Let $f:\left[0, \infty \right)\to R$ be a continuous function such that $f\left(x\right)=1-2x+{\int }_0^x{e}^{x-t}f\left(t\right)dt$ for all $x\in \left[0,\infty \right).$ Then, which of the following statement(s) is (are) TRUE?

• A

  The curve $y=f\left(x\right)$ pass through the point $\left(1,2\right)$
• Correct answer
  B

  The curve $y=f\left(x\right)$ passes through the point $\left(2,-1\right)$
• Correct answer
  C

  The area of the region $\left\{\left(x,y\right)\in \left[\mathrm{0,1}\right]\times R:f\left(x\right)\le y\le \sqrt{1-x^2}\right\} is\frac{\pi -2}{4}$
• D

  The area of the region $\left\{\left(x,y\right)\in \left[\mathrm{0,1}\right]\times R:f\left(x\right)\le y\le \sqrt{1-x^2}\right\} is\frac{\pi -1}{4}$

Solution

$f\left(x\right)=1-2x+{\int }_0^x{e}^{x-t}f\left(t\right)dt$ $\Rightarrow e^{-x}f\left(x\right)=e^{-x}\left(1-2x\right)+{\int }_0^x{e}^{-t}f\left(t\right)dt$ Differentiate w.r.t. $x$ $e^{-x}f\left(x\right)+e^{+x}{f}^{\prime }\left(x\right)={-e}^{-x}\left(1-2x\right)+e^{-x}\left(-2\right)+e^{-x}f\left(x\right)$ $\Rightarrow -f\left(x\right)+f^{\prime }\left(x\right)=-\left(1-2x\right)-2+f\left(x\right)$ $\Rightarrow f^{\prime }\left(x\right)-2f\left(x\right)=2x-3$ Integrating factor $=e^{-2x}$ $f\left(x\right).e^{-2x}=\int e^{-2x}\left(2x-3\right)dx$ $=\left(2x-3\right)\int e^{-2x}dx-\int \left(\left(2\right)\int e^{-2x}dx\right)dx$ $= \frac{\left(2x-3\right)e^{-2x}}{-2}- \frac{e^{-2x}}{2}+c$ $f\left(x\right)=\frac{2x-3}{-2}-\frac{1}{2}+c{e}^{2x}$ $f\left(0\right)=\frac{3}{2}-\frac{1}{2}+c=1\Rightarrow c=0$ $\therefore f\left(x\right)=1-x$ Area $=\frac{\pi }{4}-\frac{1}{2}=\frac{\pi -2}{4}$

Hence, the correct answer is options: 'B', 'C'.