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  • If <img alt="f(x)=\sin \left ( \cos ^{-1}\left ( \frac{1-2^{2x}}{1+2^{2x}} \right ) \right )" src="/latex-image/?f%28x%29%3D%5Csin%20%5Cleft%20%28%20%5Ccos%20%5E%7B-1%7D%5Cleft%20%28%20%5Cfrac%7B1-2%5E%7B2x%7D%7D%7B1&plus;2%5E%7B2x%7D%7D%20%5Cright

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 \left | a^{2} -b^{2}\right | is _________________.
Option: 1 381
Option: 2 481
Option: 3 581
Option: 4 681

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best_answer

 

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

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

Let 2^x=\tan\theta

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

f(x)=\sin \left(\cos ^{-1}\left(\cos{2\theta}\right)\right)

f(x)=\sin 2\theta

f(x)=\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\frac{2\cdot2^{ x}}{1+2^{2 x}}

\begin{aligned} &\therefore \quad 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 ^{\prime}(1)=\frac{20 \ln 2-32 \ln 2}{25}=-\frac{12}{25} \ln 2 \\ &\text { So, } a =25, b =12 \Rightarrow\left| a ^{2}- b ^{2}\right|=25^{2}-12^{2} \\& =625-144 \\& =481 \end{array}

 

Posted by

Suraj Bhandari

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