JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

f'\left( x \right) = {\tan ^{ - 1}}\left( {\sec \,x + \tan \,x} \right),\,\frac{\pi }{2} < x < \frac{\pi }{2}

and

f\left( 0 \right) = 0,

then

f\left( 1 \right)

is equal to

View answer and explanation

Correct option: A

\frac{{\pi + 1}}{4}

Option A ((

\pi

+ 1)/4) is correct because, after simplifying f'(x) and integrating, using the initial condition f(0) = 0, we find f(1) = (

\pi

+ 1)/4.

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