Question

A function $f:R\to R^{+}$ satisfies $f(x+y)=f\left(x\right)f\left(y\right)\forall x,y\in R.$ If $f\left(0\right)=1$ and $f^{\prime }\left(0\right)=2$, then the value of $\underset{0}{\overset{{\log }_{e}3}{\int }}\left[f\right(x\left)e^{-x}\right]dx$ is

(where $[\cdot ]$ is represents the greatest integer function)

• A. ${\log }_e\left(\frac{3}{2}\right)$
• B. ${\log }_e\left(\frac{9}{2}\right)$
• C. ${\log }_e\left(\frac{9}{4}\right)$
• D. ${\log }_{e}3$

Answer 1

The correct option is B ${\log }_e\left(\frac{9}{2}\right)$

$f(x+y)=f\left(x\right)f\left(y\right)\forall x,y\in R$
From the above equation, we can conclude that
$f\left(x\right)=a^x,(a\ne 0,1)$ where $a$ is some constant.
$\Rightarrow f^{\prime }\left(x\right)=a^x\ln a$
$\Rightarrow 2=\ln a(\because f^{\prime }(0)=2)$
$\Rightarrow a=e^2$
$\Rightarrow f\left(x\right)=a^x=(e^2{)}^x=e^{2x}$

Now,
$I=\underset{0}{\overset{{\log }_{e}3}{\int }}\left[f\right(x\left)e^{-x}\right]dx=\underset{0}{\overset{{\log }_{e}3}{\int }}\left[e^x\right]dx=\underset{{\log }_{e}1}{\overset{{\log }_{e}2}{\int }}1dx+\underset{{\log }_{e}2}{\overset{{\log }_{e}3}{\int }}2dx=0+{\log }_{e}2+2({\log }_{e}3-{\log }_{e}2)={\log }_e\left(\frac{9}{2}\right)$

Alternate method :
$f(x+y)=f\left(x\right)f\left(y\right)$
Differentiating w.r.t $x$
$\Rightarrow f^{\prime }(x+y)(1+\frac{dy}{dx})=f^{\prime }\left(x\right)f\left(y\right)+f\left(x\right)f^{\prime }\left(y\right)\frac{dy}{dx}$
Put $x=0$ and $y=x$
$\Rightarrow f^{\prime }(0+x)(1+\frac{dx}{dx})=f^{\prime }\left(0\right)f\left(x\right)+f\left(0\right)f^{\prime }\left(x\right)\frac{dx}{dx}$
$\Rightarrow 2f^{\prime }\left(x\right)=2f\left(x\right)+f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=2f\left(x\right)$
$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int 2dx$
$\Rightarrow {\log }_e|f\left(x\right)|=2x+c$

Now, put $x=0$
$\Rightarrow {\log }_e|f\left(0\right)|=2\left(0\right)+c$
$\Rightarrow {\log }_{e}1=c\Rightarrow c=0$
$\therefore {\log }_e|f\left(x\right)|=2x$
$\Rightarrow f\left(x\right)=e^{2x}$