Q84 / 180 · KCET 2009 · Mathematics
If $f(x)=\frac{g(x)+g(-x)}{2}+\frac{2}{[h(x)+h(-x)]^{-1}}$ where $g$ and $h$ are differentiable function, then $\mathrm{f}^{\prime}(0)$
- A1
- B$\frac{1}{2}$
- C$\frac{3}{2}$
- Correct answerD0
Solution
Given,
$f(x)=\frac{g(x)+g(-x)}{2}+\frac{2}{[h(x)+h(-x)]^{-1}}$ $\Rightarrow f(x)=\frac{g(x)+g(-x)}{2}+2[h(x)+h(-x)]$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(-x)}{2}+2\left[h^{\prime}(x)-h^{\prime}(-x)\right]$ $\therefore f^{\prime}(0)=\frac{g^{\prime}(0)-g^{\prime}(0)}{2}+2\left[h^{\prime}(0)-h^{\prime}(0)\right]$ $=0$
Hence, the correct answer is option: 'D'.