JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 11

Answer: (a) 31

Explanation:
1. \( [\sqrt x+2]=2+[\sqrt x] \). So \( \int_0^92\,dx=18 \).
2. \( [\sqrt x]=0 \) on \( [0,1) \), \( 1 \) on \( [1,4) \), \( 2 \) on \( [4,9) \): \( 0+3+10=13 \). Total \( 18+13=31 \).

Teacher's Note:
1. Pull the integer 2 out of the greatest integer function.
2. Split the interval where \( \sqrt x \) crosses integers.

Answer: (d) \( k^{-1}e^{-1} \)

Explanation:
1. Let \( L=\lim\left(\frac{n!}{n^n}\right)^{1/n}\cdot\frac1k \). \( \ln\left(\frac{n!}{n^n}\right)^{1/n}=\frac1n\sum\ln\frac rn\to\int_0^1\ln x\,dx=-1 \).
2. So \( L=\frac1k\cdot e^{-1}=k^{-1}e^{-1} \).

Teacher's Note:
1. Convert the logarithm of the product into a sum and use the limit of a Riemann sum.
2. \( \int_0^1\ln x\,dx=-1 \).

Answer: (a) \( A_{n+1}=A_n,\ B_{n+1}-B_n=A_{n+1} \)

Explanation:
1. \( A_{n+1}-A_n=\int\frac{\sin(2n+1)x-\sin(2n-1)x}{\sin x}=\int2\cos2nx\,dx=0 \).
2. \( B_{n+1}-B_n=\int\frac{\sin^2(n+1)x-\sin^2nx}{\sin^2x}=\int\frac{\sin(2n+1)x\sin x}{\sin^2x}=A_{n+1} \).

Teacher's Note:
1. Use \( \sin^2P-\sin^2Q=\sin(P+Q)\sin(P-Q) \).
2. Difference formulas telescope such integrals.

Answer: (a) \( -\frac{19}6 \)

Explanation:
1. \( \int_2^9f=\int_{-3}^9f-\int_{-3}^2f \).
2. \( =-\frac56-\frac73=-\frac{19}6 \).

Teacher's Note:
1. Use additivity of the limits.
2. Take care with the common denominators.

Answer: (b) \( f(p,q)=\frac p{p+q}f(p-1,q-1) \)

Explanation:
1. Integrate by parts: \( f=\left[\cos^px\frac{\sin qx}q\right]_0^{\pi/2}+\frac pq\int\cos^{p-1}x\sin x\sin qx\,dx \). The boundary term is 0 for \( p>0 \).
2. Use \( \sin x\sin qx=\cos(q-1)x-\cos x\cos qx \): this gives \( f=\frac pq\left[f(p-1,q-1)-f\right] \), so \( f(p,q)=\frac p{p+q}f(p-1,q-1) \).

Teacher's Note:
1. Integrate by parts taking \( \cos qx \) as the integrable factor.
2. Convert the product of sines using the cosine of a difference.

Answer: (d) \( -\frac\pi2\ln2 \)

Explanation:
1. With \( h(x)=x+\sin x \) odd, use \( \int_{-a}^a\frac{F(x)}{1+e^{h(x)}}dx=\int_0^aF(x)dx \) for even \( F \). Here \( F=\ln\cos x \).
2. \( \int_0^{\pi/2}\ln\cos x\,dx=-\frac\pi2\ln2 \).

Teacher's Note:
1. Add the integral to itself with \( x\to-x \); the denominators combine to 1.
2. Remember \( \int_0^{\pi/2}\ln\cos x\,dx=-\frac\pi2\ln2 \).

Answer: (c) 0

Explanation:
1. \( f(x)+f(-x) \) is even and \( g(x)-g(-x) \) is odd.
2. The product is odd, and the interval is symmetric, so the integral is 0.

Teacher's Note:
1. Build even and odd parts first.
2. Odd times even is odd.

Answer: (a) \( \frac1e\log\left(\frac{1+e}2\right) \)

Explanation:
1. \( \frac1{e^x+e}=\frac1e\left[1-\frac{e^x}{e^x+e}\right] \).
2. \( \int_0^1=\frac1e\left[1-\ln\frac{e+e}{1+e}\right]=\frac1e\left[1-\ln\frac{2e}{1+e}\right]=\frac1e\ln\frac{1+e}2 \).

Teacher's Note:
1. Write \( 1=\frac{e^x+e}{e^x+e} \) to split the integrand.
2. Then the log term integrates directly.

Answer: (c) 0

Explanation:
1. Replace \( x \) by \( \pi-x \): \( \sin2m(\pi-x)=-\sin2mx \) while \( \sin(\pi-x)=\sin x \).
2. So \( I=-I \) and \( I=0 \).

Teacher's Note:
1. Use the property \( \int_0^af(x)dx=\int_0^af(a-x)dx \).
2. Even multiples give a sign change.

Answer: (a) 3

Explanation:
1. \( I=\int_0^4f(4-x)\phi(4-x)dx=\int_0^4f(x)\phi(4-x)dx \). Adding: \( 2I=\int_0^4f(x)\{\phi(x)+\phi(4-x)\}dx=3\cdot2=6 \).
2. So \( I=3 \).

Teacher's Note:
1. Add the integral to its reflected form.
2. The sum of the two \( \phi \) values is a constant.

Answer: (d) \( \frac{\pi^2}{32} \)

Explanation:
1. Put \( t=\tan^{-1}x \), \( dt=\frac{dx}{1+x^2} \): \( \int_0^{\pi/4}t\,dt \).
2. \( =\frac12\left(\frac\pi4\right)^2=\frac{\pi^2}{32} \).

Teacher's Note:
1. The derivative of \( \tan^{-1}x \) is already in the integrand.
2. Change the limits with the substitution.

Answer: (c) 0

Explanation:
1. \( f'(x)=\frac{A\pi}2\cos\frac{\pi x}2 \); \( f'\left(\frac12\right)=\frac{A\pi}{2\sqrt2}=\sqrt2\Rightarrow A=\frac4\pi \).
2. \( \int_0^1f=\frac{2A}\pi+B=\frac{2A}\pi \), so \( B=0 \).

Teacher's Note:
1. The integral condition alone determines \( B \).
2. \( \int_0^1\sin\frac{\pi x}2dx=\frac2\pi \).

Answer: (b) \( \frac19\left[\frac14\tan^{-1}\frac14-\frac15\tan^{-1}\frac15\right] \)

Explanation:
1. \( \frac1{(x^2+16)(x^2+25)}=\frac19\left[\frac1{x^2+16}-\frac1{x^2+25}\right] \).
2. Integrating from 0 to 1 gives \( \frac19\left[\frac14\tan^{-1}\frac14-\frac15\tan^{-1}\frac15\right] \).

Teacher's Note:
1. Subtract the two denominators to get the factor \( 25-16=9 \).
2. \( \int\frac{dx}{x^2+c^2}=\frac1c\tan^{-1}\frac xc \).

Answer: (b) \( \frac{\pi^2}{12} \)

Explanation:
1. The integrand is even: \( I=2\int_0^{\pi/2}\frac x{8\cos^22x+1}dx \). The denominator is unchanged by \( x\to\frac\pi2-x \), so \( I=\frac\pi2\int_0^{\pi/2}\frac{dx}{8\cos^22x+1} \).
2. With \( u=2x \) and \( t=\tan u \): \( \int_0^{\pi/2}\frac{dx}{1+8\cos^22x}=\int_0^{\infty}\frac{dt}{t^2+9}=\frac\pi6 \). Hence \( I=\frac\pi2\cdot\frac\pi6=\frac{\pi^2}{12} \).

Teacher's Note:
1. Use the reflection \( x\to\frac\pi2-x \) to remove the factor \( x \).
2. Always check the value numerically for such integrals.

Answer: (c) \( \frac{n!}{a^{n+1}} \)

Explanation:
1. Differentiate \( \int_0^\infty e^{-ax}dx=\frac1a \) \( n \) times with respect to \( a \): \( (-1)^n\int x^ne^{-ax}dx=\frac{(-1)^nn!}{a^{n+1}} \).
2. Cancelling \( (-1)^n \): \( \frac{n!}{a^{n+1}} \).

Teacher's Note:
1. Differentiation under the integral sign generates the powers of \( x \).
2. The signs \( (-1)^n \) cancel on both sides.

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

Explanation:
1. Put \( t=x^2 \): \( \frac12\int_0^1e^tdt \).
2. \( =\frac{e-1}2 \).

Teacher's Note:
1. Use \( dt=2x\,dx \).
2. The limits stay 0 and 1.

Answer: (c) \( 2\int_0^{\pi/2}f(\cos x)dx \)

Explanation:
1. \( f(\cos(\pi-x))=f(-\cos x)=f(\cos x) \) because \( f \) is even.
2. So the integrand is symmetric about \( \frac\pi2 \) and \( \int_0^\pi=2\int_0^{\pi/2} \).

Teacher's Note:
1. Test symmetry about the midpoint \( \frac\pi2 \).
2. Even \( f \) removes the sign in \( \cos(\pi-x) \).

Answer: (a) \( e\left(\frac e2-1\right) \)

Explanation:
1. \( \int e^x\left(f+f'\right)dx=e^xf \) with \( f=\frac1x \), \( f'=-\frac1{x^2} \).
2. Value \( =\left[\frac{e^x}x\right]_1^2=\frac{e^2}2-e=e\left(\frac e2-1\right) \).

Teacher's Note:
1. Recognise the pattern \( e^x(f+f') \).
2. Evaluate at both limits.

Answer: (b) \( \frac1{n-1} \)

Explanation:
1. \( I_n+I_{n-2}=\int_0^{\pi/4}\tan^{n-2}x\sec^2x\,dx \).
2. \( =\left[\frac{\tan^{n-1}x}{n-1}\right]_0^{\pi/4}=\frac1{n-1} \).

Teacher's Note:
1. This is the standard reduction formula for powers of tangent.
2. Use \( 1+\tan^2x=\sec^2x \).

Answer: (a) \( \log_e2 \)

Explanation:
1. \( \left[2e^{-x}\right]=1 \) when \( 1\le2e^{-x}<2 \), i.e. \( 0<x\le\ln2 \), and \( 0 \) for \( x>\ln2 \).
2. So the integral \( =\int_0^{\ln2}1\,dx=\log_e2 \).

Teacher's Note:
1. Find where \( 2e^{-x} \) crosses 1.
2. Beyond that point the greatest integer is 0.

Answer: (a) \( K=3 \)

Explanation:
1. The integral equals \( K^2+6-6K+15=K^2-6K+21 \).
2. \( K^2-6K+21\le12\Rightarrow(K-3)^2\le0\Rightarrow K=3 \).

Teacher's Note:
1. A perfect square being \( \le0 \) forces equality.
2. Integrate term by term first.

Answer: (a) \( [f(0)+4f(\tfrac12)+f(1)]/6 \)

Explanation:
1. \( \int_0^1f=a+\frac b2+\frac c3 \). Also \( f(0)=a,\ f(\tfrac12)=a+\frac b2+\frac c4,\ f(1)=a+b+c \).
2. \( f(0)+4f(\tfrac12)+f(1)=6a+3b+2c \), and \( \frac{6a+3b+2c}6=a+\frac b2+\frac c3 \).

Teacher's Note:
1. This is Simpson's rule, exact for quadratics.
2. Compute both sides in terms of \( a,b,c \).

Answer: (c) \( \frac12 \)

Explanation:
1. \( F'(x)=\frac{\ln x}{1+x}+\frac{\ln(1/x)}{1+1/x}\cdot\left(-\frac1{x^2}\right)=\frac{\ln x}{1+x}+\frac{\ln x}{x(1+x)}=\frac{\ln x}x \).
2. So \( F(x)=\frac12\ln^2x+F(1) \), \( F(1)=0 \). Hence \( F(e)=\frac12 \).

Teacher's Note:
1. Differentiate \( F \) using the Leibniz rule.
2. Evaluate the constant at \( x=1 \).