TIFR-2020-Maths-A: 7

Question

Let $f\left ( x \right )=\frac{log\left ( 2+x \right )}{\sqrt{1+x}}$ for $x\geq 0$, and $a_{m}=\frac{1}{m}\int_{0}^{m}f\left ( t \right )dt$ for every positive integer $m$.

Then the sequence $\{a_{m}\} \infty_{m=1}$

• A. diverges to $+\infty$
• B. has more than one limit point
• C. converges and satisfies $\lim_{m\rightarrow \infty }a_{m}=\frac{1}{2}$log $2$
• D. converges and satisfies $\lim_{m\rightarrow \infty }a_{m}=0$

Answer 1

$f(t)= \frac{\ln (2+t)}{\sqrt{1+t}}$

$\int \frac{\ln (2+t)}{\sqrt{1+t}}\; dt$

$\int \frac{\ln (1+(\sqrt{1+t})^2)}{\sqrt{1+t}}\; dt$

$\text {Let, $\sqrt{1+t}=z$} \Rightarrow \frac{1}{\sqrt{1+t}}\;dt = 2dz$

$\text{so,}$ $\int \frac{\ln (1+(\sqrt{1+t})^2)}{\sqrt{1+t}}\; dt = 2\int \ln (1+z^2)*1\; dz$

$\text{Consider}$ $\ln (1+z^2)$ as first function $“1”$ as second function.

$\text{On applying integration by parts and substituting $z=\sqrt{1+t}$, we get}$

$\int \frac{\ln (2+t)}{\sqrt{1+t}}\; dt = 2\left [ \sqrt{1+t} \ln (2+t) - 2 (\sqrt{1+t} - \tan^{-1}(\sqrt{1+t}))\right ]$

$\text{Now, $a_m=\frac{1}{m}\int_{0}^{m} f(t)\;dt=\frac{2}{m}\left [ \sqrt{1+m} \ln (2+m) - 2 (\sqrt{1+m} - \tan^{-1}(\sqrt{1+m}))\right ]$}$

$\text{By using L’H$\hat{o}$pital’s rule,}$

$\lim_{m\rightarrow \infty} \frac{\sqrt{1+m} \ln (2+m)}{m} = 0$

$\lim_{m\rightarrow \infty} \frac{\sqrt{1+m}}{m} = 0$

$\lim_{m\rightarrow \infty} \frac{\tan^{-1}\sqrt{1+m}}{m} = 0$

$\text{Hence,$\lim_{m\rightarrow \infty}a_m = 0$} $