Q233 / 270 · NDA 2017 (Phase 2) · Mathematics

The set of all points, where the function $f(\mathrm{x})=\sqrt{1-\mathrm{e}^{-\mathrm{x}^2}}$ is differentiable, is

  • A
    $(0, \infty)$
  • B
    $(-\infty, \infty)$
  • Correct answer
    C
    $(-\infty, 0) \cup(0, \infty)$
  • D
    $(-1, \infty)$

Solution

$\begin{aligned} & f(x)=\sqrt{1-e^{-x^2}} \\ & f^{\prime}(x)=\frac{1}{2 \sqrt{1-e^{-x^2}}} \cdot\left(-e^{-x^2}\right)(-2 x)\end{aligned}$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{xe}^{-\mathrm{x}^2}}{\sqrt{1-\mathrm{e}^{-\mathrm{x}^2}}}$ $\mathrm{f}^{\prime}(\mathrm{x})$ is defined for all values of x , except 0 . $\therefore \mathrm{f}(\mathrm{x})$ is differentiable on $(-\infty, 0) \cup(0, \infty)$

Hence, the correct answer is option: 'C'.

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