Question

If $f\left(x\right)=\frac{\left[g\right(x)+g(-x\left)\right]}{2}+\frac{2}{{\left[h\right(x)+h(-x\left)\right]}^{-1}}$, where $g$ and $h$ are differentiable functions, then $f\text{'}\left(0\right)$is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option:

Find the value of $f\text{'}\left(0\right)$:

Given that,

$f\left(x\right)=\frac{\left[g\right(x)+g(-x\left)\right]}{2}+\frac{2}{{\left[h\right(x)+h(-x\left)\right]}^{-1}}$

$f\left(x\right)=\frac{\left[g\right(x)+g(-x\left)\right]}{2}+2\left[h\right(x)+h(-x\left)\right]$

Differentiate the given function with respect to $x$,

$f\text{'}\left(x\right)=\left(\frac{1}{2}\right)[g\text{'}(x)+g\text{'}(-x\left)\right(-1\left)\right]+2[h\text{'}(x)+h\text{'}(-x\left)\right(-1\left)\right]$

Now, substitute $x=0$ in the differentiated function, we get

$$\begin{gathered} f’\left(0\right)=\left(\frac{1}{2}\right)[g\text{'}(0)–g\text{'}(0\left)\right]+2[h\text{'}(0)–h\text{'}(0\left)\right] \\ =0 \end{gathered}$$

Hence, the correct option is D.