1
JEE Main 2026 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\vec{a}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{b}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{c}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\vec{v}=\vec{a} \times \vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9 p^2$ is equal to

A

9

B

6

C

4

D

12

2
JEE Main 2026 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $f(x)=\left\{\begin{array}{cc}\frac{a|x|+x^2-2(\sin |x|)(\cos |x|)}{x} & , x \neq 0 \\ b & , x=0\end{array}\right.$

is continuous at $x=0$, then $a+b$ is equal to :

A

1

B

2

C

0

D

4

3
JEE Main 2026 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The area of the region enclosed between the circles $x^2+y^2=4$ and $x^2+(y-2)^2=4$ is:

A

$\frac{2}{3}(4 \pi-3 \sqrt{3})$

B

$\frac{4}{3}(2 \pi-\sqrt{3})$

C

$\frac{4}{3}(2 \pi-3 \sqrt{3})$

D

$\frac{2}{3}(2 \pi-3 \sqrt{3})$

4
JEE Main 2026 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An equilateral triangle OAB is inscribed in the parabola $y^2=4 x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having $A B$ as a diameter from the origin is

A

$2(8-3 \sqrt{3})$

B

$4(6+\sqrt{3})$

C

$4(3-\sqrt{3})$

D

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

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