Question

Let f be a differentiable function on R and satisfying the integral equation $\int_{0}^{x}f(t)dt + \int_{0}^{x}t \cdot f(x-t)dt = -1 + e^{-x}$, for all $x \in R$, then

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

Answer 1

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

Options (A), (B), and (C) are correct.

Answer 2

1. Rewrite the given equation:

   $$\int_{0}^{x}f(t)dt + \int_{0}^{x}t\,f(x-t)dt = -1 + e^{-x}.$$

   Change variable in the second integral using $u = x-t$ so that

   $$\int_{0}^{x}t\,f(x-t)dt = \int_{0}^{x}(x-u)f(u)du.$$

   Hence, the equation becomes:

   $$\int_{0}^{x}\Big[ 1+(x-u)\Big]f(u)du = \int_{0}^{x}\big[(x+1)-u\big]f(u)du = e^{-x}-1.$$

2. Differentiate both sides with respect to $x$. Use Leibniz’s rule:

   $$\frac{d}{dx}\Big[\int_{0}^{x}\big[(x+1)-u\big]f(u)du\Big] = \int_{0}^{x}f(u)du + f(x)\quad (\text{since }(x+1)-x=1).$$

   And the derivative of the right side is:

   $$\frac{d}{dx}\Big[e^{-x}-1\Big] = -e^{-x}.$$

   So, we have:

   $$\int_{0}^{x}f(u)du + f(x) = -e^{-x}.$$

3. Let $F(x)=\int_{0}^{x}f(u)du$, then $F'(x)=f(x)$. The equation becomes:

   $$F(x)+F'(x)=-e^{-x}.$$

   This is a first-order linear ODE for $F(x)$.

4. Solve using the integrating factor $e^x$:

   $$\frac{d}{dx}\Big[e^x F(x)\Big] = -e^x e^{-x} = -1.$$

   Integrate:

   $$e^x F(x) = -x + C.$$

   Since $F(0)=0$, we have $C=0$. Thus:

   $$F(x) = -x\,e^{-x}.$$

5. Differentiate $F(x)$ to find $f(x)$:

   $$f(x) = F'(x) = \frac{d}{dx}\Big[-x\,e^{-x}\Big] = -e^{-x} + x\,e^{-x} = e^{-x}(x-1).$$

6. Now check the options:

   • $f(2) = e^{-2}(2-1)=e^{-2}$. So option (A) is correct.

   • $f(0) = e^{0}(0-1) = -1$ and $f'(x) = \frac{d}{dx}\Big[e^{-x}(x-1)\Big] = e^{-x}(2-x)$ giving $f'(0)=2$. Thus, $f(0) + f'(0)=-1+2=1$. So option (B) is correct.

   • $f'(0)=2$ confirms option (C) is correct.

   • Option (D) $f'(0)=1$ is false.