Practice MCQs for JEE Mathematics Miscellaneous Problems
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Multiple Choice Questions
(a) 1
(b) 2
(c) 0
(d) \( \frac1{\sqrt2} \)
Show Answer & Explanation
Answer: (c) 0
Explanation:
1. Let \( f(x)=\log(x+\sqrt{1+x^2}) \). Then \( f(-x)=\log(\sqrt{1+x^2}-x)=\log\frac{1}{x+\sqrt{1+x^2}}=-f(x) \).
2. f is odd, so the integral over the symmetric interval [−2, 2] is 0.
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(a) 1
(b) 2
(c) 0
(d) \( \frac1{\sqrt2} \)
Show Answer & Explanation
Answer: (b) 2
Explanation:
1. Write as \( \frac1n\sum_{r=1}^n\sqrt{\frac nr}=\frac1n\sum\left(\frac rn\right)^{-1/2} \).
2. As a Riemann sum: \( \int_0^1x^{-1/2}dx=2 \).
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(a) 1
(b) 2
(c) 0
(d) \( \frac1{\sqrt2} \)
Show Answer & Explanation
Answer: (a) 1
Explanation:
1. Combine: \( \frac{n^2+1-an^2-an}{n+1}=\frac{(1-a)n^2-an+1}{n+1} \).
2. For a finite limit the coefficient of n² must vanish: a = 1.
3. Then the limit of \( \frac{-n+1}{n+1} \) is −1, and −1 + 1 = 0 as required.
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81. .
(a) 1
(b) 2
(c) 0
(d) \( \frac1{\sqrt2} \)
Show Answer & Explanation
Answer: (d) \( \frac1{\sqrt2} \)
Explanation:
1. At x = 3π/4, cos x < 0, so \( y=-\cos x \).
2. Then \( \frac{dy}{dx}=\sin x=\sin\frac{3\pi}{4}=\frac1{\sqrt2} \).
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58. .
(a) \( -\frac38 \)
(b) \( -\frac12 \)
(c) −2
(d) −1
Show Answer & Explanation
Answer: (d) −1
Explanation:
1. Expand: \( 2\sqrt5\sin x\cos x+4\sin^2x-6=\sqrt5\sin2x+2(1-\cos2x)-6 \).
2. This is \( \sqrt5\sin2x-2\cos2x-4 \), with maximum \( \sqrt{5+4}-4=-1 \).
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63. .
(a) \( -\frac38 \)
(b) \( -\frac12 \)
(c) −2
(d) −1
Show Answer & Explanation
Answer: (c) −2
Explanation:
1. On (−2, −1): [x] = −2, so the integrand is 2x; integral \( =[x^2]_{-2}^{-1}=1-4=-3 \).
2. On (−1, 0): |[x]| = 1: \( \int_{-1}^0x\,dx=-\frac12 \). On (0, 1): 0. On (1, 2): \( \int_1^2x\,dx=\frac32 \).
3. Total \( -3-\frac12+\frac32=-2 \).
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70. .
(a) \( -\frac38 \)
(b) \( -\frac12 \)
(c) −2
(d) −1
Show Answer & Explanation
Answer: (b) \( -\frac12 \)
Explanation:
1. Series: \( \tan^{-1}x=x-\frac{x^3}{3}+\dots \) and \( \sin^{-1}x=x+\frac{x^3}{6}+\dots \).
2. Difference \( =-\frac{x^3}{2}+\dots \), so the limit is \( -\frac12 \).
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(a) \( -\frac38 \)
(b) \( -\frac12 \)
(c) −2
(d) −1
Show Answer & Explanation
Answer: (a) \( -\frac38 \)
Explanation:
1. Differentiate: \( 2xy+x^2y'+3y^2y'=0 \); at (1, 1): \( 2+4y'=0 \), so \( y'=-\frac12 \).
2. Differentiate again: \( 2y+4xy'+x^2y''+6y(y')^2+3y^2y''=0 \). At (1, 1): \( 2-2+\frac32+4y''=0 \).
3. So \( y''=-\frac38 \).
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81. .
(a) \( \frac\pi4 \)
(b) \( \frac\pi2 \)
(c) \( \frac\pi3 \)
(d) \( \frac\pi6 \)
Show Answer & Explanation
Answer: (a) \( \frac\pi4 \)
Explanation:
1. Sum \( \tan(x+y)=\frac{\frac{b}{c+a}+\frac{c}{a+b}}{1-\frac{bc}{(c+a)(a+b)}}=\frac{b(a+b)+c(c+a)}{(c+a)(a+b)-bc} \).
2. Using \( b^2+c^2=a^2 \): numerator \( =a(b+c)+a^2 \), denominator \( =a^2+a(b+c) \). The ratio is 1.
3. So the sum is π/4.
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(a) \( \frac\pi4 \)
(b) \( \frac\pi2 \)
(c) \( \frac\pi3 \)
(d) \( \frac\pi6 \)
Show Answer & Explanation
Answer: (d) \( \frac\pi6 \)
Explanation:
1. \( \tan x\tan2x=1 \) (with both positive), so \( \frac{2\tan^2x}{1-\tan^2x}=1 \) giving \( 3\tan^2x=1 \).
2. tan x = 1/√3, x = π/6 (then tan 2x = √3 > 0, valid).
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77. .
(a) \( \frac\pi4 \)
(b) \( \frac\pi2 \)
(c) \( \frac\pi3 \)
(d) \( \frac\pi6 \)
Show Answer & Explanation
Answer: (c) \( \frac\pi3 \)
Explanation:
1. \( \frac{\tan A-\tan B}{\tan A+\tan B}=\frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin(A-B)}{\sin C} \).
2. By the sine rule \( \frac{c-b}{c}=\frac{\sin C-\sin B}{\sin C} \). So \( \sin(A-B)=\sin C-\sin B=\sin(A+B)-\sin B \).
3. Hence \( \sin(A+B)-\sin(A-B)=\sin B \), i.e. \( 2\cos A\sin B=\sin B \), so \( \cos A=\frac12 \) and A = π/3.
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96. .
(a) neither continuous nor differentiable at x = 2
(b) differentiable and continuous for all x
(c) continuous but not differentiable at x = 2
(d) differentiable and continuous at x = 2
Show Answer & Explanation
Answer: (c) continuous but not differentiable at x = 2
Explanation:
1. At x = 2: left limit and right limit are both 0 and f(2) = 0, so f is continuous.
2. Left derivative: for x < 2, \( f=x(2-x) \), slope \( 2-2x=-2 \). Right derivative: for x > 2, \( f=x(x-2) \), slope \( 2x-2=2 \).
3. The one-sided derivatives differ, so it is not differentiable at x = 2.
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78. .
(a) neither continuous nor differentiable at x = 2
(b) differentiable and continuous for all x
(c) continuous but not differentiable at x = 2
(d) differentiable and continuous at x = 2
Show Answer & Explanation
Answer: (a) neither continuous nor differentiable at x = 2
Explanation:
1. Left limit at x = 2: \( x^2\cdot1\to4 \). Value and right limit: \( g(2)=4\cdot2=8 \).
2. The left limit (4) differs from g(2) = 8, so g is discontinuous at x = 2 and hence not differentiable there.
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82. .
(a) neither continuous nor differentiable at x = 2
(b) differentiable and continuous for all x
(c) continuous but not differentiable at x = 2
(d) differentiable and continuous at x = 2
Show Answer & Explanation
Answer: (d) differentiable and continuous at x = 2
Explanation:
1. At x = 2: left value \( 8 \), \( \phi(2)=8 \), right limit \( 0+8=8 \): continuous.
2. Left derivative \( 3x^2=12 \); right derivative of \( 4(x-2)+2x^2 \) is \( 4+4x=12 \): equal, so differentiable at x = 2.
3. (It is discontinuous at x = 1, so option (b) is false.)
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78. .
(a) 6
(b) 2
(c) 4
(d) 3
Show Answer & Explanation
Answer: (d) 3
Explanation:
1. \( \frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}{r}=\frac{84}{36}=\frac73 \) and \( \frac{{}^nC_{r+1}}{{}^nC_r}=\frac{n-r}{r+1}=\frac{126}{84}=\frac32 \).
2. Solving: \( 3n-10r=-3 \) and \( 2n-5r=3 \), giving n = 9 and r = 3 (check: 9C2 = 36, 9C3 = 84, 9C4 = 126).
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64. .
(a) 6
(b) 2
(c) 4
(d) 3
Show Answer & Explanation
Answer: (b) 2
Explanation:
1. \( f(x)=2^x+x^2-1 \) has f(0) = 0.
2. f(−1) = 0.5 > 0 and f(−0.5) ≈ −0.04 < 0, so there is another root between −1 and −0.5.
3. For x > 0, f is increasing and positive, so there are no more roots. Total 2.
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(a) 6
(b) 2
(c) 4
(d) 3
Show Answer & Explanation
Answer: (a) 6
Explanation:
1. \( x^2=k\left(y-\frac3k\right) \): vertex \( \left(0,\frac3k\right) \) and \( 4a=k \), so the focus is \( \left(0,\frac3k+\frac k4\right) \).
2. Setting \( \frac3k+\frac k4=2 \): \( k^2-8k+12=0 \), k = 2 or 6. Since k ≠ 2, k = 6.
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(a) 6
(b) 2
(c) 4
(d) 3
Show Answer & Explanation
Answer: (c) 4
Explanation:
1. Groups of four in A.P. from 1 to 12 that partition the set: {1,2,3,4},{5,6,7,8},{9,10,11,12}; {1,4,7,10},{2,5,8,11},{3,6,9,12}; {1,2,3,4},{5,7,9,11},{6,8,10,12}; and {9,10,11,12},{1,3,5,7},{2,4,6,8}.
2. Enumerating all partitions by computer confirms exactly 4 ways.
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(a) 1/2
(b) 3
(c) 2
(d) 4
Show Answer & Explanation
Answer: (b) 3
Explanation:
1. Three planes through the origin share a common line when the determinant of coefficients is zero.
2. The determinant \( \begin{vmatrix}2&-3&-7\\\lambda&-14&-13\\8&-31&-33\end{vmatrix}=118\lambda-354 \).
3. So \( \lambda=3 \).
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(a) 2
(b) 1
(c) 0
(d) 8
Show Answer & Explanation
Answer: (a) 2
Explanation:
1. Let the integral be I. Replacing t by 4 − t: \( I=\int_0^4\frac{f(4-t)}{f(4-t)+f(t)}dt \).
2. Adding: \( 2I=\int_0^41\,dt=4 \), so I = 2.
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(a) 2
(b) 1
(c) 0
(d) 8
Show Answer & Explanation
Answer: (d) 8
Explanation:
1. f even implies f″ even. So \( x^3f(x) \) and \( xf''(x) \) are odd functions and integrate to 0 over [−2, 2].
2. The remaining integral is \( \int_{-2}^22\,dx=8 \).
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(a) 2
(b) 1
(c) 0
(d) 8
Show Answer & Explanation
Answer: (c) 0
Explanation:
1. By the fundamental theorem \( f'(x)=|\cos x| \).
2. At x = π/2: \( |\cos\frac\pi2|=0 \).
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(a) 2
(b) 1
(c) 0
(d) 8
Show Answer & Explanation
Answer: (b) 1
Explanation:
1. Intersection: \( x-bx^2=\frac{x^2}{b} \) gives x = 0 and \( x=\frac{b}{1+b^2} \).
2. Area \( =\left(b+\frac1b\right)\frac{1}{6}\left(\frac{b}{1+b^2}\right)^3=\frac16\cdot\frac{b^2}{(1+b^2)^2} \).
3. The function \( \frac{b^2}{(1+b^2)^2}=\frac{1}{(b+1/b)^2} \) is maximum when \( b+\frac1b \) is minimum, i.e. b = 1.
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(a) (1, −2)
(b) \( \left(\frac32,-\frac32\right) \)
(c) (1, 1)
(d) (0, 3)
Show Answer & Explanation
Answer: (c) (1, 1)
Explanation:
1. Complete the square: \( (y-1)^2=-8x+24=-8(x-3) \).
2. Vertex (3, 1), \( 4a=8 \), a = 2, opening to the left, so the focus is (3 − 2, 1) = (1, 1).
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Free study material for Mathematics
Practice MCQs for JEE Mathematics Miscellaneous Problems
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