JEE Main 2021 (Online) 27th August Evening Shift | Inverse Trigonometric Functions Question 25 | Mathematics

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)

-1

Let M and m respectively be the maximum and minimum values of the function
f(x) = tan^$$-$$1 (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to :

$$2 + \sqrt 3 $$

$$2 - \sqrt 3 $$

$$3 + 2\sqrt 2 $$

$$3 - 2\sqrt 2 $$

JEE Main 2021 (Online) 27th August Morning Shift

MCQ (Single Correct Answer)

-1

If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x² $$-$$ 1 is :

$$\cos \left( {{{4a} \over \pi }} \right)$$

$$\sin \left( {{{2a} \over \pi }} \right)$$

$$\cos \left( {{{2a} \over \pi }} \right)$$

$$\sin \left( {{{4a} \over \pi }} \right)$$

JEE Main 2021 (Online) 26th August Evening Shift

MCQ (Single Correct Answer)

-1

If $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p} $$, then the value of tan p is :

$${{101} \over {102}}$$

$${{50} \over {51}}$$

100

$${{51} \over {50}}$$

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 :

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

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

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

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

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