1
JEE Main 2026 (Online) 21st January Evening Shift
Numerical
+4
-1
Change Language
If $\int\limits_0^1 4 \cot ^{-1}\left(1-2 x+4 x^2\right) \mathrm{d} x=\mathrm{a\,tan}^{-1}(2)-\mathrm{b\,log}_{\mathrm{e}}(5)$, where $\mathrm{a}, \mathrm{b} \in \mathrm{N}$, then $(2 \mathrm{a}+\mathrm{b})$ is equal to $\_\_\_\_$ .
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2
JEE Main 2026 (Online) 21st January Evening Shift
Numerical
+4
-1
Change Language
Let the maximum value of $\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2$ for $x \in\left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{\mathrm{m}}{\mathrm{n}} \pi^2$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$. Then $\mathrm{m}+\mathrm{n}$ is equal to $\_\_\_\_$。
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3
JEE Main 2026 (Online) 21st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A spherical body of radius $r$ and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_0$. Estimated maximum error in the quantity $\eta$ is : (Ignore errors associated with $\sigma$, $\rho$ and $g$, gravitational acceleration)

A

$2 \left[ \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0} \right]$

B

$2 \left[ \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0} \right]$

C

$\frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}$

D

$2 \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$

4
JEE Main 2026 (Online) 21st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Keeping the significant figures in view, the sum of the physical quantities 52.01 m, 153.2 m and 0.123 m is :

A

205.3 m

B

205 m

C

205.333 m

D

205.33 m

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