Question

If $f(x)=\sin \left({\cos }^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)$ 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

Answer: 481

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