JEE MAIN 2026 Previous Year Questions (PYQs) – Page 1 of 25
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If the area of the region ${(x,y) : 1 - 2x \le y \le 4 - x^2,; x \ge 0,; y \ge 0}$ is $\frac{\alpha}{\beta}$, $\alpha, \beta \in \mathbb{N}$, $\gcd(\alpha,\beta) = 1$, then the value of $(\alpha + \beta)$ is:
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Solution
Required area
$= \frac{2}{3}\times 8 - \frac{1}{2}\times \frac{1}{2}\times 1$
$= \frac{16}{3} - \frac{1}{4} = \frac{61}{12} = \frac{\alpha}{\beta}$
$\Rightarrow \alpha + \beta = 73$
Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2},; \ldots, \frac{a_{10}}{2^9}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \cdots + a_{10} = 62$, then $a_1$ is equal to:
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Solution
$\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \cdots = \frac{a_{10}}{2a_9} = \frac{1}{\sqrt{2}}$
$\Rightarrow a_1, a_2, a_3, \ldots, a_{10}$ are in G.P. with common ratio $\sqrt{2}$
$\sum_{i=1}^{10} a_i = a_1 \cdot \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = 62$
$\Rightarrow a_1 = 2(\sqrt{2}-1)$
Let $A = {x : |x^2 - 10| \le 6}$ and $B = {x : |x - 2| > 1}$. Then
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Solution
$|x^2 - 10| \le 6$
$-6 \le x^2 - 10 \le 6$
$4 \le x^2 \le 16$
$\Rightarrow A = [-4, -2] \cup [2, 4]$
$|x - 2| > 1$
$\Rightarrow B = (-\infty, 1) \cup (3, \infty)$
$A \cup B = (-\infty, 1) \cup (2, \infty)$
$A \cap B = [-4, -2] \cup [3, 4]$
$A - B = [2, 3]$
$B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$
For the matrices
$A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -29 & 49 \ -13 & 18 \end{bmatrix}$
if $(A^{15} + B)\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$, then among the following which one is true?
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Solution
Here
$A^n = \begin{bmatrix} 2n+1 & -4n \ n & -2n+1 \end{bmatrix}$
$\Rightarrow A^{15} = \begin{bmatrix} 31 & -60 \ 15 & -29 \end{bmatrix}$
$A^{15} + B = \begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}$
Now
$\begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$
$\Rightarrow 2x - 11y = 0$
$\Rightarrow x = 11,; y = 2$
For a triangle $ABC$, let $\vec{p} = \overrightarrow{BC},; \vec{q} = \overrightarrow{CA}$ and $\vec{r} = \overrightarrow{BA}$. If $|\vec{p}| = 2\sqrt{3},; |\vec{q}| = 2$ and $\cos\theta = \frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then
$|\vec{p}\times(\vec{q}-3\vec{r})|^2 + 3|\vec{r}|^2$ is equal to:
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Solution
$\vec{p} + \vec{q} = \vec{r}$
$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$
$-\frac{1}{\sqrt{3}} = \frac{12 + 4 - |\vec{r}|^2}{2\cdot 2\sqrt{3}\cdot 2}$
$\Rightarrow |\vec{r}|^2 = 24$
$|\vec{p}\times(\vec{q}-3\vec{r})|^2 + 3|\vec{r}|^2$
$= |\vec{p}\times(\vec{q}-3(\vec{p}+\vec{q}))|^2 + 72$
$= |\vec{p}\times(-3\vec{p}-2\vec{q})|^2 + 72$
$= | -2\vec{p}\times\vec{q} |^2 + 72$
$= 4|\vec{p}|^2 |\vec{q}|^2 \sin^2\theta + 72$
$= 4\cdot 12 \cdot 4 \cdot \frac{2}{3} + 72 = 200$
Let $y = y(x)$ be the solution of differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $y(0) = -\frac{7}{4}$. Then $y\left(\frac{\pi}{6}\right)$ is equal to:
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Solution
$\frac{dy}{dx} - 2y\cos x = 2\cos x + 3\sin x \cos x$
$I.F. = e^{-2\sin x}$
$e^{-2\sin x}y = \int e^{-2\sin x}(3\sin x \cos x + 2\cos x),dx$
$= e^{-2\sin x}\left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C$
$\Rightarrow y = -\frac{3}{2}\sin x - \frac{7}{4} + Ce^{2\sin x}$
$y(0) = -\frac{7}{4} \Rightarrow C = 0$
$y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\cdot\frac{1}{2} - \frac{7}{4} = -\frac{5}{2}$
Let $A = {2,3,5,7,9}$. Let $R$ be the relation on $A$ defined by $xRy$ if and only if $2x \le 3y$. Let $\ell$ be the number of elements in $R$, and $m$ be the minimum number of elements required to be added in $R$ to make it a symmetric relation. Then $\ell + m$ is equal to:
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Solution
$A = {2,3,5,7,9}$
$y \ge \frac{2x}{3}$
$x = 2 \Rightarrow y = 2,3,5,7,9$
$x = 3 \Rightarrow y = 2,3,5,7,9$
$x = 5 \Rightarrow y = 5,7,9$
$x = 7 \Rightarrow y = 5,7,9$
$x = 9 \Rightarrow y = 7,9$
$\ell = 18$
To make symmetric, elements to be added:
${(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)}$
$m = 7$
$\ell + m = 25$
The system of equations
$3x + y + 4z = 3$
$2x + ay - z = -3$
$x + 2y + z = 4$
has no solution, then the value of $\alpha$ is equal to:
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Solution
for no solution $\Delta = 0$
$\begin{vmatrix} 3 & 1 & 4 \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{vmatrix} = 0$
$\Rightarrow 3(\alpha + 2) + 1(-1 - 2) + 4(4 - \alpha) = 0$
$\Rightarrow 3\alpha + 6 - 3 + 16 - 4\alpha = 0$
$\Rightarrow 19 - \alpha = 0 \Rightarrow \alpha = 19$
$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \ -3 & \alpha & -1 \ 4 & 2 & 1 \end{vmatrix}$
$\neq 0$
$\therefore$ no solution for $\alpha = 19$
Let $z$ be the complex number satisfying $|z - 5| \le 3$ and having maximum positive principal argument. Then $34\left|\frac{5z - 12}{5iz + 16}\right|^2$ is equal to:
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Solution
$|z - 5| \le 3$
For $\arg(z)$ to be maximum, $z$ lies at point $P$
$z = (4, 4)$
$= \left(4\left(\frac{4}{5}\right),; 4\left(\frac{3}{5}\right)\right)$
$= \left(\frac{16}{5},; \frac{12}{5}\right)$
Now
$34\left|\frac{5z - 12}{5iz + 16}\right|^2$
$= 34\left|\frac{(16 + 12i) - 12}{(16i + 12i^2) - 12i^2}\right|^2$
$= 34\left|\frac{4 + 12i}{16i + 4}\right|^2$
$= 34\left(\frac{16 + 144}{256 + 16}\right)$
$= 34\left(\frac{160}{272}\right) = 20$
The largest $n \in \mathbb{N}$, for which $7^n$ divides $101!$, is:
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Solution
Exponent of $7$ in $101!$
$= \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{7^2} \right\rfloor + \left\lfloor \frac{101}{7^3} \right\rfloor + \cdots$
$= 14 + 2 = 16$
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