JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 09

Answer: (d) \( \log2 \)

Explanation:
1. \( \int_0^1\tan^{-1}x\,dx=\frac\pi4-\frac12\ln2 \), so \( \int_0^1\cot^{-1}\left(1-x+x^2\right)dx=\frac\pi2-\ln2 \).
2. Since \( \tan^{-1}y+\cot^{-1}y=\frac\pi2 \), the required integral is \( \int_0^1\frac\pi2\,dx-\left(\frac\pi2-\ln2\right) \).
3. This equals \( \ln2 \).

Teacher's Note:
1. Use tan^-1 y + cot^-1 y = pi/2 to link the two integrals.

Answer: (d) \( \frac12(\log x)^2 \)

Explanation:
1. In \( f\left(\frac1x\right)=\int_1^{1/x}\frac{\log t}{1+t}dt \) put \( t=\frac1u \): it becomes \( \int_1^x\frac{\log u}{u(1+u)}du \).
2. Adding, \( f(x)+f\left(\frac1x\right)=\int_1^x\log t\left[\frac1{1+t}+\frac1{t(1+t)}\right]dt=\int_1^x\frac{\log t}tdt \).
3. \( =\frac12(\log x)^2 \).

Teacher's Note:
1. Substituting t = 1/u in the second integral makes the denominators combine neatly.

Answer: (c) 100

Explanation:
1. The graph is symmetric about \( x=2 \) and about \( x=4 \), so \( f \) has period \( 2(4-2)=4 \).
2. By symmetry about \( x=2 \), \( \int_2^4f=\int_0^2f=5 \), so the integral over one period is 10.
3. \( [10,50] \) has length 40, which is 10 periods, so the integral is 100.

Teacher's Note:
1. Two lines of symmetry give a period equal to twice the distance between them.

Answer: (c) \( \frac\pi{16}\ln2 \)

Explanation:
1. Put \( u=2x \): the integral becomes \( \frac12\int_0^1\frac{\ln(1+u)}{1+u^2}du \).
2. Using \( \int_0^1\frac{\ln(1+u)}{1+u^2}du=\frac\pi8\ln2 \) (put \( u=\tan\theta \) and use the reflection \( \theta\to\frac\pi4-\theta \)).
3. The value is \( \frac12\cdot\frac\pi8\ln2=\frac\pi{16}\ln2 \).

Teacher's Note:
1. This is a standard integral worth remembering: the integral of ln(1 + tan theta) over [0, pi/4] is (pi/8) ln 2.

Answer: (c) \( -9e \)

Explanation:
1. Integrating by parts, \( P_n=\left[x(\log x)^n\right]_1^e-n\int_1^e(\log x)^{n-1}dx=e-nP_{n-1} \).
2. So \( P_{10}=e-10P_9 \) and \( P_9=e-9P_8 \).
3. \( P_{10}=e-10e+90P_8 \), hence \( P_{10}-90P_8=-9e \).

Teacher's Note:
1. Derive the reduction formula P_n = e - n P_(n-1) by parts and apply it twice.

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

Explanation:
1. On \( \left(0,\frac\pi2\right] \), \( \cos x\in[0,1) \) so \( [\cos x]=0 \).
2. On \( \left(\frac\pi2,\pi\right) \), \( \cos x\in(-1,0) \) so \( [\cos x]=-1 \).
3. The integral is \( -1\cdot\frac\pi2=-\frac\pi2 \).

Teacher's Note:
1. Find where cos x lies in each unit interval.

Answer: (a) \( e^{-a}\left[F(x+a)-F(1+a)\right] \)

Explanation:
1. Put \( u=t+a \): \( t=1\to u=1+a \) and \( t=x\to u=x+a \).
2. \( \int_1^x\frac{e^t}{t+a}dt=\int_{1+a}^{x+a}\frac{e^{u-a}}udu=e^{-a}\int_{1+a}^{x+a}\frac{e^u}udu \).
3. \( =e^{-a}\left[F(x+a)-F(1+a)\right] \).

Teacher's Note:
1. Shift the variable so that the denominator becomes u and the integrand matches the definition of F.

Answer: (d) \( \log2\sqrt2 \)

Explanation:
1. \( \sqrt{\tan^2x}=|\tan x| \). On \( \left(\frac{7\pi}4,2\pi\right) \) tangent is negative, and on \( \left(2\pi,\frac{7\pi}3\right) \) it is positive.
2. \( \int_{7\pi/4}^{2\pi}-\tan x\,dx=\left[\ln|\cos x|\right]_{7\pi/4}^{2\pi}=\ln\sqrt2 \) and \( \int_{2\pi}^{7\pi/3}\tan x\,dx=\left[-\ln|\cos x|\right]=\ln2 \).
3. The total is \( \ln\sqrt2+\ln2=\ln2\sqrt2 \).

Teacher's Note:
1. sqrt(tan^2 x) is |tan x|, so split at the point 2 pi where tan changes sign.

Answer: (d) \( \frac32 \)

Explanation:
1. \( \{2x\} \) has period \( \frac12 \), and over each period its integral is \( \int_0^{1/2}2x\,dx=\frac14 \).
2. The interval \( [-1,2] \) has length 3, which is 6 periods.
3. The integral is \( 6\times\frac14=\frac32 \).

Teacher's Note:
1. The function {2x} repeats every 1/2, so count the periods.

Answer: (a) 2

Explanation:
1. The sum is \( \frac3n\sum_{r=0}^{n-1}\frac{1}{\sqrt{1+\frac{3r}n}} \), a Riemann sum of \( \frac{1}{\sqrt{1+3x}} \) on \( [0,1] \) multiplied by 3.
2. \( 3\int_0^1\frac{dx}{\sqrt{1+3x}}=3\cdot\frac23\left[\sqrt{1+3x}\right]_0^1 \).
3. \( =2(2-1)=2 \).

Teacher's Note:
1. Write each term as 1/sqrt(1 + 3r/n) and convert the limit of the sum into an integral.

Answer: (c) \( \frac53 \)

Explanation:
1. Split at \( x=1 \): \( \int_0^1x^3dx=\frac14 \).
2. \( \int_1^2x^2(x-1)dx=\left[\frac{x^4}4-\frac{x^3}3\right]_1^2=\left(4-\frac83\right)-\left(\frac14-\frac13\right)=\frac43+\frac1{12} \).
3. Total \( =\frac14+\frac43+\frac1{12}=\frac{3+16+1}{12}=\frac53 \).

Teacher's Note:
1. Use the first piece of f on [0, 1) and the second on [1, 2].

Answer: (b) I and II only

Explanation:
1. \( \frac{x}{(1+x)(1+x^2)}=-\frac{1}{2(1+x)}+\frac{x+1}{2(1+x^2)} \), whose antiderivative is \( -\frac12\ln(1+x)+\frac14\ln(1+x^2)+\frac12\tan^{-1}x \).
2. As \( x\to\infty \) the logarithmic part tends to 0, so the value is \( \frac12\cdot\frac\pi2=\frac\pi4 \). So I is true and III is false.
3. \( \frac1{(1+x)(1+x^2)}=\frac1{2(1+x)}+\frac{1-x}{2(1+x^2)} \) similarly gives \( \frac\pi4 \), so II is true.

Teacher's Note:
1. Partial fractions show both integrals have the same value pi/4.