Question

Let f be a differentiable function on
R and satisfying the integral equation ${\int }_0^{x}f\left(t\right)dt+{\int }_0^{x}t.f(x-t)dt=-1+e^{-x}\text{for all}xϵR,$ then

• A. $f^{\prime }\left(0\right)=2$
• B. $f\left(2\right)=e^{-2}$
• C. $f\left(0\right)+f^{\prime }\left(0\right)=1$
• D. $f^{\prime }\left(0\right)=1$

Answer 1

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

$f\left(0\right)+f^{\prime }\left(0\right)=1$

Given equation is
${\int }_0^{x}f\left(t\right)dt+{\int }_0^{x}tf(x-t)dt=-1+e^{-x}$
$\Rightarrow {\int }_0^{x}f\left(t\right)dt+{\int }_0^x(x+0-t)f\{x-(x+0-t\left)\right\}dt=-1+e^{-x}$
$\Rightarrow {\int }_0^{x}f\left(t\right)dt+x{\int }_0^{x}f\left(t\right)dt-{\int }_0^{x}tf\left(t\right)dt=-1+e^{-x}$
Differentiating both sides w.r.t. x, we get
$f\left(x\right)\times 1+1\times {\int }_0^{x}f\left(t\right)dt+x\left[f\right(x)\times 1-f(0)\times 0]$
$-\left[xf\right(x\left)\frac{x}{0}f\right(t)dt+xf(x)-xf(x)=-e^{-x}]$ ...(1)
Put $x=0$ in (1), we get
$f\left(0\right)+0=-e^0=-1\Rightarrow f\left(0\right)=-1$
Again diff. both sides w.r.t. x, we get
$f^{\prime }\left(x\right)+\left[f\right(x)\times 1-f(0)\times 0]=e^{-x}$
$\Rightarrow e^x\left[f\right(x)+f^{\prime }(x\left)\right]=\varphi$
$\Rightarrow \frac{d}{dx}\left[e^{x}f\right(x\left)\right]=\varphi$
Integrating, we get $e^{x}f\left(x\right)=x+c$ ...(2)
Put $x=0$ is $\left(2\right)e^{0}f\left(0\right)=0+c\Rightarrow c=1\times (-1)=-1$
$\therefore$ From (2),
$f\left(x\right)=(x-1)e^{-x}$
$\therefore f\left(2\right)=(2-1)e^{-2}=e^{-2}\Rightarrow$ option (A) is correct
$f^{\prime }\left(x\right)=-(x-1)e^{-x}+e^{-x}$
$\Rightarrow f^{\prime }\left(0\right)=-(0-1)e^0+e^0=2$
$\therefore f\left(0\right)+f^{\prime }\left(0\right)=-1+2=1$
$\Rightarrow$ Option (B) and (C) are correct