If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2}

$${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$

JEE Main · Quantitative Aptitude

JEE Main Differentiation Of Inverse Trigonometric Function Questions with Answers and Solutions

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JEE Main Differentiation Of Inverse Trigonometric Function practice

Topic-wise practice8 questions 8 min

1Preview question 1 of 5 · instant answer reveal

Let

f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)

, 0 < x < 1. Then :

Choose one answer to reveal the solution.

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JEE Main Differentiation Of Inverse Trigonometric Function practice FAQ

Can I solve these JEE Main Differentiation Of Inverse Trigonometric Function questions for free?

Yes. The preview questions, answers and step-by-step solutions on this page are free to read without any login.

What opens after free signup?

Signup unlocks the full JEE Main Differentiation Of Inverse Trigonometric Function topic test, saved attempts, performance analysis, Maya AI doubt-chat, friend challenges and a personalised study plan.

Are the solutions explained step by step?

Yes. Every question here carries a worked solution so you understand the method, not just the final answer.

Prepare together, not alone

JEE Main Differentiation Of Inverse Trigonometric Function practice

What visitors can use on this page

JEE Main Differentiation Of Inverse Trigonometric Function practice FAQ

Prepare together, not alone

JEE Main Differentiation Of Inverse Trigonometric Function practice

What visitors can use on this page

JEE Main Differentiation Of Inverse Trigonometric Function practice FAQ