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All years 2026 2025 2024 2022

Let $ \dfrac{\pi}{2} < x < \pi $ be such that $ \cot x = -\dfrac{5}{\sqrt{11}} $. Then $ \displaystyle \left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x) $ is equal to:

Let $ \alpha=\dfrac{1}{\sin 60^\circ \sin 61^\circ}+\dfrac{1}{\sin 62^\circ \sin 63^\circ}+\cdots+\dfrac{1}{\sin 118^\circ \sin 119^\circ}. $ Then the value of $ \left(\dfrac{\text{cosec}\, 1^\circ}{\alpha}\right)^2 $ is ________.

$Let\, \alpha = \left(1 - 2\cos\left(\frac{\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{3\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{9\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{27\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{81\pi}{11}\right)\right)$
Then the value of $5 - \alpha^2$ is_______.

Let $ \alpha $ and $ \beta $ be real numbers such that $ -\dfrac{\pi}{4} < \beta < 0 < \alpha < \dfrac{\pi}{4} $. If $ \sin(\alpha+\beta)=\dfrac{1}{3} $ and $ \cos(\alpha-\beta)=\dfrac{2}{3} $, then the greatest integer less than or equal to: $\displaystyle \left( \frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha} \right)^2 $ is __________.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

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JEE Advanced 2026 Paper - 1

Match each entry in List-I to the correct entry in List-II and choose the correct option.

Consider the following lists:

The correct option is:

JEE MainsJEE AdvanceBITSATJEE Advance | PYQ

JEE Mains JEE Advance BITSAT
JEE Advance | PYQ