Question

Let $f$ be a continuous and differentiable function in $\left(x_{1} , x_{2}\right).$ If $f\left(x\right).f^{'}\left(x\right)\geq x\sqrt{1 - \left(f \left(x\right)\right)^{4}}$ and $\underset{x \rightarrow x_{1}^{+}}{lim}\left(f \left(x\right)\right)^{2}=1$ and $\underset{x \rightarrow x_{2}^{-}}{lim}\left(f \left(x\right)\right)^{2}=\frac{1}{2}$ then minimum value of $x_{1}^{2}-x_{2}^{2}$ is $\lambda $ then $\frac{\pi }{\lambda }$ equals to

Answer 1

$\frac{2 f \left(x\right) f^{'} \left(x\right)}{\sqrt{1 - \left(f^{2} \left(x\right)\right)^{2}}}\geq 2x$
$\Rightarrow \displaystyle \int _{x_{1}}^{x_{2}}\frac{2 f \left(x\right) f^{'} \left(x\right)}{\sqrt{1 - \left(f^{2} \left(x\right)\right)^{2}}}dx\geq \displaystyle \int _{x_{1}}^{x_{2}}2xdx$
$\left(\Rightarrow \left(x_{2}^{2} - x_{1}^{2}\right) \leq \left(sin\right)^{- 1} \left(f^{2} \left(x\right)\right|\right)_{x_{1}}^{x_{2}}$
$\Rightarrow x_{1}^{2}-x_{2}^{2}\geq \frac{\pi }{3}$