Let  $f:[0,\infty )\to R$ be a continuous function such that $f\left(x\right)=1-2x+\underset{0}{\overset{x}{\int }}{e}^{x-t}f\left(t\right)dt$ for all $x\in [0,\infty ).$ Then, which of the following statement(s) is (are) TRUE?

$f\left(x\right)=1-2x+\underset{0}{\overset{x}{\int }}{e}^{x-t}f\left(t\right)dt \Rightarrow f\left(x\right)=1-2x+e^x\underset{0}{\overset{x}{\int }}{e}^{-t}f\left(t\right)dt \mathrm{...}\left(i\right)$
Differentiating w.r.t.x, we get
 $f^{\text{'}}\left(x\right)=-2+e^x{e}^{-x}f\left(x\right)+e^x\underset{0}{\overset{x}{\int }}{e}^{-t}f\left(t\right)dt \mathrm{...}\left(ii\right)$
Subtracting (i) from (ii), we get
$f\text{'}\left(x\right)-f\left(x\right)=-2+f\left(x\right)-1+2x \Rightarrow f\text{'}\left(x\right)-2f\left(x\right)=2x-3$ 
 
This is linear differential equation.
      I.F. $=e^{\int -2dx}=e^{-2x}$ 
Therefore, solution is   $y.e^{-2x}=\int (2x-3).\frac{e^{-2x}}{-2}-\int 2\times \frac{e^{-2x}}{-2}dx+c$
 $\Rightarrow y.e^{-2x}=(2x-3).\frac{e^{-2x}}{-2}-\int 2\times \frac{e^{-2x}}{-2}dx+c \Rightarrow y.e^{-2x}=\frac{-(2x-3)e^{-2x}}{2}-\frac{e^{-2x}}{2}+c$
 $\Rightarrow y.e^{-2x}=(1-x)e^{-2x}+c$
 $\Rightarrow y=(1-x)+c.e^{2x} \mathrm{...}\left(iii\right)$
Putting x = 0 in (i), we get f(0) = 1.
So, from above equation, we have
 1 = 1 + c   $\Rightarrow$ c = 0
Now, (iii) reduces to y = 1 – x, which passes through point
Now,  $f\left(x\right)\le y\le \sqrt{1-x^2}$
$\Rightarrow 1-x\le y\le \sqrt{1-x^2}$
$1-x\le y$ represents the region above the line $1-x=y$ and $y\le \sqrt{1-x^2}$ represents the region inside the semicircle (lying above x-axis) $x^2+y^2=1.$ 
So, common region is as the shaded one in the following figure:
 
Area of the shaded region
 = Area of quarter of the circle – Area of triangle $OAB$  

$=\frac{1}{4}\times \pi {\left(1\right)}^2-\frac{1}{2}\times 1\times 1=\frac{\pi -2}{4}$