JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

Let

f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {\frac{{1 - x}}{x}} } \right)} \right),\,\,0 < x < 1

. Then,

View answer and explanation

Correct option: C

{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

The correct answer is C (

{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

) because it is the relationship derived after simplifying the function and finding its derivative.

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