Question

Let $f(x+y)=f\left(x\right).f\left(y\right)$ for all $xϵR$ and $f\left(x\right)=1+x\varphi \left(x\right)\log 2$ where ${lim}_{x\to 0}\varphi \left(x\right)=1$ then $f^{\prime }\left(x\right)$ is equal to

• A. $\log 2^{f\left(x\right)}$
• B. $\log \left(f\right(x){)}^2$
• C. $\log 2$
• D. none of these

Answer 1

The correct option is A $\log 2^{f\left(x\right)}$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$={lim}_{h\to 0}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}$ $[\because f(x+y)=f(x).f(y\left)\right]$
$=f\left(x\right){lim}_{h\to 0}\left(\frac{f\left(h\right)-1}{h}\right)$
$=f\left(x\right){lim}_{h\to 0}\frac{1+h\varphi \left(h\right)log2-1}{h}$ $[\because f(x)=1+x(x\left)log2\right]$
$=f\left(x\right)log2{lim}_{h\to 0}\varphi \left(h\right)$
$=f\left(x\right).log2.1[\because {lim}_{h\to 0}\varphi (h)=1]$
$\Rightarrow f^{\prime }\left(x\right)=log{2}^{f\left(x\right)}$
Hence, option 'A' is correct.