Question

Consider the non-constant differentiable function $f$ of one variable which obeys the relation $\frac{f\left(x\right)}{f\left(y\right)}=f(x-y)$. If $f^{\prime }\left(0\right)=p$ and $f^{\prime }\left(5\right)=q$, then $f^{\prime }(-5)$ is

• A. $\frac{p^2}{q}$
• B. $\frac{q}{p}$
• C. $\frac{p}{q}$
• D. $q$

Answer 1

The correct option is A $\frac{p^2}{q}$

$f\left(x\right)=a^{kx}$ is non-constant differentiable function

Also, $\frac{f\left(x\right)}{f\left(y\right)}=\frac{a^{kx}}{a^{ky}}=a^{kx}{a}^{-ky}=a^{k(x-y)}=f(x-y)$

$\therefore f\left(x\right)=a^{kx}$ is the required function

$\Rightarrow f^{\prime }\left(x\right)=k{a}^{kx}\ln a$

Given $f^{\prime }\left(0\right)=p$ and $f^{\prime }\left(5\right)=q$
$⟹k\ln a=p$ and $k{a}^{5k}\ln a=q$
$⟹a^{5k}=\frac{q}{p}$
$\therefore f^{\prime }(-5)=k.a^{-5k}\ln a=k\ln a^{-5k}=p\times \frac{p}{q}=\frac{p^2}{q}$

Hence, $f^{\prime }(-5)=\frac{p^2}{q}$