# Solve the Following Questions

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 $-\frac{b}{a} \log _{c} 2$ when $x=1$, where a and b are integers, then the minimum value of $\left|a^{2}-b^{2}\right|$ is

Solution:

$f(x)=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x}}{1+2^{2 x}}\right)\right)$ at $x=1 ; 2^{2 x}=4$

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

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

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

$\left\{\begin{array}{cc}\text { If } \quad x>1 \Rightarrow \frac{\pi}{2}>\theta>\frac{\pi}{4} \\ \therefore \quad \pi>2 \theta>\frac{\pi}{2}\end{array}\right\}$

$=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$=\frac{2 x}{1+x^{2}}$

Hence, $f(x)=\frac{2 \cdot 2^{x}}{1+2^{2 x}}$

$\therefore f^{\prime}(x)=\frac{\left(1+2^{2 x}\right)\left(2.2^{x} \ln 2\right)-2^{2 x} \cdot 2 \cdot \ln 2 \cdot 2 \cdot 2^{x}}{\left(1+2^{2 x}\right)}$

$\therefore \quad f^{1}(1)=\frac{20 \ln 2-32 \ln 2}{25}=-\frac{12}{25} \ln 2$

So, $a=25, b=12 \Rightarrow\left|a^{2}-b^{2}\right|=25^{2}-12^{2}$

$=625-144$

$=481$