If f(x) = sin ( ^ - 1 ( 1 - 2 ^ 2x 1 + 2 ^ 2x )) and its first...

Given $f\left(x\right) = \sin \left(\cos^{- 1}\left(\dfrac{1 - {2}^{2x}}{1 + {2}^{2x}}\right)\right)\text{at}x = 1; {2}^{2x} = {\left({2}^{x}\right)}^{2}$

For $\sin \left(\cos^{- 1}\left(\dfrac{1 - {\left({2}^{x}\right)}^{2}}{1 + {\left({2}^{x}\right)}^{2}}\right)\right)$

Let $\tan^{- 1}{2}^{x} = \theta; \theta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$∴ \sin \left(\cos^{- 1}\cos 2\theta\right) = \sin 2\theta$

\left\{$\begin{cases}\text{If} & x > 1 ⇒ \dfrac{\pi}{2} > \theta > \dfrac{\pi}{4} \\ ∴ & \pi > 2\theta > \dfrac{\pi}{2}\end{cases}$\right\}$</p><p>$= 2sin\theta cos\theta = \dfrac{2tan\theta}{1 + \tan^{2}\theta}$</p><p>Hence,$f\left(x\right) = \dfrac{2 \cdot {2}^{x}}{1 + {2}^{2x}}$</p><p>$∴ {f}^{'}\left(x\right) = \dfrac{\left(1 + {2}^{2x}\right)\left(2 . {2}^{x}\ln 2\right) - {2}^{2x} \cdot 2 \cdot \ln 2 \cdot 2 \cdot {2}^{x}}{\left(1 + {2}^{2x}\right)}$</p><p>$∴ {f}^{'}\left(1\right) = \dfrac{20\ln 2 - 32\ln 2}{25} = - \dfrac{12}{25}\ln 2$</p><p>So,$a = 25, b = 12$.</p><p>Then, \left|{a}^{2} - {b}^{2}\right| = {25}^{2} - {12}^{2}$

$= 625 - 144$

$= 481$