Verified Solution • 2025–26 Session moderate

QJEEM21Mad8e5: JEE Main 2021 Mathematics - Class 12 JEEM21Mad8e5 Solution: If and its first derivative with respect to x is when x = 1,…

If f(x)=sin⁡(cos⁡−1(1−22x1+22x))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 −balog⁡e2- {b \over a}{\log _e}2 when x = 1, where a and b are integers, then the minimum value of | a2 −- b2 | is ____________ .

Step-by-Step Verified Solution

f(x)=sin⁡cos⁡−1(1−(2x)21+(2x)2)f(x) = \sin {\cos ^{ - 1}}\left( {{{1 - {{({2^x})}^2}} \over {1 + {{({2^x})}^2}}}} \right)

=sin⁡(2tan⁡−12x)= \sin (2{\tan ^{ - 1}}{2^x})

f′(x)=cos⁡(2tan⁡−12x).2.11+(2x)2×2x.log⁡e2f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2

∴\therefore f′(1)=cos⁡(2tan⁡−12).21+4×2×log⁡e2f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2

⇒f′(1)=cos⁡cos⁡−1(1−221+22).45log⁡e2\Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2

=−1225log⁡e2= - {{12} \over {25}}{\log _e}2

⇒a=25,b=12\Rightarrow a = 25,b = 12

∣a2−b2∣=∣625−144∣=481|{a^2} - {b^2}| = |625 - 144| = 481

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