Question

A function $f:R\to R^{+}$ satisfying $f(x+y)=f\left(x\right)f\left(y\right)\forall x,y\in R,f\left(0\right)=1,f^{\prime }\left(0\right)=2$, then

• A. $\underset{0}{\overset{{\log }_{e}3}{\int }}\left[f\right(x\left)e^{-x}\right]dx={\log }_e\left(\frac{9}{2}\right)$
  $\left\{\right[\cdot \left]\text{is greatest integer function}\right\}$
• B. $\underset{x\to 0}{lim}\left[f\right(x\left)\right]$ does not exist $\left\{\right[\cdot \left]\text{is greatest integer function}\right\}$
• C. $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{e^2-1}{2}$
• D. $0<f\left(x\right)\le 1$ if $x\in (0,1]$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\underset{0}{\overset{{\log }_{e}3}{\int }}\left[f\right(x\left)e^{-x}\right]dx={\log }_e\left(\frac{9}{2}\right)$
$\left\{\right[\cdot \left]\text{is greatest integer function}\right\}$
B $\underset{x\to 0}{lim}\left[f\right(x\left)\right]$ does not exist $\left\{\right[\cdot \left]\text{is greatest integer function}\right\}$
C $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{e^2-1}{2}$

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

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 $y=x$
$\Rightarrow f^{\prime }(0+x)(1+\frac{dx}{dx})=f^{\prime }\left(0\right)f\left(x\right)+f\left(0\right)f^{\prime }\left(x\right)\frac{dx}{dx}$
$\Rightarrow 2f^{\prime }\left(x\right)=2f\left(x\right)+f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=2f\left(x\right)$
$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int 2dx$
$\Rightarrow \log |f\left(x\right)|=2x+c$

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

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

$RHL=\underset{x\to 0^{+}}{lim}\left[e^{2x}\right]=1LHL=\underset{x\to 0^{-}}{lim}\left[e^{2x}\right]=0\therefore LHL\ne RHL$
Limit does not exist.

$\underset{0}{\overset{1}{\int }}{e}^{2x}dx={\left[\frac{e^{2x}}{2}\right]}_0^1=\frac{e^2-1}{2}$

$x=1\Rightarrow f\left(1\right)=e^2>1$