Question

Let $f : [0, \infty) \to \mathbb R $ be a continuous function such that $f(x) = 1 - 2x + \int^x_0 e^{x -t} f(t) dt$ 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 $\{(x, y) \in [0, 1] \times \mathbb R ∶ f(x) \leq y \leq \sqrt{1 - x^2} \}$ is $\frac{\pi - 2}{4}$
• D. The area of the region $\{(x, y) \in [0, 1] \times \mathbb R ∶ f(x) \leq y \leq \sqrt{1 - x^2} \}$ is $\frac{\pi - 1}{4}$

Answer 1

Answer: (B), (C)

$f\left(x\right) = 1 - 2x + \int^{x}_{0} e^{x-t} f\left(t\right)dt $
$\Rightarrow e^{-x}f\left(x\right)=e^{-x}\left(1-2x\right)+ \int^{x}_{0} e^{-t} f\left(t\right)dt $
Differentiate w.r.t.x.
$-e^{-x}f\left(x\right)+e^{-x}f'\left(x\right) = - e^{-x}\left(1-2x\right)+e^{-x}\left(-2\right)+e^{-x}f\left(x\right) $
$\Rightarrow -f\left(x\right)+ f'\left(x\right)=-\left(1-2x\right)-2+f\left(x\right)$
$ \Rightarrow f'\left(x\right) - 2 f\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 $
$f\left(x\right) = \frac{2x-3}{-2} - \frac{1}{2}+ ce^{2x}$
$ 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} $