Question

Let $f$ be a continuous function. If $\underset{0}{\overset{x}{\int }}f\left(t\right)dt=e^x-c{e}^{2x}\underset{0}{\overset{1}{\int }}f\left(t\right)e^{-t}dt,$ where $c$ is a non-zero constant, then

• A. $f\left(x\right)=e^x+2e^x$
• B. $f\left(x\right)=e^x-2e^{2x}$
• C. $c=-\frac{1}{1+2e}$
• D. $c=\frac{1}{3-2e}$

Answer 1

Answer: (B), (D)

$c=\frac{1}{3-2e}$

$\underset{0}{\overset{x}{\int }}f\left(t\right)dt=e^x-c{e}^{2x}\underset{0}{\overset{1}{\int }}f\left(t\right)e^{-t}dt\cdots \left(1\right)$
Differentiating w.r.t. $x,$ we get
$f\left(x\right)=e^x-2c{e}^{2x}\underset{0}{\overset{1}{\int }}f\left(t\right)e^{-t}dt\cdots \left(2\right)$
Putting $x=0$ in $\left(1\right),$ we get
$\underset{0}{\overset{0}{\int }}f\left(t\right)dt=e^0-c{e}^0\underset{0}{\overset{1}{\int }}f\left(t\right)e^{-t}dt$
$\Rightarrow \underset{0}{\overset{1}{\int }}f\left(t\right)e^{-t}dt=\frac{1}{c}\cdots \left(3\right)$
From $\left(2\right)$ and $\left(3\right),$ we get
$f\left(x\right)=e^x-2e^{2x}$


From $\left(3\right),$ we have
$\underset{0}{\overset{1}{\int }}(e^t-2e^{2t})e^{-t}dt=\frac{1}{c}$
$\Rightarrow \underset{0}{\overset{1}{\int }}(1-2e^t)dt=\frac{1}{c}$
$\Rightarrow 1-2(e-1)=\frac{1}{c}$
$\therefore c=\frac{1}{3-2e}$