Question

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

• A. The curve $y=f(x)$ passes through the point (1, 2)
• B. The curve $y=f(x)$ passes through the point (2, -1)
• C. The area of the region $\{(x,y)\in [0,1]\times \mathbb{R}:f(x)\le y\le \sqrt{1-x^2}\}$ is $\frac{\pi -2}{4}$
• D. The area of the region $\{(x,y)\in [0,1]\times \mathbb{R}:f(x)\le y\le \sqrt{1-x^2}\}$ is $\frac{\pi -1}{4}$

Answer 1

Answer: (B), (C)

$f\left(x\right)=1-2x+{\int }_0^x{e}^{x-t}f\left(t\right)dt$
$\Rightarrow e^{-x}f\left(x\right)=e^{-x}\left(1-2x\right)+{\int }_0^x{e}^{-t}f\left(t\right)dt$
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 $=\frac{\pi }{4}-\frac{1}{2}=\frac{\pi -2}{4}$