UPSC NDA 2 2021 Mathematics Solved Paper

Concept:A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point are all equal.Explanation:Statement 1:As $x \to 0$, $\frac{1}{x^2} \to \infty$.The function $\sin\left(\frac{1}{x^2}\right)$ oscillates between $-1$ and $1$ without approaching a fixed value.Hence, $\lim_{x \to 0} \sin\left(\frac{1}{x^2}\right)$ does not exist.Even if $f(0) = 0$, the limit does not exist, so $f$ is not continuous at $x=0$.Thus, statement 1 is false.Statement 2:At $x = \frac{2}{\sqrt{\pi}}$, we have $\frac{1}{x^2} = \frac{1}{\left(\frac{2}{\sqrt{\pi}}\right)^2} = \frac{\pi}{4}$.So $f\left(\frac{2}{\sqrt{\pi}}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.Since $\sin$ is continuous everywhere, $\lim_{x \to \frac{2}{\sqrt{\pi}}} \sin\left(\frac{1}{x^2}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.Both the left-hand and right-hand limits equal this value, matching $f\left(\frac{2}{\sqrt{\pi}}\right)$.Therefore, $f$ is continuous at $x = \frac{2}{\sqrt{\pi}}$.Thus, statement 2 is true.Answer:Only statement 2 is correct, so the answer is option B. 2 only.