1
JEE Main 2023 (Online) 1st February Evening Shift
+4
-1

Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$.

If $$\mathrm{n(S)}$$ denotes the number of elements in $$\mathrm{S}$$ then :

A
$$\mathrm{n}(\mathrm{S})=0$$
B
$$\mathrm{n}(\mathrm{S})=1$$ and only one element in $$\mathrm{S}$$ is less than $$\frac{1}{2}$$.
C
$$\mathrm{n}(\mathrm{S})=1$$ and the elements in $$\mathrm{S}$$ is more than $$\frac{1}{2}$$.
D
$$\mathrm{n}(\mathrm{S})=1$$ and the element in $$\mathrm{S}$$ is less than $$\frac{1}{2}$$.
2
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1

Let $$S$$ be the set of all solutions of the equation $$\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$$. Then $$\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$$ is equal to

A
$$\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$$
B
$$\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$$
C
$$\frac{-2 \pi}{3}$$
D
None
3
JEE Main 2023 (Online) 31st January Evening Shift
+4
-1
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 :
A
$\frac{\pi}{16}$
B
$\frac{\pi}{48}$
C
$\frac{\pi}{8}$
D
$\frac{\pi}{12}$
4
JEE Main 2023 (Online) 31st January Morning Shift
+4
-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

A
16
B
$$\pi$$
C
16 $$-$$ 5$$\pi$$
D
0
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