Q114 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{1}{n} \sec ^2 1\right]=$
- A$\frac{1}{2} \sec (1)$
- B$\frac{1}{2} \operatorname{cosec}(1)$
- C$\tan (1)$
- Correct answerD$\frac{1}{2} \tan (1)$
Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}\right] \\
= & \lim _{n \rightarrow \infty} \sum_{r=1}^n\left[\frac{1}{n}\left(\frac{r}{n} \sec ^2\left(\frac{r}{n}\right)^2\right)\right]=\int_0^1 x \sec ^2 x^2 d x
\end{aligned}
$$
Put $x^2=t ; 2 x d x=d t$
$$
=\frac{1}{2} \int_0^1 \sec ^2 t d t=\frac{1}{2}[\tan t]_0^1=\frac{1}{2} \tan (1) .
$$
Hence, the correct answer is option: 'D'.