Question

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

Answer 1

Step 1: Determine the value of first derivative of the given function

The given function is $f\left(x\right)=\sin \left[{\cos }^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right]$ and its first derivative with respect to $x$ is $\left(-\frac{b}{a}\right){\log }_{e}2$ when $x=1$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[\sin \left({\cos }^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)\right] \\ \Rightarrow f\text{'}\left(x\right)=\cos \left({\cos }^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)·\frac{d}{dx}\left[{\cos }^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right]\left[\because \frac{d}{\mathrm{dx}}\left[f\left(g\left(x\right)\right)\right]=f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)\right] \end{gathered}$$

$$\begin{gathered} =\frac{1-2^{2x}}{1+2^{2x}}·\left(-\frac{1}{\sqrt{1-{\left({\displaystyle \frac{1-2^{2x}}{1+2^{2x}}}\right)}^2}}\right)·\frac{d}{dx}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\left[\because \frac{d}{\mathrm{dx}}{\cos }^{-1}\left(x\right)=\frac{-1}{\sqrt{1-x^2}}\right] \\ =-\frac{1-2^{2x}}{1+2^{2x}}·\frac{1}{\sqrt{{\displaystyle \frac{{\left(1+2^{2x}\right)}^2-{\left(1-2^{2x}\right)}^2}{{\left(1+2^{2x}\right)}^2}}}}·\frac{d}{dx}\left(\frac{1-2^{2x}}{1+2^{2x}}\right) \\ =-\frac{1-2^{2x}}{1+2^{2x}}·\frac{1}{\sqrt{{\displaystyle \frac{4\left(1\right)\left(2^{2x}\right)}{{\left(1+2^{2x}\right)}^2}}}}·\frac{d}{dx}\left(\frac{1-2^{2x}}{1+2^{2x}}\right) \\ =-\frac{1-2^{2x}}{1+2^{2x}}·\frac{\left(1+2^{2x}\right)}{2·2^x}·\frac{\left(1+2^{2x}\right)·{\displaystyle \frac{d}{dx}}\left(1-2^{2x}\right)-\left(1-2^{2x}\right)·{\displaystyle \frac{d}{dx}}\left(1+2^{2x}\right)}{{\left(1+2^{2x}\right)}^2}\left[\because \frac{d}{\mathrm{dx}}\left(\frac{U}{V}\right)=\frac{V·{\displaystyle \frac{\mathrm{dU}}{\mathrm{dx}}}-U·{\displaystyle \frac{\mathrm{dV}}{\mathrm{dx}}}}{V^2}\right] \\ =-\frac{1-2^{2x}}{2^{x+1}}·\frac{\left(1+2^{2x}\right)·\left(-2^{2x}\ln 2\right){\displaystyle ·\frac{d}{dx}}\left(2x\right)-\left(1-2^{2x}\right)·\left(2^{2x}\ln 2\right)·{\displaystyle \frac{d}{dx}}\left(2x\right)}{{\left(1+2^{2x}\right)}^2}\left[\because \frac{d}{\mathrm{dx}}{a}^x=a^x\ln \left(a\right)\right] \\ =-\frac{1-2^{2x}}{2^{x+1}}·\frac{-2·2^{2x}\ln 2\left(1+2^{2x}\right)-2·2^{2x}\ln 2\left(1-2^{2x}\right)}{{\left(1+2^{2x}\right)}^2} \\ =-\frac{1-2^{2x}}{2^{x+1}}·\frac{-2·2^{2x}\ln 2\left(1+2^{2x}+1-2^{2x}\right)}{{\left(1+2^{2x}\right)}^2} \\ =\frac{1-2^{2x}}{2^{x+1}}·\frac{2·2^{2x}\ln 2\left(2\right)}{{\left(1+2^{2x}\right)}^2} \\ =\frac{1-2^{2x}}{2^{x+1}}·\frac{2^{2x+2}\ln 2}{{\left(1+2^{2x}\right)}^2} \\ =\frac{1-2^{2x}}{2^{x+1}}·\frac{2^{2\left(x+1\right)}\ln 2}{{\left(1+2^{2x}\right)}^2} \\ =\frac{2^{x+1}·\ln 2·\left(1-2^{2x}\right)}{{\left(1+2^{2x}\right)}^2} \end{gathered}$$

Step 2: Determine the value of $f\text{'}\left(1\right)$

Substitute $x=1$ in $f\text{'}\left(x\right)$.

$f\text{'}\left(1\right)=\frac{2^{1+1}·\ln 2·\left(1-2^{2·1}\right)}{{\left(1+2^{2·1}\right)}^2}$

$$\begin{gathered} =\frac{2^2·\ln 2·\left(1-2^2\right)}{{\left(1+2^2\right)}^2} \\ =\frac{4·\ln 2·\left(1-4\right)}{{\left(1+4\right)}^2} \\ =\frac{4·\ln 2·\left(-3\right)}{{\left(5\right)}^2} \\ =-\frac{12·\ln 2}{25} \\ =\left(-\frac{12}{25}\right){\log }_{e}2 \end{gathered}$$

Step 3: Solve for the required minimum value

It is given that $f\text{'}\left(1\right)=\left(-\frac{b}{a}\right){\log }_{e}2$.

$$\begin{gathered} \Rightarrow \left(-\frac{12}{25}\right){\log }_{e}2=\left(-\frac{b}{a}\right){\log }_{e}2 \\ \Rightarrow \frac{b}{a}=\frac{12}{25} \end{gathered}$$

Thus $a=25$ and $b=12$.

the minimum value of $\left|a^2-b^2\right|$ can be given by, ${\left|a^2-b^2\right|}_{\min }=\left|{\left(25\right)}^2-{\left(12\right)}^2\right|$

$$\begin{gathered} \Rightarrow {\left|a^2-b^2\right|}_{\min }=\left|625-144\right| \\ \Rightarrow {\left|a^2-b^2\right|}_{\min }=\left|481\right| \\ \Rightarrow {\left|a^2-b^2\right|}_{\min }=481 \end{gathered}$$

Hence, the minimum value of $\left|a^2-b^2\right|$ is $481$.