\sum_{k=1}^n(2k-1)^2=c^2\cdot\frac{n(2n-1)(2n+1)}3=\frac{n(4n^2-1)c^2}3 \).

Teacher's Note:
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Answer: (a) 48

Explanation:
1. The left side is \( |z|^2(\bar z^2+z^2)=2(x^2+y^2)(x^2-y^2) \), so \( (x^2+y^2)(x^2-y^2)=175 \).
2. Integer solutions: x² + y² = 25 and x² − y² = 7, giving x² = 16, y² = 9 (the other factor pairs give non-squares).
3. The roots are (±4, ±3), forming a rectangle of sides 8 and 6, area 48.

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Answer: (d) \( \frac1{4\sin2^\circ} \)

Explanation:
1. \( \text{Im}(z^{2m-1})=\sin(2m-1)\theta \).
2. Sum \( \sum_{m=1}^{15}\sin(2m-1)\theta=\frac{\sin^2(15\theta)}{\sin\theta} \).
3. At θ = 2°: \( \frac{\sin^230^\circ}{\sin2^\circ}=\frac1{4\sin2^\circ} \).

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Answer: (c) 77

Explanation:
1. Subtract 1 from each digit: seven digits from {0, 1, 2} with sum 3.
2. Case 1,1,1 and four 0s: \( \binom73=35 \). Case 2,1 and five 0s: 7 × 6 = 42.
3. Total = 35 + 42 = 77.

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Answer: (d) a straight line

Explanation:
1. The vertices are A(pq, (p+1)(q+1)), B(−p, 0) and C(−q, 0).
2. The altitude from A is x = pq. The altitude from B is perpendicular to the second line, giving the orthocentre at (pq, −pq).
3. Its coordinates satisfy x + y = 0, so the locus is a straight line.

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Answer: (b) x² + y² − 4x − 10y + 19 = 0

Explanation:
1. The centre of the given circle is C(3, 2). Since ∠PAC = ∠PBC = 90°, the circumcircle of PAB has PC as a diameter.
2. Its equation is (x − 1)(x − 3) + (y − 8)(y − 2) = 0.
3. Expanding: x² + y² − 4x − 10y + 19 = 0.

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Answer: (d) \( \frac{27}{10} \)

Explanation:
1. The ellipse is \( \frac{x^2}9+y^2=1 \): A(3, 0), B(0, 1), auxiliary circle \( x^2+y^2=9 \).
2. Line AB: x + 3y = 3. Substituting x = 3 − 3y: 10y² − 18y = 0, so y = 9/5 at M (x = −12/5).
3. Area \( =\frac12\cdot OA\cdot|y_M|=\frac12\cdot3\cdot\frac95=\frac{27}{10} \).

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Answer: (c) \( \left(\pm2\sqrt3,\pm\frac17\right) \)

Explanation:
1. Take P(4 cos θ, 2 sin θ). The normal meets the x-axis at Q(3 cos θ, 0), so M is \( \left(\frac72\cos\theta,\sin\theta\right) \).
2. Eliminating θ: \( \frac{4x^2}{49}+y^2=1 \).
3. The latus rectum is x = ±2√3 (since ae = 2√3): \( y^2=1-\frac{48}{49}=\frac1{49} \), so y = ±1/7.

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Answer: (c) b, d are non-parallel

Explanation:
1. Since all are unit vectors, |a × b| ≤ 1 and |c × d| ≤ 1, so the product equals 1 only if |a × b| = |c × d| = 1 and the two cross products are the same unit vector n.
2. Then a ⊥ b, c ⊥ d and all four vectors lie in the plane perpendicular to n, so they are coplanar. This rules out (a) and (b).
3. In that plane let a be at 0°, b at 90°, c at 60° (from a·c = 1/2) and d at 150°. Then b and d are 60° apart, so they are not parallel; (d) is also false.

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Answer: (a) \( \frac14 \)

Explanation:
1. Q = (1 − 3μ, −1 + μ, 2 + 5μ), so \( \vec{PQ}=(-2-3\mu,\ -3+\mu,\ -4+5\mu) \).
2. PQ is parallel to the plane, so it is perpendicular to the normal (1, −4, 3): \( (-2-3\mu)+12-4\mu-12+15\mu=8\mu-2=0 \).
3. So μ = 1/4.

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Answer: (c) \( \sqrt3 \)

Explanation:
1. Equal angles give direction cosines \( \frac1{\sqrt3} \) each, so points are (2 + t, −1 + t, 2 + t) with t = distance/√3.
2. Substituting: 2(2 + t) + (−1 + t) + (2 + t) = 9 gives 4t = 4 and t = 1.
3. The distance PQ is √3 · 1 = √3.

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Answer: (c) f(1/2) < 1/2 and f(1/3) < 1/3

Explanation:
1. Differentiating both sides: \( \sqrt{1-f'^2}=f \), so f² + f′² = 1.
2. With f(0) = 0 and f ≥ 0, the solution is f(x) = sin x.
3. sin(1/2) ≈ 0.479 < 1/2 and sin(1/3) ≈ 0.327 < 1/3, since sin x < x for x > 0.

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Answer: (a) \( \frac23 \)

Explanation:
1. Let t = tan²x. Then sin⁴x = t²/(1+t)² and cos⁴x = 1/(1+t)².
2. The equation becomes \( \frac{t^2}{2}+\frac13=\frac15(1+t)^2 \), which simplifies to \( (3t-2)^2=0 \).
3. So t = 2/3.

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Answer: (a) \( \frac1{125} \)

Explanation:
1. From the previous step tan²x = 2/3, so sin²x = 2/5 and cos²x = 3/5.
2. sin⁸x = 16/625 and cos⁸x = 81/625.
3. The value is \( \frac{16}{625\cdot8}+\frac{81}{625\cdot27}=\frac{2}{625}+\frac{3}{625}=\frac1{125} \).

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Answer: (a) \( \frac\pi2 \)

Explanation:
1. Use \( \csc A\csc B=\frac{\cot A-\cot B}{\sin(B-A)} \) with B − A = π/4, so each term is \( \sqrt2[\cot(\ldots)-\cot(\ldots)] \).
2. The sum telescopes to \( \sqrt2\left[\cot\theta-\cot\left(\theta+\frac{3\pi}2\right)\right]=\sqrt2(\cot\theta+\tan\theta)=\frac{2\sqrt2}{\sin2\theta} \).
3. Setting this equal to 4√2 gives sin 2θ = 1/2, so θ = π/12 or 5π/12. The sum is π/2.

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48. g
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50. n
51. t
52.
53. d
54. i
55. f
56. f
57. e
58. r
59. e
60. n
61. c
62. e
63. s
64. .

Answer: (b) 2a

Explanation:
1. \( \cos B+\cos C=2\sin\frac A2\cos\frac{B-C}2 \), so \( \cos\frac{B-C}2=2\sin\frac A2 \).
2. By the sine rule \( \frac{b+c}a=\frac{\cos\frac{B-C}2}{\sin\frac A2}=2 \).
3. So b + c = 2a (A moves on an ellipse with foci B and C).

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

Answer: (a) (a, 0)

Explanation:
1. With P(at², 2at): T(−at², 0) and N(at² + 2a, 0). The centroid is \( \left(\frac{at^2+2a}3,\frac{2at}3\right) \).
2. Eliminating t: \( y^2=\frac{4a}3\left(x-\frac{2a}3\right) \), with vertex (2a/3, 0) and latus rectum 4a/3.
3. The focus is at distance a/3 from the vertex: (a, 0).

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

Answer: (a) x² + 2y² = 2

Explanation:
1. The hyperbola has eccentricity √2 and foci (±1, 0). The ellipse has eccentricity 1/√2.
2. Orthogonal intersection means the conics are confocal, so ae = 1 giving a² = 2 and b² = 2(1 − 1/2) = 1.
3. The ellipse is x²/2 + y² = 1, i.e. x² + 2y² = 2.

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

Answer: (b) 1

Explanation:
1. \( f'(x)=\cos\frac1x+\frac1x\sin\frac1x \).
2. As x → ∞, cos(1/x) → 1 and (1/x) sin(1/x) → 0, so the limit is 1.

Teacher's Note:
1. D
2. i
3. f
4. f
5. e
6. r
7. e
8. n
9. t
10. i
11. a
12. t
13. e
14.
15. b
16. y
17.
18. t
19. h
20. e
21.
22. p
23. r
24. o
25. d
26. u
27. c
28. t
29.
30. r
31. u
32. l
33. e
34.
35. a
36. n
37. d
38.
39. t
40. a
41. k
42. e
43.
44. t
45. h
46. e
47.
48. l
49. i
50. m
51. i
52. t
53.
54. t
55. e
56. r
57. m
58.
59. b
60. y
61.
62. t
63. e
64. r
65. m
66. .

Answer: (b) f(x + 2) − f(x) > 2 for all x in [1, ∞)

Explanation:
1. By the mean value theorem, f(x + 2) − f(x) = 2f′(c) for some c in (x, x + 2).
2. f′ is strictly decreasing with limit 1 as x → ∞, so f′(c) > 1 for every c.
3. Hence f(x + 2) − f(x) > 2 for all x ≥ 1.

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

Answer: (a) \( 10\pi \)

Explanation:
1. Add the integral with x replaced by −x: \( 2I_n=\int_{-\pi}^{\pi}\frac{\sin nx}{\sin x}dx \) (since \( \frac1{1+\pi^x}+\frac1{1+\pi^{-x}}=1 \)).
2. For odd n this equals 2π, so I_n = π.
3. The sum of ten such terms is 10π.

Teacher's Note:
1. T
2. h
3. e
4.
5. s
6. u
7. b
8. s
9. t
10. i
11. t
12. u
13. t
14. i
15. o
16. n
17.
18. x
19.
20. t
21. o
22.
23. −
24. x
25.
26. a
27. n
28. d
29.
30. t
31. h
32. e
33.
34. i
35. d
36. e
37. n
38. t
39. i
40. t
41. y
42.
43. 1
44. /
45. (
46. 1
47.
48. +
49.
50. a
51. ^
52. x
53. )
54.
55. +
56.
57. 1
58. /
59. (
60. 1
61.
62. +
63.
64. a
65. ^
66. −
67. x
68. )
69.
70. =
71.
72. 1
73.
74. r
75. e
76. m
77. o
78. v
79. e
80.
81. t
82. h
83. e
84.
85. a
86. w
87. k
88. w
89. a
90. r
91. d
92.
93. d
94. e
95. n
96. o
97. m
98. i
99. n
100. a
101. t
102. o
103. r
104. .

Answer: (b) 1

Explanation:
1. The curve meets y = e at x = 1.
2. Area \( =\int_0^1(e-e^x)dx=e-(e-1)=1 \).

Teacher's Note:
1. I
2. n
3. t
4. e
5. g
6. r
7. a
8. t
9. e
10.
11. t
12. h
13. e
14.
15. v
16. e
17. r
18. t
19. i
20. c
21. a
22. l
23.
24. g
25. a
26. p
27.
28. b
29. e
30. t
31. w
32. e
33. e
34. n
35.
36. t
37. h
38. e
39.
40. l
41. i
42. n
43. e
44.
45. y
46.
47. =
48.
49. e
50.
51. a
52. n
53. d
54.
55. t
56. h
57. e
58.
59. c
60. u
61. r
62. v
63. e
64.
65. f
66. r
67. o
68. m
69.
70. 0
71.
72. t
73. o
74.
75. 1
76. .

Answer: (a) \( \frac1{64} \)

Explanation:
1. Expand \( \sqrt{a^2-x^2}=a-\frac{x^2}{2a}-\frac{x^4}{8a^3}-\cdots \).
2. The numerator becomes \( \left(\frac1{2a}-\frac14\right)x^2+\frac{x^4}{8a^3}+\cdots \). For L to be finite, 1/(2a) = 1/4 so a = 2.
3. Then \( L=\frac1{8a^3}=\frac1{64} \).

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

Answer: (a) 12

Explanation:
1. A symmetric 3 × 3 matrix has 3 diagonal and 3 independent off-diagonal entries. Five 1s in total means (diagonal 1s + 2 × off-diagonal 1s) = 5.
2. Cases: 3 diagonal 1s and 1 off-diagonal 1: 3 ways; 1 diagonal 1 and 2 off-diagonal 1s: 3 × 3 = 9 ways.
3. Total = 3 + 9 = 12.

Teacher's Note:
1. C
2. o
3. u
4. n
5. t
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. t
22.
23. e
24. n
25. t
26. r
27. i
28. e
29. s
30.
31. s
32. e
33. p
34. a
35. r
36. a
37. t
38. e
39. l
40. y
41.
42. b
43. e
44. c
45. a
46. u
47. s
48. e
49.
50. e
51. a
52. c
53. h
54.
55. o
56. f
57. f
58. -
59. d
60. i
61. a
62. g
63. o
64. n
65. a
66. l
67.
68. 1
69.
70. a
71. p
72. p
73. e
74. a
75. r
76. s
77.
78. t
79. w
80. i
81. c
82. e
83. .

Answer: (b) at least 4 but less than 7

Explanation:
1. A unique solution needs a non-zero determinant.
2. Checking all 12 matrices in A, exactly 6 are non-singular.

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

Answer: (b) more than 2

Explanation:
1. Singular matrices (6 of them) give either no solution or infinitely many.
2. Comparing ranks of the coefficient and augmented matrices, 4 of the singular matrices give an inconsistent system.
3. So the number is 4, which is more than 2.

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