JEE MAIN 2026 Previous Year Questions (PYQs) – Page 12 of 25

Let the lines $L_1:\ \vec r=\hat i+2\hat j+3\hat k+\lambda(2\hat i+3\hat j+4\hat k)$ and $L_2:\ \vec r=(4\hat i+\hat j)+\mu(5\hat i+2\hat j+\hat k)$ intersect at point $R$. Let $P$ and $Q$ be points on $L_1$ and $L_2$ such that $|PR|=\sqrt{29}$ and $|OQ|=\sqrt{\frac{47}{3}}$. If $P$ lies in the first octant, then $27(OR)^2$ is equal to:

Let $729, 81, 9, 1, \ldots$ be a sequence and $P_n$ denote the product of the first $n$ terms of this sequence. If $2\sum_{n=1}^{40}(P_n)=\dfrac{3^\alpha-1}{3^\beta}$ and $\gcd(\alpha,\beta)=1$, then $\alpha+\beta$ is equal to:

Let $\vec a = 2\hat i + \hat j - 2\hat k,; \vec b = \hat i + \hat j$ and $\vec c = \vec a \times \vec b$. Let $\vec d$ be a vector such that $|\vec d - \vec a| = \sqrt{11},; |\vec c \times \vec d| = 3$ and the angle between $\vec c$ and $\vec d$ is $\frac{\pi}{4}$. Then $\vec a \cdot \vec d$ is equal to:

If the domain of the function $f(x)=\log_{(10x^2-17x+7)}(18x^2-11x+1)$ is $(-\infty,a)\cup(b,c)\cup(d,\infty)-{e}$, then $90(a+b+c+d+e)$ equals:

Let each of the two ellipses \( E_1:\ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \ (a > b) \) and \( E_2:\ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \ (A < B) \) have eccentricity \( \frac{4}{5} \). Let $\ell_1,\ell_2$ be lengths of latus rectum of $E_1,E_2$ respectively such that $2\ell_1=\ell_2$. If the distance between the foci of $E_1$ is $8$, then the distance between foci of $E_2$ is:

The value of $\dfrac{\sqrt{3}\csc20^\circ-\sec20^\circ}{\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ}$ is equal to:

Let $S={z\in\mathbb{C}:\left|\frac{z-6i}{z-2i}\right|=1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5}}$ Then $\sum_{z\in S}|z|^2$ is equal to:

Let $f(t)=\int \frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}dt,; t>1$ If $f(e^{\pi/2})=-e^{-\pi/2}$ and $f(e^{\pi/4})=\alpha e^{-\pi/4}$, then $\alpha$ equals:

Let $R$ be a relation defined on ${1,2,3,4}\times{1,2,3,4}$ by $R={((a,b),(c,d)):2a+3b=3c+4d}$ Then the number of elements in $R$ is:

Let $A(1,0),; B(2,-1)$ and $C\left(\frac{7}{3},\frac{4}{3}\right)$ be three points. If equation of bisector of angle $ABC$ is $\alpha x+\beta y=5$, then $\alpha^2+\beta^2$ is: