JEE Mathematics Previous Year Questions MCQs Set 01

Answer: (c) two

Explanation:
1. First equation: \( 2\sin^2\theta-(1-2\sin^2\theta)=0 \), so \( \sin^2\theta=\frac14 \) and \( \sin\theta=\pm\frac12 \).
2. Second equation: \( 2(1-s^2)-3s=0 \) gives \( 2s^2+3s-2=0 \), so \( s=\frac12 \) (s = −2 is impossible).
3. Common solutions: \( \sin\theta=\frac12 \), i.e. \( \theta=\frac\pi6,\frac{5\pi}{6} \). Two solutions.

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Answer: (d) \( \frac29(2p-q)(2q-p) \)

Explanation:
1. \( \alpha+\beta=p \) and \( \frac\alpha2+2\beta=q \). Doubling the second: \( \alpha+4\beta=2q \).
2. Subtract: \( 3\beta=2q-p \), so \( \beta=\frac{2q-p}{3} \) and \( \alpha=p-\beta=\frac{4p-2q}{3} \).
3. Hence \( r=\alpha\beta=\frac{2(2p-q)(2q-p)}{9} \).

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Answer: (c) two

Explanation:
1. Coplanar means the scalar triple product is zero. With \( t=\lambda^2 \): \( \begin{vmatrix}-t&1&1\\1&-t&1\\1&1&-t\end{vmatrix}=0 \).
2. The determinant equals \( (2-t)(1+t)^2 \).
3. So \( t=2 \) or \( t=-1 \). Since \( t=\lambda^2\ge0 \), t = 2 and \( \lambda=\pm\sqrt2 \): two values.

Teacher's Note:
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Answer: (c) \( \frac25 \)

Explanation:
1. Given that each American couple is adjacent, treat the 4 couples as blocks (each block has 2 internal orders) and the Indian man and his wife as 2 separate people: 6 objects in a circle, \( 5!\times2^4 \) arrangements.
2. Favourable: the Indian couple also forms a block (2 internal orders): 5 objects, \( 4!\times2\times2^4 \) arrangements.
3. Probability \( =\frac{4!\times2}{5!}=\frac25 \).

Teacher's Note:
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Answer: (a) \( \frac8\pi f(2) \)

Explanation:
1. At x = π/4, sec²x = 2, so numerator and denominator both tend to 0. Apply L'Hospital's rule.
2. Differentiate: numerator \( f(\sec^2x)\cdot2\sec^2x\tan x \), denominator \( 2x \).
3. At x = π/4: \( \frac{f(2)\cdot2\cdot2\cdot1}{\pi/2}=\frac{8f(2)}{\pi} \).

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Answer: (a) \( \frac{1}{n(n-1)}(1+nx^n)^{1-\frac1n}+K \)

Explanation:
1. Composing once more: \( f(f(x))=\frac{x}{(1+2x^n)^{1/n}} \). By induction \( g(x)=\frac{x}{(1+nx^n)^{1/n}} \).
2. So the integrand is \( x^{n-1}(1+nx^n)^{-1/n} \). Put \( u=1+nx^n \), \( du=n^2x^{n-1}dx \).
3. Integral \( =\frac{1}{n^2}\cdot\frac{u^{1-1/n}}{1-1/n}=\frac{1}{n(n-1)}(1+nx^n)^{1-\frac1n}+K \).

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Answer: (c) 96

Explanation:
1. Alphabetical order of letters: C, C, H, I, N, O. No letter is smaller than C, so every earlier word starts with C.
2. Words starting with C whose second letter is smaller than O: second letter C, H, I or N, each followed by an arrangement of the remaining 4 distinct letters: 4 × 4! = 96.
3. Words starting with CO: the smallest has the remaining letters C, H, I, N in order, which is COCHIN itself. So exactly 96 words come before it.

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Answer: (b) 2

Explanation:
1. Let CD = x, AB = 2x and AD = 2r (the diameter of the incircle). Area \( =\frac{(x+2x)}{2}\cdot2r=3xr=18 \), so \( xr=6 \).
2. BC = \( \sqrt{x^2+4r^2} \). For a tangential quadrilateral \( AB+CD=AD+BC \): \( 3x=2r+\sqrt{x^2+4r^2} \).
3. Squaring: \( 9x^2-12xr+4r^2=x^2+4r^2 \), so \( x=\frac{3r}{2} \).
4. Then \( \frac32r^2=6 \), r = 2.

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Answer: (a) \( x^2\csc^2\theta-y^2\sec^2\theta=1 \)

Explanation:
1. The ellipse \( \frac{x^2}{4}+\frac{y^2}{3}=1 \) has \( c^2=4-3=1 \), foci (±1, 0).
2. For the hyperbola a = sin θ and c = 1, so \( b^2=c^2-a^2=1-\sin^2\theta=\cos^2\theta \).
3. Equation \( \frac{x^2}{\sin^2\theta}-\frac{y^2}{\cos^2\theta}=1 \), i.e. \( x^2\csc^2\theta-y^2\sec^2\theta=1 \).

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Answer: (d) \( (3+4i)e^{i\pi/4} \)

Explanation:
1. The first displacement is \( 3e^{i\pi/4} \). The second is at 135° to the real axis: \( 4e^{3i\pi/4}=4ie^{i\pi/4} \).
2. Sum \( =(3+4i)e^{i\pi/4} \).

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Answer: (a) \( \frac{1}{3x}+\frac{2x^2}{3} \)

Explanation:
1. The limit is the derivative with respect to t at t = x: \( 2xf(x)-x^2f'(x)=1 \).
2. So \( f'-\frac2xf=-\frac1{x^2} \), a linear equation with integrating factor \( x^{-2} \): \( \frac{d}{dx}\left(\frac{f}{x^2}\right)=-\frac1{x^4} \).
3. Hence \( \frac{f}{x^2}=\frac{1}{3x^3}+C \), i.e. \( f=\frac{1}{3x}+Cx^2 \). With f(1) = 1, C = 2/3.

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Answer: (b) \( -\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3} \)

Explanation:
1. \( \frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1} \).
2. Differentiate with respect to y: \( \frac{d^2x}{dy^2}=-\left(\frac{dy}{dx}\right)^{-2}\frac{d^2y}{dx^2}\cdot\frac{dx}{dy} \).
3. Using \( \frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1} \): \( \frac{d^2x}{dy^2}=-\frac{d^2y}{dx^2}\left(\frac{dy}{dx}\right)^{-3} \).

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Answer: (c) fixed radius 1 and variable centres along the x-axis

Explanation:
1. Separate variables: \( \frac{y\,dy}{\sqrt{1-y^2}}=dx \).
2. Integrate: \( -\sqrt{1-y^2}=x+c \).
3. Squaring: \( (x+c)^2+y^2=1 \): circles of radius 1 with centre (−c, 0) on the x-axis.

Teacher's Note:
1. S
2. e
3. p
4. a
5. r
6. a
7. t
8. e
9.
10. t
11. h
12. e
13.
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15. a
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18. a
19. b
20. l
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22. s
23.
24. a
25. n
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27.
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30. t
31. e
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33. r
34. a
35. t
36. e
37. .
38.
39. S
40. q
41. u
42. a
43. r
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45.
46. t
47. h
48. e
49.
50. r
51. e
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53. u
54. l
55. t
56.
57. t
58. o
59.
60. r
61. e
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63. o
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65. n
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68. e
69.
70. t
71. h
72. e
73.
74. c
75. i
76. r
77. c
78. l
79. e
80. .

Answer: (d) the y-axis

Explanation:
1. Divide by z: \( \frac{z}{1-z^2}=\frac{1}{\bar z-z} \) since \( 1/z=\bar z \).
2. With \( z=e^{i\theta} \): \( \bar z-z=-2i\sin\theta \), so the value is \( \frac{i}{2\sin\theta} \).
3. This is purely imaginary, so it lies on the y-axis.

Teacher's Note:
1. O
2. n
3.
4. t
5. h
6. e
7.
8. u
9. n
10. i
11. t
12.
13. c
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20. 1
21. /
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23.
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34.
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41. a
42. t
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47.
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49. .
50.
51. T
52. h
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56. i
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58. f
59. e
60. r
61. e
62. n
63. c
64. e
65.
66. o
67. f
68.
69. c
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71. n
72. j
73. u
74. g
75. a
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79.
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89.
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91. m
92. a
93. g
94. i
95. n
96. a
97. r
98. y
99. .

Answer: (c) \( \left(3,\frac43\right) \)

Explanation:
1. A point that divides the triangle into three equal-area triangles is the centroid.
2. Centroid \( =\left(\frac{0+3+6}{3},\frac{0+4+0}{3}\right)=\left(3,\frac43\right) \).

Teacher's Note:
1. E
2. q
3. u
4. a
5. l
6.
7. a
8. r
9. e
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11. s
12.
13. o
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23. e
24. e
25.
26. s
27. u
28. b
29. -
30. t
31. r
32. i
33. a
34. n
35. g
36. l
37. e
38. s
39.
40. i
41. d
42. e
43. n
44. t
45. i
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47. y
48.
49. t
50. h
51. e
52.
53. c
54. e
55. n
56. t
57. r
58. o
59. i
60. d
61. .

Answer: (d) Statement 1 is false, Statement 2 is true

Explanation:
1. \( F(x)=\frac x2-\frac{\sin2x}{4}+C \).
2. Then \( F(x+\pi)=F(x)+\frac\pi2\neq F(x) \), so Statement 1 is false.
3. Statement 2 is true since sin² has period π.

Teacher's Note:
1. A
2. n
3.
4. i
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6. t
7. e
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9. r
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12.
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15.
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17.
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24. i
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26.
27. f
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30. c
31. t
32. i
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34. n
35.
36. n
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40.
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42. o
43. t
44.
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51. i
52. o
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56. ,
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59. e
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65.
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67. h
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69.
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75. p
76. e
77. r
78.
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80. e
81. r
82. i
83. o
84. d
85.
86. i
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88.
89. n
90. o
91. n
92. -
93. z
94. e
95. r
96. o
97. .

Answer: (d) Statement 1 is false, Statement 2 is true

Explanation:
1. Direction of the line = cross product of normals: \( (3,-6,-2)\times(2,1,-2)=(14,2,15) \), so Statement 2 is true.
2. Check the point (3, 1, 0) on the plane 3x − 6y − 2z = 15: 9 − 6 = 3, not 15. So the stated line does not lie on the planes and Statement 1 is false.

Teacher's Note:
1. T
2. h
3. e
4.
5. d
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7. r
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9. c
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25.
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28.
29. r
30. i
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32. h
33. t
34.
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42.
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52.
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54. h
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56. s
57. e
58. n
59.
60. i
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62.
63. w
64. r
65. o
66. n
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68. .
69.
70. A
71. l
72. w
73. a
74. y
75. s
76.
77. t
78. e
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80. t
81.
82. t
83. h
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85.
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91.
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93. n
94.
95. b
96. o
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98. h
99.
100. p
101. l
102. a
103. n
104. e
105. s
106. .

Answer: (a) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Explanation:
1. Write \( y=-\frac12(x-1)^2+\frac32 \). The vertex is at x = 1 and the axis is the line x = 1.
2. So the parabola is symmetric about x = 1 because a parabola is symmetric about its axis. Both statements are true and the second explains the first.

Teacher's Note:
1. C
2. o
3. m
4. p
5. l
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7. t
8. e
9.
10. t
11. h
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20.
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25. i
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27. d
28.
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30. h
31. e
32.
33. a
34. x
35. i
36. s
37. .
38.
39. S
40. y
41. m
42. m
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44. t
45. r
46. y
47.
48. a
49. b
50. o
51. u
52. t
53.
54. t
55. h
56. e
57.
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60. i
61. s
62.
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65. p
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70. s
71.
72. t
73. h
74. e
75.
76. s
77. y
78. m
79. m
80. e
81. t
82. r
83. y
84. .

Answer: (a) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Explanation:
1. The director circle of \( x^2+y^2=a^2 \) is \( x^2+y^2=2a^2 \), here \( 2\times169=338 \).
2. For (17, 7): \( 289+49=338 \), so the point lies on it and the tangents are perpendicular. Statement 2 explains Statement 1.

Teacher's Note:
1. T
2. h
3. e
4.
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7. r
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9. c
10. t
11. o
12. r
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20.
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23.
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27.
28. l
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31. u
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33.
34. o
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36.
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43.
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46. o
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50. h
51. i
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68.
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70. a
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72. g
73. e
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77.
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81.
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88.
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90. u
91. b
92. s
93. t
94. i
95. t
96. u
97. t
98. e
99.
100. t
101. h
102. e
103.
104. p
105. o
106. i
107. n
108. t
109. .

Answer: (c) \( G_1=G_2=G_3=\dots \)

Explanation:
1. Since \( G_1^2=A_1H_1 \), the geometric mean of \( A_1 \) and \( H_1 \) is \( G_2=\sqrt{A_1H_1}=G_1 \).
2. Repeating the argument for each n, all the geometric means are equal: \( G_n=\sqrt{ab} \).

Teacher's Note:
1. F
2. o
3. r
4.
5. a
6. n
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8.
9. t
10. w
11. o
12.
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20. e
21.
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25. b
26. e
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30.
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33. e
34.
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36. r
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41. t
42.
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45.
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49. i
50. r
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53. M
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58.
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60. M
61.
62. e
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68.
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70. h
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72.
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74. q
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79.
80. o
81. f
82.
83. t
84. h
85. e
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90. M
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92.
93. T
94. h
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96.
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98. M
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100. t
101. h
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103. r
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106. o
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109.
110. n
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112. v
113. e
114. r
115.
116. c
117. h
118. a
119. n
120. g
121. e
122. s
123. .

Answer: (c) 1 : 4

Explanation:
1. Intersection: \( x^2+8x-9=0 \) gives x = 1, so \( P(1,2\sqrt2),\ Q(1,-2\sqrt2) \) and \( PQ=4\sqrt2 \).
2. Tangent to the circle at P: \( x+2\sqrt2y=9 \), meeting the x-axis at R(9, 0). Tangent to the parabola at P: \( 2\sqrt2y=4(x+1) \), meeting the axis at S(−1, 0).
3. Heights from S and R to PQ (the line x = 1): 2 and 8. Areas are in the ratio 2 : 8 = 1 : 4.

Teacher's Note:
1. B
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26. b
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69. a
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73.
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93. n
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96. R
97.
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102.
103. t
104. h
105. e
106.
107. l
108. i
109. n
110. e
111.
112. P
113. Q
114. .

Answer: (b) \( 3\sqrt3 \)

Explanation:
1. As before \( P(1,2\sqrt2),\ R(9,0),\ S(-1,0) \).
2. Sides: \( RS=10,\ PS=\sqrt{4+8}=2\sqrt3,\ PR=\sqrt{64+8}=6\sqrt2 \). Area \( =\frac12\cdot10\cdot2\sqrt2=10\sqrt2 \).
3. Circumradius \( R=\frac{abc}{4K}=\frac{10\cdot2\sqrt3\cdot6\sqrt2}{40\sqrt2}=3\sqrt3 \).

Teacher's Note:
1. F
2. i
3. n
4. d
5.
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10. t
11. h
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24. ,
25.
26. t
27. h
28. e
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30.
31. u
32. s
33. e
34.
35. R
36.
37. =
38.
39. a
40. b
41. c
42. /
43. (
44. 4
45.
46. ×
47.
48. a
49. r
50. e
51. a
52. )
53. .

Answer: (d) 2

Explanation:
1. \( P(1,2\sqrt2),\ Q(1,-2\sqrt2),\ R(9,0) \).
2. Sides: \( PQ=4\sqrt2,\ PR=QR=6\sqrt2 \). Area \( =\frac12\cdot4\sqrt2\cdot8=16\sqrt2 \).
3. Semi-perimeter \( s=\frac{4\sqrt2+12\sqrt2}{2}=8\sqrt2 \), so \( r=\frac{16\sqrt2}{8\sqrt2}=2 \).

Teacher's Note:
1. I
2. n
3. r
4. a
5. d
6. i
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9.
10. e
11. q
12. u
13. a
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21.
22. d
23. i
24. v
25. i
26. d
27. e
28. d
29.
30. b
31. y
32.
33. s
34. e
35. m
36. i
37. -
38. p
39. e
40. r
41. i
42. m
43. e
44. t
45. e
46. r
47. .

Answer: (b) \( \frac1{12}n(n+1)(3n^2+n+2) \)

Explanation:
1. \( V_r=\frac r2[2r+(r-1)(2r-1)]=\frac{r(2r^2-r+1)}{2} \).
2. Summing with the standard sums of r, r² and r³ gives \( \sum V_r=\frac{n(n+1)(3n^2+n+2)}{12} \).

Teacher's Note:
1. F
2. i
3. n
4. d
5.
6. V
7. _
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66. a
67. l
68.
69. n
70. u
71. m
72. b
73. e
74. r
75. s
76. .

Answer: (d) a composite number

Explanation:
1. With \( V_r=\frac{r(2r^2-r+1)}{2} \) we get \( T_r=3r^2+2r-1=(3r-1)(r+1) \).
2. For every natural number r, both factors are at least 2, so T_r is composite (e.g. 4, 15, 32, 55).

Teacher's Note:
1. F
2. a
3. c
4. t
5. o
6. r
7. i
8. s
9. e
10.
11. T
12. _
13. r
14. .
15.
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17. _
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19.
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22.
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24. d
25. d
26.
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28. o
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78. t
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83.
84. f
85. a
86. i
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88. .

Answer: (b) \( Q_1,Q_2,Q_3,\dots \) are in A.P. with common difference 6

Explanation:
1. \( T_r=3r^2+2r-1 \), so \( Q_r=T_{r+1}-T_r=6r+5 \).
2. The difference of consecutive Q's is 6, so they are in A.P. with common difference 6.

Teacher's Note:
1. Q
2. _
3. r
4.
5. i
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7.
8. a
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11. i
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29. r
30.
31. w
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35.
36. s
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40. e
41.
42. 6
43. ,
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48. c
49. e
50.
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52. n
53.
54. A
55. .
56. P
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58.
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61.
62. d
63. i
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65. f
66. e
67. r
68. e
69. n
70. c
71. e
72.
73. 6
74. .