1
JEE Main 2022 (Online) 26th July Morning Shift
+4
-1

$$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$$ is equal to :

A
1
B
2
C
$$\frac{1}{4}$$
D
$$\frac{5}{4}$$
2
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1

Let m and M respectively be the minimum and the maximum values of $$f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]$$. Then m + M is equal to :

A
$$1 + \sqrt 2 + \pi$$
B
$$\left( {1 + \sqrt 2 } \right)\pi$$
C
$$\pi + \sqrt 2$$
D
$$1 + \pi$$
3
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1

Let $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$ and $$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$ where the inverse trigonometric functions take principal values. Then, the equation whose roots are $$\alpha$$ and $$\beta$$ is :

A
$$15{x^2} - 8x - 7 = 0$$
B
$$5{x^2} - 12x + 7 = 0$$
C
$$25{x^2} - 18x - 7 = 0$$
D
$$25{x^2} - 32x + 7 = 0$$
4
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

The domain of the function $${\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right)$$ is :

A
$$R - \left\{ { - {1 \over 2},{1 \over 2}} \right\}$$
B
$$( - \infty , - 1] \cup [1,\infty ) \cup \{ 0\}$$
C
$$\left( { - \infty ,{{ - 1} \over 2}} \right) \cup \left( {{1 \over 2},\infty } \right) \cup \{ 0\}$$
D
$$\left( { - \infty ,{{ - 1} \over {\sqrt 2 }}} \right] \cup \left[ {{1 \over {\sqrt 2 }},\infty } \right) \cup \{ 0\}$$
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