Question

Let $f:R\to (0,\infty )$ be a real valued differentiable function satisfying $\underset{0}{\overset{x}{\int }}tf(x-t)dt=e^{2x}-1.$ Then which of the following is/are correct?

• A. $\left(f^{-1}{)}^{\prime }\right(4)=\frac{1}{8}$
• B. Derivative of $f\left(x\right)$ w.r.t. $e^x$ at $x=0$ is equal to $8$.
• C. $\underset{x\to 0}{lim}\frac{f\left(x\right)-4}{x}=4$
• D. $f\left(0\right)=4$

Answer 1

Answer: (A), (B), (D)

$f\left(0\right)=4$

Given : $\underset{0}{\overset{x}{\int }}tf(x-t)dt=e^{2x}-1$
Putting $t\to x-t$
$\underset{0}{\overset{x}{\int }}(x-t)f\left(t\right)dt=e^{2x}-1\Rightarrow x\underset{0}{\overset{x}{\int }}f\left(t\right)dt-\underset{0}{\overset{x}{\int }}tf\left(t\right)dt=e^{2x}-1$

Differentiating both sides w.r.t. $x$, we get
$xf\left(x\right)+\underset{0}{\overset{x}{\int }}f\left(t\right)dt-xf\left(x\right)=2e^{2x}\Rightarrow \underset{0}{\overset{x}{\int }}f\left(t\right)dt=2e^{2x}\Rightarrow f\left(x\right)=4e^{2x}\therefore f\left(0\right)=4$

Let $g\left(x\right)=f^{-1}\left(x\right)$, so
$g\left(f\right(x\left)\right)=x\Rightarrow g^{\prime }\left(f\right(x\left)\right)\times f^{\prime }\left(x\right)=1$
Putting $x=0$, we get
$g^{\prime }\left(4\right)=\frac{1}{8}\therefore \left(f^{-1}{)}^{\prime }\right(4)=\frac{1}{8}$

Now, $\frac{df\left(x\right)}{d{e}^x}=\frac{\frac{d\left(4e^{2x}\right)}{dx}}{\frac{d\left(e^x\right)}{dx}}=\frac{8e^{2x}}{e^x}$
$\Rightarrow \frac{df\left(x\right)}{dx}{|}_{x=0}=8$

$\underset{x\to 0}{lim}\frac{f\left(x\right)-4}{x}=\underset{x\to 0}{lim}\frac{4e^{2x}-4}{x}=\underset{x\to 0}{lim}\frac{8(e^{2x}-1)}{2x}=8$