JEE Main 202126 Aug, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002Let f(x)=cos(2tan−1sin(cot−11−xx))f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)f(x)=cos(2tan−1sin(cot−1x1−x)), 0 < x < 1. Then :AOption A. (1−x)2f′(x)−2(f(x))2=0{(1 - x)^2}f'(x) - 2{(f(x))^2} = 0(1−x)2f′(x)−2(f(x))2=0BOption B. (1+x)2f′(x)+2(f(x))2=0{(1 + x)^2}f'(x) + 2{(f(x))^2} = 0(1+x)2f′(x)+2(f(x))2=0COption C. (1−x)2f′(x)+2(f(x))2=0{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0(1−x)2f′(x)+2(f(x))2=0DOption D. (1+x)2f′(x)−2(f(x))2=0{(1 + x)^2}f'(x) - 2{(f(x))^2} = 0(1+x)2f′(x)−2(f(x))2=0CheckSkip · show solution
