JEE Mathematics Definite Integral MCQs Set 07

Answer: (c) \( \pi \)

Explanation:
1. Using \( x\to\pi-x \): \( \int_0^\pi xf(\sin x)dx=\frac\pi2\int_0^\pi f(\sin x)dx \).
2. By symmetry about \( \frac\pi2 \), \( \int_0^\pi f(\sin x)dx=2\int_0^{\pi/2}f(\sin x)dx \).
3. So \( A=\pi \).

Teacher's Note:
1. Apply the property of definite integrals f(x) to f(a - x).
2. Sin is symmetric about pi/2.

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

Explanation:
1. Adding the integral with \( x\to-x \) (since \( \frac1{1+\pi^x}+\frac1{1+\pi^{-x}}=1 \)) gives \( I_n=\int_0^\pi\frac{\sin nx}{\sin x}dx \).
2. \( I_{n+2}-I_n=\int_0^\pi2\cos(n+1)x\,dx=0 \) (i).
3. So \( I_{odd}=I_1=\pi \) and \( I_{even}=I_0=0 \): (ii) and (iii) hold; (iv) fails since \( I_1\ne I_2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( 2g(1) \)

Explanation:
1. \( \int_{-1}^1g'(x)dx=g(1)-g(-1) \).
2. Replacing \( (x,y) \) by \( (-x,-y) \) leaves \( y^3-3y+x=0 \) unchanged, so \( g \) is odd: \( g(-1)=-g(1) \) and the value is \( 2g(1) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac\pi8(1+\sqrt2) \)

Explanation:
1. Here \( a=0,b=\frac\pi2,c=\frac\pi4 \).
2. Value \( =\frac{\pi/2}{4}\left\{0+1+2\cdot\frac{\sqrt2}2\right\}=\frac\pi8(1+\sqrt2) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) an odd function

Explanation:
1. The integrand \( g(t)=\frac{t^2}{e^t+e^{-t}} \) is even, since \( g(-t)=g(t) \).
2. Then \( f(-x)=\int_0^{-x}g(t)dt=-\int_0^xg(u)du=-f(x) \), so \( f \) is odd.

Teacher's Note:
1. The integral from 0 of an even function is odd.
2. Substitute t = -u to check.

Answer: (b) \( e-1 \)

Explanation:
1. The sum is \( \frac1n\sum_{k=1}^{n}e^{k/n} \), a Riemann sum.
2. Its limit is \( \int_0^1e^x\,dx=e-1 \).

Teacher's Note:
1. Rewrite each n-th root as e to the power k/n.
2. Recognise the limit of a sum as a definite integral.

Answer: (d) 2

Explanation:
1. The curves meet at \( x=\pm1 \) (where \( |x|=1 \)).
2. Area \( =\int_{-1}^{1}\left[(1-|x|)-(|x|-1)\right]dx=2\int_{-1}^1(1-|x|)dx=2 \) (a square of diagonal 2).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \left(\frac34,3\right) \)

Explanation:
1. Here \( t\in(\frac12,\frac34) \) (approximately 0.606) and \( f>0 \) on \( [0,t] \).
2. Area \( =\int_0^tf=t+t^2+t^3+t^4\approx1.33 \), which lies in \( (\frac34,3) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac\pi2 \)

Explanation:
1. Using \( x\to-x \): \( I=\int\frac{a^x\cos^2x}{1+a^x}dx \). Adding, \( 2I=\int_{-\pi}^{\pi}\cos^2x\,dx=\pi \).
2. So \( I=\frac\pi2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac\pi{12} \)

Explanation:
1. Put \( u=x^3 \), \( du=3x^2dx \).
2. The integral becomes \( \frac13\int_0^1\frac{du}{1+u^2}=\frac13\cdot\frac\pi4=\frac\pi{12} \).

Teacher's Note:
1. Substitute u = x cubed to reach the arctan form.
2. The factor x squared is the derivative of x cubed divided by 3.

Answer: (c) \( \frac12\log\left(\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+c \)

Explanation:
1. \( J=\int\frac{e^{3x}}{1+e^{2x}+e^{4x}}dx \), so \( J-I=\int\frac{e^{3x}-e^x}{e^{4x}+e^{2x}+1}dx \).
2. Put \( u=e^x \): \( \int\frac{u^2-1}{u^4+u^2+1}du=\int\frac{1-u^{-2}}{u^2+1+u^{-2}}du \); with \( w=u+\frac1u \) this is \( \int\frac{dw}{w^2-1}=\frac12\log\frac{w-1}{w+1} \).
3. Back-substituting: \( \frac12\log\frac{u^2-u+1}{u^2+u+1}+c \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 1

Explanation:
1. Integrating by parts: \( \int\cos^{-1}x\,dx=x\cos^{-1}x-\sqrt{1-x^2} \).
2. Between 0 and 1: \( (0-0)-(0-1)=1 \).

Teacher's Note:
1. Use integration by parts or the area interpretation.
2. The area under arccos on [0,1] equals the area under cos on [0, pi/2], which is 1.

Answer: (a) 1

Explanation:
1. Write \( \int_0^2f=\int_0^1f(x)dx+\int_0^1f(x+1)dx=\int_0^1\left[f(x)+f(x+1)\right]dx=1 \).

Teacher's Note:
1. Work from the given condition to find the function or quantity, then substitute.
2. Verify the answer on a simple case.

Answer: (d) \( \int_0^{2a}f(x)\,dx \)

Explanation:
1. Put \( t=2a-x \) in the second integral: \( \int_0^af(2a-x)dx=\int_a^{2a}f(t)dt \).
2. Adding gives \( \int_0^{2a}f(x)dx \).

Teacher's Note:
1. Substitute 2a - x to shift the limits.
2. The two pieces join end to end.

Answer: (c) \( \frac{f(b)-f(a)}{b-a} \)

Explanation:
1. \( F'(c)=\frac12\{f(a)-f(b)\}+\frac{b-a}2f'(c) \) (the terms in \( f(c) \) cancel).
2. Setting \( F'(c)=0 \): \( f'(c)=\frac{f(b)-f(a)}{b-a} \) (the mean value point; \( F''<0 \) gives a maximum as \( f''<0 \)).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac{1}{2x^2}\sqrt{2x^4-2x^2+1}+c \)

Explanation:
1. Put \( t=\frac1{x^2} \), \( dt=-\frac{2}{x^3}dx \). The integrand becomes \( -\frac12\cdot\frac{1-t}{\sqrt{t^2-2t+2}}dt=\frac12\cdot\frac{t-1}{\sqrt{t^2-2t+2}}dt \).
2. Integral \( =\frac12\sqrt{t^2-2t+2}+c=\frac{1}{2x^2}\sqrt{2x^4-2x^2+1}+c \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( -\frac12 \)

Explanation:
1. \( [x]=-1 \) on \( \left[-\frac12,0\right) \) and \( 0 \) on \( \left[0,\frac12\right] \), so \( \int[x]\,dx=-\frac12 \).
2. The logarithm term is an odd function, so its integral over a symmetric interval is 0.
3. Total \( -\frac12 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. By induction \( f^{(k)}(x)=\frac{x}{(1+kx^n)^{1/n}} \), so \( g(x)=\frac{x}{(1+nx^n)^{1/n}} \).
2. Then \( \int x^{n-2}g\,dx=\int x^{n-1}(1+nx^n)^{-1/n}dx \). Put \( u=1+nx^n \), \( du=n^2x^{n-1}dx \).
3. Result \( \frac1{n^2}\cdot\frac{u^{1-1/n}}{1-1/n}=\frac{1}{n(n-1)}(1+nx^n)^{1-\frac1n}+K \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( 3I \)

Explanation:
1. Put \( u=2x \): \( \int_3^{3+3T}f(2x)dx=\frac12\int_6^{6+6T}f(u)\,du \).
2. The interval has length \( 6T \), i.e. 6 periods, so the integral is \( 6I \) and the result is \( \frac12\cdot6I=3I \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( 2x\sin x \)

Explanation:
1. By the Leibniz rule, \( \frac{dy}{dx}=\sin\sqrt{x^2}\cdot\frac{d}{dx}(x^2) \).
2. This is \( \sin x\cdot2x=2x\sin x \) (taking x positive).

Teacher's Note:
1. Differentiate under an integral with a variable upper limit: substitute the limit and multiply by its derivative.
2. Don't forget the chain-rule factor 2x.

Answer: (d) 1

Explanation:
1. Let \( A=\int_0^1(e^{x^2}-1)dx \) (area between \( y=1 \) and \( y=e^{x^2} \)). The rectangle \( [0,1]\times[1,e] \) has area \( e-1 \) and is split by the curve into \( A \) and \( \int_1^e\sqrt{\ln y}\,dy \) (the curve is \( x=\sqrt{\ln y} \)).
2. So \( \int_0^1e^{x^2}dx+\int_1^e\sqrt{\ln x}\,dx=1+A+(e-1-A)=e \).
3. Also \( \int_1^e(-1)\,dx=-(e-1) \). Total \( e-(e-1)=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 2

Explanation:
1. \( f(a)+f(-a)=1 \), so the limits are symmetric about \( \frac12 \).
2. Replacing \( x \) by \( 1-x \) leaves \( g\{x(1-x)\} \) unchanged: \( I_1=\int(1-x)g\,dx=I_2-I_1 \).
3. So \( I_2=2I_1 \) and the ratio is 2.

Teacher's Note:
1. Check that the limits add up to 1.
2. Use the substitution x to 1 - x.

Answer: (c) \( \frac12 \)

Explanation:
1. Since \( f(t+n)=f(t) \), \( \phi(x)=\int_0^xf(t)\,dt \) and \( \phi'(x)=f(x) \).
2. So \( \phi'\left(\tfrac32\right)=f\left(\tfrac32\right)=f\left(\tfrac12\right)=\tfrac12 \).

Teacher's Note:
1. Periodicity removes the shift n inside the integral.
2. Reduce 3/2 modulo 1 to 1/2.

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

Explanation:
1. \( \frac{n}{n^2+kn+k^2}=\frac1n\cdot\frac1{1+\frac kn+\frac{k^2}{n^2}} \), so both sums approximate \( \int_0^1\frac{dx}{1+x+x^2}=\frac{\pi}{3\sqrt3} \).
2. The integrand is decreasing: the right-endpoint sum \( S_n \) is below the integral and the left-endpoint sum \( T_n \) is above it.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) Statement 1 is false and Statement 2 is true

Explanation:
1. \( F(x)=\frac x2-\frac{\sin2x}{4}+C \), so \( F(x+\pi)=F(x)+\frac\pi2\ne F(x) \): Statement 1 is false.
2. Statement 2 is true because \( \sin(x+\pi)=-\sin x \) and squaring removes the sign.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \frac{\pi}{2\sqrt2}\log_e(\sqrt2+1) \)

Explanation:
1. \( \sqrt{1+\sin2x}=\sin x+\cos x \) on the interval.
2. Using \( x\to\frac\pi2-x \): \( I=\frac\pi4\int_0^{\pi/2}\frac{dx}{\sin x+\cos x} \).
3. \( \int_0^{\pi/2}\frac{dx}{\sin x+\cos x}=\frac1{\sqrt2}\int_0^{\pi/2}\sec\left(x-\frac\pi4\right)dx=\sqrt2\ln(\sqrt2+1) \).
4. Hence \( I=\frac{\pi}{4}\cdot\sqrt2\ln(\sqrt2+1)=\frac{\pi}{2\sqrt2}\ln(\sqrt2+1) \).

Teacher's Note:
1. Use the property of definite integrals f(x) to f(a - x) and average the two forms.
2. Numerically the value is about 0.979.

Answer: (b) 1

Explanation:
1. The bracket times \( (\sin x)^{\cos x} \) is the derivative of \( (\sin x)^{\cos x} \).
2. So the integral is \( \left[(\sin x)^{\cos x}\right]_0^{\pi/2}=1-\lim_{x\to0^+}(\sin x)^{\cos x}=1-0=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 4

Explanation:
1. Differentiate: \( 2x\,f(x^2)=2x+3x^2 \), so \( f(x^2)=1+\frac{3x}{2} \).
2. Put \( x=2 \): \( f(4)=1+3=4 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac43 \)

Explanation:
1. They meet where \( -4y^2=1-5y^2 \): \( y=\pm1 \).
2. Area \( =\int_{-1}^{1}\left[(1-5y^2)-(-4y^2)\right]dy=\int_{-1}^1(1-y^2)dy=\frac43 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.