Question

If $f\left(x\right)$ is invertible and twice differentiable function satisfying $f^{\prime }\left(x\right)=\stackrel{f\left(x\right)}{\underset{0}{\int }}{f}^{-1}\left(t\right)dt,\forall x\in R$ and $f^{\prime }\left(0\right)=1$, then $f^{\prime }\left(1\right)$ can be

• A. $e$
• B. $e^2$
• C. $\frac{1}{e}$
• D. $\sqrt{e}$

Answer 1

The correct option is D $\sqrt{e}$

$f^{\prime }\left(x\right)=\stackrel{f\left(x\right)}{\underset{0}{\int }}{f}^{-1}\left(t\right)dt$
Differentiating both sides, we get
$f^{\prime \prime }\left(x\right)=f^{-1}\left(f\right(x\left)\right).f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime \prime }\left(x\right)=x{f}^{\prime }\left(x\right)\Rightarrow \int \frac{f^{\prime \prime }\left(x\right)}{f^{\prime }\left(x\right)}dx=\int xdx$
$\Rightarrow \ln |f^{\prime }\left(x\right)|=\frac{x^2}{2}+c$
$f^{\prime }\left(0\right)=1\Rightarrow c=0$
$\therefore \ln |f^{\prime }\left(x\right)|=\frac{x^2}{2}$
$\Rightarrow |f^{\prime }\left(x\right)|=e^{x^2/2}$
$\Rightarrow |f^{\prime }\left(1\right)|=e^{1/2}=\sqrt{e}$