Consider the non-constant differentiable func | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the functional equation $\frac{f(x)}{f(y)}=f(x-y)$.Setting $y=0$, we get $\frac{f(x)}{f(0)}=f(x)$, which implies $f(0)=1$.Differentiating bo

Vedclass · Accepted Answer

$\frac{p^{2}}{q}$

Vedclass · Answer

(A) Given the functional equation $\frac{f(x)}{f(y)}=f(x-y)$.Setting $y=0$, we get $\frac{f(x)}{f(0)}=f(x)$, which implies $f(0)=1$.Differentiating both sides with respect to $x$, we get $\frac{f^{\prime}(x)}{f(y)}=f^{\prime}(x-y)$.Setting $x=0$, we have $\frac{f^{\prime}(0)}{f(y)}=f^{\prime}(-y)$.Since $f^{\prime}(0)=p$, we get $f^{\prime}(-y) = \frac{p}{f(y)}$.Also, differentiating the original equation with respect to $y$, we get $f(x) \cdot (-\frac{f^{\prime}(y)}{(f(y))^2}) = f^{\prime}(x-y) \cdot (-1)$.This simplifies to $\frac{f(x) f^{\prime}(y)}{(f(y))^2} = f^{\prime}(x-y)$.At $y=0$, $\frac{f(x) f^{\prime}(0)}{(f(0))^2} = f^{\prime}(x)$, so $f^{\prime}(x) = p f(x)$.This is a linear differential equation with solution $f(x) = e^{px}$.Then $f^{\prime}(x) = p e^{px}$.Given $f^{\prime}(5) = q$, we have $p e^{5p} = q$, so $e^{5p} = \frac{q}{p}$.We need $f^{\prime}(-5) = p e^{-5p} = \frac{p}{e^{5p}} = \frac{p}{q/p} = \frac{p^2}{q}$.

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

Similar Questions

If $ \cos y = x \cos (a+y) $ with $ \cos a \neq \pm 1 $,then $ \frac{d y}{d x} $ is equal to

If $\log (x+y)=\log (x y)+a$,where $a$ is a constant,then $\frac{dy}{dx}$ at $x=2$ and $y=4$ is

If ${x^y} = {y^x},$ then $\frac{dy}{dx} = $

If $y=y(x)$ is an implicit function of $x$ such that $\log _{e}(x+y)=4 x y$,then $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is equal to .... .

If $x^2+y^2=4$, then $y \frac{dy}{dx}+x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module