Q81 / 225 · JEE Main 2005 · Mathematics
$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots .+\frac{1}{n^2} \sec ^2 1\right]$ equals
- A$\frac{1}{2} \sec 1$
- B$\frac{1}{2} \operatorname{cosec} 1$
- C$\tan 1$
- Correct answerD$\frac{1}{2} \tan 1$
Solution
$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots .+\frac{1}{n} \sec ^2 1\right]$ is equal to $\lim _{n \rightarrow \infty} \frac{r}{n^2} \sec ^2 \frac{r^2}{n^2}=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{r}{n} \sec ^2 \frac{r^2}{n^2}$
$\Rightarrow$ Given limit is equal to value of integral $\int_0^1 x \sec ^2 x^2 d x$
or $\frac{1}{2} \int_0^1 2 x \sec x^2 d x=\frac{1}{2} \int_0^1 \sec ^2 t d t$
[put $\left.x^2=t\right]$ $=\frac{1}{2}(\tan t)_0^1=\frac{1}{2} \tan 1$.
Hence, the correct answer is option: 'D'.