JEE Mathematics Definite Integration MCQs Set 37

Answer: (d) \( \log_e1 \)

Explanation:
1. Here \( g(f(x))=\log_e\dfrac{2-x\cos x}{2+x\cos x} \) and \( f(-x)=\dfrac{2+x\cos x}{2-x\cos x}=\dfrac1{f(x)} \).
2. So \( g(f(-x))=-g(f(x)) \): the integrand is odd, and the integral over a symmetric interval is \( 0=\log_e1 \). Answer: (d)

Teacher's Note:
1. Test the integrand for odd symmetry.

Answer: (c) \( \dfrac{\pi-1}{4} \)

Explanation:
1. Use \( \int_0^a f(x)dx=\int_0^a f(a-x)dx \): the integral also equals \( \int_0^{\pi/2}\dfrac{\cos^3x}{\sin x+\cos x}dx \).
2. Adding, \( 2I=\int_0^{\pi/2}(1-\sin x\cos x)\,dx=\dfrac\pi2-\dfrac12 \), so \( I=\dfrac{\pi-1}{4} \). Answer: (c)

Teacher's Note:
1. Add the integral to its reflected form.

Answer: (a) \( \dfrac\pi4-\dfrac12\log_e2 \)

Explanation:
1. Put \( t=x^2 \): \( I=\frac12\int_0^1\cot^{-1}(1-t+t^2)\,dt=\frac12\int_0^1\tan^{-1}\dfrac{1}{1-t+t^2}\,dt \).
2. Since \( \dfrac1{1-t+t^2}=\dfrac{t-(t-1)}{1+t(t-1)} \), the integrand is \( \tan^{-1}t-\tan^{-1}(t-1) \), and \( \int_0^1\tan^{-1}(t-1)dt=-\int_0^1\tan^{-1}t\,dt \).
3. So \( I=\int_0^1\tan^{-1}t\,dt=\dfrac\pi4-\dfrac12\log_e2 \). Answer: (a)

Teacher's Note:
1. Write the inverse cotangent as a difference of two inverse tangents.

Answer: (c) \( \dfrac34(2)^{4/3}-\dfrac34 \)

Explanation:
1. The sum is \( \dfrac1n\sum_{r=1}^n\left(1+\dfrac rn\right)^{1/3} \), which tends to \( \int_0^1(1+x)^{1/3}dx \).
2. \( \int_0^1(1+x)^{1/3}dx=\dfrac34\left[2^{4/3}-1\right] \). Answer: (c)

Teacher's Note:
1. Recognise a Riemann sum.

Answer: (b) \( 3^{7/6}-3^{5/6} \)

Explanation:
1. As in the indefinite case the antiderivative is \( -3\tan^{-1/3}x \).
2. Value \( =-3\left[(\sqrt3)^{-1/3}-(1/\sqrt3)^{-1/3}\right]=3\left(3^{1/6}-3^{-1/6}\right)=3^{7/6}-3^{5/6} \). Answer: (b)

Teacher's Note:
1. Substitute \( t=\tan x \) and change the limits.

Answer: (a) \( -2 \)

Explanation:
1. Partial fractions: \( \dfrac1{x+\alpha}-\dfrac1{x+\alpha+1} \), so the integral is \( \log_e\dfrac{(2\alpha+1)^2}{2\alpha(2\alpha+2)} \).
2. Setting \( \dfrac{(2\alpha+1)^2}{4\alpha(\alpha+1)}=\dfrac98 \) gives \( \alpha^2+\alpha-2=0 \), so \( \alpha=1 \) or \( -2 \); only \( -2 \) is in the options. Answer: (a)

Teacher's Note:
1. Use partial fractions.

Answer: (d) \( \dfrac43 \)

Explanation:
1. \( |\cos x| \) is symmetric about \( x=\pi/2 \), so the integral is \( 2\int_0^{\pi/2}\cos^3x\,dx \).
2. With \( \cos^3x=\frac14(\cos3x+3\cos x) \): \( 2\cdot\frac14\left[\frac{\sin3x}3+3\sin x\right]_0^{\pi/2}=\frac12\left(-\frac13+3\right)=\dfrac43 \). Answer: (d)

Teacher's Note:
1. Use symmetry about \( \pi/2 \).

Answer: (d) \( (-\sqrt2,\sqrt2) \)

Explanation:
1. \( x^4-2x^2=x^2(x-\sqrt2)(x+\sqrt2) \) is negative exactly on \( (-\sqrt2,\sqrt2) \) (except at \( 0 \)).
2. The integral is smallest when the interval covers all of the region where the integrand is negative, so \( (a,b)=(-\sqrt2,\sqrt2) \). Answer: (d)

Teacher's Note:
1. Find where the integrand is negative.

Answer: (a) \( \dfrac{24}{25} \)

Explanation:
1. Differentiate: \( f(x)=2x-x^2f(x) \), so \( f(x)=\dfrac{2x}{1+x^2} \).
2. \( f'(x)=\dfrac{2(1-x^2)}{(1+x^2)^2} \), and at \( x=\frac12 \) this is \( \dfrac{2\cdot\frac34}{\frac{25}{16}}=\dfrac{24}{25} \). Answer: (a)

Teacher's Note:
1. Differentiate both sides with respect to \( x \).

Answer: (d) \( 0 \)

Explanation:
1. For non-integer \( y \), \( [-y]=-1-[y] \). So with \( f(x)=\dfrac{\sin^2x}{[x/\pi]+1/2} \), \( f(-x)=\dfrac{\sin^2x}{-1-[x/\pi]+1/2}=-f(x) \) (apart from isolated points).
2. The integrand is odd and the interval is symmetric, so the integral is \( 0 \). Answer: (d)

Teacher's Note:
1. Test the integrand for odd or even symmetry.

Answer: (c) \( 2\displaystyle\int_0^af(x)\,dx \)

Explanation:
1. Replace \( x \) by \( a-x \): \( I=\int_0^af(x)g(a-x)\,dx=\int_0^af(x)[4-g(x)]\,dx=4\int_0^af-I \).
2. So \( 2I=4\int_0^af(x)\,dx \) and \( I=2\int_0^af(x)\,dx \). Answer: (c)

Teacher's Note:
1. Use the reflection property of definite integrals.

Answer: (d) \( \dfrac\pi4 \)

Explanation:
1. Replacing \( x \) by \( -x \) gives \( I=\int\dfrac{2^x\sin^2x}{2^x+1}dx \). Adding the two forms, \( 2I=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=2\int_0^{\pi/2}\sin^2x\,dx \).
2. So \( I=\int_0^{\pi/2}\sin^2x\,dx=\dfrac\pi4 \). Answer: (d)

Teacher's Note:
1. Add the integral to its form with \( x\to-x \).

Answer: (b) \( \left(\dfrac15,0\right) \)

Explanation:
1. \( I_n+I_{n+2}=\int\tan^nx(1+\tan^2x)\,dx=\int\tan^nx\sec^2x\,dx=\dfrac{\tan^{n+1}x}{n+1}+C \).
2. For \( n=4 \): \( I_4+I_6=\dfrac{\tan^5x}5+C \), so \( a=\dfrac15 \) and \( b=0 \). Answer: (b)

Teacher's Note:
1. Combine \( I_n \) and \( I_{n+2} \).

Answer: (c) \( 1 \)

Explanation:
1. The integrand is \( \dfrac{\log x}{\log x+\log(6-x)} \). Using \( \int_2^4f(x)dx=\int_2^4f(6-x)dx \) gives the companion \( \dfrac{\log(6-x)}{\log(6-x)+\log x} \).
2. Adding, \( 2I=\int_2^41\,dx=2 \), so \( I=1 \). Answer: (c)

Teacher's Note:
1. Use the property with \( a+b=6 \).

Answer: (d) A is false, but R is true.

Explanation:
1. Using the property, \( I=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx \) as well. Adding, \( 2I=\int_{\pi/6}^{\pi/3}dx=\dfrac\pi6 \), so \( I=\dfrac\pi{12} \) and the Assertion is false.
2. The Reason is a standard true property. Answer: (d)

Teacher's Note:
1. Add the integral to its reflected form.

Answer: (c) \( g(x)-g(\pi) \)

Explanation:
1. \( g(x+\pi)=g(x)+\int_x^{x+\pi}\cos4t\,dt \), and the integral over a full period of \( \cos4t \) is \( \int_0^\pi\cos4t\,dt=0 \).
2. Also \( g(\pi)=\dfrac{\sin4\pi}4=0 \), so \( g(x+\pi)=g(x)=g(x)-g(\pi) \). The other options give an undefined value, \( g(x)+1 \), or \( 0 \). Answer: (c) Answer: (c)

Teacher's Note:
1. Use the periodicity of \( \cos4t \).

Answer: (d) \( \dfrac34 \)

Explanation:
1. \( [x^2]=0 \) on \( [0,1) \), \( 1 \) on \( [1,\sqrt2) \) and \( 2 \) on \( [\sqrt2,1.5] \).
2. Integral \( =\int_1^{\sqrt2}x\,dx+2\int_{\sqrt2}^{1.5}x\,dx=\dfrac12+(2.25-2)=\dfrac34 \). Answer: (d)

Teacher's Note:
1. Find where \( [x^2] \) takes each integer value.

Answer: (d) \( -\dfrac\pi2 \)

Explanation:
1. Replacing \( x \) by \( \pi-x \) gives \( I=\int_0^\pi[-\cot x]\,dx \). Adding, \( 2I=\int_0^\pi([\cot x]+[-\cot x])dx \).
2. For non-integer values \( [t]+[-t]=-1 \), and \( \cot x \) is an integer only at isolated points, so \( 2I=-\pi \) and \( I=-\dfrac\pi2 \). Answer: (d)

Teacher's Note:
1. Use the reflection \( x\to\pi-x \).

Answer: (c) \( I\lt\dfrac23 \) and \( J\lt2 \)

Explanation:
1. For \( x\in(0,1) \), \( \sin x\lt x \), so \( I\lt\int_0^1\sqrt x\,dx=\dfrac23 \).
2. Also \( \cos x\lt1 \), so \( J\lt\int_0^1x^{-1/2}dx=2 \). Answer: (c)

Teacher's Note:
1. Bound the integrands using \( \sin x\lt x \) and \( \cos x\lt1 \).