JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 19)
If $$f(x) = \sin \left( {{{\cos }^{ - 1}}\left( {{{1 - {2^{2x}}} \over {1 + {2^{2x}}}}} \right)} \right)$$ and its first derivative with respect to x is $$ - {b \over a}{\log _e}2$$ when x = 1, where a and b are integers, then the minimum value of | a2 $$-$$ b2 | is ____________ .
Answer
481
Explanation
$$f(x) = \sin {\cos ^{ - 1}}\left( {{{1 - {{({2^x})}^2}} \over {1 + {{({2^x})}^2}}}} \right)$$
$$ = \sin (2{\tan ^{ - 1}}{2^x})$$
$$f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$$
$$ \therefore $$ $$f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$$
$$ \Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$$
$$ = - {{12} \over {25}}{\log _e}2$$
$$ \Rightarrow a = 25,b = 12$$
$$|{a^2} - {b^2}| = |625 - 144| = 481$$
$$ = \sin (2{\tan ^{ - 1}}{2^x})$$
$$f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$$
$$ \therefore $$ $$f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$$
$$ \Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$$
$$ = - {{12} \over {25}}{\log _e}2$$
$$ \Rightarrow a = 25,b = 12$$
$$|{a^2} - {b^2}| = |625 - 144| = 481$$
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