Let f : R → R be a function given by f(x + y) = f(x) f(y) for all x, y ∈ R. If $f(x)=1+x g(x) . \log _e 2$, where $\lim\limits_{x \rightarrow 0} g(x)=1$. Then, f'(x) =
Answer & explanation
Correct answer: option 1
We have,
$f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h}$ [∵ f(x + y) = f(x) f(y)]
$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} f(x)\left(\frac{f(h)-1}{h}\right)$
$\Rightarrow f'(x)=f(x) \lim\limits_{h \rightarrow 0} \frac{1+h g(h) . \log _e 2-1}{h}$ [∵ f(h)= 1 + hg(h) loge2]
$f'(x)=f(x)\log_e 2$ $\left[∵ \lim\limits_{h \rightarrow 0} g(h)=1\right]$