Q218 / 270 · NDA 2017 (Phase 2) · Mathematics
A function is defined in $(0, \infty)$ by
$f(x)=\left\{\begin{array}{ccc}1-x^2 & \text { for } & 0 \lt x \leq 1 \\\text { In } & x & \text { for } 1 \lt x \leq 2 \\\ln 2-1+0.5 x & \text { for } & 2 \lt x \lt \infty\end{array}\right.$
Which one of the following is correct in respect of the derivative of the function, i.e., $f^{\prime}(\mathrm{x})$ ?
- A$f^{\prime}(x)=2 x$ for $0 \lt x \leq 1$
- B$f^{\prime}(\mathrm{x})=-2 \mathrm{x}$ for $0 \lt \mathrm{x} \leq 1$
- Correct answerC$f^{\prime}(x)=-2 x$ for $0 \lt x \lt 1$
- D$f^{\prime}(\mathrm{x})=0$ for $0 \lt x \lt \infty$
Solution
$f(x)=\left\{\begin{array}{l}1-x^2 \text { for } 0 \lt x \leq 1 \\ \ln x \text { for } 1 \lt x \leq 2 \\ \ln 2-1+0.5 x \text { for } 2 \lt x \lt \infty\end{array}\right.$
$f^{\prime}(x)=-2 x$ for $0 \lt x \lt 1$
Hence, the correct answer is option: 'C'.