Question

Let $f$ be a continuous and differentiable function in $\left(x_1,x_2\right).$ If $f\left(x\right).f^{{}^{\prime }}\left(x\right)\ge x\sqrt{1-{\left(f\left(x\right)\right)}^4}$ and $\underset{x\to x_1^{+}}{\lim }{\left(f\left(x\right)\right)}^2=1$ and $\underset{x\to x_2^{-}}{\lim }{\left(f\left(x\right)\right)}^2=\frac{1}{2}$ then minimum value of $x_1^2-x_2^2$ is $\lambda$ then $\frac{\pi }{\lambda }$ equals to

Answer 1

Answer: 3

$\frac{2f\left(x\right)f^{{}^{\prime }}\left(x\right)}{\sqrt{1-{\left(f^2\left(x\right)\right)}^2}}\ge 2x$
$\Rightarrow {\int }_{x_1}^{x_2}\frac{2f\left(x\right)f^{{}^{\prime }}\left(x\right)}{\sqrt{1-{\left(f^2\left(x\right)\right)}^2}}dx\ge {\int }_{x_1}^{x_2}2xdx$
${\left(\Rightarrow \left(x_2^2-x_1^2\right)\le {\left(sin\right)}^{-1}\left(f^2\left(x\right)\right|\right)}_{x_1}^{x_2}$
$\Rightarrow x_1^2-x_2^2\ge \frac{\pi }{3}$