Question

If $f ( x )=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x }}{1+2^{2 x }}\right)\right)$ and its first derivative with respect to $x$ is $-\frac{b}{a} \log _{e} 2$ when $x =1$, where $a$ and $b$ are integers, then the minimum value of $\mid a ^{2}- b ^{2} \mid$ is ______.

Answer 1

$f ( x )=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x }}{1+2^{2 x }}\right)\right)$ at $x =1 ; 2^{2 x }=4$
for $\sin \left(\cos ^{-1}\left(\frac{1- x ^{2}}{1+ x ^{2}}\right)\right) ;$
Let $\tan ^{-1} x =\theta ;\,\, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\therefore \sin \left(\cos ^{-1} \cos 2 \theta\right)=\sin 2 \theta$
$\begin{Bmatrix}If &x>1 \Rightarrow \frac{\pi}{2} > \theta > \frac{\pi}{4}\\ \therefore &\pi > 2\theta > \frac{\pi}{2}\end{Bmatrix}$
$=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$
$=\frac{2x}{1+x^{2}}$
Hence, $f(x)=\frac{2 \cdot 2^{x}}{1+2^{2 x}}$
$\therefore f'(x)=\frac{\left(1+2^{2 x}\right)\left(2.2^{x} \ln 2\right)-2^{2 x} \cdot 2 \cdot \ln 2 \cdot 2 \cdot 2^{x}}{\left(1+2^{2 x}\right)}$
$\therefore f^{1}(1)=\frac{20 \ln 2-32 \ln 2}{25}=-\frac{12}{25} \ln 2$
So, a $=25, b=12$
$\Rightarrow \left|a^{2}-b^{2}\right|=25^{2}-12^{2}$
$=625-144$
$=481$