1
JEE Main 2026 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{A}_1$ be the bounded area enclosed by the curves $y=x^2+2, x+y=8$ and $y$-axis that lies in the first quadrant. Let $\mathrm{A}_2$ be the bounded area enclosed by the curves $y=x^2+2, y^2=x, x=2$, and $y$-axis that lies in the first quadrant. Then $\mathrm{A}_1-\mathrm{A}_2$ is equal to

A

$\frac{2}{3}(2 \sqrt{2}+1)$

B

$\frac{2}{3}(3 \sqrt{2}+1)$

C

$\frac{2}{3}(\sqrt{2}+1)$

D

$\frac{2}{3}(4 \sqrt{2}+1)$

2
JEE Main 2026 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let a circle of radius 4 pass through the origin O , the points $\mathrm{A}(-\sqrt{3} a, 0)$ and $\mathrm{B}(0,-\sqrt{2} b)$, where $a$ and $b$ are real parameters and $a b \neq 0$. Then the locus of the centroid of $\triangle \mathrm{OAB}$ is a circle of radius

A

$\frac{7}{3}$

B

$\frac{11}{3}$

C

$\frac{5}{3}$

D

$\frac{8}{3}$

3
JEE Main 2026 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The number of the real solutions of the equation: $x|x+3|+|x-1|-2=0$ is

A

3

B

2

C

5

D

4

4
JEE Main 2026 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The value of $\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}}$ is equal to

A

32

B

64

C

12

D

16

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