Q114 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics
If $\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}=\mathrm{ae}^{\mathrm{b}}$, then $a+b=$
- A$\pi-2$
- B$\pi$
- C$\pi+2$
- Correct answerD$\frac{\pi}{2}$
Solution
Let $y=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}$
$\begin{aligned} & \Rightarrow \log y=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left[\log \left(1+\frac{r^2}{n^2}\right)\right] \\ & \Rightarrow \log y=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left[1+\left(\frac{r}{n}\right)^2\right]\end{aligned}$
$\Rightarrow \log y=\int_0^1 \log \left(1+x^2\right)=\left[\log \left(1+x^2\right) \cdot x\right]_0^1-\int_0^1 \frac{2 x^2}{1+x^2} d x$
$\begin{aligned}
& =\log 2-2(x-\tan x)=\log 2-2+\frac{\pi}{2} \\
& \Rightarrow y=e^{\log 2-2+\frac{\pi}{2}}=a e^b
\end{aligned}$
So, $\dot{a}=2, b=-2+\frac{\pi}{2} \Rightarrow a+b=\frac{\pi}{2}$.
Hence, the correct answer is option: 'D'.