JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 10

Answer: (c) \( \frac14\log2 \)

Explanation:
1. \( \sum\frac{r^3}{r^4+n^4}=\frac1n\sum\frac{(r/n)^3}{(r/n)^4+1} \), a Riemann sum.
2. \( \int_0^1\frac{x^3}{x^4+1}dx=\frac14\ln\left(1+x^4\right)\Big|_0^1 \).
3. The limit is \( \frac14\ln2 \).

Teacher's Note:
1. Divide the numerator and denominator by n^4 to expose the Riemann sum.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. By the Reason with \( t=5.5 \): \( [t]=5 \) and \( \{t\}=0.5 \), so the integral is \( \frac52+\frac{0.25}2=\frac52+\frac18=\frac{21}8 \). The Assertion is true.
2. The Reason is true: each full unit interval contributes \( \int_0^1x\,dx=\frac12 \), and the last partial interval contributes \( \frac{\{t\}^2}2 \).
3. The Reason gives exactly the computation used for the Assertion, so it is a correct explanation.

Teacher's Note:
1. Each unit interval contributes 1/2 and the leftover part contributes {t}^2/2.

Answer: (d) \( 4n \)

Explanation:
1. \( 0\le|\sin x|\le1 \), so \( 0\le\frac{|\sin x|}2\le\frac12 \) and \( \left[\frac{|\sin x|}2\right]=0 \).
2. The integral reduces to \( \int_0^{2n\pi}|\sin x|dx \).
3. Each period of length \( \pi \) contributes 2, and there are \( 2n \) of them, giving \( 4n \).

Teacher's Note:
1. The greatest integer of a number below 1/2 is zero.

Answer: (b) 3

Explanation:
1. Integrating by parts twice, \( \int f''\sin x\,dx=f'\sin x-f\cos x-\int f\sin x\,dx \).
2. So \( \int_0^\pi(f+f'')\sin x\,dx=\left[f'\sin x-f\cos x\right]_0^\pi=f(\pi)+f(0) \).
3. Therefore \( 2+f(0)=5 \), so \( f(0)=3 \).

Teacher's Note:
1. The sum (f + f'') sin x is the derivative of f' sin x - f cos x.

Answer: (a) I and II only

Explanation:
1. By parts, \( I_n=\left[\frac x{(1+x^2)^n}\right]_0^1+2n\int_0^1\frac{x^2}{(1+x^2)^{n+1}}dx=2^{-n}+2n\left(I_n-I_{n+1}\right) \).
2. Rearranged: \( 2nI_{n+1}=2^{-n}+(2n-1)I_n \), so I is true.
3. With \( n=1 \): \( 2I_2=\frac12+I_1=\frac12+\frac\pi4 \), so \( I_2=\frac\pi8+\frac14 \): II true, III false. With \( n=2 \): \( 4I_3=\frac14+3I_2=1+\frac{3\pi}8 \), so \( I_3=\frac14+\frac{3\pi}{32} \): IV false.

Teacher's Note:
1. Integrate 1/(1 + x^2)^n by parts and write x^2 as (1 + x^2) - 1.

Answer: (c) \( \sin(3x+4)+3(x-1)\cos(3x+4) \)

Explanation:
1. Given \( \int_1^bf(x)dx=(b-1)\sin(3b+4) \).
2. Differentiate both sides with respect to \(b\): \( f(b)=\sin(3b+4)+3(b-1)\cos(3b+4) \).
3. So \( f(x)=\sin(3x+4)+3(x-1)\cos(3x+4) \).

Teacher's Note:
1. Differentiate the area function with respect to its upper limit.

Answer: (a) TFTT

Explanation:
1. S1: \( \cos^{-1}(\cos x)=x \) on \( [0,\pi] \) and \( 2\pi-x \) on \( [\pi,2\pi] \), so the integral is \( 2\int_0^\pi x\,dx=\pi^2 \): true.
2. S2: the curve is a square (diamond) with diagonals of length 2, so its area is \( \frac{2\cdot2}2=2 \), not 3: false. S3: \( \int fg\,dx=\int ff'dx=\frac{f^2}2 \): true.
3. S4: \( \int_0^1\left(x-x^2\right)dx=\frac16 \): true. The order is T, F, T, T.

Teacher's Note:
1. Evaluate each statement separately; the diamond |x| + |y| = 1 has area 2.

Answer: (b) \( (0,\infty) \)

Explanation:
1. \( f'(x)=2x\left[e^{-(x^2+1)^2}-e^{-x^4}\right] \).
2. Since \( (x^2+1)^2>x^4 \), the bracket is always negative.
3. So \( f'(x)<0 \) exactly when \( x>0 \): \( f \) decreases on \( (0,\infty) \).

Teacher's Note:
1. Differentiate with both limits variable and check the sign of the bracket.

Answer: (a) \( \frac{4\pi}{\sqrt3}\tan^{-1}\frac12 \)

Explanation:
1. The part \( \frac{4x^3}{2-\cos\left(|x|+\frac\pi3\right)} \) is odd and integrates to 0 over the symmetric interval.
2. The remaining part is even: \( 2\pi\int_0^{\pi/3}\frac{dx}{2-\cos\left(x+\frac\pi3\right)}=2\pi\int_{\pi/3}^{2\pi/3}\frac{du}{2-\cos u} \).
3. With \( \int\frac{du}{2-\cos u}=\frac2{\sqrt3}\tan^{-1}\left(\sqrt3\tan\frac u2\right) \), the value is \( \frac{4\pi}{\sqrt3}\left(\tan^{-1}3-\tan^{-1}1\right)=\frac{4\pi}{\sqrt3}\tan^{-1}\frac12 \).

Teacher's Note:
1. Split the numerator into an odd and an even part.

Answer: (c) 3

Explanation:
1. Differentiate both sides with respect to \(x\): \( -\sin^2x\,f(\sin x)\cos x=-\cos x \).
2. So \( f(\sin x)=\frac1{\sin^2x} \), i.e. \( f(t)=\frac1{t^2} \).
3. \( f\left(\frac1{\sqrt3}\right)=3 \).

Teacher's Note:
1. Differentiate with respect to x; the lower limit gives a minus sign.

Answer: (c) 4

Explanation:
1. \( x^3+3x^2+3x+3=(x+1)^3+2 \). Put \( u=x+1 \), so the limits become \( -1 \) and 1.
2. The integrand is \( u^3+2+u\cos u \). \( u^3 \) and \( u\cos u \) are odd, so they integrate to 0.
3. The value is \( \int_{-1}^12\,du=4 \).

Teacher's Note:
1. Recognise (x + 1)^3 in the polynomial and shift the variable.

Answer: (a) \( \frac14\ell n\frac32 \)

Explanation:
1. Put \( u=x^2 \): \( x\,dx=\frac{du}2 \), and the integral is \( \frac12\int_{\ln2}^{\ln3}\frac{\sin u}{\sin u+\sin(\ln6-u)}du \).
2. The limits satisfy \( \ln2+\ln3=\ln6 \), so by the property \( \int_a^bf(u)du=\int_a^bf(a+b-u)du \) the last integral equals \( \frac12(\ln3-\ln2) \).
3. The value is \( \frac12\cdot\frac12\ln\frac32=\frac14\ln\frac32 \).

Teacher's Note:
1. The two limits add up to ln 6, matching the term sin(ln 6 - x^2).

Answer: (d) Local maximum at \( \pi \) and local minimum at \( 2\pi \)

Explanation:
1. \( f'(x)=\sqrt x\sin x \). On \( (0,\pi) \) it is positive, on \( (\pi,2\pi) \) negative, and on \( \left(2\pi,\frac{5\pi}2\right) \) positive.
2. The sign of \( f' \) changes from + to - at \( \pi \): local maximum.
3. It changes from - to + at \( 2\pi \): local minimum.

Teacher's Note:
1. Follow the sign of sin x, since the square root factor is positive.

Answer: (c) 21

Explanation:
1. Integrate \( p'(x)=p'(1-x) \): \( p(x)=-p(1-x)+C \).
2. Putting \( x=0 \): \( 1=-41+C \), so \( C=42 \) and \( p(x)+p(1-x)=42 \).
3. Hence \( 2\int_0^1p(x)dx=\int_0^1\left[p(x)+p(1-x)\right]dx=42 \), giving 21.

Teacher's Note:
1. Integrate the relation between the derivatives to link p(x) and p(1 - x).

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

Explanation:
1. Let \( I=\int_0^\pi[\cot x]dx \). Replacing \(x\) by \( \pi-x \): \( I=\int_0^\pi[-\cot x]dx \).
2. For non-integer \(y\), \( [y]+[-y]=-1\); the exceptional points are finitely many and do not affect the integral.
3. Adding: \( 2I=\int_0^\pi(-1)dx=-\pi \), so \( I=-\frac\pi2 \).

Teacher's Note:
1. Use the identity [y] + [-y] = -1 for non-integers together with x to pi - x.

Answer: (d) \( \pi\log2 \)

Explanation:
1. \( \int_0^1\frac{\log(1+x)}{1+x^2}dx=\frac\pi8\log2 \) (put \( x=\tan\theta \) and use \( \theta\to\frac\pi4-\theta \)).
2. Multiplying by 8 gives \( \pi\log2 \).
3. So the value is \( \pi\log2 \).

Teacher's Note:
1. Remember this standard integral: it equals (pi/8) ln 2.

Answer: (c) 19

Explanation:
1. Split \( [-2,4] \) into unit intervals starting at \( n=-2,-1,0,1,2,3 \).
2. The integral is \( (-2)^2+(-1)^2+0^2+1^2+2^2+3^2 \).
3. \( =4+1+0+1+4+9=19 \).

Teacher's Note:
1. Each unit interval [n, n + 1] contributes n^2.

Answer: (d) 4

Explanation:
1. \( f'(1)=2A+B=8 \). \( f(2)+f''(2)=4A+2B+C+2A=33 \), i.e. \( 6A+2B+C=33 \). \( \int_0^1f=\frac A3+\frac B2+C=\frac73 \).
2. From the first, \( B=8-2A \). The second gives \( C=17-2A \). Substituting in the third: \( \frac A3+4-A+17-2A=\frac73 \), so \( -\frac{8A}3=-\frac{56}3 \) and \( A=7 \).
3. Then \( B=-6 \) and \( C=3 \), so \( A+B+C=4 \).

Teacher's Note:
1. Form three linear equations in A, B, C and solve them step by step.

Answer: (b) 4

Explanation:
1. The limit has the form \( \frac00 \). By L'Hopital's rule it equals \( \lim\frac{2x\left(f(x^2)-x^2\right)}{2(x-1)} \), again \( \frac00 \) since \( f(1)=1 \).
2. Apply the rule again: the numerator's derivative at \( x=1 \) is \( 2\left(f(1)-1\right)+2\left(2f'(1)-2\right)=0+2(6-2)=8 \), and the denominator's derivative is 2.
3. The limit is \( \frac82=4 \).

Teacher's Note:
1. Apply L'Hopital's rule twice, differentiating the integral with the Leibniz rule.

Answer: (b) 2

Explanation:
1. Differentiate both sides with respect to \(x\): \( \phi(x)=1-x\phi(x) \).
2. So \( (1+x)\phi(x)=1 \) and \( \phi(x)=\frac1{1+x} \).
3. \( 10\phi(4)=\frac{10}5=2 \).

Teacher's Note:
1. The right side has a variable lower limit, so its derivative carries a minus sign.