Question

If $f:R\to R$ be a function defined as $f\left(x\right)=e^x+\underset{0}{\overset{1}{\int }}(x+y{e}^x)f\left(y\right)dy,$ then

• A. $f\left(x\right)$ is decreasing function
• B. $f\left(x\right)$ is one-one function
• C. $f\left(x\right)$ is onto function
• D. $f\left(x\right)=0$ has only one solution

Answer 1

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

$f\left(x\right)=0$ has only one solution

$f\left(x\right)=e^x+\underset{0}{\overset{1}{\int }}(x+y{e}^x)f\left(y\right)dy\Rightarrow f\left(x\right)=e^x+x\underset{0}{\overset{1}{\int }}f\left(y\right)dy+e^x\underset{0}{\overset{1}{\int }}yf\left(y\right)dy\Rightarrow f\left(x\right)=e^x(1+\underset{0}{\overset{1}{\int }}yf\left(y\right)dy)+x\underset{0}{\overset{1}{\int }}f\left(y\right)dy\Rightarrow f\left(x\right)=a{e}^x+bx$
$\Rightarrow f\left(y\right)=a{e}^y+by...\left(1\right)$

Now,
$a=1+\underset{0}{\overset{1}{\int }}yf\left(y\right)dy$
From eqn $\left(1\right),$
$a=1+\underset{0}{\overset{1}{\int }}y(a{e}^y+by)dy$
$\Rightarrow a=1+a\underset{0}{\overset{1}{\int }}y{e}^{y}dy+b\underset{0}{\overset{1}{\int }}{y}^{2}dy$
Using by parts
$\Rightarrow a=1+a[e^y\cdot y-e^y{]}_0^1+\frac{b}{3}\Rightarrow a=1+a+\frac{b}{3}\Rightarrow b=-3$

Now,
$b=\underset{0}{\overset{1}{\int }}f\left(y\right)dy\Rightarrow b=\underset{0}{\overset{1}{\int }}(a{e}^y+by)dy\Rightarrow b={[a{e}^y+\frac{b{y}^2}{2}]}_0^1\Rightarrow b=a(e-1)+\frac{b}{2}\Rightarrow a=\frac{-3}{2(e-1)}$


$\therefore f\left(x\right)=-3[\frac{e^x}{2(e-1)}+x]$

When
$x\to -\infty \Rightarrow f\left(x\right)\to \infty x\to \infty \Rightarrow f\left(x\right)\to -\infty$

Differentiating the function w.r.t. $x$,
$f^{\prime }\left(x\right)=-3[\frac{e^x}{2(e-1)}+1]\Rightarrow f^{\prime }\left(x\right)<0,\forall x\in R$
$\therefore f\left(x\right)$ is decreasing function.
Hence, $f\left(x\right)$ is both one-one and onto.
$f\left(x\right)=0$ has only one solution.