Question

Let $f:R\to R$ be a differentiable function such that $f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-t}f(x-t)dt.$ Then, $y=f\left(x\right)$ is

• A. injective but not surjective
• B. surjective but not injective
• C. bijective
• D. neither injective nor surjective

Answer 1

The correct option is B surjective but not injective

$f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-t}f(x-t)dt\cdots \left(i\right)$
Using the property:
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-(x-t)}f(x-(x-t))dt$
$=x^2+e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt\cdots \left(ii\right)$
Differentiating w.r.t. $x,$ we get
$\Rightarrow f^{\prime }\left(x\right)=2x-e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt+e^{-x}{e}^{x}f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=2x-e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt+f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=2x+x^2\text{[Using equation}\left(ii\right)]$
$\Rightarrow f\left(x\right)=\frac{x^3}{3}+x^2+c$
Also, $f\left(0\right)=0\text{[From equation}\left(i\right)]$
$\Rightarrow f\left(x\right)=\frac{x^3}{3}+x^2$
$\Rightarrow f\left(x\right)=x^2\left(\frac{x+3}{3}\right)$
Clearly, $f\left(0\right)=f(-3)=0$
$\Rightarrow f\left(x\right)$ is not injective function.
and range $=R=$ co-domain,
hence $f\left(x\right)$ is surjective.