UPSC NDA 2 2017 Mathematics Solved Paper

Concept:A function is differentiable at a point if its derivative exists and is finite. The derivative must be defined without any discontinuity or indeterminate form.Explanation:The function is $f(x) = \sqrt{1 - e^{-x^2}}$.Its domain is all real numbers because $e^{-x^2} \le 1$ for every $x$, so the radicand is non‑negative.Differentiate using the chain rule:$f'(x) = \frac{1}{2\sqrt{1-e^{-x^2}}} \cdot \frac{d}{dx}(1-e^{-x^2})$$= \frac{1}{2\sqrt{1-e^{-x^2}}} \cdot (-e^{-x^2} \cdot (-2x))$$= \frac{x e^{-x^2}}{\sqrt{1-e^{-x^2}}}$.Now the derivative exists wherever the denominator is non‑zero and the whole expression is finite.The denominator $\sqrt{1-e^{-x^2}}$ becomes zero when $1-e^{-x^2}=0 \Rightarrow x=0$.At $x=0$, the derivative formula gives $\frac{0}{0}$, an indeterminate form, so $f'(0)$ does not exist.For any $x \neq 0$, the denominator is positive and the numerator is well‑defined, so $f'(x)$ exists and is finite.Thus the function is differentiable at all points except $x=0$.Answer:$(-\infty,0) \cup (0,\infty)$, which corresponds to option C.