JEE Mains JEE Advance BITSAT
JEE Mains | PYQ

Year

All years 2026 2025 2024 2023 2022 2021

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

If the domain of the function \( f(x) = \log_e\left(4x^2+11x+6\right)+\sin^{-1}(4x+3)+\cos^{-1}\left(\dfrac{10x+6}{3}\right) \) is \( (\alpha,\beta] \), then \( 36|\alpha+\beta| \) is equal to:

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

Let $S = \left\{ x \in R : 0 < x < 1 \text{ and } 2\tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right\}$. If $n(S)$ denotes the number of elements in $S$ then:

The angle of elevation of the top \( P \) of a tower from the feet of one person standing due South of the tower is \( 45^\circ \) and from the feet of another person standing due west of the tower is \( 30^\circ \). If the height of the tower is \( 5 \, \text{meters} \), then the distance (in meters) between the two persons is equal to:

\( \max\limits_{0 \le x \le \pi} \left\{ x - 2 \sin x \cos x + \dfrac{1}{3} \sin 3x \right\} = \)

If \( S=\left\{x\in \mathbb{R}:\sin^{-1}\!\left(\dfrac{x+1}{\sqrt{x^{2}+2x+2}}\right)-\sin^{-1}\!\left(\dfrac{x}{\sqrt{x^{2}+1}}\right)=\dfrac{\pi}{4}\right\} \), then \( \sum\limits_{x\in S}\left(\sin\!\left(\left(x^{2}+x+5\right)\dfrac{\pi}{2}\right)-\cos\!\left(\left(x^{2}+x+5\right)\pi\right)\right) \) is equal to _____.

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2},0\right)$ and directrix $y=-\frac{1}{2}$. Then $S=\left\{x \in R: \tan^{-1}\left(\sqrt{f(x)}\right)+\sin^{-1}\left(\sqrt{f(x)+1}\right)=\frac{\pi}{2}\right\}$:

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

If the domain of the function \( f(x)=\sec^{-1}\!\left(\dfrac{2x}{5x+3}\right) \) is \( [\alpha,\beta)\cup(\gamma,\delta] \), then \( \left|3\alpha+10(\beta+\gamma)+21\delta\right| \) is equal to ______.

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

If the sum of all the solutions of $\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}, -1 < x < 1, x \neq 0,$ is $a - \frac{4}{\sqrt{3}}$, then $a$ is equal to _____.

Let \( a_{1}=1,\, a_{2}, a_{3}, a_{4}, \ldots \) be consecutive natural numbers.
Then \( \tan^{-1}\!\left(\dfrac{1}{1+a_{1}a_{2}}\right)+\tan^{-1}\!\left(\dfrac{1}{1+a_{2}a_{3}}\right)+\ldots+\tan^{-1}\!\left(\dfrac{1}{1+a_{2021}a_{2022}}\right) \) is equal to:

Let \( g(x) = f(x) + f(1-x) \) and \( f''(x) > 0, \; x \in (0,1) \). If \( g \) is decreasing in the interval \( (0, \alpha) \) and increasing in the interval \( (\alpha, 1) \), then \( \tan^{-1}(2\alpha) + \tan^{-1}\!\left(\dfrac{1}{\alpha}\right) + \tan^{-1}\!\left(\dfrac{\alpha + 1}{\alpha}\right) \) is equal to:

For \( x\in(-1,1] \), the number of solutions of the equation \( \sin^{-1}x=2\tan^{-1}x \) is equal to __________.

← PreviousQuestion 10 of 15Next →

JEE Main 2023 10th April Evening Shift

If the domain of the function \( f(x)=\sec^{-1}\!\left(\dfrac{2x}{5x+3}\right) \) is \( [\alpha,\beta)\cup(\gamma,\delta] \), then \( \left|3\alpha+10(\beta+\gamma)+21\delta\right| \) is equal to ______.