Question

Consider non-constant differentiable function f(x) such that f(xy)=f(x).f(y)\forall x,y \in R-{0}$f^{\prime }\left(1\right)=1,thenForX>0,lnf\left(x\right)=kf\left(x\right)$ has only one positive real solution if k is equal to

• A. $\frac{1}{e}$
• B. e
• C. $e^2$
• D. none

Answer 1

Answer: (A)

$\frac{1}{e}$

Given that,

$f\left(x\right)=f\left(x\right)f\left(y\right)\forall x,yinR-\left|0\right|$

$f^{\prime }\left(x\right)=1thenforX>0inf\left(x\right)=kf\left(x\right)$

$\therefore k=\frac{1}{e}$

Hence, Option $\left(A\right)$ is correct answer.