Let f be a differentiable function satisfying the condition: $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ for all x, y(≠ 0) ∈ R and f(y) ≠ 0. If f'(1) = 2, then f'(x) is equal to
Answer & explanation
Correct answer: option 4
We have,
$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ for all $x, y \in R, y \neq 0$ and $f(y) \neq 0$
Replacing x and y both by 1 , we get
$f(1)=\frac{f(1)}{f(1)} \Rightarrow f(1)=1$
Now,
$f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\Rightarrow f'(x)=f(x) \lim\limits_{h \rightarrow 0}\left\{\frac{\frac{f(x+h)}{f(x)}-1}{h}\right\}$
$\Rightarrow f'(x)=f(x) \lim _{h \rightarrow 0}\left\{\frac{f\left(\frac{x+h}{x}\right)-1}{h}\right\}~~~~~~~\left[∵ f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}\right]$
$\Rightarrow f'(x)=f(x) \lim\limits_{h \rightarrow 0} \frac{f\left(1+\frac{h}{x}\right)-f(1)}{h}$ [∵ f(1) = 1]
$\Rightarrow f'(x)=\frac{f(x)}{x} \lim\limits_{h \rightarrow 0} \frac{f\left(1+\frac{h}{x}\right)-f(1)}{h / x}$
$\Rightarrow f'(x)=\frac{f(x)}{x} f'(1)=\frac{2 f(x)}{x}$ [∵ f'(1) = 2]