JEE Main 2021 (Online) 26th August Morning Shift | Inverse Trigonometric Functions Question 28 | Mathematics

JEE Main 2021 (Online) 26th August Morning Shift

MCQ (Single Correct Answer)

-1

Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :

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

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

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

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

JEE Main 2021 (Online) 20th July Evening Shift

MCQ (Single Correct Answer)

-1

The value of $$\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right)$$ is equal to :

$${{ - 181} \over {69}}$$

$${{220} \over {21}}$$

$${{ - 291} \over {76}}$$

$${{151} \over {63}}$$

JEE Main 2021 (Online) 20th July Morning Shift

MCQ (Single Correct Answer)

-1

The number of real roots of the equation $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$ is :

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

The number of solutions of the equation

$${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$$, for x$$\in$$[$$-$$1, 1], and [x] denotes the greatest integer less than or equal to x, is :

Infinite

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