JEE Mathematics Definite Integration MCQs Set 33

Answer: (c) \( \left(2e^{\pi/2-2}\right)^m \)

Explanation:
1. The logarithm of the expression is \( m\cdot\frac1n\sum_{r=1}^n[\ln(n^2+r^2) - 2\ln n] \).
2. By the given limit this tends to \( m(\ln2 + \pi/2 - 2) \), so the value is \( (2e^{\pi/2-2})^m \).

Teacher's Note:
1. The logarithm turns the product into the given sum.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac\pi2 \)

Explanation:
1. \( f(x)f(\pi/2 - x) = \frac{\sin2x}{2x(\pi/2 - x)} = \frac{\sin2x}{\pi}\left(\frac1x + \frac{2}{\pi - 2x}\right) \).
2. Each of \( \int_0^{\pi/2}\frac{\sin2x}{x}dx \) and \( \int_0^{\pi/2}\frac{2\sin2x}{\pi-2x}dx \) equals \( J = \int_0^\pi\frac{\sin t}{t}dt \).
3. So the integral is \( 2J/\pi \), and \( k(2J/\pi) = J \) gives \( k = \pi/2 \).

Teacher's Note:
1. Partial fractions split the product into two copies of \( J \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) (i) and (ii) only

Explanation:
1. The numerator is about \( \frac34n^{4/3} \); the bracket is about \( \frac1n\int_0^1\frac{dx}{(a+x)^2} = \frac1{n\,a(a+1)} \).
2. So the limit is \( \frac34a(a+1) = 54 \), i.e. \( a(a+1) = 72 \), giving a = \( 8 \) or a = \( -9 \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (b) \( \frac13\left(16\sqrt2-20\right) \)

Explanation:
1. A point \( (x, y) \) with \( 0 \le x \le y \) is nearer to the origin than to the edge \( y = 1 \) when \( \sqrt{x^2+y^2} \lt 1 - y \), i.e. \( y \lt (1 - x^2)/2 \); this part of the region needs also \( x \le y \), so \( x \le \sqrt2 - 1 \).
2. Octant area = \( \int_0^{\sqrt2-1}\left(\frac{1-x^2}2 - x\right)dx \). By the eight-fold symmetry, area = \( 8 \) times this = \( \frac13(16\sqrt2 - 20) \).

Teacher's Note:
1. Use symmetry and the parabola equidistant from the origin and an edge.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 13.93 \)

Explanation:
1. \( C \) is \( y = \sqrt{1 + x^2} \); the line meets it at \( x = \sqrt3, y = 2 \).
2. Area = \( \int_0^{\sqrt3}(\sqrt{1+x^2} - \frac{2x}{\sqrt3})dx = [\frac x2\sqrt{1+x^2} + \frac12\sinh^{-1}x - \frac{x^2}{\sqrt3}]_0^{\sqrt3} = \frac12\ln(2+\sqrt3) \).
3. \( \frac12\ln(2+\sqrt3) = \frac14\ln(2+\sqrt3)^2 = \frac14\ln(7+4\sqrt3) \), so \( k = 7 + 4\sqrt3 \approx 13.93 \).

Teacher's Note:
1. \( k = (2 + sqrt3)^2 = 7 + 4 \) sqrt3.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 10\left(\frac\pi2\right)^9 \)

Explanation:
1. Integrating by parts twice: \( u_n = n(\pi/2)^{n-1} - n(n-1)u_{n-2} \).
2. With \( n = 10: u_{10} + 90u_8 = 10(\pi/2)^9 \).

Teacher's Note:
1. The recurrence \( u_n + n(n-1)u_{n-2} = n(pi/2)^{n-1} \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 3e \)

Explanation:
1. Put \( t = x^3 \) in \( I_2: I_2 = (1/3)\int_0^1\frac{e^{-t}}{2-t}dt \).
2. Now put \( s = 1 - t: I_2 = (1/3)\int_0^1\frac{e^{s-1}}{1+s}ds = I_1/(3e) \).
3. So \( I_1/I_2 = 3e \).

Teacher's Note:
1. \( I_2 \) equals \( I_1 \) divided by \( 3e \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 3+2\sqrt2 \)

Explanation:
1. \( f'(x) = \sec x - \csc x = 0 \) gives \( x = \pi/4 \), and \( f' \) changes from negative to positive there, so it is the minimum.
2. \( f(\pi/4) = \ln(\sec x + \tan x)|_0^{\pi/4} + \ln\tan\frac x2|_{\pi/4}^{\pi/2} = \ln(\sqrt2+1) - \ln(\sqrt2-1) = \ln\frac{\sqrt2+1}{\sqrt2-1} = \ln(3+2\sqrt2) \).

Teacher's Note:
1. Differentiate to find the minimum point \( x = pi/4 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{\pi^2}2-4 \)

Explanation:
1. The function \( \ln\frac{\pi+x}{\pi-x}\cos x \) is odd, so its integral is \( 0 \).
2. \( \int_{-\pi/2}^{\pi/2}x^2\cos x\,dx = 2[x^2\sin x + 2x\cos x - 2\sin x]_0^{\pi/2} = 2(\pi^2/4 - 2) = \pi^2/2 - 4 \).

Teacher's Note:
1. Odd and even parts separate.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \int_{-2}^2f(x)\,dx\ge12 \)

Explanation:
1. Convexity gives \( f(x) + f(-x) \ge 2f(0) = 6 \), so \( \int_{-2}^2 f = \int_0^2[f(x) + f(-x)]dx \ge 12 \).
2. Equality holds for a linear \( f (for \) example \( f = 3 + x) \), so strict inequality fails.
3. \( f \) is increasing, so on \( [-1, 0] f \lt 3 \) and \( \int_{-1}^0 f \lt 3 \); the other options are false.

Teacher's Note:
1. Use \( f(x) + f(-x) \gt = 6 \) and test a linear function.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 2\pi e^{3/4}\lt I\lt 2\pi e^3 \)

Explanation:
1. Let \( s = sin x \) in \( [-1, 1]. s^2 + s + 1 \) has minimum \( 3/4 (at s = -1/2) \) and maximum \( 3 (at s = 1) \).
2. So \( 2\pi e^{3/4} \lt \) I < \( 2\pi e^3 \).

Teacher's Note:
1. Find the range of the exponent.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 25 \)

Explanation:
1. Put \( t = x - \) a - \( \pi \): the integral is \( \int_{-2\pi}^{2\pi}|t|\cos\frac{t+a}{2}dt = 2\cos\frac a2\int_0^{2\pi}t\cos\frac t2dt = -16\cos\frac a2 \).
2. This equals \( -16 \) only when \( \cos(a/2) = 1 \), i.e. a = \( 4k\pi \).
3. \( 0 \le 4k\pi \le 314 \) gives \( k = 0, 1 \), ..., \( 24 \), so there are \( 25 \) values.

Teacher's Note:
1. Count multiples of \( 4 \) pi up to \( 314 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) (ii) and (iii) only

Explanation:
1. \( g(x) = \frac{\sin4x}4 \), so \( g(\pi) = 0 \) and \( g(x + \pi) = \frac{\sin(4x + 4\pi)}4 = g(x) \).
2. Since \( g(\pi) = 0, g(x) = g(x) + g(\pi) = g(x) - g(\pi) \); the quotient is undefined and the product is \( 0 \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (d) \( \frac1{2\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{11}-1} \)

Explanation:
1. Replace \( x \) by \( 1 - x: 5 + 2x - 2x^2 \) is unchanged and \( \frac1{1+e^{2-4x}} + \frac1{1+e^{4x-2}} = 1 \). So I = \( \frac12\int_0^1\frac{dx}{5+2x-2x^2} \).
2. \( 5 + 2x - 2x^2 = \frac{11}2 - 2(x - \tfrac12)^2 \), and \( \int_0^1\frac{dx}{5+2x-2x^2} = \frac1{\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{11}-1} \).
3. Hence I = \( \frac1{2\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{11}-1} \).

Teacher's Note:
1. Use the \( x -\gt 1 - x \) pairing trick.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Reduction: \( I_n = \frac{n-1}nI_{n-2} \), with \( I_0 = \pi \) and \( I_1 = 2 \). Odd \( n \) gives rational numbers \( (a \) rational multiple of \( 2) \); even \( n \) gives a rational multiple of \( \pi \), which is irrational.
2. Since \( 0 \le \sin x \le 1, (\sin x)^{n+1} \le (\sin x)^n \), so the sequence is decreasing.

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (b) \( 2g''=3g^2 \)

Explanation:
1. Since \( f'(x) = (1 + x^3)^{-1/2}, g'(x) = 1/f'(g(x)) = \sqrt{1 + g^3} \).
2. \( g'' = \frac{3g^2g'}{2\sqrt{1+g^3}} = \frac{3g^2}{2} \), so \( 2g'' = 3g^2 \).

Teacher's Note:
1. Differentiate the inverse function twice.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac\pi4 \)

Explanation:
1. Pairing \( x \) and \( -x: \frac1{1+2^x} + \frac1{1+2^{-x}} = 1 \), so I = \( \int_0^{\pi/2}\sin^2x\,dx = \frac\pi4 \).

Teacher's Note:
1. Use the \( 1/(1+2^x) \) pairing trick.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 4 \)

Explanation:
1. \( x^2 - 4x + 3 = (x-1)(x-3) \) is positive on \( [0,1) \) and \( (3,4] \), negative on \( (1,3) \).
2. Areas: \( \frac43 \) on \( [0,1], \frac43 \) on \( [1,3], \frac43 \) on \( [3,4] \).
3. Total = \( 4 \).

Teacher's Note:
1. Factorise and find the sign on each subinterval.

Answer: (a) (2, 4, 8, 0)

Explanation:
1. \( P: f = ax^2 + \) bx with \( a/3 + b/2 = 1 \), i.e. \( 2a + 3b = 6 \), so \( (a, b) = (3, 0) \) or \( (0, 2): P = 2 \).
2. \( Q \): the maximum \( \sqrt2 \) occurs when \( x^2 = \pi/4 + 2k\pi \); with \( x^2 \le 13 \) this gives \( x^2 = \pi/4 \) and \( 9\pi/4 \), so four values of \( x: Q = 4 \).
3. \( R: \int_{-2}^2\frac{3x^2}{1+e^x}dx = \int_0^2 3x^2dx = 8. S \): the numerator integrand is even times odd = odd, so it integrates to \( 0 \) and \( S = 0 \).

Teacher's Note:
1. Evaluate each of the four parts separately.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) (i) and (ii) only

Explanation:
1. \( f(x) = (7\tan^6x - 3\tan^2x)\sec^2x \), so \( \int_0^{\pi/4}f\,dx = \int_0^1(7t^6 - 3t^2)dt = 1 - 1 = 0 \), and an antiderivative is \( h = \tan^7x - \tan^3x \).
2. \( \int_0^{\pi/4}xf = [xh]_0^{\pi/4} - \int_0^{\pi/4}h = -(I_7 - I_3) \), where \( I_n = \int_0^{\pi/4}\tan^nx\,dx \) and \( I_n + I_{n-2} = \frac1{n-1} \).
3. \( I_7 - I_3 = (\frac16 - I_5) - I_3 = \frac16 - \frac14 = -\frac1{12} \), so the integral is \( 1/12 \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (c) \( 0 \)

Explanation:
1. \( f'(x) = \pi/\sqrt{1-x^2} \) is never zero and is defined for every \( |x| \lt 1 \).
2. So there are no critical points.

Teacher's Note:
1. \( f(x) = \) pi \( sin^{-1}x \) is strictly increasing.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \ln\sqrt{\frac32} \)

Explanation:
1. The sum is \( (1/n)\sum 1/((r/n)^2 - 1) \), giving \( \int_2^3\frac{dx}{x^2-1} = \frac12\ln\frac{x-1}{x+1}\Big|_2^3 \).
2. = \( (1/2)[\ln(1/2) - \ln(1/3)] = (1/2)\ln(3/2) \).

Teacher's Note:
1. The limits of \( x = r/n \) are \( 2 \) and \( 3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 7 \)

Explanation:
1. For \( x \ne 0, 2\int_1^2f(tx)dt = (2/x)\int_x^{2x}f(u)du = x + 2 \), so \( \int_x^{2x}f = x(x+2)/2 \).
2. \( \int_0^1 8f(8x)dx = \int_0^8f \), and \( \int_0^8f - \int_0^1f = \int_1^2f + \int_2^4f + \int_4^8f = 3/2 + 4 + 12 = 35/2 \).
3. Subtract \( \int_0^121x\,dx = 21/2: 35/2 - 21/2 = 7 \).

Teacher's Note:
1. Break \( [0,8] \) into doubling intervals.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 8 \)

Explanation:
1. \( f \) and its inverse are mirror images in \( y = x \), so the area between them is twice the area between \( f \) and \( y = x \).
2. The area between \( f \) and \( y = x \) is \( \int_0^{2\pi}|\sin \) x|dx = \( 4 \), so the answer is \( 8 \).

Teacher's Note:
1. Reflect in the line \( y = x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac43 \)

Explanation:
1. Differentiating the functional equations: \( f(x) = 8x - 4x^2 \) and \( g(x) = x^3 - 4x \).
2. On \( [0, 2], |g(x)| = x(4 - x^2) \) and \( f(x) - |g(x)| = x(x-2)^2 \ge 0 \).
3. Area = \( \int_0^2x(x-2)^2dx = \frac43 \).

Teacher's Note:
1. Find \( f \) and \( g \) explicitly first.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac12+\ln3 \)

Explanation:
1. The limits add up to \( 5 \), and replacing \( x \) by \( 5 - x \) swaps \( 4 + x \) with \( 9 - x \).
2. So I = \( (b - a)/2 = (1 + 2\ln3)/2 = 1/2 + \ln 3 \).

Teacher's Note:
1. Use the property that I equals half the interval length.
2. Verify the final value with a quick numerical or limiting check.