JEE Mathematics Definite Integration MCQs Set 30

Answer: (d) \( 3\pi+1 \)

Explanation:
1. \( x sin(\pi x) \ge 0 \) on \( [-1, 1] \) and \( \le 0 \) on \( [1, 3/2] \).
2. An antiderivative is \( -x cos(\pi x)/\pi + sin(\pi x)/\pi^2: \int_{-1}^1 = 2/\pi \) and \( \int_1^{3/2} = -1/\pi - 1/\pi^2 \).
3. Total = \( 2/\pi + 1/\pi + 1/\pi^2 = (3\pi + 1)/\pi^2 \).

Teacher's Note:
1. Take absolute values piece by piece.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( -1 \)

Explanation:
1. \( \int_0^{\pi/2}x\cos x\,dx = \frac\pi2 - 1 = \frac{\pi-2}2 \), so the second term is \( 2a \).
2. \( \int_0^{\pi/2}(\sin x + a\cos x)^3dx = \frac23 + \) a + \( a^2 + \frac23a^3 \). The equation becomes \( 2a^3 + 3a^2 - 3a - 4 = 0 \), i.e. \( (a+1)(2a^2 + \) a - \( 4) = 0 \).
3. The only integer root is a = \( -1 \).

Teacher's Note:
1. Reduce to a cubic and factor.
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. Replacing \( x \) by \( x + 2: f(x+2) + f(x+4) = 10 \), so \( f(x+4) = f(x): f \) has period \( 4 \).
2. \( \int_{-1}^{1}f + \int_1^3f = \int_{-1}^1(f(x) + f(x+2))dx = 20 \) for each block of length \( 4 \), and \( [-1, 7] \) holds two blocks: \( 40 \).

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

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

Explanation:
1. \( sin^2 x \) has period \( \pi \) and is symmetric about \( \pi/2 \), so \( (i) \) and \( (ii) \) hold; \( \int_0^{2\pi}\cos^2x\,dx = \int_0^{2\pi}\sin^2x\,dx = \pi \), so \( (iii) \) holds.
2. \( \int_0^{\pi/4}\sin^2x\,dx = \pi/8 - 1/4 \), so \( 8 \) times it is \( \pi - 2 \ne \pi (iv \) is \( false) \).

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

Answer: (d) (i), (ii) and (iii) only

Explanation:
1. \( f'(x) = F(x) + xF'(x) \). Since \( F \) decreases from \( F(1) = 0, F(x) \lt 0 \) on \( (1, 3] \), and \( F' \lt 0 \), so \( f'(x) \lt 0 \) on \( [1, 3) \).
2. Hence \( f'(1) = F'(1) \lt 0, f(2) = 2F(2) \lt 0 \), and \( f' \) never vanishes on \( (1, 3) \).

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

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

Explanation:
1. \( L = \int_0^1\frac{dx}{(x+a)(x+b)} \).
2. For a \( \ne b: = \frac1{b-a}\ln\frac{(1+a)b}{(1+b)a} = \frac1{a-b}\ln\frac{a(b+1)}{b(a+1)} \). For a = \( b: \int_0^1\frac{dx}{(x+a)^2} = \frac1a - \frac1{1+a} = \frac1{a(1+a)} \).

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

Answer: (b) \( \pi \)

Explanation:
1. Complete the square: \( x^2 + 2x + 2 = (x+1)^2 + 1 \).
2. The integral is \( [\tan^{-1}(x+1)] \) from \( -\infty \) to \( \infty \).
3. = \( \pi/2 - (-\pi/2) = \pi \).

Teacher's Note:
1. Shift the variable by \( 1 \); the limits stay infinite.

Answer: (c) \( f(x)+2f\left(\frac\pi2\right) \)

Explanation:
1. The integrand \( 2 + sin^2 3t \) has period \( \pi/3 \), so \( f(x + \pi) = f(x) + f(\pi) \).
2. \( f(\pi) = 2\pi + \pi/2 = 5\pi/2 \) and \( f(\pi/2) = \pi + \pi/4 = 5\pi/4 \), so \( f(\pi) = 2f(\pi/2) \).

Teacher's Note:
1. The extra piece \( f(pi) \) equals \( 2 f(pi/2) \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \pi \)

Explanation:
1. \( g(x) = (x + x^3)/(1 + x^4) \) is odd and \( \cot^{-1}(-u) = \pi - \cot^{-1}u \).
2. So \( \cot^{-1}g(x) + \cot^{-1}g(-x) = \pi \), and I = \( (1/2)\int_{-1}^1\pi\,dx = \pi \).

Teacher's Note:
1. Pairing \( x \) and \( -x \) gives pi every time.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 3 \)

Explanation:
1. As \( x \to \pm\infty, f(x) \to 1 \), giving one horizontal asymptote \( y = 1 \).
2. The denominator vanishes at \( x = 1 \) and \( x = 2 (numerator non-zero) \), giving two vertical asymptotes. Total \( 3 \).

Teacher's Note:
1. One horizontal and two vertical.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 2\pi \)

Explanation:
1. On \( (\pi/2, \pi], f(x) = 2 - \sin(\pi - x) = 2 - \sin x \).
2. \( \int_0^{\pi/2}\sin x\,dx = 1 \) and \( \int_{\pi/2}^{\pi}(2 - \sin x)dx = \pi - 1 \), so \( [0, \pi] \) gives \( \pi \).
3. The second half is its mirror image, so the total is \( 2\pi \).

Teacher's Note:
1. One half contributes pi.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 22 \)

Explanation:
1. For an increasing \( f: \int_4^{10}f^{-1}(x)dx = 6\cdot10 - 2\cdot4 - \int_2^6f(x)dx (areas \) of rectangles minus the area under \( f) \).
2. = \( 60 - 8 - 30 = 22 \).

Teacher's Note:
1. Use the rectangle method for inverse functions.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 0 \)

Explanation:
1. \( f(x^2/4) \) and \( g(x^2/4) \) are even; \( f(x) - f(-x) \) is odd and \( g(x) + g(-x) \) is even.
2. The integrand is odd and the limits are symmetric, so the integral is \( 0 \).

Teacher's Note:
1. An odd integrand over symmetric limits gives \( 0 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \ln2-1 \)

Explanation:
1. The limit is \( \int_0^{\pi/2}\cos x\ln\cos x\,dx \). Integrating by parts \( (or \) with \( u = \sin x): = \int_0^1\frac12\ln(1-u^2)du \).
2. \( \int_0^1\ln(1-u)du = -1 \) and \( \int_0^1\ln(1+u)du = 2\ln2 - 1 \), so the value is \( \frac12(2\ln2 - 2) = \ln2 - 1 \).

Teacher's Note:
1. Put \( u = sin x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{\pi}{2} \)

Explanation:
1. Divide numerator and denominator by \( \cos^4\theta \); with \( t = \tan^2\theta, \sin2\theta\,d\theta/\cos^4\theta = \sec^2\theta\,d(\tan^2\theta) \).
2. The integral becomes \( \int_0^\infty\frac{dt}{t^2+1} \).
3. = \( \pi/2 \).

Teacher's Note:
1. Introduce \( t = tan^2(theta) \), so \( dt = 2 tan(theta) sec^2(theta) d(theta) \).

Answer: (c) a local maximum at \( \pi \) and a local minimum at \( 2\pi \)

Explanation:
1. \( f'(x) = \sqrt x\sin x \): positive on \( (0, \pi) \), negative on \( (\pi, 2\pi) \), positive on \( (2\pi, 5\pi/2) \).
2. So \( f \) has a local maximum at \( \pi \) and a local minimum at \( 2\pi \).

Teacher's Note:
1. Look at the sign of \( f' \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) Statement I is false; Statement II is true

Explanation:
1. Statement II is a standard property and is true.
2. By it, \( 2I = \int_{\pi/6}^{\pi/3}\left[\frac1{1+\sqrt{\tan \) x}} + \( \frac1{1+\sqrt{\cot x}}\right]dx = \frac\pi3 - \frac\pi6 = \frac\pi6 \), so I = \( \pi/12 \ne \pi/6 \); Statement I is false.

Teacher's Note:
1. The integral is half the interval length, \( pi/12 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 3+\cos1+\cos2+\cos3 \)

Explanation:
1. \( \cos^{-1}x \) takes values in \( [0,\pi] \); its integer part is \( 3 \) for \( x \le \cos3, 2 \) for \( \cos3 \lt x \le \cos2, 1 \) for \( \cos2 \lt x \le \cos1, 0 \) above.
2. Integral = \( 3(\cos3+1) + 2(\cos2-\cos3) + 1(\cos1-\cos2) \).
3. = \( 3 + \cos1 + \cos2 + \cos3 \).

Teacher's Note:
1. Remember \( cos \) is decreasing on \( [0, pi] \), so large integer parts correspond to small \( x \).

Answer: (b) All of (i), (ii), (iii) and (iv)

Explanation:
1. The curves meet at \( (1/\sqrt2, \pi/4) \). Integrating with respect to \( x \) under the lower curve gives \( (i) \); integrating horizontally between \( x = \sin y \) and \( x = \cos y \) gives \( (iii) = \sqrt2 - 1 \).
2. \( \int_{\pi/4}^{\pi/2}(\sin y - \cos y)dy = \sqrt2 - 1 \) as well, and \( (i) \) evaluates to \( (\pi/(4\sqrt2) + 1/\sqrt2 - 1) + (1/\sqrt2 - \pi/(4\sqrt2)) = \sqrt2 - 1 \).

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

Answer: (c) \( \frac{2\sin x^2}{x\cos y^2} \)

Explanation:
1. Differentiate both sides: \( \cos(y^2)\,y' = (\sin x^2/x^2)\cdot 2x \).
2. So \( y' = 2\sin x^2/(x\cos y^2) \).

Teacher's Note:
1. Leibnitz rule on both sides.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac\pi3 \)

Explanation:
1. Divide by \( x^2: (1 + 1/x^2)/(x^2 + 1/x^2 + 7) \). Put \( t = x - 1/x, dt = (1 + 1/x^2)dx \).
2. \( x^2 + 1/x^2 = t^2 + 2 \), so the integral is \( \int_{-\infty}^{\infty}dt/(t^2 + 9) = \pi/3 \).

Teacher's Note:
1. \( t = x - 1/x \) converts it to the integral of \( 1/(t^2+9) \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 1 \)

Explanation:
1. Use \( 1/(t(1+t^2)) = 1/t - t/(1+t^2) \). The sum becomes \( \frac12\ln(1+\tan^2x) - \frac12\ln(1+\cot^2x) + \ln(\cot x) + 1 \).
2. \( \frac12\ln\sec^2x = -\ln\cos x \) and \( \frac12\ln\csc^2x = -\ln\sin x \), so the logarithms cancel, leaving \( 1 \).

Teacher's Note:
1. All the x-dependence cancels.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{27}{16} \)

Explanation:
1. For large \( n, {}^{3n}C_n \sim (27/4)^n \) and \( {}^{2n}C_n \sim 4^n (up \) to polynomial factors, which vanish after taking the n-th \( root) \).
2. The limit is \( \frac{27/4}{4} = \frac{27}{16} \).

Teacher's Note:
1. Use Stirling's approximation.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 4 \)

Explanation:
1. On \( [n, n+1) \) put \( x = n + s \). For odd \( n, f = s \) and \( \cos\pi x = -\cos\pi s \); for even \( n, f = 1 - s \) and \( \cos\pi x = \cos\pi s \).
2. \( \int_0^1s\cos\pi s\,ds = -2/\pi^2 \), so each odd interval gives \( 2/\pi^2 \) and each even interval gives \( \int_0^1(1-s)\cos\pi s\,ds = 2/\pi^2 \).
3. There are \( 20 \) intervals: \( \int = 40/\pi^2 \), and \( \frac{\pi^2}{10}\cdot\frac{40}{\pi^2} = 4 \).

Teacher's Note:
1. Each unit interval contributes \( 2/pi^2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac12 \)

Explanation:
1. \( R_1 = [1 - (1-b)^3]/3 \) and \( R_2 = (1-b)^3/3 \).
2. \( R_1 - R_2 = [1 - 2(1-b)^3]/3 = 1/4 \), so \( (1-b)^3 = 1/8 \) and \( b = 1/2 \).

Teacher's Note:
1. Write both areas with \( (1-b)^3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac{3\pi a^2}{8} \)

Explanation:
1. The curve is the astroid; by symmetry the area is \( 4\int_0^{\pi/2}y\left|\frac{dx}{dt}\right|dt = 4\int_0^{\pi/2}3a^2\sin^4t\cos^2t\,dt \).
2. \( \int_0^{\pi/2}\sin^4t\cos^2t\,dt = \pi/32 \), so the area is \( 12a^2\pi/32 = 3\pi a^2/8 \).

Teacher's Note:
1. Four equal quadrants.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( g\text{ is decreasing on }(1,\infty) \)

Explanation:
1. \( g'(x) = h(x)f(x) \), where \( h(x) = \frac{2(x-1)}{x+1} - \ln x \) and \( f(x) \gt 0 \).
2. \( h(1) = 0 \) and \( h'(x) = \frac4{(x+1)^2} - \frac1x = -\frac{(x-1)^2}{x(x+1)^2} \lt 0 \), so \( h \lt 0 \) for \( x \gt 1 \).
3. Hence \( g' \lt 0 \) and \( g \) is decreasing on \( (1, \infty) \).

Teacher's Note:
1. The sign of \( g' \) is the sign of \( h \).
2. Verify the final value with a quick numerical or limiting check.