JEE Mathematics Definite Integration MCQs Set 29

Answer: (a) \( 3+\ln2 \)

Explanation:
1. The square \( [-1, 1]^2 \) has area \( 4 \). The part with xy > \( 1/2 \) lies in quadrants I and III: \( 2\int_{1/2}^1(1 - \frac1{2x})dx = 1 - \ln2 \).
2. Area = \( 4 - (1 - \ln 2) = 3 + \ln 2 \).

Teacher's Note:
1. Remove two corner regions bounded by the hyperbola.
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. \( T_n = \frac1n\sum_{r=1}^n\frac1{(r/n-1)^2+1} \) is a right-end Riemann sum of the increasing function \( 1/((x-1)^2+1) \) on \( [0, 1]; S_n \) is the left-end sum.
2. So \( S_n \lt \int_0^1 = \pi/4 \lt T_n, T_n - S_n = \frac1{2n} \gt 0 \), and both sums tend to \( \pi/4 \).

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

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

Explanation:
1. Pairing \( x \) and \( -x: \frac{1}{1+e^x} + \frac1{1+e^{-x}} = 1 \), so I = \( \int_0^{\pi/2}x^2\cos x\,dx \).
2. = \( [x^2\sin x + 2x\cos x - 2\sin x]_0^{\pi/2} = \frac{\pi^2}4 - 2 \).

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

Answer: (d) \( 9 \)

Explanation:
1. The line \( x = 2y + 3 \) meets \( y^2 = x \) where \( y^2 - 2y - 3 = 0 \), i.e. \( y = 3 (point (9, 3)) \).
2. Area = \( \int_0^3(2y + 3 - y^2)dy = 9 + 9 - 9 = 9 \).

Teacher's Note:
1. Integrate with respect to \( y \).
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. The area is \( \int_1^t(f(x) - x)dx \), so \( f(t) - t = \frac{d}{dt}\left[t + \sqrt{1+t^2}\right] = 1 + \frac{t}{\sqrt{1+t^2}} \).
2. \( f(x) = x + 1 + \frac{x}{\sqrt{1+x^2}} \).

Teacher's Note:
1. Differentiate the area with respect to \( t \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 0 \)

Explanation:
1. \( \sin(\sin x) \) is odd about \( x = \pi \), so its integral over \( [0, 2\pi] \) is \( 0 \).
2. \( \sin(\cos x) \) changes sign under \( x \to \pi - x (cos x \) becomes -cos \( x) \), so \( \int_0^\pi = 0 \), and similarly on \( [\pi, 2\pi] \).
3. The total is \( 0 \).

Teacher's Note:
1. Test the symmetry \( x -\gt \) pi - \( x \) on \( [0, pi] \) and \( x -\gt 3pi - x \) on \( [pi, 2pi] \).

Answer: (c) \( \pi\ln 2 \)

Explanation:
1. Put \( x = \tan\theta: \int_0^{\pi/2}\ln(\tan\theta + \cot\theta)d\theta \).
2. \( \tan\theta + \cot\theta = \frac{2}{\sin2\theta} \), so the integral is \( \int_0^{\pi/2}(\ln2 - \ln\sin2\theta)d\theta \).
3. Since \( \int_0^{\pi/2}\ln\sin2\theta\,d\theta = -\frac\pi2\ln2 \), the value is \( \frac\pi2\ln2 + \frac\pi2\ln2 = \pi\ln2 \).

Teacher's Note:
1. The substitution \( x = tan(theta) \) absorbs the factor \( 1/(1+x^2) \).

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

Explanation:
1. For large \( x \) the integrand behaves like \( (\lambda - 1/2)/x \), so convergence needs \( \lambda = 1/2 \).
2. The antiderivative is \( \frac14\ln(x^2+1) - \frac12\ln(2x+1) = \frac14\ln\frac{x^2+1}{(2x+1)^2} \), which tends to \( \frac14\ln\frac14 \) at infinity and equals \( \frac14\ln\frac15 \) at \( 2 \).
3. I = \( \frac14\ln\frac14 - \frac14\ln\frac15 = \frac14\ln\frac54 \).

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 (ii) only

Explanation:
1. Mean value theorem for integrals applied to \( h(x) = \sqrt xf(x) \) on \( [1, 9] (length 8) \): I = \( 8h(c) \), so \( (i) \) holds. \( (iii) \) would need length \( 9 \), which is not the interval length.
2. With \( x = t^2 \), I = \( 2\int_1^3t^2f(t^2)dt = 2\cdot2\cdot(\text{average \) of \( }h(t) = t^2f(t^2)) \), and the average of \( h \) lies between two values \( h(p), h(q) (take p = q) \), so \( (ii) \) holds.
3. If \( f \ge 0 \) we only get I \( \ge 0 (f \) may be \( 0) \), so \( (iv) \) 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: (b) \( 4 \)

Explanation:
1. \( I_1 = \int(\sec^2x + 2\tan x)dx = 1 + \ln 2 \).
2. Partial fractions: \( \frac{1}{(1+x)^2(1+x^2)} = \frac{1}{2(1+x)} + \frac{1}{2(1+x)^2} - \frac{x}{2(1+x^2)} \), so \( I_2 = \frac12\ln2 + \frac14 - \frac14\ln2 = \frac{1+\ln2}{4} \).
3. \( I_1/I_2 = 4 \).

Teacher's Note:
1. The two integrals are related by a factor of \( 4 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 0 \)

Explanation:
1. \( dy/dx = 4x^3\ln x^4 - 3x^2\ln x^3 = 16x^3\ln x - 9x^2\ln x \).
2. Both \( x^3\ln x \) and \( x^2\ln x \) tend to \( 0 \) as \( x \to 0^+ \), so the limit is \( 0 \).

Teacher's Note:
1. Powers of \( x \) beat the logarithm.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 2\pi \)

Explanation:
1. Use \( \int_0^\pi xf(x)dx = (\pi/2)\int_0^\pi f(x)dx \) for \( f \) symmetric about \( \pi/2 \).
2. \( \int_0^\pi\sqrt{1+|\cos \) x|}dx = \( 2\int_0^{\pi/2}\sqrt2\cos(x/2)dx = 2\sqrt2\cdot\sqrt2 = 4 \).
3. I = \( (\pi/2)(4) = 2\pi \).

Teacher's Note:
1. The integrand \( f(x) = sqrt(1+|cos x|) \) satisfies \( f(pi - x) = f(x) \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac32-e-\frac1{2e^2} \)

Explanation:
1. Let \( u = (x/e)^x \). Then \( u' = u\ln x \), so \( (x/e)^{2x}\ln x = \frac12(u^2)' \) and \( (e/x)^x\ln x = -(u^{-1})' \).
2. The integral is \( [\frac12u^2 + u^{-1}]_1^e = (\frac12 + 1) - (\frac1{2e^2} + e) = \frac32 - e - \frac1{2e^2} \).

Teacher's Note:
1. Recognise derivatives of \( u^2 \) and \( 1/u \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac12 \)

Explanation:
1. The square \( |x| + |y| \le 1 \) has area \( 2 \). The strip \( |x| \lt 1/2 \) inside it has area \( \int_{-1/2}^{1/2}2(1 - |x|)dx = 3/2 \).
2. Required area = \( 2 - 3/2 = 1/2 \).

Teacher's Note:
1. Subtract the central strip from the whole square.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 2 \)

Explanation:
1. \( \frac1{1+\cos \) x} = \( \frac12\sec^2\frac x2 \), with antiderivative \( \tan\frac x2 \).
2. \( [\tan\frac x2]_{\pi/4}^{3\pi/4} = \tan\frac{3\pi}8 - \tan\frac\pi8 = (\sqrt2+1) - (\sqrt2-1) = 2 \).

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

Answer: (d) \( \frac1{2n+1}-a_n \)

Explanation:
1. \( a_{n+1} + a_n = \int_0^{\pi/4}\tan^{2n}x(\tan^2x + 1)dx = \int_0^{\pi/4}\tan^{2n}x\sec^2x\,dx = \frac1{2n+1} \).
2. So \( a_{n+1} = \frac1{2n+1} - a_n \).

Teacher's Note:
1. Combine consecutive terms using \( sec^2 = 1 + tan^2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac{n}{n^2-1} \)

Explanation:
1. Put \( x = \sinh t \), so \( x + \sqrt{1+x^2} = e^t \) and \( dx = \cosh t\,dt \).
2. \( \int_0^\infty e^{-nt}\cosh t\,dt = \frac12\left(\frac1{n-1} + \frac1{n+1}\right) = \frac{n}{n^2-1} \).

Teacher's Note:
1. Hyperbolic substitution.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( -\frac{10}{21} \)

Explanation:
1. The integrand simplifies to \( x^{1/6} - x^{-1/4} \).
2. Integrating: \( (6/7)x^{7/6} - (4/3)x^{3/4} \) from \( 0 \) to \( 1 \) gives \( 6/7 - 4/3 \).
3. \( 6/7 - 4/3 = (18 - 28)/21 = -10/21 \).

Teacher's Note:
1. Write every root as a power of \( x \) before integrating.
2. Both terms are integrable at \( 0 \) because the exponents are greater than \( -1 \).

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

Explanation:
1. \( x^2 - 4x + 3 = (x-1)(x-3) \) is negative only between \( 1 \) and \( 3 \).
2. \( f(a, 3) \) decreases while the integrand is positive \( (a \lt 1) \) and increases for \( 1 \lt \) a < \( 3 \), so the least value is at a = \( 1 \). For \( b \ge 4 \) the integrand is positive, so \( f(4, b) \) increases.
3. \( f(0, b) \) is least at \( b = 3 (not 2) \). The overall minimum is \( \int_1^3 = -4/3 \).

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

Answer: (d) \( \frac12 \)

Explanation:
1. Write the integrand as \( \frac{3\sec^2\theta}{(1+\sqrt{\tan\theta})^5} \) and put \( u = \sqrt{\tan\theta}: \sec^2\theta\,d\theta = 2u\,du \).
2. I = \( \int_0^\infty\frac{6u}{(1+u)^5}du = 6B(2,3) = 6\cdot\frac{1!\,2!}{4!} = \frac12 \).

Teacher's Note:
1. With \( u = sqrt(tan) \) the integral becomes a beta integral.
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. Integrate \( I_n \) by parts: \( I_n = 2^{-n} + 2n(I_n - I_{n+1}) \), which rearranges to \( 2nI_{n+1} = 2^{-n} + (2n-1)I_n \).
2. With \( I_1 = \pi/4: 2I_2 = 1/2 + \pi/4 \), so \( I_2 = \pi/8 + 1/4 \). Then \( 4I_3 = 1/4 + 3I_2 \) gives \( I_3 = 3\pi/32 + 1/4 \).

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

Answer: (b) \( -\frac{3e^x}{2(e-1)}-3x \)

Explanation:
1. Let A = \( \int_0^1f \) and \( B = \int_0^1yf(y)dy \). Then \( f(x) = (1 + B)e^x + \) Ax.
2. \( B = (1+B)\int_0^1ye^ydy + A/3 = (1+B) + A/3 \), so A = \( -3 \). Also A = \( (1+B)(e - 1) + A/2 \) gives \( 1 + B = -\frac{3}{2(e-1)} \).
3. \( f(x) = -\frac{3e^x}{2(e-1)} - 3x \).

Teacher's Note:
1. \( f \) has the form \( (constant)e^x + (constant)x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{23}6 \)

Explanation:
1. \( x^2 + 1 \le x + 1 \) for \( 0 \le x \le 1 \) and the opposite for \( x \ge 1 \).
2. Area = \( \int_0^1(x^2+1)dx + \int_1^2(x+1)dx = \frac43 + \frac52 = \frac{23}6 \).

Teacher's Note:
1. The smaller of the two curves bounds the region.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 0 \)

Explanation:
1. \( g'(a) = -\pi^2\cos(\pi a)/\sin^2(\pi a) \).
2. At a = \( 1/2, \cos(\pi/2) = 0 \), so \( g'(1/2) = 0 \).

Teacher's Note:
1. The function pi \( / sin(pi a) \) has its minimum at a = \( 1/2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 1 \)

Explanation:
1. Mean value = \( (1/2)\ln((2 + c)/c) = (1/2)\ln 3 \).
2. \( (2 + c)/c = 3 \), so \( c = 1 \).

Teacher's Note:
1. Set \( (2+c)/c = 3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \ln 4 \)

Explanation:
1. Put \( e^x - 1 = u^2 \): the integral is \( 2\tan^{-1}\sqrt{e^x-1} + \) const.
2. \( 2[\tan^{-1}\sqrt{e^x-1} - \pi/4] = \pi/6 \) gives \( \tan^{-1}\sqrt{e^x-1} = \pi/3 \).
3. \( e^x - 1 = 3 \), so \( x = \ln 4 \).

Teacher's Note:
1. The antiderivative is \( 2 tan^{-1} sqrt(e^x - 1) \).
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. Partial fractions: \( x/((1+x)(1+x^2)) = -1/(2(1+x)) + (1+x)/(2(1+x^2)) \); the logarithmic parts cancel at infinity, leaving \( \frac12\tan^{-1}x \), so \( J = \pi/4 \).
2. Putting \( x = 1/t \) turns \( J \) into \( \int_0^\infty dt/((1+t)(1+t^2)) \), so \( (iii) \) is true.

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