JEE Main 2023 (Online) 31st January Evening Shift | Inverse Trigonometric Functions Question 15 | Mathematics

Let (a, b) $\subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$, holds.

If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :

$\frac{\pi}{16}$

$\frac{\pi}{48}$

$\frac{\pi}{8}$

$\frac{\pi}{12}$

JEE Main 2023 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-1

If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$$, then $${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$$ is equal to :

$$\pi$$

16 $$-$$ 5$$\pi$$

JEE Main 2023 (Online) 24th January Morning Shift

MCQ (Single Correct Answer)

-1

$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$ is equal to :

$${\pi \over 2}$$

$${\pi \over 3}$$

$${\pi \over 6}$$

$${\pi \over 4}$$

JEE Main 2022 (Online) 29th July Evening Shift

MCQ (Single Correct Answer)

-1

The domain of the function $$f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$$ is :

$$[1, \infty)$$

$$[-1,2]$$

$$[-1, \infty)$$

$$(-\infty, 2]$$