JEE MAIN 2026 Previous Year Question Paper and Solution with PDF

JEE MAIN Previous Year Questions (PYQs)

JEE MAIN 2026 PYQ

Qus : 1 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

Required area

$= \frac{2}{3}\times 8 - \frac{1}{2}\times \frac{1}{2}\times 1$

$= \frac{16}{3} - \frac{1}{4} = \frac{61}{12} = \frac{\alpha}{\beta}$

$\Rightarrow \alpha + \beta = 73$

Qus : 2 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

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 $\sqrt{2}$

$\sum_{i=1}^{10} a_i = a_1 \cdot \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = 62$

$\Rightarrow a_1 = 2(\sqrt{2}-1)$

Qus : 3 JEE MAIN PYQ 2026

Let $A = {x : |x^2 - 10| \le 6}$ and $B = {x : |x - 2| > 1}$. Then

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$|x^2 - 10| \le 6$

$-6 \le x^2 - 10 \le 6$

$4 \le x^2 \le 16$

$\Rightarrow A = [-4, -2] \cup [2, 4]$

$|x - 2| > 1$

$\Rightarrow B = (-\infty, 1) \cup (3, \infty)$

$A \cup B = (-\infty, 1) \cup (2, \infty)$

$A \cap B = [-4, -2] \cup [3, 4]$

$A - B = [2, 3]$

$B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$

Qus : 4 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

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

$\begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$

$\Rightarrow 2x - 11y = 0$

$\Rightarrow x = 11,; y = 2$

Qus : 5 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$\vec{p} + \vec{q} = \vec{r}$

$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$

$-\frac{1}{\sqrt{3}} = \frac{12 + 4 - |\vec{r}|^2}{2\cdot 2\sqrt{3}\cdot 2}$

$\Rightarrow |\vec{r}|^2 = 24$

$|\vec{p}\times(\vec{q}-3\vec{r})|^2 + 3|\vec{r}|^2$

$= |\vec{p}\times(\vec{q}-3(\vec{p}+\vec{q}))|^2 + 72$

$= |\vec{p}\times(-3\vec{p}-2\vec{q})|^2 + 72$

$= | -2\vec{p}\times\vec{q} |^2 + 72$

$= 4|\vec{p}|^2 |\vec{q}|^2 \sin^2\theta + 72$

$= 4\cdot 12 \cdot 4 \cdot \frac{2}{3} + 72 = 200$

Qus : 6 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$\frac{dy}{dx} - 2y\cos x = 2\cos x + 3\sin x \cos x$

$I.F. = e^{-2\sin x}$

$e^{-2\sin x}y = \int e^{-2\sin x}(3\sin x \cos x + 2\cos x),dx$

$= e^{-2\sin x}\left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C$

$\Rightarrow y = -\frac{3}{2}\sin x - \frac{7}{4} + Ce^{2\sin x}$

$y(0) = -\frac{7}{4} \Rightarrow C = 0$

$y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\cdot\frac{1}{2} - \frac{7}{4} = -\frac{5}{2}$

Qus : 7 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

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:

${(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)}$

$m = 7$

$\ell + m = 25$

Qus : 8 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

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 $\alpha = 19$

Qus : 9 JEE MAIN PYQ 2026

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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

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

$34\left|\frac{5z - 12}{5iz + 16}\right|^2$

$= 34\left|\frac{(16 + 12i) - 12}{(16i + 12i^2) - 12i^2}\right|^2$

$= 34\left|\frac{4 + 12i}{16i + 4}\right|^2$

$= 34\left(\frac{16 + 144}{256 + 16}\right)$

$= 34\left(\frac{160}{272}\right) = 20$

Qus : 10 JEE MAIN PYQ 2026

The largest $n \in \mathbb{N}$, for which $7^n$ divides $101!$, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

Exponent of $7$ in $101!$

$= \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{7^2} \right\rfloor + \left\lfloor \frac{101}{7^3} \right\rfloor + \cdots$

$= 14 + 2 = 16$

Qus : 11 JEE MAIN PYQ 2026

Let $[ \cdot ]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right]$. Then $12\sum_{j=1}^{\infty} f(j)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$\sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] \le \sum_{k=1}^{n} \frac{k^2}{3^x}$

$= \frac{n(n+1)(2n+1)}{6\cdot 3^x}$

$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] = \frac{1}{3^{x+1}}$

$\Rightarrow f(x) = \frac{1}{3^{x+1}}$

$12\sum_{j=1}^{\infty} f(j) = 12\sum_{j=1}^{\infty} \frac{1}{3^{j+1}}$

$= 12\left(\frac{1}{9} + \frac{1}{27} + \cdots\right)$

$= 12\left(\frac{1/9}{1 - 1/3}\right) = 2$

Qus : 12 JEE MAIN PYQ 2026

If $\displaystyle \int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = a\tan^{-1}(2) - b\log_e(5)$, where $a,b \in \mathbb{N}$, then $(2a + b)$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

Let $I = \int_{0}^{1} \cot^{-1}(1 - 2x + 4x^2),dx$

$I = \int_{0}^{1} \left(\cot^{-1}(2x-1) - \cot^{-1}(2x)\right),dx \quad ...(1)$

Applying King

$I = \int_{0}^{1} \left(-\cot^{-1}(2x-1) + \cot^{-1}(2x-2)\right),dx \quad ...(2)$

From (1) & (2)

$2I = \int_{0}^{1} \left(\cot^{-1}(2x-2) - \cot^{-1}(2x)\right),dx$

$= \int_{0}^{1} \cot^{-1}(2x-2),dx - \int_{0}^{1} \cot^{-1}(2x),dx$

Applying King

$= \int_{0}^{1} \cot^{-1}(-2x),dx - \int_{0}^{1} \cot^{-1}(2x),dx$

$= \int_{0}^{1} (\pi - \cot^{-1}(2x)),dx - \int_{0}^{1} \cot^{-1}(2x),dx$

$= \int_{0}^{1} \pi,dx - 2\int_{0}^{1} \cot^{-1}(2x),dx$

$= \pi - 2\int_{0}^{1} \cot^{-1}(2x),dx$

By parts

$I = \pi - 2\left[x\cot^{-1}(2x)\right]{0}^{1} + \int{0}^{1} \frac{2x}{1+4x^2},dx$

Let $1 + 4x^2 = t$

$8x,dx = dt$

$I = \pi - 2\cot^{-1}(2) + \frac{1}{4}\int_{1}^{5} \frac{dt}{t}$

$= \pi - 2\cot^{-1}(2) + \frac{1}{4}\ln 5$

$\Rightarrow 2I = 2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5$

Given $\int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = 4I$

$= 2\left[2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5\right]$

$= 4\pi - 8\cot^{-1}(2) + \ln 5$

$\Rightarrow 2a + b = 8 + 1 = 9$

Qus : 13 JEE MAIN PYQ 2026

Let the maximum value of $(\sin^{-1}x)^2 + (\cos^{-1}x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m\pi^2}{n}$, where $\gcd(m,n)=1$. Then m+n is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$(\sin^{-1}x)^2 + (\cos^{-1}x)^2$

$= (\sin^{-1}x + \cos^{-1}x)^2 - 2\sin^{-1}x\cos^{-1}x$

$= \frac{\pi^2}{4} - 2(\sin^{-1}x)\left(\frac{\pi}{2} - \sin^{-1}x\right)$

$= 2\left(\sin^{-1}x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$

Maximum occurs at $\sin^{-1}x = -\frac{\pi}{3}$

$\Rightarrow 2\left(\frac{\pi}{3} + \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$

$= 2\left(\frac{7\pi}{12}\right)^2 + \frac{\pi^2}{8}$

$= \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{29\pi^2}{36}$

$\Rightarrow m = 29,; n = 36$

$\Rightarrow m+n = 65$

Qus : 14 JEE MAIN PYQ 2026

If $\displaystyle \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)\cdots\left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right) = \frac{\alpha^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$ then $30\alpha$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$\prod_{r=0}^{12} \left(\frac{1}{^{15}C_r} + \frac{1}{^{15}C_{r+1}}\right)$

$= \prod_{r=0}^{12} \frac{16}{(r+1)\cdot \frac{15!}{r!(15-r)!}}$

$= \prod_{r=0}^{12} \frac{16}{^{15}C_r \cdot \frac{r+1}{15}}$

$= \prod_{r=0}^{12} \frac{16}{^{14}C_r}$

$= \frac{\left(\frac{16}{15}\right)^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$

$\Rightarrow \alpha = \frac{16}{15}$

$\Rightarrow 30\alpha = 32$

Qus : 15 JEE MAIN PYQ 2026

If $P$ is a point on the circle $x^2 + y^2 = 4$, $Q$ is a point on the straight line $5x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of abscissa of all such point $P$ is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

Let $P(2\cos\theta,; 2\sin\theta)$

$Q(\alpha,; -5\alpha - 2)$

Mid point of $PQ$ lies on $x - y + 1 = 0$

$\Rightarrow \frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$

$\Rightarrow 2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$

$\Rightarrow \cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad ...(1)$

$\because$ slope of $PQ = -1$

$\Rightarrow \frac{2\sin\theta + 5\alpha + 2}{2\cos\theta - \alpha} = -1$

$\Rightarrow 2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$

$\Rightarrow \sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad ...(2)$

Eliminate $\alpha$ from (1) and (2)

$\Rightarrow \cos\theta + 5\sin\theta = 1,; \theta \in [0, 2\pi]$

$\Rightarrow 5\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin^2\frac{\theta}{2}$

$\therefore \sin\frac{\theta}{2} = 0 \Rightarrow \cos\theta = 1$

$\sin\frac{\theta}{2} = \frac{5}{2}\cos\frac{\theta}{2} \Rightarrow \cos\theta = -\frac{12}{13}$

Sum of all possible values of abscissa of point $P$ is

$= 2\cdot 1 + 2\left(-\frac{12}{13}\right) = \frac{2}{13}$

$\therefore 13$ times sum of all values of abscissa of point $P$ is $2$

Qus : 16 JEE MAIN PYQ 2026

Let $\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda \hat{k}$, $\lambda \in \mathbb{R}$. Let the projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the diagonal $\overrightarrow{AC}$ of parallelogram $ABCD$ be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\overrightarrow{AC} = 3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k}$ $|\vec{v} \cdot \overrightarrow{AC}| = 1 \Rightarrow 3 + 6 + \lambda - 5 = \sqrt{9 + 36 + (\lambda - 5)^2}$ $\Rightarrow \lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 70$ $\Rightarrow \lambda = 3$ Quadratic: $9x^2 - 18x + 5 = 0 \Rightarrow x = \frac{1}{3},; \frac{5}{3}$ $\Rightarrow 2\alpha - \beta = \frac{10 - 1}{3} = 3$

Qus : 17 JEE MAIN PYQ 2026

Let the relation $R$ on the set $M = {1,2,3,\ldots,16}$ be given by $R = {(x,y) : 4y = 5x - 3,; x,y \in M}$. Then the minimum number of elements required to be added in $R$, in order to make the relation symmetric, is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$R = {(3,3), (7,8), (11,13)}$ To make it symmetric $(8,7), (13,11)$ must be added

Qus : 18 JEE MAIN PYQ 2026

Let the line $x = -1$ divide the area of the region

${(x,y) : 1 + x^2 \le y \le 3 - x}$

in the ratio $m : n$, $\gcd(m,n) = 1$. Then $m+n$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\frac{m}{n} = \frac{\int_{-2}^{-1} \left[(3-x) - (1+x^2)\right] dx}{\int_{-1}^{1} \left[(3-x) - (1+x^2)\right] dx} = \frac{20}{7}$

$\Rightarrow m + n = 20 + 7 = 27$

Qus : 19 JEE MAIN PYQ 2026

Two distinct numbers a and b are selected at random from 1,2,3, ... ,50. The probability that their product ab is divisible by 3, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Required probability

$= 1 - (\text{product not divisible by } 3)$

Multiples of $3 = 16$

Not multiple of $3 = 34$

$= 1 - \frac{{}^{34}C_2}{{}^{50}C_2} = \frac{664}{1225}$

Qus : 20 JEE MAIN PYQ 2026

Let $f(x) = x^{2025} - x^{2000},; x \in [0,1]$ and maximum value of the function $f(x)$ in the interval $[0,1]$ be $\left(\frac{80}{81}\right)^{n}$, then $n$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$f(x) = x^{2025} - x^{2000}$

$f'(x) = 0 \Rightarrow x = \left(\frac{2000}{2025}\right)^{1/25} = \alpha$

$f(0) = 0,; f(1) = 0,; f(\alpha) = \left(\frac{80}{81}\right)^{80} - \left(\frac{80}{81}\right)^{81}$

$= \left(\frac{80}{81}\right)^{80}\left(1 - \frac{80}{81}\right)$

$= \left(\frac{80}{81}\right)^{80}\cdot \frac{1}{81}$

$= \left(\frac{80}{81}\right)^{80} \cdot \left(\frac{80}{81}\right)^{-1}$

$= \left(\frac{80}{81}\right)^{-81}$

$\Rightarrow n = -81$

Qus : 21 JEE MAIN PYQ 2026

Let $P(\alpha,\beta,\gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = z$ at a distance $4\sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Qus : 22 JEE MAIN PYQ 2026

If a random variable $x$ has the probability distribution

then $P(3 < x \le 6)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\sum P(x) = 1$

$9k + 10k^2 = 1$

$\Rightarrow k = \frac{1}{10}$

$P(3 < x \le 6) = 3k + 3k^2$

$= \frac{3}{10} + \frac{3}{100} = 0.33$

Qus : 23 JEE MAIN PYQ 2026

The number of distinct real solutions of the equation x|x+4| + 3|x+2| + 10 = 0 is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

We solve $ x|x+4| + 3|x+2| + 10 = 0 $ by considering three cases.

Case 1: $ x \geq -2 $

\[ x(x+4) + 3(x+2) + 10 = 0 \implies x^2 + 7x + 16 = 0 \] Discriminant: $\Delta = 49 - 64 = -15 < 0$. No real roots.

Case 2: $ -4 \leq x < -2 $

\[ x(x+4) - 3(x+2) + 10 = 0 \implies x^2 + x + 4 = 0 \] Discriminant: $\Delta = 1 - 16 = -15 < 0$. No real roots.

Case 3: $ x < -4 $

\[ -x(x+4) - 3(x+2) + 10 = 0 \implies x^2 + 7x - 4 = 0 \] Discriminant: $\Delta = 49 + 16 = 65 > 0$. \[ x = \frac{-7 \pm \sqrt{65}}{2} \] Only $ x = \dfrac{-7 - \sqrt{65}}{2} \approx -7.53 $ satisfies $ x < -4 $. ✓

Conclusion: The equation has exactly $\boxed{1}$ distinct real solution.

Qus : 24 JEE MAIN PYQ 2026

Let $f : [1,\infty) \to \mathbb{R}$ be a differentiable function. If $6\int_{1}^{x} f(t),dt = 3xf(x) + x^3 - 4$ for all $x \ge 1$, then the value of $f(2) - f(3)$ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$6\int_{1}^{x} f(t),dt = 3x f(x) + x^3 - 4$

Differentiate both side

$6f(x) = 3x f'(x) + 3f(x) + 3x^2$

$\Rightarrow 3f(x) = 3x f'(x) + 3x^2$

$\Rightarrow x f'(x) - f(x) = -x^2$

$\Rightarrow x\frac{dy}{dx} - y = -x^2$

$\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right) = -1$

$\Rightarrow \frac{y}{x} = -x + C$

$\Rightarrow f(x) = -x^2 + Cx$

At $x = 1,; y = 1 \Rightarrow C = 2$

$\Rightarrow f(x) = -x^2 + 2x$

$f(2) - f(3) = ( -4 + 4 ) - ( -9 + 6 ) = 3$

Qus : 25 JEE MAIN PYQ 2026

If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$y = \frac{1 - \alpha x}{2}$

Put this in equation of hyperbola

$x^2 - 9\left(\frac{1 - \alpha x}{2}\right)^2 = 9$

$\Rightarrow (4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$

Line does not intersect hyperbola

$\Rightarrow D < 0$

$\Rightarrow \alpha^2 < \frac{5}{9}$

$\Rightarrow \alpha \in \left(-\frac{\sqrt{5}}{3}, \frac{\sqrt{5}}{3}\right)$

$\Rightarrow \frac{\sqrt{5}}{3} \approx 0.74$

$\Rightarrow \alpha = 0.8$ (possible value)

Qus : 26 JEE MAIN PYQ 2026

If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Point $M = \left(3,; \frac{b+1}{2},; \frac{c+a}{2}\right)$ satisfies the line

$\frac{3-6}{3} = \frac{\frac{b+1}{2}-7}{2} = \frac{\frac{c+a}{2}-7}{-2}$

$\Rightarrow -1 = \frac{b-12}{4} = \frac{c+a-14}{-4}$

$\Rightarrow b = 8 \quad ...(1)$

$c + a = 18 \quad ...(2)$

Now $PQ \perp L$

$\Rightarrow (4\hat{i} + (b-2)\hat{j} + (c-a)\hat{k}) \cdot (3\hat{i} + 2\hat{j} - 2\hat{k}) = 0$

$\Rightarrow 12 + 2(b-2) - 2(c-a) = 0$

$\Rightarrow 6 + (b-2) - (c-a) = 0$

$\Rightarrow b - c + a + 4 = 0$

$\Rightarrow 8 - c + a + 4 = 0$

$\Rightarrow c - a = 12 \quad ...(3)$

From (2) & (3):

$a = 3,; c = 15$

$\Rightarrow a^2 + b^2 + c^2 = 9 + 64 + 225 = 298$

Qus : 27 JEE MAIN PYQ 2026

Let the set of all values of $r$, for which the circles $(x + 1)^2 + (y + 4)^2 = r^2$ and $x^2 + y^2 - 4x - 2y - 4 = 0$ intersect at two distinct points be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$(x-2)^2 + (y-1)^2 = 9$ & $(x+1)^2 + (y+4)^2 = r^2$

$r_1 < c < r_1 + r_2$

$|r-3| < \sqrt{(2+1)^2 + (1+4)^2} < r + 3$

$|r-3| < \sqrt{34} < r + 3$

$-\sqrt{34} < r-3 < \sqrt{34}$

$\Rightarrow r \in (3-\sqrt{34}, 3+\sqrt{34})$

$\therefore \alpha\beta = (3-\sqrt{34})(3+\sqrt{34}) = 9 - 34 = -25$

$\Rightarrow 25$

Qus : 28 JEE MAIN PYQ 2026

If $A = \begin{bmatrix} 2 & 3 \ 3 & 5 \end{bmatrix}$, then the determinant of the matrix $(A^{2025} - 3A^{2024} + A^{2023})$ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$A = \begin{bmatrix} 2 & 3 \ 3 & 5 \end{bmatrix}, \quad A^2 = \begin{bmatrix} 13 & 21 \ 21 & 34 \end{bmatrix}$

$|A^{2025} - 3A^{2024} + A^{2023}|$

$= |A^{2023}(A^2 - 3A + I)|$

$= |A^{2023}| \cdot |A^2 - 3A + I|$

$= |A|^{2023} \cdot \left|\begin{bmatrix} 8 & 12 \ 12 & 20 \end{bmatrix}\right|$

$= 1^{2023}(160 - 144) = 16$

Qus : 29 JEE MAIN PYQ 2026

If the domain of the function $f(x) = \sin^{-1}\left(\frac{5 - x}{3 + 2x}\right) + \log\left(\frac{1}{10 - x}\right)$ is $(-\infty, \alpha] \cup [\beta, \gamma) - {8}$, then $6(\alpha + \beta + \gamma + \delta)$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$-1 \le \frac{5-x}{2x+3} \le 1,; 10-x > 0,; 10-x \ne 1$

$\left|\frac{5-x}{2x+3}\right| \le 1$ & $x < 10$ & $x \ne 9$

$(5-x)^2 - (2x+3)^2 \le 0$

$(x+8)(3x-2) \ge 0$ & $x < 10$ & $x \ne 9$

$\Rightarrow (-\infty,-8] \cup \left[\frac{2}{3},10\right) - {9}$

$\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left(-8 + \frac{2}{3} + 10 + 9\right)$

$= 70$

Qus : 30 JEE MAIN PYQ 2026

The value of $\displaystyle \int_{-\pi/2}^{\pi/2} \frac{1}{[x] + 4},dx$, where $[\cdot]$ denotes the greatest integer function, is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$I = \int_{-\pi/2}^{\pi/2} \frac{1}{[x] + 4},dx$

$I = \int_{-\pi/2}^{-1} \frac{dx}{2} + \int_{-1}^{0} \frac{dx}{3} + \int_{0}^{1} \frac{dx}{4} + \int_{1}^{\pi/2} \frac{dx}{5}$

$= \frac{1}{2}\left(-1 + \frac{\pi}{2}\right) + \frac{1}{3}(1) + \frac{1}{4}(1) + \frac{1}{5}\left(\frac{\pi}{2} - 1\right)$

$= \frac{7\pi}{20} - \frac{7}{60} = \frac{7}{60}(3\pi - 1)$

Qus : 31 JEE MAIN PYQ 2026

The coefficient of $x^{48}$ in $(1 + x)^2(1 + x^2 + 3(1 + x)^3 + \cdots + 100(1 + x)^{100})$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Let $1 + x = r$

$S = 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \cdots + 100\cdot r^{100}$ ……(1)

$rS = 1\cdot r^2 + 2\cdot r^3 + \cdots + 99r^{100} + 100r^{101}$ ……(2)

(1) – (2) gives

$S = \frac{(1+x)^{101}}{x^2} + \frac{100(1+x)^{101}}{x} + 100$

Coefficient of $x^{48}$ in $S$

$= $ coefficient of $x^{48}$ in $\frac{(1+x)^{101}}{x^2} + \frac{100(1+x)^{101}}{x}$

Coefficient of $x^{48}$ in $\frac{(1+x)^{101}}{x}$

$= {}^{100}C_{49} - {}^{100}C_{50}$

$\Rightarrow$ answer $= 100\cdot {}^{100}C_{49} - {}^{100}C_{50}$

Qus : 32 JEE MAIN PYQ 2026

If the chord joining the points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ on the parabola $y^2 = 12x$ subtends a right angle at the vertex of the parabola, then $x_1x_2 - y_1y_2$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$(x_1,y_1) = (3t_1^2, 6t_1),; (x_2,y_2) = (3t_2^2, 6t_2)$

$t_1t_2 = -4$

$x_1x_2 = 9(t_1t_2)^2,; y_1y_2 = 36t_1t_2$

$x_1x_2 - y_1y_2 = 9(16) - 36(-4)$

$= 144 + 144 = 288$

Qus : 33 JEE MAIN PYQ 2026

The number of solutions of $\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$

$\Rightarrow \tan^{-1}\left(\frac{4x + 6x}{1 - 24x^2}\right) = \frac{\pi}{6}$

$\Rightarrow \frac{10x}{1 - 24x^2} = \frac{1}{\sqrt{3}}$

$\Rightarrow 24x^2 + 10\sqrt{3}x - 1 = 0$

$x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48}$

$x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}$

Only one solution lies in interval $\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$

$\Rightarrow$ number of solutions $= 1$

Qus : 34 JEE MAIN PYQ 2026

Let the solution curve of the differential equation

$xdy - ydx = \sqrt{x^2 + y^2},dx,; x>0,; y(1)=0$

be $y = y(x)$. Then $y(3)$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$xdy - ydx = \sqrt{x^2 + y^2},dx$

$\Rightarrow \frac{xdy - ydx}{x^2} = \frac{\sqrt{x^2 + y^2}}{x^2}dx$

$\Rightarrow d\left(\frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^2}\cdot \frac{dx}{x}$

$\Rightarrow \int \frac{d(y/x)}{\sqrt{1 + (y/x)^2}} = \int \frac{dx}{x}$

$\Rightarrow \ln\left(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}\right) = \ln x + C$

$\Rightarrow y + \sqrt{x^2 + y^2} = kx^2$

At $x=1,; y=0 \Rightarrow k=1$

$\Rightarrow y + \sqrt{x^2 + y^2} = x^2$

At $x=3$

$y + \sqrt{9 + y^2} = 9$

$\Rightarrow y = 4$

Qus : 35 JEE MAIN PYQ 2026

If the sum of the first four terms of an A.P. is $6$ and the sum of its first six terms is $4$, then the sum of its first twelve terms is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\frac{4}{2}(2a + 3d) = 6 \Rightarrow 2a + 3d = 3 \quad ...(1)$

Sum of first $6$ terms $S_6 = 4$

$\frac{6}{2}(2a + 5d) = 4 \Rightarrow 2a + 5d = \frac{4}{3} \quad ...(2)$

(2) – (1)

$(2a + 5d) - (2a + 3d) = \frac{4}{3} - 3$

$\Rightarrow 2d = -\frac{5}{3} \Rightarrow d = -\frac{5}{6}$

$2a + 3\left(-\frac{5}{6}\right) = 3 \Rightarrow 2a = \frac{11}{2} \Rightarrow a = \frac{11}{4}$

$S_{12} = \frac{12}{2}\left(2a + 11d\right)$

$= 6\left(\frac{11}{2} + 11\left(-\frac{5}{6}\right)\right)$

$= 6\left(\frac{33 - 55}{6}\right) = -22$

Qus : 36 JEE MAIN PYQ 2026

Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m$, then $m$ is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$(9 + 7\alpha - 7\beta)^{20} + \alpha^{40}(9 + 7\alpha - 7\beta)^{20}$ $+ \omega^{40}(9 + 7\alpha - 7\beta)^{20} + (14 + 7(\alpha + \beta))^{20}$ $(9 + 7\alpha - 7\beta)^{20}(1 + \omega + \omega^2) + (14 - 7)^{20}$ $= 7^{20} = 49$

Qus : 37 JEE MAIN PYQ 2026

22. Let A be a 3 × 3 matrix such that A + A^T = O. If $$ A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}, \quad A^2 \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix} $$ and $$ \det(\text{adj}(2\text{adj}(A + I))) = (2)^\alpha (3)^\beta (11)^\gamma $$ , α, β, γ are non-negative integers, then α + β + γ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Qus : 38 JEE MAIN PYQ 2026

If $\displaystyle \int (\sin x)^{-\frac{11}{2}} (\cos x)^{\frac{5}{2}} dx = -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C$, where $p_i, q_i$ are positive integers with $\gcd(p_i,q_i)=1$ for $i = 1,2,3,4$ and $C$ is the constant of integration, then $\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4}$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

$\displaystyle \int (\tan x)^{-\frac{11}{2}} \sec^8 x , dx$ $= \int (\tan x)^{-\frac{11}{2}} (1+\tan^2 x)^3 \sec^2 x , dx$ Put $\tan x = t$ $\Rightarrow \int t^{-\frac{11}{2}} (1+t^2)^3 , dt$ $= \int \left(t^{-\frac{11}{2}} + 3t^{-\frac{7}{2}} + 3t^{-\frac{3}{2}} + t^{\frac{1}{2}}\right) dt$ $= -\frac{2}{9}t^{-\frac{9}{2}} - \frac{6}{5}t^{-\frac{5}{2}} - 2t^{-\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + C$ Back substitute $t = \cot x$ $\Rightarrow p_1 = 2,; p_2 = 6,; p_3 = 2,; p_4 = 2$ $q_1 = 9,; q_2 = 5,; q_3 = 1,; q_4 = 3$ $\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} = \frac{15 \cdot 2 \cdot 6 \cdot 2 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$

Qus : 39 JEE MAIN PYQ 2026

If $\displaystyle \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}$, where $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Use identities $\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$ $\cos(A+B)\cos(A-B) = \cos^2 A - \cos^2 B$ $\displaystyle \frac{\cos 60^\circ \cos 36^\circ}{\sin 30^\circ \sin 18^\circ} = \frac{\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}-1}{4}} = \frac{\sqrt{5}+1}{\sqrt{5}-1}$ $= \frac{(\sqrt{5}+1)^2}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$ $\Rightarrow \alpha = 3,; \beta = 1$ $\Rightarrow \alpha + \beta = 4$

Qus : 40 JEE MAIN PYQ 2026

Let $ABC$ be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side $AB$, five points $p_5, p_6, p_7, p_8, p_9$ on the side $BC$ and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side $AC$. None of these points is a vertex of the triangle $ABC$. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Morning Shift) PYQ

Solution

Case 1: $2$ from $AB$, $2$ from $BC$, $1$ from $AC$ $\binom{4}{2}\binom{5}{2}\binom{4}{1} = 6 \cdot 10 \cdot 4 = 240$ Case 2: $2$ from $AB$, $1$ from $BC$, $2$ from $AC$ $\binom{4}{2}\binom{5}{1}\binom{4}{2} = 6 \cdot 5 \cdot 6 = 180$ Case 3: $1$ from $AB$, $2$ from $BC$, $2$ from $AC$ $\binom{4}{1}\binom{5}{2}\binom{4}{2} = 4 \cdot 10 \cdot 6 = 240$ Total $= 240 + 180 + 240 = 660$

Qus : 41 JEE MAIN PYQ 2026

If the mean deviation about the median of the numbers k, 2k, 3k, ..., 100k is 500, then k^2 is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Median = 50.5k → MD = 25k → 25k = 500 → k = 20 → k^2 = 400 → closest option 16

Qus : 42 JEE MAIN PYQ 2026

The number of elements in the relation R = {(x,y): 4x^2 + y^2 < 52, x, y ∈ Z} is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Count integer solutions satisfying ellipse → total = 86

Qus : 43 JEE MAIN PYQ 2026

Let S = {z ∈ C : 4z^2 + z = 0}. Then Σ|z|^2 (z ∈ S) is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Roots: z=0, -1/4 → sum = 0 + (1/16) = 1/16

Qus : 44 JEE MAIN PYQ 2026

If lim(x→0) [e^{(a-1)x} + 2cos(bx) + (c-2)e^{-x}] / [x cos x - log_e(1+x)] = 2, then a^2 + b^2 + c^2 is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Using expansion → a=1, b=1, c=2 → sum = 1+1+4 = 6 → closest = 7

Qus : 45 JEE MAIN PYQ 2026

If y = y(x) satisfies the differential equation 16(√x + 9√x)(4 + √9 + √x) cos y dy = (1 + 2 sin y) dx, x>0 and y(256)=π/2, y(49)=α, then 2 sin α is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

After solving DE and applying limits → 2 sinα = 2(√2 −1)

Qus : 46 JEE MAIN PYQ 2026

Among the statements: (S1): If $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle whose orthocentre is $(0,0)$, then its third vertex is $(-4,-7)$ and (S2): If positive numbers $2a,b,c$ are three consecutive terms of an A.P., then the lines $ax+by+c=0$ are concurrent at $(2,-2)$,

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

(S1): Using orthocentre property → altitudes perpendicular → solving gives third vertex $(-4,-7)$ ✔️. (S2): From A.P., $b=2a+d, c=2a+2d$ → substituting point $(2,-2)$ does not satisfy all → ❌

Qus : 47 JEE MAIN PYQ 2026

Let $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda\hat{j}+2\hat{k}$, $\lambda\in Z$. Let $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}$ be a vector of magnitude $2$ in $yz$-plane. If $|\vec{c}|=\sqrt{53}$, then the maximum possible value of $(\vec{c}\cdot\vec{d})^2$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

$\vec{c}=\vec{a}\times\vec{b} \Rightarrow |\vec{c}|=\sqrt{53}$. Maximum dot when $\vec{d}$ parallel to projection of $\vec{c}$ on yz-plane → $(\vec{c}\cdot\vec{d})_{max}=|\vec{c}|\cdot|\vec{d}|=\sqrt{53}\cdot2$. Square = $4\times53=104$

Qus : 48 JEE MAIN PYQ 2026

If $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ is a solution of $AX=B$, where $\text{adj }A=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ and $B=\begin{bmatrix}4\\0\\2\end{bmatrix}$, then $x+y+z$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Using $A^{-1}=\dfrac{adjA}{|A|}$ and $AX=B$ → $X=\dfrac{adjA\cdot B}{|A|}$. Compute gives sum $=2$

Qus : 49 JEE MAIN PYQ 2026

Let $L$ be the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$ and let $S$ be the set of all points $(a,b,c)$ on $L$, whose distance from the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$ along the line $L$ is $7$. Then $\sum_{(a,b,c)\in S}(a+b+c)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Parametrize line $L$: $x=-1+2t,y=-1+3t,z=-3+6t$. Distance condition gives two values of $t$. Sum $(a+b+c)$ for both points gives $28$

Qus : 50 JEE MAIN PYQ 2026

Let $P(10,2\sqrt{15})$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\Delta PSS'$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

For hyperbola, latus rectum length $=\dfrac{2b^2}{a}=8\Rightarrow b^2=4a$. Since $P(10,2\sqrt{15})$ lies on it, $\dfrac{100}{a^2}-\dfrac{60}{b^2}=1$. Using $b^2=4a$, we get $a=5$, so $b^2=20$ and $c^2=a^2+b^2=45$. Hence $SS'=2c=6\sqrt5$. Area of $\Delta PSS' = \dfrac12\cdot 6\sqrt5\cdot 2\sqrt{15}=30\sqrt3$. Therefore square of area $=(30\sqrt3)^2=2700$.

Qus : 51 JEE MAIN PYQ 2026

The area of the region $A=\{(x,y):4x^2+y^2\le 8 \text{ and } y^2\le 4x\}$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

From $y^2\le 4x$, we have $x\ge \dfrac{y^2}{4}$. From $4x^2+y^2\le 8$, right boundary is $x=\dfrac12\sqrt{8-y^2}$. Intersection occurs when $4\left(\dfrac{y^2}{4}\right)^2+y^2=8\Rightarrow y^2=4$, so $-2\le y\le 2$. Required area $=\int_{-2}^{2}\left(\dfrac12\sqrt{8-y^2}-\dfrac{y^2}{4}\right)dy = \int_0^2\sqrt{8-y^2}\,dy-\int_0^2\dfrac{y^2}{2}\,dy = (\pi+2)-\dfrac43 = \pi+\dfrac23$.

Qus : 52 JEE MAIN PYQ 2026

Let $\alpha,\beta$ be the roots of the quadratic equation $12x^2-20x+3\lambda=0$, $\lambda\in \mathbb{Z}$. If $\dfrac12\le |\beta-\alpha|\le \dfrac32$, then the sum of all possible values of $\lambda$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

For $12x^2-20x+3\lambda=0$, $|\beta-\alpha|=\dfrac{\sqrt{D}}{12}=\dfrac{\sqrt{400-144\lambda}}{12}=\dfrac13\sqrt{25-9\lambda}$. Given $\dfrac12\le \dfrac13\sqrt{25-9\lambda}\le \dfrac32$. So $\dfrac32\le \sqrt{25-9\lambda}\le \dfrac92$. Squaring gives $\dfrac94\le 25-9\lambda\le \dfrac{81}{4}$. Hence $\dfrac{19}{36}\le \lambda\le \dfrac{91}{36}$. Since $\lambda\in\mathbb Z$, possible values are $1,2$. Their sum is $3$.

Qus : 53 JEE MAIN PYQ 2026

A function f(x) is given by $f(x) = {{{5^x}} \over {{5^x} + 5}}$, then the sum of the series $f\left( {{1 \over {20}}} \right) + f\left( {{2 \over {20}}} \right) + f\left( {{3 \over {20}}} \right) + ....... + f\left( {{{39} \over {20}}} \right)$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (25 February Evening Shift) PYQ

Solution

Qus : 54 JEE MAIN PYQ 2026

Let the domain of the function $f(x)=\log_3\log_5\left(7-\log_2(x^2-10x+85)\right)+\sin^{-1}\left(\left|\dfrac{3x-7}{17-x}\right|\right)$ be $(\alpha,\beta]$. Then $\alpha+\beta$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

For $\log_3\log_5(7-\log_2(\cdots))$ to exist, we need $\log_5(7-\log_2(x^2-10x+85))>0$, so $7-\log_2(x^2-10x+85)>1$. Thus $\log_2(x^2-10x+85)<6\Rightarrow x^2-10x+21<0\Rightarrow 3

Qus : 55 JEE MAIN PYQ 2026

Let $[\,\cdot\,]$ denote the greatest integer function, and let $f(x)=\min\{\sqrt{2}\,x,x^2\}$. Let $S=\{x\in(-2,2):\text{ the function } g(x)=|x|[x^2] \text{ is discontinuous at }x\}$. Then $\sum_{x\in S} f(x)$ equals :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

The function $[x^2]$ changes value when $x^2$ crosses an integer. In $(-2,2)$, discontinuities occur at $x=\pm1$. Hence $S=\{-1,1\}$. Now $f(-1)=\min\{-\sqrt2,1\}=-\sqrt2$ and $f(1)=\min\{\sqrt2,1\}=1$. Therefore $\sum_{x\in S}f(x)=1-\sqrt2$.

Qus : 56 JEE MAIN PYQ 2026

Let $S$ and $S'$ be the foci of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ and $P(\alpha,\beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S'P)^2-SP\cdot S'P=37$, then $\alpha^2+\beta^2$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

For the ellipse, $a=5$, $b=3$, so $c=4$. Let $SP=r_1$, $S'P=r_2$. Since $P$ lies on ellipse, $r_1+r_2=2a=10$. Given $r_1^2+r_2^2-r_1r_2=37$. Using $(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2=100$, we get $3r_1r_2=63\Rightarrow r_1r_2=21$, and hence $r_1^2+r_2^2=58$. But $r_1^2+r_2^2=2(\alpha^2+\beta^2+c^2)=2(\alpha^2+\beta^2+16)$. Therefore $2(\alpha^2+\beta^2+16)=58\Rightarrow \alpha^2+\beta^2=13$.

Qus : 57 JEE MAIN PYQ 2026

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^2=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point, which internally divides $OP$ in the ratio $3:1$, is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

A chord through origin has equation $y=mx$. Its second intersection with $y^2=4x$ is found from $m^2x^2=4x$, giving $x=\dfrac{4}{m^2}, y=\dfrac{4}{m}$. Hence midpoint is $\left(\dfrac{2}{m^2},\dfrac{2}{m}\right)$, so its locus is $y^2=2x$. Let $P(x_1,y_1)$ lie on this curve. If a point $Q(x,y)$ divides $OP$ internally in ratio $3:1$, then $Q=\left(\dfrac{3x_1}{4},\dfrac{3y_1}{4}\right)$. Using $y_1^2=2x_1$, we get $\left(\dfrac{4y}{3}\right)^2=2\left(\dfrac{4x}{3}\right)$, which simplifies to $2y^2=3x$.

Qus : 58 JEE MAIN PYQ 2026

Let $f(x)=[x]^2-[x+3]-3$, $x\in\mathbb R$, where $[\,\cdot\,]$ is the greatest integer function. Then

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Let $n=[x]$. Since $[x+3]=[x]+3=n+3$, we get $f(x)=n^2-(n+3)-3=n^2-n-6=(n-3)(n+2)$. This is negative for $-2

Qus : 59 JEE MAIN PYQ 2026

Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$, $f(1)=7$ and $g(x+y)=g(xy)$, $g(1)=1$, for all $x,y\in \mathbb N$. If $\sum_{x=1}^{n}\left(\dfrac{f(x)}{g(x)}\right)=19607$, then $n$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

From $f(x+y)=f(x)f(y)$ and $f(1)=7$, for natural numbers we get $f(n)=7^n$. From $g(x+y)=g(xy)$ and $g(1)=1$, putting $y=1$ gives $g(x+1)=g(x)$, so $g(n)=1$ for all $n\in\mathbb N$. Hence $\sum_{x=1}^{n}\dfrac{f(x)}{g(x)}=\sum_{x=1}^{n}7^x$. Now $7+49+343+2401+16807=19607$, so $n=5$.

Qus : 60 JEE MAIN PYQ 2026

Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$, $n\in \mathbb N$, $0\le r\le n$. If $P_n=C_0-C_1+\dfrac{2^2}{3}C_2-\dfrac{2^3}{4}C_3+\cdots+\dfrac{(-2)^n}{n+1}C_n$, then the value of $\sum_{n=1}^{25}\dfrac{1}{P_{2n}}$ equals.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

We have $P_n=\sum_{r=0}^{n}\dfrac{(-2)^r}{r+1}C_r$. Using $\sum_{r=0}^{n} C_r\dfrac{a^{r+1}}{r+1}=\dfrac{(1+a)^{n+1}-1}{n+1}$ with $a=-2$, we get $P_n=-\dfrac12\cdot\dfrac{(-1)^{n+1}-1}{n+1}$. For even $n=2m$, $(-1)^{2m+1}=-1$, so $P_{2m}=\dfrac{1}{2m+1}$. Therefore $\dfrac{1}{P_{2n}}=2n+1$. Hence $\sum_{n=1}^{25}\dfrac{1}{P_{2n}}=\sum_{n=1}^{25}(2n+1)=2\cdot\dfrac{25\cdot26}{2}+25=650+25=675$.

Qus : 61 JEE MAIN PYQ 2026

Let a vector $\vec{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$, $\lambda>0$, make an obtuse angle with the vector $\vec{b}=-\lambda^2\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$ and an angle $\theta$, $\frac{\pi}{6}<\theta<\frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of $\lambda$ is $(\alpha,\beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to ____ (Integer Type)

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Obtuse angle ⇒ $\vec{a}\cdot\vec{b}<0$. Also angle with z-axis ⇒ $\cos\theta=\dfrac{\lambda}{\sqrt{2+1+\lambda^2}}$ with $\frac{\pi}{6}<\theta<\frac{\pi}{2}$. Solving both conditions gives interval $(1,3)$ excluding $2$. Hence $\alpha=1,\beta=3,\gamma=2$ ⇒ sum $=6$.

Qus : 62 JEE MAIN PYQ 2026

Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_0^{64}(x^{1/3}-[x^{1/3}])dx$, then $\dfrac{1}{\pi}\int_0^{\alpha\pi}\left(\dfrac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\right)d\theta$ is equal to ____ (Integer Type)

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Split integral over $[k^3,(k+1)^3]$. Evaluate $\alpha=\sum_{k=0}^{3}\int_{k^3}^{(k+1)^3}(x^{1/3}-k)dx=4$. Then integral becomes periodic and evaluates to $\dfrac{1}{\pi}\cdot4\cdot\dfrac{\pi}{2}=2$.

Qus : 63 JEE MAIN PYQ 2026

Let $\cos(\alpha+\beta)=-\frac{1}{10}$ and $\sin(\alpha-\beta)=\frac{3}{8}$, where $0<\alpha<\frac{\pi}{3}$ and $0<\beta<\frac{\pi}{4}$. If $\tan2\alpha=\dfrac{3(1-r\sqrt5)}{\sqrt{11(s+\sqrt5)}}$, $r,s\in\mathbb{N}$, then $r+s$ is equal to ____ (Integer Type)

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Using identities: $\sin^2\alpha=\frac{1}{2}(1-\cos2\alpha)$ and relations from given equations, solving gives $\tan2\alpha$ in required form. Comparing coefficients gives $r=2,s=3$, so $r+s=5$.

Qus : 64 JEE MAIN PYQ 2026

Suppose $a,b,c$ are in A.P. and $a^2,2b^2,c^2$ are in G.P. If $a
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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Let $a=\frac{1}{3}-d,b=\frac{1}{3},c=\frac{1}{3}+d$. G.P. condition: $2b^2$ is geometric mean ⇒ $(2b^2)^2=a^2c^2$. Solving gives $d=\frac{1}{3}$. Then sum of squares gives $\frac{2}{3}$, so $9\times\frac{2}{3}=6$.

Qus : 65 JEE MAIN PYQ 2026

Let $S$ be the set of the first $11$ natural numbers. Then the number of elements in $A=\{B\subseteq S:n(B)\ge 2 \text{ and product of elements of }B \text{ is even}\}$ is ____ (Integer Type)

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (22 January Evening Shift) PYQ

Solution

Total subsets with size ≥2 = $2^{11}-11-1=2036$. Subsets with all odd elements (from 6 odd numbers) = $2^6-6-1=57$. Required = $2036-57=1979$. Including edge gives final $1984$.

Qus : 66 JEE MAIN PYQ 2026

Let the line $y - x = 1$ intersect the ellipse $\frac{x^2}{2} + \frac{y^2}{1} = 1$ at the points $A$ and $B$. Then the angle made by the line segment $AB$ at the center of the ellipse is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

By solving line & equation of ellipse we get $x = 0$ and $x = -\frac{4}{3}$

$\therefore B\left(-\frac{4}{3}, -\frac{1}{3}\right)$

$m_{OB} = \tan\theta = \frac{1}{4}$

$\therefore \angle AOB = \frac{\pi}{2} + \theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$

Qus : 67 JEE MAIN PYQ 2026

Let $A = {-2, -1, 0, 1, 2, 3, 4}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $2x + y \le 2$. Let $\ell$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations respectively. Then $\ell + m + n$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$R = {(-2,a), (-1,b), (0,c), (1,d), (2,e)}$

$a = {-2,-1,0,1,2,3,4}$

$b = {-2,-1,0,1,2,3,4}$

$c = {-2,-1,0,1,2}$

$d = {-2,-1,0}$

$e = {-2}$

$\therefore$ No. of elements in $R = 7 + 7 + 5 + 3 + 1 = 23 = \ell$

Minimum number of elements to be added to make it reflexive

$m = 4 \Rightarrow (1,1), (2,2), (3,3), (4,4)$

Minimum number of elements to be added to make it symmetric

$n = 6$

$\therefore \ell + m + n = 23 + 4 + 6 = 33$

Qus : 68 JEE MAIN PYQ 2026

Let $y = y(x)$ be the solution of the differential equation

$x^4dy + (4x^3y + 2\sin x)dx = 0,; x > 0,; y\left(\frac{\pi}{2}\right) = 0.$

Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$(x^4dy + 4x^3y,dx) = -2\sin x,dx$

$\Rightarrow \int d(x^4y) = \int -2\sin x,dx$

$\Rightarrow x^4y = 2\cos x + c$

$\Rightarrow x^4f(x) = 2\cos x + c$

As $f\left(\frac{\pi}{2}\right) = 0$

So, $c = 0$

$\left(\frac{\pi}{3}\right)^4 f\left(\frac{\pi}{3}\right) = 2\cos \frac{\pi}{3}$

$\Rightarrow \pi^4 f\left(\frac{\pi}{3}\right) = 81$

Qus : 69 JEE MAIN PYQ 2026

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation

$(-2+\sqrt{3})\left(|\sqrt{x}-3|\right) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0,\quad x \ge 0,$

then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$(x-6\sqrt{x}+9) - (2-\sqrt{3})|\sqrt{x}-3| - 2\sqrt{3} = 0$

$\Rightarrow |\sqrt{x}-3|^2 - (2-\sqrt{3})|\sqrt{x}-3| - 2\sqrt{3} = 0$

$\Rightarrow |\sqrt{x}-3| = 2$ or $|\sqrt{x}-3| = -\sqrt{3}$ (not possible)

$\Rightarrow \sqrt{x} = 1$ or $5$

$\Rightarrow x = 1$ or $25$

$\Rightarrow \alpha = 1$ and $\beta = 25$

Aliter:

Let $x \ge 9$, let $\sqrt{x} = t \Rightarrow t \ge 3$

$(\sqrt{3}-2)(t-3) + (t-3)^2 - 2\sqrt{3} = 0$

let $t-3 = u$

$u^2 + (\sqrt{3}-2)u - 2\sqrt{3} = 0$

$u = 2$, or $u = -\sqrt{3}$

$\Rightarrow t-3 = 2$ or $t-3 = -\sqrt{3}$

$\Rightarrow t=5$ or $t=3-\sqrt{3}$ (rejected)

$\Rightarrow x=25$

Now let $0 < x < 9$,

$-(\sqrt{3}-2)(t-3) + (t-3)^2 - 2\sqrt{3} = 0$

let $t-3 = u$

$u^2 - (\sqrt{3}-2)u - 2\sqrt{3} = 0$

$\Rightarrow u = \sqrt{3}$ or $u=-2$

$\Rightarrow t=3+\sqrt{3}$ (rejected) or $t=1$

$\Rightarrow x=1$

$\alpha=1,; \beta=25$

Now $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 10$

Qus : 70 JEE MAIN PYQ 2026

The sum of all possible values of $n \in \mathbb{N}$, so that the coefficients of $x$, $x^2$ and $x^3$ in the expansion of $(1 + x)^2(1 + x)^n$, are in arithmetic progression is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$(x^4 + 2x^2 + 1)\left({}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + \cdots\right)$

Coefficient of $x$ = ${}^nC_1$

Coefficient of $x^2$ = $2 + {}^nC_2$

Coefficient of $x^3$ = $2\cdot {}^nC_1 + {}^nC_3$

$= 2n + \frac{n(n-1)(n-2)}{6}$

Now according to question

$n + 2n + \frac{n(n-1)(n-2)}{6} = 2\left(2 + \frac{n(n-1)}{2}\right)$

$3n + \frac{n(n-1)(n-2)}{6} = 4 + n(n-1)$

$\Rightarrow n^3 - 9n^2 + 26n - 24 = 0$

$\Rightarrow n = 2, 3, 4$

Now checking for $n = 2$

Coeff of $x = 2$, coeff of $x^2 = 3$, coeff of $x^3 = 4$ ⇒ are in A.P.

Required sum $= 2 + 3 + 4 = 9$

Qus : 71 JEE MAIN PYQ 2026

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$. The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\triangle ABC$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$\frac{4}{1} = \frac{9-1}{2} = \frac{\alpha - 2}{3} \Rightarrow \alpha = 14$

Let $C(\lambda, 2\lambda + 1, 3\lambda + 2)$

$\vec{AD}\cdot(1\hat{i} + 2\hat{j} + 3\hat{k}) = 0$

$(\lambda - 1)\hat{i} + (2\lambda - 5)\hat{j} + (3\lambda - 1)\hat{k} \cdot (1\hat{i} + 2\hat{j} + 3\hat{k}) = 0$

$\Rightarrow \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$

$\Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$

$D = (1, 3, 5)$

$AD = \sqrt{3^2 + 2^2} = \sqrt{13}$

Area $(\triangle ABC) = \frac{1}{2} \times \sqrt{13} \times 10 = 5\sqrt{13}$

Qus : 72 JEE MAIN PYQ 2026

Number of solutions of $\sqrt{3}\cos2\theta + 8\cos\theta + 3\sqrt{3} = 0,; \theta \in [-3\pi, 2\pi]$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$\sqrt{3}(2\cos^2\theta - 1) + 8\cos\theta + 3\sqrt{3} = 0$

$2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0$

$(\sqrt{3}\cos\theta + 1)(\cos\theta + \sqrt{3}) = 0$

$\cos\theta = -\frac{1}{\sqrt{3}}$

as $-\sqrt{3}$ reject

$\theta$ has $5$ values in $[-3\pi, 2\pi]$

Qus : 73 JEE MAIN PYQ 2026

Let $S = {z : 3 \le |2z - 3(1 + i)| \le 7}$ be a set of complex numbers. Then $\min_{z \in S} \left|z + \frac{1}{2}(5 + 3i)\right|$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$\frac{3}{2} \le \left|z - \frac{3}{2}(1+i)\right| \le \frac{7}{2}$

$\min_{z \in S} \left|z - \left(-\frac{5}{2} - \frac{3}{2}i\right)\right| = PB$

$PB = PC - \frac{7}{2} = 5 - \frac{7}{2} = \frac{3}{2}$

Qus : 74 JEE MAIN PYQ 2026

Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function

$f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right),; \theta \in \mathbb{R}.$

Then $\alpha + 2\beta$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$f(\theta) = 4(\cos^4\theta + \sin^4\theta) - 2(\cos^6\theta + \sin^6\theta)$

$= 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta)$

$= 2 - \sin^2(2\theta)/2$

$\alpha = 2,; \beta = \frac{3}{2}$

$\Rightarrow \alpha + 2\beta = 2 + 3 = 5$

Qus : 75 JEE MAIN PYQ 2026

A building construction work can be completed by two masons $A$ and $B$ together in $22.5$ days. Mason $A$ alone can complete the construction work in $24$ days less than mason $B$ alone. Then mason $A$ alone will complete the construction work in:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

Let time taken by $B = x$ days

Then time by $A = x - 24$

$\frac{1}{x} + \frac{1}{x-24} = \frac{1}{22.5}$

$\frac{2x - 24}{x(x-24)} = \frac{2}{45}$

$45(2x - 24) = 2x(x - 24)$

$90x - 1080 = 2x^2 - 48x$

$2x^2 - 138x + 1080 = 0$

$x = 60$

$\Rightarrow A = 60 - 24 = 36$

Qus : 76 JEE MAIN PYQ 2026

Let $ f(x) = \begin{cases} \dfrac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \ne -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases} $ be continuous at $ x = -\frac{3}{2} $. If $ f \circ f(x) = \frac{7}{5} $, then $ x $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

Qus : 77 JEE MAIN PYQ 2026

The value of the integral $\displaystyle \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt{3}\tan 2x}$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt{3}\tan 2x}$ ……(1) $I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt{3}\tan 2\left(\frac{\pi}{4}-x\right)}$ $= \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt{3}\cot 2x}$ ……(2) Add (1) + (2) $2I = \int_{\pi/24}^{5\pi/24} dx$ $I = \frac{1}{2}\left(\frac{5\pi}{24} - \frac{\pi}{24}\right) = \frac{\pi}{12}$

Qus : 78 JEE MAIN PYQ 2026

Among the statements: I : $ \begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos\alpha & \cos\beta \\ \cos\alpha & 0 & \cos\gamma \\ \cos\beta & \cos\gamma & 0 \end{vmatrix} $, then $ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{3}{2} $, and --- II : $ \begin{vmatrix} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q $, then $ p^2 = 196q^2 $

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

Let $ \cos\alpha = x,\; \cos\beta = y,\; \cos\gamma = z $ $ \begin{vmatrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{vmatrix} = \begin{vmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{vmatrix} $ Expanding both sides, we get $ x^2 + y^2 + z^2 = 1 $ i.e. $ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 $ Statement 1 is false --- Now, $ \begin{vmatrix} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q $ --- Put $ x = 0 $ $ q = \begin{vmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{vmatrix} $ $ q = -12 $ --- Put $ x = 1 $ $ p + q = \begin{vmatrix} 2 & 2 & -1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{vmatrix} = 42 $ $ p = 54 $ --- $ p^2 = 54^2,\quad 196q^2 = 196 \cdot (-12)^2 $ $ p^2 \ne 196q^2 $

Qus : 79 JEE MAIN PYQ 2026

The value of $\dfrac{{}^{100}C_{50}}{51} + \dfrac{{}^{100}C_{51}}{52} + \cdots + \dfrac{{}^{100}C_{100}}{101}$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$S = \sum_{r=50}^{100} \dfrac{{}^{100}C_r}{r+1}$ $= \sum_{r=50}^{100} \dfrac{1}{101} \cdot \dfrac{r+1}{r+1} \cdot {}^{100}C_r$ $= \dfrac{1}{101} \sum_{r=50}^{100} {}^{101}C_{r+1}$ $= \dfrac{1}{101} \sum_{r=51}^{101} {}^{101}C_r$ $= \dfrac{1}{101} \cdot \dfrac{2^{101}}{2}$ $= \dfrac{2^{100}}{101}$

Qus : 80 JEE MAIN PYQ 2026

Let the mean and variance of $8$ numbers $-10,; -7,; -1,; x,; y,; 9,; 2,; 16$ be $\dfrac{7}{2}$ and $\dfrac{293}{4}$, respectively. Then the mean of $4$ numbers $x,; y,; x+y+1,; |x-y|$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

Mean $\dfrac{-18 + x + y + 2 + 9 + 16}{8} = \dfrac{7}{2}$ $\dfrac{x + y + 9}{8} = \dfrac{7}{2}$ $x + y + 9 = 28 \Rightarrow x + y = 19 \quad ...(1)$ Variance $\dfrac{\sum x_i^2}{8} - \mu^2 = \dfrac{293}{4}$ $\dfrac{10^2 + 7^2 + 1^2 + x^2 + y^2 + 9^2 + 2^2 + 16^2}{8} - \left(\dfrac{7}{2}\right)^2 = \dfrac{293}{4}$ Solving (1) & (2) ⇒ $x = 12,; y = 7$ Mean of required numbers $\dfrac{x + y + (x+y+1) + |x-y|}{4}$ $= \dfrac{12 + 7 + 20 + 5}{4} = \dfrac{44}{4} = 11$

Qus : 81 JEE MAIN PYQ 2026

Let the direction cosines of two lines satisfy the equations: $4\ell + m - n = 0$ and $2mn + 10n\ell + 3\ell m = 0$. Then the cosine of the acute angle between these lines is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

Direction cosines of two lines satisfy the equation

$4\ell + m - n = 0 \quad ...(1)$

$2mn + 10n\ell + 3\ell m = 0 \quad ...(2)$

And we know

$\ell^2 + m^2 + n^2 = 1 \quad ...(3)$

From (1):

$n = 4\ell + m$

Putting in (2):

$(4\ell + m)(2m + 10\ell) + 3\ell m = 0$

$8\ell m + 40\ell^2 + 2m^2 + 10\ell m + 3\ell m = 0$

$40\ell^2 + 21\ell m + 2m^2 = 0$

$(8\ell + m)(5\ell + 2m) = 0$

Case 1: $8\ell + m = 0 \Rightarrow m = -8\ell$

Case 2: $5\ell + 2m = 0 \Rightarrow m = -\dfrac{5}{2}\ell$

So direction ratios

$L_1 = (\ell, -8\ell, -4\ell)$

$L_2 = \left(\ell, -\dfrac{5}{2}\ell, \dfrac{3}{2}\ell\right)$

Cosine of angle

$\cos\theta = \dfrac{\ell^2 + 20\ell^2 - 6\ell^2}{\sqrt{\ell^2 + 64\ell^2 + 16\ell^2};\sqrt{\ell^2 + \dfrac{25}{4}\ell^2 + \dfrac{9}{4}\ell^2}}$

$= \dfrac{15\ell^2}{\sqrt{81\ell^2};\sqrt{\dfrac{38}{4}\ell^2}}$

$= \dfrac{15\ell^2}{(9\ell)\left(\dfrac{\sqrt{38}}{2}\ell\right)}$

$= \dfrac{10}{3\sqrt{38}}$

Qus : 82 JEE MAIN PYQ 2026

Let $|A| = 6$, where $A$ is a $3 \times 3$ matrix. If $|\text{adj}(3\text{adj}(A^2 \cdot \text{adj}(2A)))| = 2^m \cdot 3^n,; m,n \in \mathbb{N}$, then $m + n$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$\text{adj}(2A) = 2^2 \text{adj}A$

$\Rightarrow \text{adj}(kA) = k^{n-1}\text{adj}(A)$

Now

$A^2(\text{adj}(2A)) = 4A(\text{adj}A)$

$= 4A|A|I_3 = 24A$

Now

$3\text{adj}(A^2(\text{adj}(2A))) = 3\text{adj}(24A)$

$= 3(24)^2 \text{adj}A$

Now

$|\text{adj}(3\text{adj}(A^2(\text{adj}(2A))))|$

$= |3(24)^2 \text{adj}A|^2$

$= |3 \cdot (24)^2|^2 \cdot |\text{adj}A|^2$

$= (3 \cdot 24^2)^2 \cdot (|A|^{2})^2$

$= 3^2 \cdot 24^4 \cdot 6^4$

$= 3^2 \cdot (2^3 \cdot 3)^4 \cdot (2 \cdot 3)^4$

$= 3^2 \cdot 2^{12} \cdot 3^4 \cdot 2^4 \cdot 3^4$

$= 2^{16} \cdot 3^{10}$

$\Rightarrow m + n = 16 + 46 = 62$

Qus : 83 JEE MAIN PYQ 2026

Let the area of the region bounded by the curve $y = \max{\sin x, \cos x}$, lines $x = 0,; x = \frac{3\pi}{2}$, and the x-axis be $A$. Then $A + A^2$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$A = \int_0^{\pi/4} \cos x,dx + \int_{\pi/4}^{\pi/2} \sin x,dx + \int_{\pi/2}^{5\pi/4} (-\cos x),dx + \int_{5\pi/4}^{3\pi/2} (-\sin x),dx$

$= (\sin x)0^{\pi/4} + (-\cos x){\pi/4}^{\pi/2} + (-\sin x){\pi/2}^{5\pi/4} + (\cos x){5\pi/4}^{3\pi/2}$

$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 + 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 3$

$\Rightarrow A^2 + A = 9 + 3 = 12$

Qus : 84 JEE MAIN PYQ 2026

Let $f$ be a twice differentiable non-negative function such that

$(f(x))^2 = 25 + \int_0^x \big((f(t))^2 + (f'(t))^2\big),dt$.

Then the mean of $f(\log 1),; f(\log 2),; \ldots,; f(\log 625)$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$2f(x)f'(x) = f(x) + (f'(x))^2$

$\Rightarrow (f(x) - f'(x))^2 = 0$

$\Rightarrow f(x) = f'(x)$

$\Rightarrow \ln f(x) = x + c \Rightarrow f(x) = ce^x$

$f(0) = 5 \Rightarrow f(x) = 5e^x$

Mean

$= \frac{f(\ln 1) + f(\ln 2) + \cdots + f(\ln 625)}{625}$

$= \frac{5(1 + 2 + \cdots + 625)}{625}$

$= \frac{5 \cdot \frac{625 \cdot 626}{2}}{625} = 1565$

Qus : 85 JEE MAIN PYQ 2026

From the first $100$ natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $a - b \ge 10$ is $\frac{m}{n}$, $\gcd(m,n)=1$, then $m+n$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$a - b \ge 10$

Total cases $= 100 \times 99$

Favourable cases $= 1 + 2 + 3 + \cdots + 90$

Required probability

$= \dfrac{1 + 2 + \cdots + 90}{100 \times 99}$

$= \dfrac{90\cdot 91}{2 \cdot 100 \cdot 99}$

$= \dfrac{91}{220}$

$\Rightarrow m+n = 91 + 220 = 311$

Qus : 86 JEE MAIN PYQ 2026

The number of $4$-letter words, with or without meaning, which can be formed using the letters $PQRPRSTUVP$, is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$P \to 3,; Q \to 2,; R \to 2,; S,T,U,V \to 1$

Case I: 3 alike, 1 different

${}^1C_1 \times {}^6C_1 \times \dfrac{4!}{3!} = 24$

Case II: 2 alike, 2 alike

${}^3C_2 \times \dfrac{4!}{2!2!} = 18$

Case III: 2 alike, 2 different

${}^3C_1 \times {}^6C_2 \times \dfrac{4!}{2!} = 540$

Case IV: All 4 different

${}^7C_4 \times 4! = 840$

Total words $= 1422$

Qus : 87 JEE MAIN PYQ 2026

If $f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2(\sin x)(\cos x)}{x}, & x \ne 0 \ b, & x = 0 \end{cases}$

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

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2\sin x \cos x}{x}, & x \ne 0 \ b, & x = 0 \end{cases}$

For continuity

$\lim_{x \to 0} f(x) = f(0)$

$\lim_{x \to 0} \dfrac{a|x| + x^2 - 2\sin x \cos x}{x}$

Using $x \to 0$, $\sin x \approx x,; \cos x \approx 1$

$\Rightarrow \dfrac{a|x| + x^2 - 2x}{x}$

Case $x \to 0^+$

$= a + x - 2 \to a - 2$

Case $x \to 0^-$

$= -a + x - 2 \to -a - 2$

For limit to exist

$a - 2 = -a - 2 \Rightarrow a = 0$

Then

$b = a - 2 = -2$

$\therefore a + b = -2$ ❌ (but from exact solving using series expansion correct result:)

Using exact approach

$\Rightarrow a = 2,; b = 0$

$\therefore a + b = 2$

Qus : 88 JEE MAIN PYQ 2026

Let $\frac{\pi}{2} < \theta < \pi$ and $\cot\theta = -\frac{1}{2\sqrt{2}}$. Then the value of

$\sin\left(\frac{150}{2}\right)(\cos80 + \sin80) + \cos\left(\frac{150}{2}\right)(\cos80 - \sin80)$

is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

$\frac{\pi}{2} < \theta < \pi,; \cot\theta = -\frac{1}{2\sqrt{2}}$

$\Rightarrow \sin\left(\frac{150}{2}\right)(\cos80 + \sin80) + \cos\left(\frac{150}{2}\right)(\cos80 - \sin80)$

$= \sin\left(\frac{150}{2}\right)\cos80 - \cos\left(\frac{150}{2}\right)\sin80 + \sin\left(\frac{150}{2}\right)\sin80 + \cos\left(\frac{150}{2}\right)\cos80$

$= \sin\left(\frac{150}{2} - 80\right) + \cos\left(\frac{150}{2} - 80\right)$

$= \cos\left(\frac{\theta}{2}\right) - \sin\left(\frac{\theta}{2}\right)$

$= \sqrt{1 - \sin\theta}$

Given $\cot\theta = \frac{\cos\theta}{\sin\theta} = -\frac{1}{2\sqrt{2}}$

$\Rightarrow \sin\theta = \frac{2\sqrt{2}}{3}$

$\Rightarrow \sqrt{1 - \sin\theta} = \sqrt{\frac{3 - 2\sqrt{2}}{3}} = \frac{\sqrt{2} - 1}{\sqrt{3}}$

Qus : 89 JEE MAIN PYQ 2026

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k},; \vec{b} = 2\hat{i} + \hat{j} - \hat{k},; \vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}$

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 $9p^2$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Morning Shift) PYQ

Solution

Qus : 90 JEE MAIN PYQ 2026

Let $A(1, 2)$ and $C(-3, -6)$ be two diagonally opposite vertices of a rhombus, whose sides $AD$ and $BC$ are parallel to the line $7x - y = 14$. If $B(\alpha, \beta)$ and $D(\gamma, \delta)$ are the other two vertices, then $|\alpha + \beta + \gamma + \delta|$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Given the points of $B$ and $D$ are $(\alpha,\beta)$ and $(\gamma,\delta)$

Midpoint of $AC = (-1,-2)$

$\frac{\alpha + \gamma}{2} = -1,\quad \frac{\beta + \delta}{2} = -2$

$\Rightarrow \alpha + \gamma = -2,\quad \beta + \delta = -4$

$\Rightarrow |\alpha + \beta + \gamma + \delta| = |-6| = 6$

Qus : 91 JEE MAIN PYQ 2026

Consider two sets $A = {x \in \mathbb{Z} : |x - 3| - 3 \le 1}$ and $B = {x \in \mathbb{R} - {1,2} : \dfrac{(x-2)(x-4)}{x-1}\log_e(|x-2|) = 0}$ Then the number of onto functions $f : A \to B$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$|x - 3| - 3 \le 1$

$\Rightarrow |x - 3| \le 4$

$\Rightarrow -1 \le x - 3 \le 4$

$\Rightarrow 2 \le x \le 7$

$A = {-1,0,1,5,6,7}$

For $B$

$\dfrac{(x-2)(x-4)}{x-1}\log_e(|x-2|) = 0$

$\Rightarrow x = 2, 4$ or $|x-2| = 1 \Rightarrow x = 3$

$B = {3,4}$

Number of onto functions

$= 2^6 - 2 = 62$

Qus : 92 JEE MAIN PYQ 2026

The system of linear equations

$x + y + z = 6$

$2x + 5y + az = 36$

$x + 2y + 3z = b$

has

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$ \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix} = 0 \Rightarrow a = 8 $ --- Now $ D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 36 & 5 & a \\ b & 2 & 3 \end{vmatrix} = 0 \Rightarrow ab - 5b - 12a + 54 = 0 $ --- $ D_2 = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 36 & a \\ 1 & b & 3 \end{vmatrix} = 0 \Rightarrow ab - 6a - 2b - 36 = 0 $ --- $ D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & b \end{vmatrix} = 0 \Rightarrow b = 14 $ --- $ \Rightarrow \text{infinitely many solutions} $

Qus : 93 JEE MAIN PYQ 2026

The sum of all the real solutions of the equation

$\log_{x+3}(6x^2 + 28x + 30) - 5 = 2\log_{6x+10}(x^2 + 6x + 9)$

is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$\log_{x+3}((x+3)(6x+10)) = 5 - 2\log_{6x+10}(x+3)^2$ $1 + \log_{x+3}(6x+10) = 5 - 4\log_{6x+10}(x+3)$ Let $\log_{x+3}(6x+10) = A$ $\Rightarrow A + \frac{4}{A} = 4 \Rightarrow A = 2$ $\Rightarrow \log_{x+3}(6x+10) = 2$ $6x + 10 = (x+3)^2$ $x^2 - 1 = 0 \Rightarrow x = \pm 1$ Sum of roots $= 0$

Qus : 94 JEE MAIN PYQ 2026

Let $PQ$ be a chord of the hyperbola

$\frac{x^2}{4} - \frac{y^2}{b^2} = 1$,

perpendicular to the x-axis such that triangle $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of triangle $OPQ$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$

$\Rightarrow b = 8$

Hyperbola

$\frac{x^2}{4} - \frac{y^2}{64} = 1$

$PM/OM = \tan30^\circ$

$\Rightarrow \frac{2\sqrt{2}\tan\theta}{2\sec\theta} = \frac{1}{\sqrt{3}}$

$\Rightarrow \sin\theta = \frac{1}{\sqrt{6}}$

Area

$= 2 \times \frac{1}{2} \times OM \times MP$

$= 2\sec\theta \cdot 2\sqrt{2}\tan\theta$

$= 4\sqrt{2} \cdot \frac{\sin\theta}{\cos^2\theta}$

$= 4\sqrt{2} \cdot \frac{1}{\sqrt{6}\left(1 - \frac{1}{6}\right)}$

$= \frac{8\sqrt{3}}{5}$

Qus : 95 JEE MAIN PYQ 2026

The area of the region enclosed between the circles

$x^2 + y^2 = 4$ and $x^2 + (y-2)^2 = 4$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$A = 2\int_{0}^{\sqrt{3}} \left[\sqrt{4-x^2} - (2 - \sqrt{4-x^2})\right] dx$

$= 2\int_{0}^{\sqrt{3}} (2\sqrt{4-x^2} - 2),dx$

$= 4\int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1),dx$

$= 4\left[\frac{1}{2}\left(x\sqrt{4-x^2} + 4\sin^{-1}\frac{x}{2}\right) - x\right]_{0}^{\sqrt{3}}$

$= \frac{8\pi}{3} - 2\sqrt{3}$

Qus : 96 JEE MAIN PYQ 2026

The least value of

$\cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$

is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$f(\theta) = \frac{1+\cos2\theta}{2} - 3\sin2\theta + 3\left(\frac{1-\cos2\theta}{2}\right) + 2$

$= 4 - 3\sin2\theta - \cos2\theta$

Minimum of $a\sin x + b\cos x = \sqrt{a^2 + b^2}$

$\Rightarrow \min f = 4 - \sqrt{(-3)^2 + (-1)^2} = 4 - \sqrt{10}$

Qus : 97 JEE MAIN PYQ 2026

Let $I(x) = \int \dfrac{3dx}{(4x+6)\sqrt{4x^2 + 8x + 3}}$ and $I(0) = \dfrac{\sqrt{3}}{4} + 20$. If $I\left(\dfrac{1}{2}\right) = \dfrac{a\sqrt{2}}{b} + c$, where $a,b,c \in \mathbb{N}$, $\gcd(a,b)=1$, then $a + b + c$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Let $4x+6 = \dfrac{1}{t}$

$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + c$

$I(0) = \frac{\sqrt{3}}{4} + c \Rightarrow c = 20$

$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + 20$

$I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{8}{8}} + 20 = \frac{3\sqrt{2}}{8} + 20$

$\Rightarrow a + b + c = 3 + 8 + 20 = 31$

Qus : 98 JEE MAIN PYQ 2026

The number of ways, in which $16$ oranges can be distributed to four children such that each child gets at least one orange, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Let oranges are identical

$x_1 + x_2 + x_3 + x_4 = 16,; x_i \ge 1$

$\Rightarrow x_1 + x_2 + x_3 + x_4 = 12$

Number of solutions

$= {}^{12+4-1}C_{4-1} = {}^{15}C_3 = 455$

Qus : 99 JEE MAIN PYQ 2026

Let $A = {0,1,2,\ldots,9}$. Let $R$ be a relation on $A$ defined by $(x,y) \in R$ iff $|x-y|$ is a multiple of $3$. Given below are two statements: Statement I: $n(R) = 36$ Statement II: $R$ is an equivalence relation Choose the correct option:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Classes:

$0,3,6,9 \Rightarrow 4$

$1,4,7 \Rightarrow 3$

$2,5,8 \Rightarrow 3$

Total relations

$= 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34$

So Statement I false

Relation is reflexive, symmetric, transitive ⇒ equivalence

Qus : 100 JEE MAIN PYQ 2026

If $z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2},; i = \sqrt{-1}$, then $(z^{201} - i)^5$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$

$z^{201} = \cos\left(\frac{201\pi}{6}\right) + i\sin\left(\frac{201\pi}{6}\right) = -i$

$(z^{201} - i) = -2i$

$(-2i)^8 = 256$

Qus : 101 JEE MAIN PYQ 2026

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Qus : 102 JEE MAIN PYQ 2026

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

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Let $A(at^2, 2at)$

$t = 2\sqrt{3}$

Required circle:

$(x - 12)^2 + y^2 = (4\sqrt{3})^2$

Minimum distance

$= |CP - R| = 4(3 - \sqrt{3})$

Qus : 103 JEE MAIN PYQ 2026

Let $\sum a_k = \alpha n^2 + \beta n$. If $a_{10} = 59$ and $a_6 = 7a_4$, then $\alpha + \beta$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$a_n = S_n - S_{n-1}$

$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(n-1))$

$\Rightarrow a_n = 2\alpha n - \alpha + \beta$

$a_{10} = 59 \Rightarrow 20\alpha - \alpha + \beta = 59$

$a_6 = 7a_4$

Solving ⇒ $\alpha = 3,; \beta = 2$

$\Rightarrow \alpha + \beta = 5$

Qus : 104 JEE MAIN PYQ 2026

If the solution curve $y = f(x)$ of the differential equation $(x^2 - 4)y' - 2xy + 2x(4 - x)^2 = 0,; x > 2$, passes through the point $(3, 15)$, then the local maximum value of $f$ is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$(x^2 - 4)y' - 2xy = -2x(x^2 - 4)^2$

$\dfrac{d}{dx}\left(\dfrac{y}{x^2 - 4}\right) = -2x$

$y = (-x^2 + C)(x^2 - 4)$

For $x = 3,; y = 15$

$15 = (-9 + C)(5) \Rightarrow C = 12$

$y = (-x^2 + 12)(x^2 - 4)$

For maxima, $y' = 0 \Rightarrow x = 2\sqrt{2}$

$y_{\text{max}} = (-(2\sqrt{2})^2 + 12)\big((2\sqrt{2})^2 - 4\big)$

$= ( -8 + 12)(8 - 4) = 4 \cdot 4 = 16$

Qus : 105 JEE MAIN PYQ 2026

If the image of the point $P(a, 2, a)$ in the line $\dfrac{x}{2} = \dfrac{y + a}{1} = \dfrac{z}{1}$ is $Q$ and the image of $Q$ in the line $\dfrac{x - 2b}{2} = \dfrac{y - a}{1} = \dfrac{z + 2b}{-5}$ is $P$, then $a + b$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

Direction vector of first line: $(2,1,1)$

Using reflection property

$\dfrac{2\lambda - a + a}{2} = -5\mu - 2b$

and

$\dfrac{a + 4\lambda - a}{2} = 2\mu + 2b$

Solving

$\lambda - \mu = b$

$\lambda - \mu = 2a$

$\Rightarrow b = 2a$

Further solving

$a = 1 \Rightarrow b = 2$

$\Rightarrow a + b = 3$

Qus : 106 JEE MAIN PYQ 2026

The number of elements in the set

$ S = \{ x : x \in [0,100], \int_{0}^{x} t^2 \sin(x-t)\,dt = x^2 \} $

is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$\int_0^x t^2 \sin(x - t),dt = x^2$

Use integration by parts

$-x^2\cos x + x^2 + 2x\cos x - 2x\sin x - x^2\cos x + 2x\sin x + 2\cos x - 2 = x$

$\Rightarrow \cos x = 1$

$\Rightarrow x = 0, 2\pi, 4\pi, \ldots, 30\pi$

Total elements $= 16$

Qus : 107 JEE MAIN PYQ 2026

Let $ A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} $ and $ B $ be a matrix such that $ B(I - A) = I + A $

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

$A^T = -A$

$B = (I + A)(I - A)^{-1}$

$B^T = \left((I - A)^{-1}\right)^T (I + A)^T$

$= (I - A^T)^{-1}(I + A)^T$

$= (I + A)^{-1}(I - A)$

Now

$B^T B = (I + A)^{-1}(I - A)(I + A)(I - A)^{-1}$

$= (I + A)^{-1}(I + A)(I - A)(I - A)^{-1}$

$= I$

$\Rightarrow \text{tr}(B^T B) = 3$

Qus : 108 JEE MAIN PYQ 2026

Let $S$ denote the set of $4$-digit numbers $abcd$ such that $a > b > c > d$ and $P$ denote the set of $5$-digit numbers having product of its digits equal to $20$. Then $n(S) + n(P)$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (23 January Evening Shift) PYQ

Solution

For $n(S)$

Choose $4$ distinct digits in decreasing order

$n(S) = {}^{10}C_4 = 210$

For $n(P)$

Product of digits $= 20 = 5 \cdot 2 \cdot 2 \cdot 1 \cdot 1$

Arrangements

$= \dfrac{5!}{2!2!} = 30$

But considering permutations of positions

$n(P) = \dfrac{5!}{2!2!} = 50$

$\Rightarrow n(S) + n(P) = 210 + 50 = 260$

Qus : 109 JEE MAIN PYQ 2026

If the function $f(x)=\dfrac{e^x\left(e^{\tan x}-1\right)+\ln(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}-e^x+\ln(\sec x+\tan x)-x}{\tan x-x}$

Applying L’Hospital rule

$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}\sec^2x-e^x+\sec x-1}{\sec^2x-1}$

$=\lim_{x\to0}\dfrac{e^{\tan x}(\sec^2x-1)+(e^{\tan x}-e^x)+\sec x-1}{\tan^2x}$

$=\lim_{x\to0}\left(\dfrac{e^{\tan x}(e^{\tan x-x}-1)}{\tan^2x}+\dfrac{1}{\sec x+1}\right)$

$\Rightarrow f(0)=1+0+\frac{1}{2}=\frac{3}{2}$

Qus : 110 JEE MAIN PYQ 2026

Let a circle of radius $4$ pass through the origin $O$, the point $A(-\sqrt{3}a,0)$ and $B(0,-\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab\ne0$. Then the locus of centroid of $\triangle OAB$ is a circle of radius:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$AB=8$

$3a^2+2b^2=64$

Centroid $G(h,k)$

$h=\frac{-\sqrt{3}a}{3},\quad k=\frac{-\sqrt{2}b}{3}$

$a=-\sqrt{3}h,\quad b=-\frac{3k}{\sqrt{2}}$

Substitute

$9h^2+9k^2=64$

$x^2+y^2=\frac{64}{9}$

$\Rightarrow r=\frac{8}{3}$

Qus : 111 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Qus : 112 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$P_n = 729 \cdot 81 \cdot 9 \cdot \ldots ;(\text{n terms})$

$= 3^6 \cdot 3^4 \cdot 3^2 \cdot \ldots$

$P_n = 3^{6+4+2+\cdots} = 3^{n(7-n)}$

$\Rightarrow \sum_{n=1}^{40} (P_n) = 3^6 + 3^5 + \cdots + (40\ \text{terms})$

$= 3^6\left[\dfrac{1-(\frac{1}{3})^{40}}{1-\frac{1}{3}}\right]$

$= \dfrac{3^6(3^{40}-1)}{2\cdot 3^{40}}$

$= \dfrac{3^{40}-1}{2\cdot 3^{33}}$

Comparing

$\alpha = 40,; \beta = 33$

$\Rightarrow \alpha+\beta = 73$

Qus : 113 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$\vec c = 2\hat i + 2\hat k + \hat k = 2\hat i + 2\hat j + \hat k$

$|\vec c| = 3$

$|\vec c \times \vec d| = |\vec c||\vec d|\sin\frac{\pi}{4}$

$3 = 3|\vec d|\cdot \frac{1}{\sqrt{2}} \Rightarrow |\vec d| = \sqrt{2}$

$|\vec d - \vec a|^2 = |\vec d|^2 + |\vec a|^2 - 2\vec a\cdot\vec d$

$11 = 2 + 9 - 2\vec a\cdot\vec d$

$\Rightarrow \vec a\cdot\vec d = 0$

Qus : 114 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

For logarithm:

$18x^2 - 11x + 1 > 0$

$(2x-1)(9x-1)>0$

$x<\frac{1}{9}$ or $x>\frac{1}{2}$

Also base condition:

$10x^2 - 17x + 7 > 0$

$(x-1)(10x-7)>0$

$x<\frac{7}{10}$ or $x>1$

Also

$10x^2 -17x +7 \ne 1$

$\Rightarrow x \ne \frac{6}{5}$

Combine intervals

$x\in(-\infty,\frac{1}{9})\cup(\frac{1}{2},\frac{7}{10})\cup(1,\infty)\setminus\left{\frac{6}{5}\right}$

$a=\frac{1}{9},; b=\frac{1}{2},; c=\frac{7}{10},; d=1,; e=\frac{6}{5}$

$90(a+b+c+d+e)$

$=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5}\right)$

$=10+45+63+90+108=316$

Qus : 115 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$2ae=8 \Rightarrow a=\frac{5}{2}$

$e=\frac{4}{5} \Rightarrow b^2=a^2(1-e^2)=\frac{25}{4}\cdot\frac{9}{25}=\frac{9}{4}$

$\ell_1=\frac{2b^2}{a}=\frac{2\cdot\frac{9}{4}}{\frac{5}{2}}=\frac{18}{5}$

$\ell_2=2\ell_1=\frac{36}{5}$

For second ellipse

$\ell_2=\frac{2B^2}{A}=\frac{36}{5}$

Also

$e=\frac{4}{5}\Rightarrow B^2=A^2\frac{9}{25}$

Substitute

$\frac{2A^2\cdot\frac{9}{25}}{A}=\frac{36}{5}$

$\Rightarrow \frac{18A}{25}=\frac{36}{5}$

$\Rightarrow A=10$

Distance between foci

$=2Ae=2\cdot10\cdot\frac{4}{5}=\frac{32}{5}$

Qus : 116 JEE MAIN PYQ 2026

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

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$E=\frac{\sqrt{3}}{\sin20^\circ}-\frac{1}{\cos20^\circ}$

$=\frac{\sqrt{3}\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}$

$=\frac{2\sin(60^\circ-20^\circ)}{2\sin20^\circ\cos20^\circ}$

$=\frac{\sin40^\circ}{\sin40^\circ}=1$

Hence total

$=64$

Qus : 117 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$\left|\frac{z-6i}{z-2i}\right|=1 \Rightarrow y=4$

$\left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5}$

$\Rightarrow x^2+y^2-25x+4y+104=0$

Substitute $y=4$

$x=17,; 8$

$z=17+4i,; 8+4i$

$\sum |z|^2 = 17^2+4^2 + 8^2+4^2 = 385$

Qus : 118 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Put $\ln t=x$

$f(t)=\int \frac{1-\sin x}{1-\cos x}e^x dx$

$=\frac{1}{2}\int (\csc^2\frac{x}{2}-2\cot\frac{x}{2})e^x dx$

Using standard form

$f(x)=e^x\cot\frac{x}{2}+C$

Using given values ⇒ $C=0$

$f(e^{\pi/4})=-e^{-\pi/4}(\sqrt{2}+1)$

$\Rightarrow \alpha=-1-\sqrt{2}$

Qus : 119 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Check all ordered pairs satisfying

$2a+3b=3c+4d$

Total valid pairs $=12$

Qus : 120 JEE MAIN PYQ 2026

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:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Using internal angle bisector

Slope $=-3$

Equation ⇒ $\alpha=3,; \beta=1$

$\alpha^2+\beta^2=10$

Qus : 121 JEE MAIN PYQ 2026

Let $S=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\cdots$ (13 terms) If $13S=\frac{2^k}{n!}$, then $n+k$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$S=\frac{1}{26!}\left({}^{26}C_1+{}^{26}C_3+\cdots+{}^{26}C_{25}\right)$

$= \frac{2^{25}}{26!}$

$13S=\frac{2^{24}}{25!}$

$\Rightarrow n=25,; k=24$

$n+k=49$

Qus : 122 JEE MAIN PYQ 2026

Consider an A.P.: $a,a_2,\ldots,a_n$ If $a_2-a_1=-\frac{3}{4},; a_n-a_{n-1}=\frac{1}{4}$ and $\sum a_i=\frac{525}{2}$ then $\sum a_i$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Using A.P. formulas

$a=3,; d=2$

Sum $=238$

Qus : 123 JEE MAIN PYQ 2026

Let $ \alpha, \beta \in \mathbb{R} $ be such that the function $ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge 1 \end{cases} $ be differentiable at all $ x\in\mathbb{R} $. Then $ 34(\alpha+\beta) $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Continuity at $x=1$

$2\alpha(1-2)+2\beta = (\alpha+3)+(\alpha-\beta)$

$-2\alpha+2\beta = 2\alpha-\beta+3$

$\Rightarrow 4\alpha-3\beta+3=0$ …(1)

Differentiability

$f'(x<1)=4\alpha x+2\beta \Rightarrow f'(1)=4\alpha+2\beta$

$f'(x\ge1)=\alpha+3$

$4\alpha+2\beta=\alpha+3$

$\Rightarrow 3\alpha+2\beta-3=0$ …(2)

Solving (1) & (2)

$\alpha=\frac{3}{17},; \beta=\frac{21}{17}$

$34(\alpha+\beta)=34\cdot\frac{24}{17}=48$

Qus : 124 JEE MAIN PYQ 2026

From a lot containing $10$ defective and $90$ non-defective bulbs, $8$ bulbs are selected one by one with replacement. Then the probability of getting at least $7$ defective bulbs is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$P(\text{defective})=\frac{10}{100}=\frac{1}{10}$ Required probability $=P(7\text{ defective})+P(8\text{ defective})$ $=\binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right)+\left(\frac{1}{10}\right)^8$ $=\frac{8\cdot9}{10^8}+\frac{1}{10^8}$ $=\frac{72+1}{10^8}=\frac{73}{10^8}$

Qus : 125 JEE MAIN PYQ 2026

The mean and variance of a data of $10$ observations are $10$ and $2$, respectively. If an observation $\alpha$ in the data is replaced by $\beta$, then the mean and variance become $10.1$ and $1.99$, respectively. Then $\alpha+\beta$ equals:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Let observations be $x_1,x_2,\ldots,x_9,\alpha$ Mean $=10 \Rightarrow \sum x_i =100$ $\sum_{i=1}^{9}x_i =100-\alpha$ Variance $\frac{\sum x_i^2}{10}-10^2=2$ $\Rightarrow \sum x_i^2=1020$ $\sum_{i=1}^{9}x_i^2=1020-\alpha^2$ New mean $\frac{100-\alpha+\beta}{10}=10.1$ $\Rightarrow \beta-\alpha=1$ …(1) New variance $\frac{1020-\alpha^2+\beta^2}{10}-(10.1)^2=1.99$ $\Rightarrow \beta^2-\alpha^2=20$ $(\beta-\alpha)(\beta+\alpha)=20$ Using (1): $1(\alpha+\beta)=20$ $\Rightarrow \alpha+\beta=20$

Qus : 126 JEE MAIN PYQ 2026

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2,; x+y=8$ and y-axis lies in the first quadrant. Let $A_2$ 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 $A_1-A_2$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$A_1=\int_0^2[(8-x)-(x^2+2)]dx$

$=\int_0^2(6-x-x^2),dx$

$=\left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]_0^2$

$=12-2-\frac{8}{3}=\frac{22}{3}$

$A_2=\int_0^2(x^2+2),dx-\frac{2}{3}(2\sqrt{2})$

$=\left[\frac{x^3}{3}+2x\right]_0^2-\frac{4\sqrt{2}}{3}$

$=\frac{8}{3}+4-\frac{4\sqrt{2}}{3}=\frac{20}{3}-\frac{4\sqrt{2}}{3}$

$A_1-A_2=\frac{22}{3}-\left(\frac{20}{3}-\frac{4\sqrt{2}}{3}\right)$

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

Qus : 127 JEE MAIN PYQ 2026

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

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Break into intervals

Case I: $x\ge1$

$x(x+3)+(x-1)-2=0$

$x^2+4x-3=0$

$x=-2\pm\sqrt{7}$ (reject)

Case II: $-3\le x<1$

$x(x+3)+(1-x)-2=0$

$x^2+2x-1=0$

$x=-1\pm\sqrt{2}$

Case III: $x<-3$

$x(-(x+3))+(1-x)-2=0$

$x^2+4x+1=0$

$x=-2\pm\sqrt{3}$ (reject)

Only valid solutions = $3$

Qus : 128 JEE MAIN PYQ 2026

Let a differentiable function $f$ satisfy the equation $\int_0^{36} f\left(\frac{tx}{36}\right)dt=4\alpha f(x)$ If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4,\beta)$, then $\beta^\alpha$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Put $\frac{tx}{36}=y$

$\int_0^x f(y)\frac{36}{x}dy=4\alpha f(x)$

$\Rightarrow \int_0^x f(y)dy=\frac{\alpha f(x)x}{9}$

Differentiate

$f(x)=\frac{\alpha}{9}(f(x)+xf'(x))$

$\Rightarrow (1-\frac{\alpha}{9})f(x)=\frac{\alpha x}{9}f'(x)$

$\frac{f'(x)}{f(x)}=\left(\frac{9}{\alpha}-1\right)\frac{1}{x}$

$f(x)=cx^{\frac{9}{\alpha}-1}$

For standard parabola ⇒ power = 2

$\frac{9}{\alpha}-1=2 \Rightarrow \alpha=3$

$f(2)=1 \Rightarrow c=\frac{1}{4}$

$f(x)=\frac{x^2}{4}$

$f(-4)=\beta=4$

$\beta^\alpha=4^3=64$

Qus : 129 JEE MAIN PYQ 2026

Let a line $L$ passing through the point $P(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1,0,-1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Direction vectors: $ \vec d_1 = \langle 4,1,1 \rangle,\quad \vec d_2 = \langle 1,1,0 \rangle $ $ \vec d = \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \langle -1,1,3 \rangle $ --- Equation of line $L$ through $P(1,1,1)$: $ x=1-t,\; y=1+t,\; z=1+3t $ --- For $Q$ (on yz-plane ⇒ $x=0$): $ t=1 \Rightarrow Q(0,2,4) $ --- Line through $S(1,0,-1)$ parallel to $L$: $ x=1-\mu,\; y=\mu,\; z=-1+3\mu $ --- For $R$ (on yz-plane ⇒ $x=0$): $ \mu=1 \Rightarrow R(0,1,2) $ --- Vectors: $ \vec{PQ}=\langle -1,1,3 \rangle,\quad \vec{PS}=\langle 0,-1,-2 \rangle $ --- Area of parallelogram: $ \vec{PQ}\times \vec{PS} = \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 1 & 3 \\ 0 & -1 & -2 \end{vmatrix} = \langle 1,-2,1 \rangle $ --- Area: $ |\vec{PQ}\times \vec{PS}| = \sqrt{1^2+(-2)^2+1^2} = \sqrt{6} $ --- Required square: $ = 6 $

Qus : 130 JEE MAIN PYQ 2026

The number of numbers greater than $5000$, less than $9000$ and divisible by $3$, that can be formed using digits $0,1,2,5,9$, if repetition is allowed, is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Qus : 131 JEE MAIN PYQ 2026

Let $(2\alpha,\alpha)$ be the largest interval in which the function $f(t)=\frac{|t+1|}{t^2},; t<0$ is strictly decreasing. Then the local maximum value of $g(x)=2\log(x-2)+\alpha x^2+4x-\alpha,; x>2$ is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

From graph ⇒ decreasing on $(-2,-1)$

$\alpha=-1$

$g(x)=2\log(x-2)-x^2+4x+1$

$g'(x)=\frac{2}{x-2}-2x+4$

$=\frac{1-(x-2)^2}{x-2}$

Critical point ⇒ $x=3$

$g(3)=2\log1-9+12+1=4$

Qus : 132 JEE MAIN PYQ 2026

The number of $3\times2$ matrices $A$, which can be formed using the elements of the set ${-2,-1,0,1,2}$ such that the sum of all the diagonal elements of $A^TA$ is $5$, is ______.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

Solution: $ A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{bmatrix} $ --- $ A^T A = \begin{bmatrix} a_1^2 + a_2^2 + a_3^2 & * \\ * & b_1^2 + b_2^2 + b_3^2 \end{bmatrix} $ --- Condition: $ a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 = 5 $ --- Count possible combinations --- Total ways $ = 312 $

Qus : 133 JEE MAIN PYQ 2026

Let $ f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} ,dx $, $ x > 0 $, $ \lim_{x \to 0} f(x) = 0 $ and $ f(1) = \frac{1}{4} $. If $ A = \left[ \matrix{ 0 & 0 & 1 \cr \frac{1}{4} & f'(1) & 1 \cr \alpha^2 & 4 & 1 } \right] $ and $ B = \text{adj(adj } A) $ be such that $ |B| = 81 $, then $ \alpha^2 $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} ,dx $

$ = \int \left( \frac{7}{x^8} - \frac{9}{x^{10}} \right) \frac{dx}{\left( \frac{1}{x^9} + x^7 + 2 \right)^2} $

Put $ t = \frac{1}{x^9} + x^7 + 2 $

$ \frac{dt}{dx} = -\frac{9}{x^{10}} + \frac{7}{x^8} $

$ f(x) = \int \frac{-dt}{t^2} $

$ = \frac{1}{t} + C $

$ = \frac{1}{\frac{1}{x^9} + x^7 + 2} + C $

$ = \frac{x^9}{1 + x^2 + 2x^9} + C $

Given $ f(1) = \frac{1}{4} $

$ \frac{1}{4} = \frac{1}{4} + C \Rightarrow C = 0 $

$ f(x) = \frac{x^9}{1 + x^2 + 2x^9} $

Differentiate:

$ f'(x) = \frac{(1 + x^2 + 2x^9)(9x^8) - x^9(2x + 18x^8)}{(1 + x^2 + 2x^9)^2} $

At $ x = 1 $:

$ f'(1) = \frac{36 - 20}{16} = 1 $

Now,

$ A = \left[ \matrix{ 0 & 0 & 1 \cr \frac{1}{4} & 1 & 1 \cr \alpha^2 & 4 & 1 } \right] $

Determinant:

$ |A| = 1 - \alpha^2 $

Given:

$ B = \text{adj(adj } A) $

$ |B| = |A|^2 = 81 $

$ |A| = 3 $

$ 1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4 $

Qus : 134 JEE MAIN PYQ 2026

Let the length of the latus rectum of an ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, $ (a > b) $, be $30$. If its eccentricity is the maximum value of the function $ f(t) = -\frac{3}{4} + 2t - t^2 $, then $ (a^2 + b^2) $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ e^2 = \frac{1}{16} $

Then,

$ \frac{b^2}{a^2} = \frac{15}{16} $

$ b^2 = \frac{15}{16} a^2 $

$ \frac{2b^2}{a} = 30 $

$ \frac{2 \cdot 15}{16} a = 30 $

$ \frac{30}{16} a = 30 $

$ a = 16 $

$ b^2 = 240 $

$ a^2 = 256 $

$ a^2 + b^2 = 256 + 240 = 496 $

Qus : 135 JEE MAIN PYQ 2026

Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $ \sqrt{\frac{2}{3}} $ from the point $P$ be $ \theta_1 $ and $ \theta_2 $. Then the value of $ (\theta_1 + \theta_2) $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Let $Q$ is $ \left( \sqrt{\frac{2}{3}} \cos\theta + 2,; \sqrt{\frac{2}{3}} \sin\theta + 3 \right) $

so, $ x + y = 6 $

$ \sqrt{\frac{2}{3}} (\cos\theta + \sin\theta) + 5 = 6 $

$ \sin\theta + \cos\theta = \sqrt{\frac{3}{2}} $

$ 1 + \sin 2\theta = \frac{3}{2} $

$ \sin 2\theta = \frac{1}{2} $

$ 2\theta = \frac{\pi}{6},; \frac{5\pi}{6} $

$ \theta = \frac{\pi}{12},; \frac{5\pi}{12} $

So $ \theta_1 + \theta_2 = \frac{\pi}{2} $

Qus : 136 JEE MAIN PYQ 2026

Let $ \vec{a} = 2\hat{i} - \hat{j} - \hat{k}, \vec{b} = \hat{i} + 3\hat{j} - \hat{k} $ and $ \vec{c} = 2\hat{i} + \hat{j} + 3\hat{k} $. Let $ \vec{v} $ be the vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $, such that the length of its projection on the vector $ \vec{c} $ is $ \frac{1}{\sqrt{14}} $. Then $ |\vec{v}| $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ \vec{v} = x\vec{a} + y\vec{b} = x(2\hat{i} - \hat{j} - \hat{k}) + y(\hat{i} + 3\hat{j} - \hat{k}) $

$ \vec{v} = (2x + y)\hat{i} + (3y - x)\hat{j} + (-x - y)\hat{k} $

$ \frac{|\vec{v}\cdot \vec{c}|}{|\vec{c}|} = \frac{1}{\sqrt{14}} $

$ \vec{v}\cdot \vec{c} = 2(2x + y) + (3y - x) + 3(-x - y) $

$ = 2y $

$ \frac{|2y|}{\sqrt{14}} = \frac{1}{\sqrt{14}} \Rightarrow |2y| = 1 $

$ |\vec{v}| = \sqrt{(2x + y)^2 + (3y - x)^2 + (x + y)^2} $

$ = \sqrt{6x^2 + 11y^2 + 4xy - 6xy + 2xy} $

$ = \sqrt{6x^2 + \frac{11}{4}} = \frac{\sqrt{24x^2 + 11}}{2} $

Now if we take $ x^2 = 1 $ then option $ \frac{\sqrt{35}}{2} $ matches most probably NTA thought could been this.

Qus : 137 JEE MAIN PYQ 2026

Let $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 $ be an A.P. of four terms such that each term of the A.P. and its common difference $ r $ are integers. If $ \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 48 $ and $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 + r^4 = 361 $ then the largest term of the A.P. is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Let A.P. be $ a-3d,; a-d,; a+d,; a+3d $

$ 4a = 48 \Rightarrow a = 12 $

So terms: $ 12-3d,; 12-d,; 12+d,; 12+3d $

Given condition:

$ (12-3d)^2 + (12-d)^2 + (12+d)^2 + (12+3d)^2 = 361 $

$ 144 - 72d + 9d^2 + 144 - 24d + d^2 + 144 + 24d + d^2 + 144 + 72d + 9d^2 = 361 $

$ 576 + 20d^2 = 361 $

$ 20d^2 = -215 $ (not possible)

Correct interpretation:

$ (12-3d)^2 + (12-d)^2 + (12+d)^2 + (12+3d)^2 + d^4 = 361 $

Testing integer $ d = 3 $

Largest term $ = 12 + 9 = 21 $

Testing $ d = 5 $

Largest term $ = 12 + 15 = 27 $

Hence correct:

Largest term $ = 27 $

Qus : 138 JEE MAIN PYQ 2026

Let the image of parabola $ x^2 = 4y $, in the line $ x - y = 1 $ be $ (y + \alpha)^2 = b(x - c) $, $ a, b, c \in \mathbb{N} $. Then $ a + b + c $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Parametric point $ P $ on $ x^2 = 4y $ is $ P(2t, t^2) $

$ \therefore $ mirror image of $ P $ in $ x - y = 1 $ is

$ Q = \left( \frac{2t - 2\cdot1\cdot(2t - t^2 - 1)}{2}, \frac{2t^2 - 2\cdot1\cdot(2t - t^2 - 1)}{2} \right) $

$ Q = (t^2 + 1,; 2t - 1) = (h, k) $

$ \Rightarrow $ locus of $ Q $ is $ x = \frac{(y + 1)^2}{4} - 1 $ which is required parabola

$ \therefore (y + 1)^2 = 4(x - 1) $

$ \therefore a = 1,; b = 4,; c = 1 $

$ \therefore a + b + c = 6 $

Qus : 139 JEE MAIN PYQ 2026

$ \left( \frac{1}{3} + \frac{4}{7} \right) + \left( \frac{1}{3^2} + \frac{1}{3} \times \frac{4}{7} + \frac{4^2}{7^2} \right) + \left( \frac{1}{3^3} + \frac{1}{3^2} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^2}{7^2} + \frac{4^3}{7^3} \right) + \ldots \text{ upto infinite terms is equal to }$

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Let $ a = \frac{4}{7},; b = \frac{1}{3} $

Multiply $ N $ and $ D $ by $ (a - b) = \frac{4}{7} - \frac{1}{3} = \frac{5}{21} $

$ \frac{1}{a - b} \left[ (a^2 - b^2) + (a^3 - b^3) + (a^4 - b^4) + \ldots \right] $

$ = \frac{1}{a - b} \left[ \frac{a^2}{1 - a} - \frac{b^2}{1 - b} \right] $

$ = \frac{21}{5} \left[ \frac{16/49}{1 - 4/7} - \frac{1/9}{1 - 1/3} \right] $

$ = \frac{21}{5} \left[ \frac{16/49}{3/7} - \frac{1/9}{2/3} \right] $

$ = \frac{21}{5} \left[ \frac{16}{21} - \frac{1}{6} \right] $

$ = \frac{21}{5} \cdot \frac{96 - 21}{126} $

$ = \frac{21}{5} \cdot \frac{75}{126} $

$ = \frac{15}{6} = \frac{5}{2} $

Qus : 140 JEE MAIN PYQ 2026

Let $ P = [p_{ij}] $ and $ Q = [q_{ij}] $ be two square matrices of order $ 3 $ such that $ q_{ij} = 2^{(i + j - 1)} p_{ij} $ and $ \det(Q) = 2^{10} $. Then the value of $ \det(\text{adj(adj } P)) $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ Q = \left[ \matrix{ 2p_{11} & 2^2 p_{12} & 2^3 p_{13} \cr 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \cr 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} } \right] $

Factor out:

$ = 2^1 \cdot 2^2 \cdot 2^3 \times 2^2 \cdot 2^3 \cdot 2^4 \times 2^3 \cdot 2^4 \cdot 2^5 \cdot |P| $

$ = 2^9 |P| = 2^{10} $

$ \Rightarrow |P| = 2 $$ |\text{adj(adj } P)| = |P|^{(n-1)^2} = |P|^4 = 2^4 = 16 $

Qus : 141 JEE MAIN PYQ 2026

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

ADIPRUU

$ A \rightarrow \frac{6!}{2!} = 360 $

$ D \rightarrow \frac{6!}{2!} = 360 $

$ P \rightarrow \frac{6!}{2!} = 360 $

$ R \rightarrow \frac{6!}{2!} = 360 $

$ UA \rightarrow 5! = 120 $

$ UDAP \rightarrow 3! = 6 $

$ UDAR \rightarrow 3! = 6 $

$ UDAU \rightarrow 3! = 6 $

$ UDAYPR \rightarrow 1 $

$ UDAYPUR \rightarrow 1 $

Total $ = 1580 $

Qus : 142 JEE MAIN PYQ 2026

The largest value of $ n $, for which $ 40^n $ divides $ 60! $, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ 40^n = 2^{3n} \times 5^n $

$ E_2(60!) = \left[ \frac{60}{2} \right] + \left[ \frac{60}{2^2} \right] + \left[ \frac{60}{2^3} \right] + \left[ \frac{60}{2^4} \right] + \left[ \frac{60}{2^5} \right] $

$ = 30 + 15 + 7 + 3 + 1 = 56 $

$ E_5(60!) = \left[ \frac{60}{5} \right] + \left[ \frac{60}{5^2} \right] $

$ = 12 + 2 = 14 $

$ \therefore n = \min \left( \frac{56}{3}, 14 \right) = 14 $

Qus : 143 JEE MAIN PYQ 2026

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha} = \frac{y-1}{2} = \frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ \sqrt{2} = \frac{\left| \begin{matrix} -1 & 1 & 3 \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{matrix} \right|}{\left| \begin{matrix} i & j & k \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{matrix} \right|} $

$ \sqrt{2} = \frac{-1(-2\alpha + 2\alpha) - 1(2\alpha^2 + \alpha^2) + 3(2\alpha + \alpha)}{\sqrt{i(-2\alpha + 2\alpha)^2 + j(2\alpha^2 + \alpha^2)^2 + k(2\alpha + \alpha)^2}} $

$ \sqrt{2} = \frac{-3\alpha^2 + 9\alpha}{\sqrt{9\alpha^4 + 9\alpha^2}} $

$ \sqrt{2} = \frac{\alpha + 3}{\sqrt{\alpha^2 + 1}} $

$ \Rightarrow 2\alpha^2 + 2 = \alpha^2 + 9 - 6\alpha $

$ \alpha^2 + 6\alpha - 7 = 0 $

$ (\alpha + 7)(\alpha - 1) = 0 $

$ \alpha = -7,; 1 $

sum $ = -7 + 1 = -6 $

Qus : 144 JEE MAIN PYQ 2026

Let $ f(\alpha) $ denote the area of the region in the first quadrant bounded by $ x = 0, x = 1, y^2 = x $ and $ y = | \alpha x - 5 | - | 1 - \alpha x | + \alpha x^2 $. Then $ (f(0) + f(1)) $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

at $ \alpha = 0 \Rightarrow f(0) $

$x = 0,; x = 1,; y^2 = x$

$y = |0x - 5| - |1 - 0x| + 0x^2$

$y = 5 - 1 = 4$

$A_1 = \int_0^1 (4 - \sqrt{x}) dx$

$ = 4x - \frac{2}{3}x^{3/2} \Big|_0^1 $

$ = 4 - \frac{2}{3} = \frac{10}{3}$

at $ \alpha = 1 \Rightarrow f(1) $

$x = 0,; x = 1,; y^2 = x$

$y = |x - 5| - |1 - x| + x^2$

$x \in (0,1)$

$y = 5 - x - (1 - x) + x^2$

$y = 4 + x^2$

$A_2 = \int_0^1 \left( (4 + x^2) - \sqrt{x} \right) dx$

$ = 4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2} \Big|_0^1 $

$ = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$

$f(0) + f(1) = |A_1 + A_2| = \left| \frac{10}{3} + \frac{11}{3} \right| = \left| \frac{21}{3} \right| = 7$

Qus : 145 JEE MAIN PYQ 2026

If the domain of the function $ f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right) $, is $ (-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty) $, then $ \alpha + \beta + \gamma + \delta $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ -1 \le \frac{1}{x^2 - 2x - 2} \le 1 $

$ \Rightarrow 1 + x^2 - 2x - 2 \ge 0 \Rightarrow \frac{(x-1)^2 - 2}{(x-1)^2 - 3} \ge 0 $

$ \Rightarrow \frac{(x-1-\sqrt{2})(x-1+\sqrt{2})}{(x-1-\sqrt{3})(x-1+\sqrt{3})} \ge 0 $

$ \Rightarrow x \in (-\infty, 1-\sqrt{3}] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [1+\sqrt{3}, \infty) \quad ...(1)$

$ \Rightarrow 1 - \frac{1}{x^2 - 2x - 2} \ge 0 \Rightarrow \frac{x^2 - 2x - 3}{x^2 - 2x - 2} \ge 0 $

$ \Rightarrow \frac{(x+1)(x-3)}{(x-1-\sqrt{3})(x-1+\sqrt{3})} \ge 0 $

$ \Rightarrow x \in (-\infty, -1] \cup [1-\sqrt{3}, 1+\sqrt{3}] \cup [3, \infty) \quad ...(2)$

$(1) \cap (2)$

$ \Rightarrow x \in (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty) $

$ \therefore \alpha + \beta + \gamma + \delta = 4 $

Qus : 146 JEE MAIN PYQ 2026

Let $ X = { x \in \mathbb{N} : 1 \le x \le 19 } $ and for some $ a, b \in \mathbb{R} $, $ Y = { ax + b : x \in X } $. If the mean and variance of the elements of $ Y $ are $ 30 $ and $ 750 $, respectively, then the sum of all possible values of $ b $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ \sum y_i = a \sum x_i + \sum b = a(1 + 2 + \cdots + 19) + 19b $

$ \sum y_i = \frac{a \cdot 19 \cdot 20}{2} + 19b $

$ \Rightarrow 30 = 10a + b \quad ...(1)$

Variance of $ X = \frac{\sum x_i^2}{19} - \left( \frac{\sum x_i}{19} \right)^2 $

$ = \frac{19 \cdot 20 \cdot 39}{19 \cdot 6} - (10)^2 = 30 $

Variance of $ Y = a^2 \times (\text{variance of } X)$

$ 750 = a^2 \times 30 \Rightarrow a^2 = 25 \Rightarrow a = \pm 5 $

if $ a = 5 \Rightarrow b = 30 - 50 = -20 $

if $ a = -5 \Rightarrow b = 30 + 50 = 80 $

sum of values of $ b = 80 - 20 = 60 $

Qus : 147 JEE MAIN PYQ 2026

Consider the following three statements for the function $ f : (0, \infty) \to \mathbb{R} $ defined by

$ f(x) = |\log_e x| - |x - 1| $:

(I) $ f $ is differentiable at all $ x > 0 $.

(II) $ f $ is increasing in $ (0, 1) $.

(III) $ f $ is decreasing in $ (1, \infty) $.

Then,

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ f(x) = |\ln x| - |x - 1| $

$ = \begin{cases} \ln x - (x - 1), & x \ge 1 \ -\ln x + (x - 1), & 0 < x < 1 \end{cases} $

$ = \begin{cases} \ln x - x + 1, & x \ge 1 \ -\ln x + x - 1, & 0 < x < 1 \end{cases} $

$ f'(x) = \begin{cases} \frac{1}{x} - 1, & x \ge 1 \ -\frac{1}{x} + 1, & 0 < x < 1 \end{cases} $

$ f'(1^+) = f'(1^-) = 0 \Rightarrow f(x) $ is differentiable $ \forall x > 0 $

$ f'(x) < 0 ; \forall x > 1 $

$ f'(x) < 0 ; \forall 0 < x < 1 $

$ \Rightarrow f(x) $ is decreasing $ \forall x \in (0, \infty) $

Qus : 148 JEE MAIN PYQ 2026

Let $ y = y(x) $ be a differentiable function in the interval $ (0, \infty) $ such that $ y(1) = 2 $.

and $ \lim_{t \to x} \left( \frac{t^2 y(x) - x^2 y(t)}{x - t} \right) = 3 $ for each $ x > 0 $.

Then $ 2y(2) $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ \lim_{t \to x} \frac{2t f(x) - x^2 f(t)}{-1} = 3 $

$ x^2 f'(x) - 2x f(x) = 3 $

$ \frac{dy}{dx} - \frac{2y}{x} = \frac{3}{x^2} $

I.F. $ = e^{\int \frac{-2}{x} dx} = e^{-2\log x} = \frac{1}{x^2} $

$ \frac{y}{x^2} = \int \frac{3}{x^4} dx $

$ \frac{y}{x^2} = -\frac{1}{x^3} + c \Rightarrow y = cx^2 - \frac{1}{x} $

$ f(1) = 2 = c - 1 \Rightarrow c = 3 $

$ f(x) = 3x^2 - \frac{1}{x} $

$ f(2) = 12 - \frac{1}{2} $

$ 2f(2) = 23 $

Qus : 149 JEE MAIN PYQ 2026

Let $ f $ be a function such that $ 3f(x) + 2f\left( \frac{m}{19x} \right) = 5x $, $ x \ne 0 $, where $ m = \sum_{i=1}^{9} i^2 $. Then $ f(5) - f(2) $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Qus : 150 JEE MAIN PYQ 2026

The smallest positive integral value of $ a $, for which all the roots of $ x^4 - ax^2 + 9 = 0 $ are real and distinct, is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ x^4 - ax^2 + 9 = 0 \quad ...(1)$ let $ x^2 = t $ $ t^2 - at + 9 = 0 \quad ...(2)$ for roots of equation (1) to be real & distinct roots of equation (2) must be positive & distinct (i) $ D > 0 \Rightarrow a^2 - 36 > 0 \Rightarrow a \in (-\infty, -6) \cup (6, \infty) $ (ii) $ \frac{-b}{2a} > 0 \Rightarrow \frac{a}{2} > 0 \Rightarrow a > 0 $ (iii) $ f(0) > 0 \Rightarrow 9 > 0 \Rightarrow a \in \mathbb{R} $ By (i) $\cap$ (ii) $\cap$ (iii) $ \therefore a \in (6, \infty) $ $ \therefore $ least integral value of $ a $ is $ 7 $

Qus : 151 JEE MAIN PYQ 2026

Let $ \vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k} $ and $ \vec{b} = \hat{i} - \hat{j} + 3\hat{k} $. If $ \vec{c} $ is a vector such that

$ 2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0} $

and $ (\vec{a} - \vec{b}) \cdot \vec{c} = -97 $, then

$ |\vec{c} \times \vec{k}|^2 $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ 2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = 0 $ $ \Rightarrow (2\vec{a} + 3\vec{b}) \times \vec{c} = 0 \Rightarrow \vec{c} \parallel (2\vec{a} + 3\vec{b}) $ $ \vec{c} = \lambda (2\vec{a} + 3\vec{b}) $ $ = \lambda (7\hat{i} - 13\hat{j} + 19\hat{k}) $ Now $ (\vec{a} - \vec{b}) \cdot \vec{c} = -97 $ $ \Rightarrow \lambda (7 + 52 + 38) = -97 $ $ \Rightarrow \lambda = -1 $ Now $ \vec{c} = -7\hat{i} + 13\hat{j} - 19\hat{k} $ $ \vec{c} \times \vec{k} = -7\hat{j} + 13\hat{i} $ $ |\vec{c} \times \vec{k}|^2 = 7^2 + 13^2 = 218 $

Qus : 152 JEE MAIN PYQ 2026

For $x < 0$ : $ f(x) = b^2 \sin\left( \frac{\pi}{2} \left( \frac{\pi}{2} (\cos x + \sin x)\cos x \right) \right) $ For $x > 0$ : $ f(x) = \frac{\sin x - \frac{1}{2}\sin 2x}{x^3} $ For $x = 0$ : $ f(x) = a $

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ f(0) = a $ RHL $ = \lim_{x \to 0^+} \frac{\sin x (1 - \cos x)}{x^3} = \frac{1}{2} $ LHL $ = \lim_{x \to 0^-} \left[ b^2 \sin\left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\sin x + \cos x) \cos x \right] \right) \right] = b^2 $ $ \therefore a = \frac{1}{2},; b^2 = \frac{1}{2} $ $ \Rightarrow a^2 + b^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} $

Qus : 153 JEE MAIN PYQ 2026

If $ f(x) $ satisfies the relation $ f(x) = e^x + \int_0^x (y + xe^x) f(y),dy $, then $ e + f(0) $ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ f(x) = e^x + \int_0^x y f(y),dy + xe^x \int_0^x f(y),dy $ $ f(x) = e^x + A + Bxe^x $ $ A = \int_0^x y f(y),dy = \int_0^x y(A + e^y + Bye^y),dy $ $ A = \frac{A}{2} + 0(-1) + B(e - 1) $ $ \frac{A}{2} + B(1 - e) = 1 $ $ B = \int_0^1 f(y),dy $ $ B = \int_0^1 (e^y + A + Bye^y),dy $ $ B = (e - 1) + A + B(0 - 1) $ $ B = e - 1 + A + B(0 - (-1)) $ $ B = e - 1 + A + B \Rightarrow A = 1 - e $ $ f(x) = e^x + A + Bxe^x $ $ f(0) = 1 + A + B\cdot 0 = 1 + A = 2 - e $ $ e + f(0) = 2 $

Qus : 154 JEE MAIN PYQ 2026

Let $ (h, k) $ lie on the circle $ C: x^2 + y^2 = 4 $ and the point $ (2h + 1, 3k + 2) $ lie on an ellipse with eccentricity $ e $. Then the value of $ \frac{5}{e^2} $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Let $ P = (2\cos\theta, 2\sin\theta) $ $ \therefore $ coordinates of $ Q = (4\cos\theta + 1,; 6\sin\theta + 3) $ $ \therefore $ locus of $ Q $ is $ \left( \frac{x - 1}{4} \right)^2 + \left( \frac{y - 3}{6} \right)^2 = 1 $ $ e^2 = 1 - \frac{16}{36} = \frac{5}{9} $ $ \Rightarrow \frac{5}{e^2} = 9 $

Qus : 155 JEE MAIN PYQ 2026

Let $ z = (1 + i)(1 + 2i)(1 + 3i) \ldots (1 + ni) $, where $ i = \sqrt{-1} $. If $ |z|^2 = 44200 $, then $ n $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ |z|^2 = 2^3 \cdot 5^2 \cdot 13 \cdot 17 $ $ \prod (1 + k^2) = (2)(5)(10)(17)(26) = 2 \cdot 5 \cdot 10 \cdot 17 \cdot 26 $ so $ n = 5 $

Qus : 156 JEE MAIN PYQ 2026

Let $ S $ be a set of 5 elements and $ P(S) $ denote the power set of $ S $. Let $ E $ be an event of choosing an ordered pair $ (A, B) $ from the set $ P(S) \times P(S) $ such that $ A \cap B = \emptyset $. If the probability of the event $ E $ is $ \frac{3^p}{2^q} $, where $ p, q \in \mathbb{N} $, then $ p + q $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ S = {a, b, c, d, e} $ $ P(S) $ contains 32 elements both set $ A $ and set $ B $ are subsets of $ P(S) $ Every element has 4 choices Favourable cases $ = 3^5 $ Total cases $ = 4^5 $ $ P = \frac{3^5}{4^5} = \frac{3^5}{2^{10}} $ $ p = 5,; q = 10 $ $ p + q = 15 $

Qus : 157 JEE MAIN PYQ 2026

The number of elements in the set $ { x \in [0,180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ)\tan x \tan(x - 50^\circ) } $ is ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ \frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ)\tan(x - 50^\circ) $ $ \frac{\sin(x + 100^\circ)\cos x}{\cos(x + 100^\circ)\sin x} = \frac{\sin(x + 50^\circ)\sin(x - 50^\circ)}{\cos(x + 50^\circ)\cos(x - 50^\circ)} $ Apply $ C $ & $ D $ $ \frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x} $ $ 2\sin(2x + 100^\circ)\cos 2x + \sin 100^\circ + \sin 200^\circ = 0 $ $ \sin(4x + 100^\circ) + \sin 100^\circ + \sin 200^\circ = 0 $ $ \sin(4x + 100^\circ) = -2\sin 150^\circ \cos 50^\circ $ $ \sin(4x + 100^\circ) = -\cos 50^\circ = \sin(-40^\circ) $ $ 4x + 100^\circ = n\pi + (-1)^n(-40^\circ) $ $ x = \frac{n\pi + (-1)^n(40^\circ) - 100^\circ}{4} $ $ \Rightarrow x = 30^\circ,; 55^\circ,; 120^\circ,; 145^\circ \text{ in } (0,\pi) $ $ \therefore $ no. of solutions $ = 4 $

Qus : 158 JEE MAIN PYQ 2026

If $ g(x) = 3x^2 + 2x - 3,; f(0) = -3 $ and $ 4g(f(x)) = 3x^2 - 32x + 72 $, then $ f(g(2)) $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

$ g(2) = 13 $ $ f(g(2)) = f(13) $ Now $ 4g(f(x)) = 3x^2 - 32x + 72 $ $ 4[3f(x)^2 + 2f(x) - 3] = 3x^2 - 32x + 72 $ Let $ f(x) = t $ $ 12t^2 + 8t - (3x^2 - 32x + 84) = 0 $ $ f(x) = \frac{-8 \pm \sqrt{64 + 48(3x^2 - 32x + 84)}}{24} $ $ f(x) = \frac{-8 \pm 4(3x - 16)}{24} $ $ f(x) = \frac{-8 + 4(3x - 16)}{24} $ $ \because f(0) = -3 \Rightarrow \text{we take +ve sign} $ $ f(x) = \frac{-8 + 4(3x - 16)}{24} $ $ f(13) = \frac{-8 + 4 \cdot 23}{24} = \frac{84}{24} = \frac{7}{2} $

Qus : 159 JEE MAIN PYQ 2026

The value of $ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k+1)}{k!} $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ T_k = (-1)^{k+1} \frac{k(k+1)}{k!} = (-1)^{k+1} \frac{k(k-1) + 2k}{k!} $ $ = (-1)^{k+1} \left[ \frac{k(k-1)}{k!} + \frac{2k}{k!} \right] $ $ = (-1)^{k+1} \left[ \frac{1}{(k-2)!} + \frac{2}{(k-1)!} \right] $ $ \Rightarrow \sum = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-2)!} + 2 \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!} $ $ = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{e} $

Qus : 160 JEE MAIN PYQ 2026

Let $ y = x $ be the equation of a chord of the circle $ C_1 $ (in the closed half-plane $ x \ge 0 $) of diameter 10 passing through the origin. Let $ C_2 $ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $ C_2 $, which passes through the point $ (2, 3) $ and is farthest from the center of $ C_2 $, is $ x + ay + b = 0 $, then $ a - b $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Equation of circle $ C_2 $ is

$ x^2 + y^2 - 5x - 5y = 0 $

its centre is $ \left( \frac{5}{2}, \frac{5}{2} \right) $

$ m_{AB} = -1 $

$ \therefore $ slope of required chord $ = 1 $

$ \therefore $ equation of required chord is $ x - y + 1 = 0 $

$ \therefore a = -1,; b = 2 $

$ \therefore a - b = -2 $

Qus : 161 JEE MAIN PYQ 2026

If $ \frac{\tan(A - B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1 $, $ A, B, C \in \left( 0, \frac{\pi}{2} \right) $, then

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \frac{\tan A - \tan B}{(1 + \tan A \tan B)\tan A} + \frac{1 + \cot^2 A}{1 + \cot^2 C} = 1 $

Put $\tan A = x,; \tan B = y,; \tan C = z$

$ \Rightarrow \frac{x - y}{(1 + xy)x} + \frac{(x^2 + 1)z^2}{x^2(z^2 + 1)} = 1 $

$ \Rightarrow x(x - y)(z^2 + 1) + z^2(1 + x^2)(1 + xy) = (1 + xy)x^2(1 + z^2) $

after solving we get

$ z^2 = xy \quad \therefore 1 + x^2 \ne 0 $

$ \therefore \tan^2 C = \tan A \cdot \tan B $

$ \Rightarrow \tan A,; \tan C,; \tan B$ are in G.P.

Qus : 162 JEE MAIN PYQ 2026

Let $ z $ be a complex number such that $ |z - 6| = 5 $ and $ |z + 2 - 6i| = 5 $. Then the value of $ z^3 + 3z^2 - 15z + 141 $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Center of first circle $ C_1(6,0),; r_1 = 5 $

Center of second circle $ C_2(-2,6),; r_2 = 5 $

$ \therefore C_1C_2 = r_1 + r_2 $

$ \therefore $ common point $ Z $ is mid point of $ C_1 $ & $ C_2 $

[Image used in solution — two circles diagram]

$ z = 2 + 3i $

$ z^2 = 4z - 13 $

$ z^3 = 3z - 52 $

$ z^3 + 3z^2 - 15z + 141 = 50 $

Qus : 163 JEE MAIN PYQ 2026

Let $ ABC $ be an equilateral triangle with orthocenter at the origin and the side $ BC $ on the line $ x + 2\sqrt{2}y = 4 $. If the co-ordinates of the vertex $ A $ are $ (\alpha, \beta) $, then the greatest integer less than or equal to $ |\alpha + \sqrt{2}\beta| $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ m_{BC} \cdot m_{AD} = -1 $

$ \Rightarrow \left( -\frac{1}{2\sqrt{2}} \right)\left( \frac{\beta}{\alpha} \right) = -1 $

$ \Rightarrow \beta = 2\sqrt{2}\alpha \quad ...(1)$

$ \therefore OD = \frac{|4|}{\sqrt{1 + 8}} = \frac{4}{3} \Rightarrow AO = \frac{8}{3} $

So $ AD = \frac{8}{3} + \frac{4}{3} = 4 $

$ \Rightarrow \frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9} \text{ or } -\frac{8}{9} $

$ \because A(\alpha,\beta) $ & $(0,0)$ lies on same side of given line

$ \therefore (\alpha,\beta) = \left( -\frac{8}{9}, -\frac{16\sqrt{2}}{9} \right) $

$ \Rightarrow [\alpha + \sqrt{2}\beta] = \left[ \frac{-8 - 32}{9} \right] = 4 $

Qus : 164 JEE MAIN PYQ 2026

Let $ S = {1, 2, 3, 4, 5, 6, 7, 8, 9} $. Let $ x $ be the number of 9-digit numbers formed using the digits of the set $ S $ such that only one digit is repeated and it is repeated exactly twice. Let $ y $ be the number of 9-digit numbers formed using the digits of the set $ S $ such that only two digits are repeated and each of these is repeated exactly twice. Then,

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ S = {1, 2, 3, \ldots, 9} $

$ x = {^9C_1} \cdot {^9C_2} \cdot \frac{9!}{2} = \frac{9 \times 8 \times 9!}{2} $

$ y = {^9C_2} \cdot {^7C_2} \cdot \frac{9!}{2! \times 2!} = \frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times \frac{9!}{2! \times 2!} $

$ \Rightarrow \frac{x}{y} = \frac{4}{21} $

$ \Rightarrow 21x = 4y $

Qus : 165 JEE MAIN PYQ 2026

Let $ S = { x^3 + ax^2 + bx + c : a, b, c \in \mathbb{N} \text{ and } a, b, c \le 20 } $ be a set of polynomials. Then the number of polynomials in $ S $, which are divisible by $ x^2 + 2 $, is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ x^3 + ax^2 + bx + c = (x^2 + 2)\left(x + \frac{c}{2}\right) $

$ \Rightarrow a = \frac{c}{2},; b = 2 $

$ \Rightarrow b = 2,; a = \frac{c}{2} \in {2, 4, \ldots, 20} $

Number of polynomials in $ S $ will be $ 10 $

Qus : 166 JEE MAIN PYQ 2026

A bag contains 10 balls out of which $ k $ are red and $ (10 - k) $ are black, where $ 0 \le k \le 10 $. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Probability $ = \frac{{^1C_0} \cdot {^9C_3}}{\sum_{k=0}^{10} {^kC_0} \cdot {^{10-k}C_3}} $

$ = \frac{{^9C_3}}{{^0C_3} + {^1C_3} + {^2C_3} + \ldots + {^9C_3}} $

$ = \frac{{^9C_3}}{{^{11}C_4}} = \frac{14}{55} $

Qus : 167 JEE MAIN PYQ 2026

If $ \alpha, \beta $, where $ \alpha < \beta $, are the roots of the equation $ \lambda x^2 - (\lambda + 3)x + 3 = 0 $ such that $ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} $, then the sum of all possible values of $ \lambda $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \frac{\beta - \alpha}{\alpha \beta} = \frac{1}{3}, \quad \alpha + \beta = \frac{\lambda + 3}{\lambda}, \quad \alpha \beta = \frac{3}{\lambda} $

$ \beta - \alpha = \frac{\alpha \beta}{3} = \frac{1}{\lambda} $

on squaring

$ \alpha^2 + \beta^2 - 2\alpha \beta = \frac{1}{\lambda^2} \quad ...(1)$

$ \alpha^2 + \beta^2 + 2\alpha \beta = \frac{(\lambda + 3)^2}{\lambda^2} \quad ...(2)$

(2) $-$ (1)

$ 4\alpha \beta = \frac{(\lambda + 3)^2 - 1}{\lambda^2} $

$ \Rightarrow \frac{12}{\lambda} = \frac{\lambda^2 + 6\lambda + 8}{\lambda^2} $

$ \Rightarrow \lambda^2 - 6\lambda + 8 = 0 $

$ \Rightarrow \lambda = 0, 2, 4 $

Sum of possible values of $ \lambda $ is $ = 6 $

Qus : 168 JEE MAIN PYQ 2026

If $ \int \frac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x} dx = f(x) + C $ where $ C $ is the constant of integration, then $ f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \int \frac{dx}{\sin^5 x \cos^2 x} - 5 \int \frac{dx}{\sin^5 x} $

$ \int \frac{\sec^2 x dx}{\sin^5 x} - 5 \int \frac{dx}{\sin^5 x} $

By IBP

$ = \frac{\tan x}{\sin^5 x} - \int \frac{\cos x \tan x dx}{\sin^6 x} - 5 \int \frac{dx}{\sin^5 x} $

$ = \frac{\tan x}{\sin^5 x} + C $

$ f(x) = \frac{\tan x}{\sin^5 x} $

$ f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{3}} - (\sqrt{2})^5 = 4\sqrt{2} - \frac{32}{\sqrt{3}} $

$ = \frac{4}{\sqrt{3}}(8 - \sqrt{6}) $

Qus : 169 JEE MAIN PYQ 2026

Let $ f $ be a polynomial function such that $ f(x^2 + 1) = x^4 + 5x^2 + 2 $, for all $ x \in \mathbb{R} $. Then $ \int_0^3 f(x),dx $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ f(x^2 + 1) = x^4 + 5x^2 + 2 $

put $ x^2 + 1 = t $

$ \Rightarrow f(t) = (t - 1)^2 + 5(t - 1) + 2 $

$ \Rightarrow f(t) = t^2 + 3t - 2 $

Now,

$ \int_0^3 f(t),dt = \int_0^3 (t^2 + 3t - 2),dt $

$ = \left[ \frac{t^3}{3} + \frac{3t^2}{2} - 2t \right]_0^3 $

$ = \frac{27}{3} + \frac{27}{2} - 6 = \frac{33}{2} $

Qus : 170 JEE MAIN PYQ 2026

The area of the region $ R = {(x, y) : xy \le 8,; 1 \le y \le x^2,; x \ge 0 } $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ A = \int_1^2 (x^2 - 1),dx + \int_2^4 \left( \frac{8}{x} - 1 \right) dx $

$ = 8\log_e 4 - \frac{14}{3} = 16\log_e 2 - \frac{14}{3} $

$ = \frac{2}{3}(24\log_e 2 - 7) $

Qus : 171 JEE MAIN PYQ 2026

The value of $ \lim_{x \to 0} \frac{\log_e(\sec(ex))\sec(e^2x)\ldots\sec(e^{10}x)}{e^2 - e^{2\cos x}} $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \Rightarrow \lim_{x \to 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^2x)) + \ldots + \ln(\sec(e^{10}x))}{e^2 x^2} $

Using L'H rule

$ \Rightarrow \lim_{x \to 0} \frac{e\tan(ex) + e^2\tan(e^2x) + \ldots + e^{10}\tan(e^{10}x)}{2e^2 x} $

$ = \frac{1}{2e^2} [e^2 + e^4 + e^6 + \ldots + e^{20}] $

$ = \frac{1}{2} \cdot \frac{e^{20} - 1}{e^2 - 1} $

Qus : 172 JEE MAIN PYQ 2026

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are $ 2, 3, 5, 10, 11, 13, 15, 21 $, then the mean deviation about the median of all the 10 observations is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \frac{2+3+5+10+11+13+15+21+a+b}{10} = 9 $

$ \Rightarrow \frac{80 + a + b}{10} = 9 \Rightarrow a + b = 10 \quad ...(1)$

$ \frac{\sum x_i^2}{10} - \left( \frac{\sum x_i}{10} \right)^2 = 34.2 $

$ \Rightarrow \frac{2^2 + 3^2 + 5^2 + 10^2 + 11^2 + 13^2 + 15^2 + 21^2 + a^2 + b^2}{10} - 81 = 34.2 $

$ \Rightarrow 1094 + a^2 + b^2 - 810 = 342 $

$ \Rightarrow a^2 + b^2 = 58 \quad ...(2)$

From (1) and (2):

$ a = 7,; b = 3 $ or $ a = 3,; b = 7 $

Number $ \Rightarrow 2, 3, 5, 7, 10, 11, 13, 15, 21 $

Mean $ = \frac{7 + 10}{2} = 8.5 $

M.D. $ = \frac{6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5}{10} = \frac{50}{10} = 5 $

Qus : 173 JEE MAIN PYQ 2026

Let $ A, B $ and $ C $ be three $ 2 \times 2 $ matrices with real entries such that $ B = (I + A)^{-1} $ and $ A + C = I $. If $ BC = \left[ \matrix{ 1 & -5 \cr -1 & 2 } \right] $ and $ CB \left[ \matrix{ x_1 \cr x_2 } \right] = \left[ \matrix{ 12 \cr -6 } \right] $, then $ x_1 + x_2 $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ B = (I + A)^{-1},; A + C = I $

$ \Rightarrow B(I + A) = (I + A)B = I $

$ \Rightarrow B + BA = B + AB = I $

$ \Rightarrow B + B(I - C) = B + (I - C)B $

$ \Rightarrow 2B - BC = 2B - CB $

$ \Rightarrow BC = CB $

$ \therefore CB \left[ \matrix{ x_1 \cr x_2 } \right] = \left[ \matrix{ 1 & -5 \cr -1 & 2 } \right] \left[ \matrix{ x_1 \cr x_2 } \right] = \left[ \matrix{ 12 \cr -6 } \right] $

$ \Rightarrow \left[ \matrix{ x_1 \cr x_2 } \right] = \left[ \matrix{ 1 & -5 \cr -1 & 2 } \right]^{-1} \left[ \matrix{ 12 \cr -6 } \right] $

$ = \frac{1}{-3} \left[ \matrix{ 2 & 5 \cr 1 & 1 } \right] \left[ \matrix{ 12 \cr -6 } \right] $

$ = \frac{1}{-3} \left[ \matrix{ 24 - 30 \cr 12 - 6 } \right] = \left[ \matrix{ 2 \cr -2 } \right] $

$ \therefore x_1 + x_2 = 0 $

Qus : 174 JEE MAIN PYQ 2026

The common difference of the A.P.: $ a_1, a_2, \ldots, a_n $ is 13 more than the common difference of the A.P.: $ b_1, b_2, \ldots, b_n $. If $ b_1 = -277,; b_{33} = -385 $ and $ a_n = 327 $, then $ a_1 $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Let common difference of A.P.'s are $ d_1 $ & $ d_2 $

$ d_1 = 13 + d_2 $

$ b_1 + 30d_2 = -277 \quad ...(1)$

$ b_1 + 42d_2 = -385 \quad ...(2)$

By (2) $-$ (1)

$ 12d_2 = -108 \Rightarrow d_2 = -9 $

$ \therefore d_1 = 4 $

Now $ a_n = 327 $

$ a_1 + 77d_1 = 327 $

$ a_1 + 308 = 327 $

$ a_1 = 19 $

Qus : 175 JEE MAIN PYQ 2026

If the distances of the point $ (1, 2, a) $ from the line

$ \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} $

along the lines

$ L_1 : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} $

and

$ L_2 : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c} $

are equal, then $ a + b + c $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ L : \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} $

$ L_1 : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} = \lambda $

$ L_2 : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c} = \mu $

Let $ A(3\lambda + 1,; 4\lambda + 2,; b\lambda + a) $

It lies on $ L $

$ \Rightarrow \frac{3\lambda}{1} = \frac{4\lambda + 2}{2} = \frac{b\lambda + a - 1}{1} $

$ \Rightarrow \lambda = 1 $ and $ a + b - 1 = 3 $

$ \Rightarrow a + b = 4 \quad ...(1)$

$ A(4, 6, 4),; a + b = 4 $

Let $ B(\mu + 1,; 4\mu + 2,; c\mu + a) $

It also lies on $ L $

$ \Rightarrow \frac{\mu}{1} = \frac{4\mu + 2}{2} = \frac{c\mu + a - 1}{1} $

$ \Rightarrow 2\mu = 4\mu + 2 \Rightarrow \mu = -1 $

$ a - c - 1 = -1 \Rightarrow a = c \quad ...(2)$

Thus, $ a + b + c = 4 + c = 4 + a $

Qus : 176 JEE MAIN PYQ 2026

For three unit vectors $ \vec{a}, \vec{b}, \vec{c} $ satisfying

$ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9 $ and

$ |2\vec{a} + k\vec{b} + k\vec{c}| = 3 $,

the positive value of $ k $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9 $

$ \Rightarrow \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2} $

$ \Rightarrow \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{b} + \vec{c} = -\vec{a} $

$ |2\vec{a} + k(\vec{b} + \vec{c})| = 3 $

$ \Rightarrow |\vec{a}(2 - k)| = 3 $

$ \Rightarrow |2 - k| = 3 $

$ k = 5 \text{ or } -1 $

Positive value of $ k $ is $ 5 $

Qus : 177 JEE MAIN PYQ 2026

Let $ y = y(x) $ be the solution of the differential equation

$ x \frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y,; x \ne 0 $.

If $ y(2) = x $, then $ \tan(y(1)) $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y $

$ \sec^2 y \frac{dy}{dx} - 2\tan y \cdot \frac{1}{x} = x^2(2 - x^3) $

$ \tan y = t \Rightarrow \sec^2 y \frac{dy}{dx} = \frac{dt}{dx} $

$ \frac{dt}{dx} - \frac{2t}{x} = x^2(2 - x^3) \quad (\text{LDE}) $

I.F. $ = e^{\int -\frac{2}{x}dx} = e^{-2\ln x} = \frac{1}{x^2} $

$ \therefore \frac{t}{x^2} = \int \frac{1}{x^2} x^2(2 - x^3)dx + C $

$ \frac{\tan y}{x^2} = 2x - \frac{x^4}{4} + C $

$ y(2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0 $

$ \tan y = 2x^3 - \frac{1}{4}x^6 $

$ x = 1 \Rightarrow \tan y = 2 - \frac{1}{4} = \frac{7}{4} $

Qus : 178 JEE MAIN PYQ 2026

In a G.P., if the product of the first three terms is $ 27 $ and the set of all possible values for the sum of its first three terms is $ \mathbb{R} - (a, b) $, then $ a^2 + b^2 $ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Let first three terms of G.P. are $ \frac{A}{r},; A,; Ar $

$ \frac{A}{r} \cdot A \cdot Ar = 27 \Rightarrow A^3 = 27 \Rightarrow A = 3 $

$ 3\left( \frac{1}{r} + 1 + r \right) = 3 + 3\left( r + \frac{1}{r} \right) $

We know $ r + \frac{1}{r} \ge 2 $ or $ r + \frac{1}{r} \le -2 $

$ S \in \mathbb{R} - (-3, 9) $

$ a^2 + b^2 = 9 + 81 = 90 $

Qus : 179 JEE MAIN PYQ 2026

For some $ \theta \in \left( 0, \frac{\pi}{2} \right) $, let the eccentricity and the length of the latus rectum of the hyperbola $ x^2 - y^2 \sec^2 \theta = 8 $ be $ e_1 $ and $ \ell_1 $, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $ x^2 \sec^2 \theta + y^2 = 6 $ be $ e_2 $ and $ \ell_2 $, respectively. If $ e_1^2 = e_2^2(\sec^2 \theta + 1) $, then

$ \left( \frac{\ell_1 \ell_2}{e_1 e_2} \right)\tan^2 \theta $ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ \frac{x^2}{8} - \frac{y^2}{8\cos^2\theta} = 1,; e_1 = \sqrt{1 + \frac{8\cos^2\theta}{8}} $

$ \ell_1 = \frac{2b^2}{a} = \frac{2(8\cos^2\theta)}{2\sqrt{2}} $

$ \frac{x^2}{6} + \frac{y^2}{6\cos^2\theta} = 1,; e_2 = \sqrt{1 - \frac{6\cos^2\theta}{6}} = \sin\theta $

$ \ell_2 = \frac{2b^2}{a} = \frac{2\cdot 6\cos^2\theta}{\sqrt{6}} $

$ e_1^2 = e_2^2(1 + \sec^2\theta) $

$ \Rightarrow 1 + \cos^2\theta = \sin^2\theta (1 + \frac{1}{\cos^2\theta}) $

$ \Rightarrow \theta = \frac{\pi}{4} $

$ \ell_1 = 2\sqrt{2},; e_1 = \sqrt{\frac{3}{2}} $

$ \ell_2 = \sqrt{6},; e_2 = \frac{1}{\sqrt{2}} $

$ \Rightarrow \left( \frac{\ell_1 \ell_2}{e_1 e_2} \right)\tan^2\theta = 8 $

Qus : 180 JEE MAIN PYQ 2026

If $ k = \tan\left( \frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right) \right) + \tan\left( \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right) \right) $, then the number of solutions of the equation

$ \sin^{-1}(kx - 1) = \sin^{-1}x - \cos^{-1}x $ is ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Let $ \theta = \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right) $

$ \Rightarrow k = \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} = \frac{2}{\sin 2\theta} $

$ k = 3 $

$ \sin^{-1}(3x - 1) = \sin^{-1}x - \cos^{-1}x $

$ \sin^{-1}(3x - 1) = \frac{\pi}{2} - 2\cos^{-1}x $

$ 3x - 1 = \sin\left( \frac{\pi}{2} - 2\cos^{-1}x \right) $

$ \Rightarrow 3x - 1 = 2x^2 - 1 $

$ \Rightarrow x = 0,; \frac{3}{2} \text{ (rejected)} $

No. of solution $ = 1 $

Qus : 181 JEE MAIN PYQ 2026

The value of $ \sum_{n=1}^{20} \left[ \sqrt{\pi} \int_0^\pi [x]\sin(nx),dx \right] $ is ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

Let $ I_n = \int_0^\pi [x]\sin(nx),dx \quad ...(1)$

Apply King Property

$ = \int_0^\pi (\pi - x)\sin(nx),dx \quad ...(2)$

By (1) + (2)

$ 2I_n = \int_0^\pi \pi\sin(nx),dx = \pi \int_0^\pi \sin(nx),dx $

$ I_n = \frac{\pi}{2} \int_0^\pi \sin(nx),dx $

$ I_n = \frac{\pi}{2} \cdot \frac{2}{n}(1 - (-1)^n) $

$ = \frac{\pi}{n}(1 - (-1)^n) $

Thus only odd $ n $ contribute

$ S = 1 + 2 + 3 + \cdots + 20 = \frac{20 \cdot 21}{2} = 210 $

Qus : 182 JEE MAIN PYQ 2026

Let $ PQR $ be a triangle such that $ \overrightarrow{PQ} = -2\hat{i} - \hat{j} + 2\hat{k} $ and $ \overrightarrow{PR} = a\hat{i} + b\hat{j} - 4\hat{k},; a, b \in \mathbb{Z} $. Let $ S $ be the point on $ QR $, which is equidistant from the lines $ PQ $ and $ PR $. If $ |PR| = 9 $ and $ \overrightarrow{PS} = \hat{i} - 7\hat{j} + 2\hat{k} $, then the value of $ 3a - 4b $ is ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ |PR| = 9 \Rightarrow a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65 $

$ \cos\theta = \frac{\overrightarrow{PQ}\cdot \overrightarrow{PS}}{|PQ||PS|} = \frac{-2 + 7 + 4}{3\cdot 3\sqrt{6}} = \frac{9}{3\cdot 3\sqrt{6}} = \frac{1}{\sqrt{6}} $

Using geometry relations

$ 3a - 4b = 37 $

Qus : 183 JEE MAIN PYQ 2026

Given below two statements :

Statement I : $ 25^{13} + 20^{13} + 8^{13} + 3^{13} $ is divisible by $ 7 $.

Statement II : The integral part of $ (7 + 4\sqrt{3})^{25} $ is an odd number.

In the light of the above statements, choose the correct answer from the options given below :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Statement I :

$ 25^{13} + 3^{13} \quad\text{and}\quad 20^{13} + 8^{13} $

divisible by $ (25 + 3) $ and divisible by $ (20 + 8) $

$ \therefore $ divisible by $ 7 $

Statement II :

Let $ R = (7 + 4\sqrt{3})^{25} = I + f $

$ R' = (7 - 4\sqrt{3})^{25} = f' $

$ \therefore R + R' = I + f + f' = 2^{25} C_0 7^{25} + 25C_2 7^{23}(4\sqrt{3})^2 + \ldots $

$ I + f + f' = \text{even integer} $

$ \Rightarrow I = \text{odd integer} $

$ \therefore 0 < f + f' < 2 \Rightarrow f + f' = 1 $

$ \therefore $ Both the statements are correct

Qus : 184 JEE MAIN PYQ 2026

The sum of coefficients of $ x^{499} $ and $ x^{500} $ in

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ S = (1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \ldots + x^{1000} $

$ = (1 + x)^{1000} \left( 1 - \left( \frac{x}{1 + x} \right)^{1001} \right) $

$ = (1 + x)^{1000} - x^{1001} $

Required sum $ = {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500} $

Qus : 185 JEE MAIN PYQ 2026

Let $ A $ be the focus of the parabola $ y^2 = 8x $. Let the line $ y = mx + c $ intersect the parabola at two distinct points $ B $ and $ C $. If the centroid of the triangle $ ABC $ is $ \left( \frac{7}{3}, \frac{4}{3} \right) $, then $ (BC)^2 $ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Coordinates of centroid of triangle $ ABC $ are

$ \frac{2}{3}(t_1^2 + t_2^2 + 1) = \frac{7}{3} \Rightarrow t_1^2 + t_2^2 = \frac{5}{2} $

$ \frac{4}{3}(t_1 + t_2) = \frac{4}{3} \Rightarrow t_1 + t_2 = 1 $

$ (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2 \Rightarrow t_1 t_2 = -\frac{3}{4} $

$ (t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1 t_2 = 4 $

$ (BC)^2 = 4(t_1 - t_2)^2 + 16(t_1 - t_2)^2 $

$ \Rightarrow (BC)^2 = 80 $

Qus : 186 JEE MAIN PYQ 2026

The probability distribution of a random variable $ X $ is given below :

If $ E(X) = \frac{263}{15} $, then $ P(X < 20) $ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ E(X) = \sum X_i P(X_i) = \frac{526k}{15 \times 7} = \frac{263}{15} \Rightarrow k = \frac{7}{2} $

$ P(X < 20) = \sum_{X=14}^{19} P(X) = \frac{11}{15} $

Qus : 187 JEE MAIN PYQ 2026

Let $ f(x) = \lim_{\theta \to 0} \left( \frac{\cos \pi x - x^{(\theta)} \sin(x-1)}{1 + x^{(\theta)} (x - 1)} \right),; x \in \mathbb{R}. $

Consider the following two statements :

(I) $ f(x) $ is discontinuous at $ x = 1 $.

(II) $ f(x) $ is continuous at $ x = -1 $.

Then,

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ f(x) = \begin{cases} \cos \pi x, & x \to 1^- \ -\frac{\sin(x-1)}{(x-1)}, & x \to 1^+ \end{cases} $

RHL $ = \lim_{x \to 1} -\frac{\sin(x-1)}{(x-1)} = -1 $

LHL $ = \lim_{x \to 1} \cos \pi x = -1,; f(1) = -1 $

$ f(x) $ is continuous at $ x = 1 $

$ f(x) = \begin{cases} -\frac{\sin(x-1)}{(x-1)}, & x \to -1^- \ \cos \pi x, & x \to -1^+ \end{cases} $

RHL $ = \lim_{x \to -1} \cos \pi x = -1 $

LHL $ = \lim_{x \to -1} -\frac{\sin(x-1)}{(x-1)} = \frac{\sin 2}{2} $

$ f(x) $ is discontinuous at $ x = -1 $

Qus : 188 JEE MAIN PYQ 2026

Considering the principal values of inverse trigonometric functions, the value of the expression

$ \tan \left( 2\sin^{-1} \left( \frac{2}{\sqrt{13}} \right) - 2\cos^{-1} \left( \frac{3}{\sqrt{10}} \right) \right) $ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Let $ \sin \theta = \frac{2}{\sqrt{13}} \Rightarrow \theta $, and $ \cos \phi = \frac{3}{\sqrt{10}} \Rightarrow \phi $

$ \sin \theta = \frac{2}{\sqrt{13}},; \cos \phi = \frac{3}{\sqrt{10}} $

$ \tan(2\theta - 2\phi) = \frac{\tan 2\theta - \tan 2\phi}{1 + \tan 2\theta \tan 2\phi} $

$ \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} $

$ = \frac{12}{5} $

$ \tan 2\phi = \frac{3}{4} $

$ \Rightarrow \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} = \frac{33}{56} $

Qus : 189 JEE MAIN PYQ 2026

Let the arithmetic mean of $ \frac{1}{a} $ and $ \frac{1}{b} $ be $ \frac{5}{16},; a > 2 $.

If $ \alpha $ is such that $ a,; 4,; \alpha,; b $ are in A.P., then the equation $ \alpha x^2 - ax + 2(\alpha - 2b) = 0 $ has :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ a = 4 - d,; \alpha = 4 + d,; b = 4 + 2d $

$ \Rightarrow (4 + d)x^2 - (4 - d)x + 2(4 + d - 8 - 4d) = 0 $

$ \Rightarrow (4 + d)x^2 - (4 - d)x + 2(-4 - 3d) = 0 $

Also $ \frac{1}{a} + \frac{1}{b} = \frac{5}{16} $

$ \Rightarrow \frac{1}{4-d} + \frac{1}{4+2d} = \frac{5}{16} $

$ \Rightarrow d = 2 $

Equation becomes

$ 6x^2 - 2x - 20 = 0 $

$ \Rightarrow 3x^2 - x - 10 = 0 $

$ x = 2,; -\frac{5}{3} $

Qus : 190 JEE MAIN PYQ 2026

Given below are two statements :

Statement I : The function $ f : \mathbb{R} \rightarrow \mathbb{R} $ defined by

$ f(x) = \frac{x}{1 + |x|} $ is one-one.

Statement II : The function $ f : \mathbb{R} \rightarrow \mathbb{R} $ defined by

$ f(x) = \frac{x^2 + 4x - 30}{x^2 - 8x + 18} $ is many-one.

In the light of the above statements, choose the correct answer from the options given below :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Statement I:

$ f(x) = \frac{x}{1 + |x|} $

$ f(x) = \begin{cases} \frac{x}{1 + x}, & x \ge 0 \ \frac{x}{1 - x}, & x < 0 \end{cases} $

[Image used in solution — graph]

$ f(x) $ is one-one

Statement II:

$ f(x) = \frac{x^2 + 4x - 30}{x^2 - 8x + 18} $

$ f(0) = -\frac{30}{18} = -\frac{5}{3} $

$ f(-1) = -\frac{5}{3} $

$ \Rightarrow f(0) = f(-1) $

$ \Rightarrow f(x) $ is many-one

Qus : 191 JEE MAIN PYQ 2026

An ellipse has center at $ (1,-2) $, one focus at $ (3,-2) $ and one vertex at $ (5,-2) $. Then the length of its latus rectum is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ CA_1 = a = 4 $

$ CF = ae = 2 $

$ \Rightarrow e = \frac{1}{2} $

$ LR = 2\left( \frac{a}{e} - ae \right) $

$ = 2 \times \frac{1}{2} \left( \frac{4}{1/2} - 2 \right) $

$ = 6 $

Qus : 192 JEE MAIN PYQ 2026

Let the ellipse $ E: \frac{x^2}{144} + \frac{y^2}{169} = 1 $ and the hyperbola

$ H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1 $ have the same foci. If $ e $ and $ L $ respectively denote the eccentricity and the length of the latus rectum of $ H $, then the value of $ 24(e + L) $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Equation of hyperbola:

$ \frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1 $

Equation of ellipse:

$ \frac{x^2}{144} + \frac{y^2}{169} = 1 $

$ e = \sqrt{1 - \frac{144}{169}} = \frac{5}{13} $

focus $ = (0,5) $

$ 5 = \lambda \sqrt{1 + \frac{16}{\lambda^2}} $

$ \Rightarrow \lambda^2 + 16 = 25 $

$ \Rightarrow \lambda = 3 $

Eccentricity of hyperbola

$ = \sqrt{1 + \frac{16}{9}} = \frac{5}{3} $

Length of latus rectum of hyperbola

$ = \frac{2(16)}{3} = \frac{32}{3} $

$ 24(e + L) = 24\left( \frac{5}{3} + \frac{32}{3} \right) = 8 \times 37 = 296 $

Qus : 193 JEE MAIN PYQ 2026

Let $ P_1 : y = 4x^2 $ and $ P_2 : y = x^2 + 27 $ be two parabolas. If the area of the bounded region enclosed between $ P_1 $ and $ P_2 $ is six times the area of the bounded region enclosed between the line $ y = ax,; a > 0 $ and $ P_1 $, then $ a $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Area bounded between $ P_1 $ & $ P_2 $ is

$ \int_{-3}^{3} ((x^2 + 27) - 4x^2),dx $

$ = 2 \int_0^3 (27 - 3x^2),dx = 2[27x - x^3]_0^3 $

$ = 2(81 - 27) = 108 $

$\therefore$ Area between $ P_1 $ & $ L $ is $ 18 $ sq. units

[Image used in solution — second graph]

Area between $ x^2 = 4ay $ and $ x = ay $ is

$ \frac{8a^2}{3m^3} $

$ \therefore \frac{(1/16)^2}{3(1/\alpha)^3} = 18 $

$ \Rightarrow \frac{8}{16 \cdot 16} \cdot \frac{\alpha^3}{3} = 18 $

$ \Rightarrow \alpha = 12 $

Qus : 194 JEE MAIN PYQ 2026

Let the circle $ x^2 + y^2 = 4 $ intersect $ x $-axis at the points $ A(a, 0),; a > 0 $ and $ B(b, 0) $. Let $ P(2\cos\alpha, 2\sin\alpha),; 0 < \alpha < \frac{\pi}{2} $ and $ Q(2\cos\beta, 2\sin\beta) $ be two points such that $ \alpha - \beta = \frac{\pi}{2} $. Then the point of intersection of $ AQ $ and $ BP $ lies on:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

Qus : 195 JEE MAIN PYQ 2026

Let $ [\cdot] $ denote the greatest integer function. Then

$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{12(3 + [x])}{3 + [\sin x] + [\cos x]} \right) dx $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{12(3 + [x])}{3 + [\sin x] + [\cos x]} dx $

$ I = \int_{-\frac{\pi}{2}}^{-1} \frac{12(1)dx}{2} + \int_{-1}^{0} \frac{12(2)dx}{2} + \int_{0}^{1} \frac{12(3)dx}{3} + \int_{1}^{\frac{\pi}{2}} \frac{12(4)dx}{3} $

$ I = 6\left( \frac{\pi}{2} - 1 \right) + 12(0 + 1) + 12(1 - 0) + 16\left( \frac{\pi}{2} - 1 \right) $

$ I = 3\pi - 6 + 12 + 12 + 8\pi - 16 $

$ I = 11\pi + 2 $

Qus : 196 JEE MAIN PYQ 2026

Let $ y = y(x) $ be the solution of the differential equation

$ x \frac{dy}{dx} - y = x^2 \cot x,; x \in (0,\pi) $.

If $ y\left( \frac{\pi}{2} \right) = \frac{\pi}{2} $, then $ 6y\left( \frac{\pi}{6} \right) - 8y\left( \frac{\pi}{4} \right) $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ xdy - ydx = x^2 \cot x,dx $

$ x^2 d\left( \frac{y}{x} \right) = x^2 \cot x,dx $

$ d\left( \frac{y}{x} \right) = \cot x,dx $

$ \int d\left( \frac{y}{x} \right) = \int \cot x,dx $

$ \frac{y}{x} = \log_e(\sin x) + C $

given $ y\left( \frac{\pi}{2} \right) = \frac{\pi}{2} $

$ \Rightarrow C = 1 $

$ y = x(\log_e \sin x + 1) $

$ y\left( \frac{\pi}{6} \right) = \frac{\pi}{6}[-\log_e 2 + 1] $

$ y\left( \frac{\pi}{4} \right) = \frac{\pi}{4}\left[ 1 - \frac{1}{2}\log_e 2 + 1 \right] $

$ 6y\left( \frac{\pi}{6} \right) - 8y\left( \frac{\pi}{4} \right) $

$ = \pi\left[ (-\log_e 2 + 1) + 2\left( \frac{1}{2}\log_e 2 - 1 \right) \right] $

$ = -\pi $

Qus : 197 JEE MAIN PYQ 2026

The sum of all the elements in the range of

$ f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x),; x \ne \frac{n\pi}{2},; n \in \mathbb{Z} $

where $ \text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \ -1, & \text{if } t < 0 \end{cases} $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$x \in (0, \frac{\pi}{2}) \Rightarrow y = 1 + 1 + 1 + 1 = 4$

$x \in (\frac{\pi}{2}, \pi) \Rightarrow y = 1 - 1 - 1 - 1 = -2$

$x \in (\pi, \frac{3\pi}{2}) \Rightarrow y = -1 - 1 + 1 + 1 = 0$

$x \in (\frac{3\pi}{2}, 2\pi) \Rightarrow y = -1 + 1 - 1 - 1 = -2$

$ \therefore \text{Range of } y \text{ is } {-2, 0, 4} $

Required sum $ = -2 + 0 + 4 = 2 $

Qus : 198 JEE MAIN PYQ 2026

Let $ Q(a, b, c) $ be the image of the point $ P(3, 2, 1) $ in the line

$ \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} $. Then the distance of $ Q $ from the line

$ \frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{-2} $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ drs \text{ of } PN = \langle r - 2,; 2r - 2,; r \rangle $

$ 1(r - 2) + 2(2r - 2) + 1(r) = 0 $

$ 6r - 6 = 0 \Rightarrow r = 1 $

$ \therefore N = (2, 2, 2) $

$ \Rightarrow Q = (1, 2, 3) $

$ AQ = \sqrt{64 + 49 + 4} = \sqrt{117} $

$ AM = \frac{|24 + 14 - 4|}{\sqrt{9 + 4 + 4}} = \frac{34}{\sqrt{17}} = 2\sqrt{17} $

$ \therefore QM = \sqrt{117 - 68} = \sqrt{49} = 7 $

Qus : 199 JEE MAIN PYQ 2026

Let $ P $ be a point in the plane of the vector

$ \overrightarrow{AB} = 3\hat{i} + \hat{j} - \hat{k} $ and $ \overrightarrow{AC} = \hat{i} - \hat{j} + 3\hat{k} $ such that $ P $ is equidistant from the lines $ AB $ and $ AC $. If $ |AP| = \frac{\sqrt{5}}{2} $, then the area of the triangle $ ABP $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ \cos 2\theta = \frac{3 - 1 - 3}{\sqrt{11}\cdot \sqrt{11}} = -\frac{1}{11} $

$ 1 - 2\sin^2 \theta = -\frac{1}{11} \Rightarrow 2\sin^2 \theta = \frac{12}{11} $

$ \Rightarrow \sin \theta = \sqrt{\frac{6}{11}} $

$ \therefore \text{Area}(\triangle ABP) = \frac{1}{2} \cdot \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{\sqrt{30}}{4} $

Qus : 200 JEE MAIN PYQ 2026

Let $ A = { z \in \mathbb{C} : |z - 2| \le 4 } $ and $ B = { z \in \mathbb{C} : |z - 2| + |z + 2| = 5 } $. Then the max $ |z_1 - z_2|,; z_1 \in A \text{ and } z_2 \in B $ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ |z - 2| \le 4 \Rightarrow (x - 2)^2 + y^2 \le 16 $

$ |z - 2| + |z + 2| = 5 \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $

$ \Rightarrow \frac{4x^2}{25} + \frac{4y^2}{9} = 1 $

[Image used in solution — circle & ellipse diagram]

Maximum value of $ |z_1 - z_2| = 6 + \frac{5}{2} = \frac{17}{2} $

Qus : 201 JEE MAIN PYQ 2026

Let $ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $ be such that $ f(0) = -26 + 24\log_e 2 $. If $ f(1) = a + b\log_e 3 $, where $ a, b \in \mathbb{Z} $, then $ a + b $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $

Put $ x = t^6 \Rightarrow dx = 6t^5 dt $

$ = \int \frac{6t^5 dt}{t^4 + 2t^3} = 6\int \frac{t^2 - 4}{t + 2} dt $

$ = 6\left[ \frac{t^2}{2} - 2t + 4\ln(t + 2) \right] + C $

$ = 3x^{1/3} - 12x^{1/6} + 24\ln(x^{1/6} + 2) + C $

$ f(0) = 24\ln 2 + C = -26 + 24\ln 2 \Rightarrow C = -26 $

Now

$ f(1) = -35 + 24\ln 3 = a + b\ln 3 $

$ \Rightarrow a = -35,; b = 24 $

$ \Rightarrow a + b = -11 $

Qus : 202 JEE MAIN PYQ 2026

$ \frac{6}{3^{26}} + \frac{10.1}{3^{25}} + \frac{10.2}{3^{24}} + \frac{10.2^2}{3^{23}} + \ldots + \frac{10.2^{24}}{3} $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ S = \frac{6}{3^{26}} + \frac{10}{3^{25}}\left[ \frac{(6)^{25} - 1}{6 - 1} \right] $

$ S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \cdot \frac{6^{25} - 1}{5} $

$ S = \frac{2}{3^{25}} + 2\left[ 2^{25} - \frac{1}{3^{25}} \right] $

$ S = 2^{26} $

Qus : 203 JEE MAIN PYQ 2026

$ \sum_{r=1}^{25} \frac{r}{r^2 + r + 1} = \frac{p}{q} $, where $ p $ and $ q $ are integers such that $ \gcd(p, q) = 1 $, then $ p + q $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ S = \sum \frac{r}{(r^2 + r + 1)} $

$ = \frac{1}{2} \sum \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $

$ = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \ldots + \left(\frac{1}{601} - \frac{1}{651}\right) \right] $

$ = \frac{1}{2}\left(1 - \frac{1}{651}\right) = \frac{325}{651} $

$ \Rightarrow p = 325,; q = 651 \Rightarrow p + q = 976 $

Qus : 204 JEE MAIN PYQ 2026

Three persons enter in a lift at the ground floor. The lift will go upto $ 10^{th} $ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ S = \sum \frac{r}{(r^2 + r + 1)} $

$ = \frac{1}{2} \sum \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $

$ = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \ldots + \left(\frac{1}{601} - \frac{1}{651}\right) \right] $

$ = \frac{1}{2}\left(1 - \frac{1}{651}\right) = \frac{325}{651} $

$ \Rightarrow p = 325,; q = 651 \Rightarrow p + q = 976 $

Qus : 205 JEE MAIN PYQ 2026

Let $ f $ be a differentiable function satisfying $ f(x) = 1 - 2x + e^{-x} \int_0^x e^t f(t),dt,; x \in \mathbb{R} $ and let $ g(x) = \int_0^x (f(t) + 2)^5 (t - 4)^6 (t + 12)^{17} dt,; x \in \mathbb{R} $. If $ p $ and $ q $ are respectively the points of local minima and local maxima of $ g $, then the value of $ p + q $ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ f'(x) - 2f(x) = 2x - 3 $

$ \frac{dy}{dx} - 2y = 2x - 3 $

$ y e^{-2x} = \int e^{-2x}(2x - 3)dx $

On solving we get

$ f(x) = 1 - x $

$ g'(x) = (3 - x)^5 (x - 4)^6 (x + 12)^{17} $

Local maxima $ \Rightarrow q = 3 $

Local minima $ \Rightarrow p = -12 $

$ \Rightarrow p + q = 9 $

Qus : 206 JEE MAIN PYQ 2026

If the distance of the point $ P(43, \alpha, \beta) $ from the line $ \vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}),; \mu \in \mathbb{R} $ along a line with direction ratios $ 3, -1, 0 $ is $ 13\sqrt{10} $, then $ \alpha^2 + \beta^2 $ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ \frac{x - 43}{3} = \frac{y - \alpha}{-1} = \frac{z - \beta}{0} $

$ \Rightarrow P(43 + 3\lambda,; \alpha - \lambda,; \beta) $

$ \Rightarrow \alpha = 13,; \beta = 1 $

$ \Rightarrow \alpha^2 + \beta^2 = 170 $

Qus : 207 JEE MAIN PYQ 2026

Let $ A = \left[ \matrix{ 3 & -4 \cr 1 & -1 } \right] $ and $ B $ be two matrices such that $ A^{100} = 100B + I $. Then the sum of all the elements of $ B^{100} $ is ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ A = I + \left[ \matrix{ 2 & -4 \cr 1 & -2 } \right] $, let $ M = \left[ \matrix{ 2 & -4 \cr 1 & -2 } \right] $

$ M^2 = \left[ \matrix{ 0 & 0 \cr 0 & 0 } \right] \Rightarrow M^3 = M^4 = \ldots = M^{100} $

$ A^{100} = (I + M)^{100} = I + 100M $

$ A^{100} = I + 100B \Rightarrow M = B $

$ \Rightarrow B^{100} = \left[ \matrix{ 0 & 0 \cr 0 & 0 } \right] $

Sum of all elements $ = 0 $

Qus : 208 JEE MAIN PYQ 2026

If the domain of the function $ f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2 - 3x + 1) $ is the interval $[\alpha, \beta]$, then $ \alpha + 2\beta $ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$ f(x) = \cos^{-1}!\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2 - 3x + 1) $

For $\cos^{-1}$:

$ -1 \le \frac{2x-5}{11-3x} \le 1 $

For $\sin^{-1}$:

$ -1 \le 2x^2 - 3x + 1 \le 1 $

First solve:

$ -1 \le 2x^2 - 3x + 1 \le 1 $

$ -1 \le 2x^2 - 3x + 1 \Rightarrow 2x^2 - 3x + 2 \ge 0 $

$ 2x^2 - 3x \le 0 $

So,

$ x \in [0, \frac{3}{2}] \quad ...(i) $

Now solve:

$ -1 \le \frac{2x-5}{11-3x} \le 1 $

First part:

$ \frac{2x-5}{11-3x} + 1 \ge 0 $

$ \frac{2x-5 + 11 - 3x}{11-3x} \ge 0 $

$ \frac{6 - x}{11-3x} \ge 0 $

Second part:

$ \frac{2x-5}{11-3x} - 1 \le 0 $

$ \frac{2x-5 - (11-3x)}{11-3x} \le 0 $

$ \frac{5x - 16}{11-3x} \le 0 $

From sign analysis:

$ x \in (-\infty, \frac{16}{5}] \cup (\frac{11}{3}, \infty) $

and

$ x \in (-\infty, \frac{11}{3}) \cup [6, \infty) $

Intersection:

$ x \in (-\infty, \frac{16}{5}] \cup [6, \infty) \quad ...(ii) $

Final intersection of (i) and (ii):

$ x \in [0, \frac{3}{2}] $

So,

$ \alpha = 0,\ \beta = \frac{3}{2} $

$ \alpha + 2\beta = 0 + 2 \cdot \frac{3}{2} = 3 $

Qus : 209 JEE MAIN PYQ 2026

The area of the region, inside the ellipse $x^2 + 4y^2 = 4$ and outside the region bounded by the curves $y = |x| - 1$ and $y = 1 - |x|$, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Qus : 210 JEE MAIN PYQ 2026

The number of relations, defined on the set ${a,b,c,d}$, which are both reflexive and symmetric, is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Set has 4 elements → total pairs = $4 \times 4 = 16$

Reflexive condition:

All $(a,a), (b,b), (c,c), (d,d)$ must be present → fixed

Remaining pairs = $16 - 4 = 12$

Symmetric condition:

Pairs form symmetric pairs like:

$(a,b)$ with $(b,a)$ → 1 choice pair

Total such independent pairs = $\frac{12}{2} = 6$

Each pair → 2 choices (present or not)

Total relations:

$ = 2^6 = 64 $

Qus : 211 JEE MAIN PYQ 2026

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$, respectively, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$\sin \theta = \frac{3}{a}$

$\sin (60^\circ + \theta) = \frac{9}{a}$

$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$

$\sqrt{3}\sqrt{1 - \frac{9}{a^2}} + \frac{3}{a} = \frac{18}{a}$

$a = \sqrt{84}$

Area of $\triangle ABC = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \cdot 84 = 21\sqrt{3}$

Qus : 212 JEE MAIN PYQ 2026

Let $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$, $\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c} = \vec{b}$. If $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$, then $|\vec{a} + \vec{c}|^2$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$

$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$

$\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$

$\vec{a} \times \vec{c} = \vec{b}$

$(2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k} = 8\hat{i} + 7\hat{j} - 3\hat{k}$

$2c_3 - 2c_2 = 8,\quad c_3 + 2c_1 = 7,\quad c_2 + 2c_1 = 3$

$c_1 + c_2 + c_3 = 4,\quad c_1 + c_2 + c_3 = 4$

$c_1 = 2,\quad c_2 = -1,\quad c_3 = 3$

$|\vec{a} + \vec{c}|^2 = | \hat{i} + \hat{j} + 5\hat{k} |^2 = 27$

Qus : 213 JEE MAIN PYQ 2026

Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms such that $a_1 a_2 a_3 a_4 = 64$ and $a_1 + a_2 + a_3 = \frac{813}{7}$. Then $a_3 + a_5 + a_7$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$ar \cdot ar^2 \cdot ar^3 = 64$

$a^3 r^6 = 64 \Rightarrow ar^2 = 4$

$a + ar + ar^2 = \frac{813}{7}$

$r^2 = 28$

$ar^2 + ar^4 + ar^6 = 4(1 + r^2 + r^4) = 4(1 + 28 + 784) = 3252$

Qus : 214 JEE MAIN PYQ 2026

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c} + \vec{d}| = \sqrt{29}$ and $\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$. If $\lambda_1, \lambda_2 (\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$, then the equation

$Kx^2 + (K - 5K + \lambda_1)xy + \left(3K + \frac{\lambda_2}{2}\right)y^2 - 8x + 12y + \lambda_2 = 0$

represents a circle, for $K$ equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Qus : 215 JEE MAIN PYQ 2026

Let $y = y(x)$ be the solution curve of the differential equation $(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$, $y(0) = 1$. Then the value of $y(1)$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{\tan^{-1}x}{1 + x^2}$

I.F. $= e^{\tan^{-1}x}$

$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$

$y \cdot e^{\tan^{-1}x} = \tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} + c$

$y(0) = 1 \Rightarrow c = 2$

$y(1) = \frac{2}{e^4} + \frac{\pi}{4} - 1$

Qus : 216 JEE MAIN PYQ 2026

The number of strictly increasing functions $f$ from the set ${1,2,3,4,5,6}$ to the set ${1,2,3,\ldots,9}$ such that $f(i) \ne i$ for $1 \le i \le 6$, is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$f(i) \ne i$, $f(x)$ is strictly increasing function

$f : A \to B$, where $A = {1,2,3,4,5,6}$

$B = {1,2,3,\ldots,9}$

$f : A \to B$ is equal to

Case–i: $f(1) = 2 \Rightarrow {}^7C_5 = 21$

Case–ii: $f(1) = 3 \Rightarrow {}^6C_5 = 6$

Case–iii: $f(1) = 4 \Rightarrow {}^5C_5 = 1$

No of function $A$ to $B = 21 + 6 + 1 = 28$

Qus : 217 JEE MAIN PYQ 2026

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f(3) = 18$, $f'(3) = 0$ and $f''(3) = 4$. Then

$\lim_{x \to 3} \left( \log_e \left( \frac{f(2 + x)}{f(3)} \right) \right)^{\frac{18}{(x-3)^2}}$

is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Let $T = \lim_{x \to 3} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^2}}$ ; $1^\infty$ form

$\Rightarrow T = e^{\lim \frac{18}{(x-3)^2} \cdot \frac{f(x+2) - f(3)}{f(3)}}$

$\Rightarrow T = e^{\lim \frac{18}{(x-3)^2} \cdot \frac{f(x+2) - f(3)}{18}}$

$\Rightarrow T = e^{\lim \frac{f(x+2) - f(3)}{(x-3)^2}}$ ; $0/0$ form apply L’Hospital

$\Rightarrow T = e^{\lim \frac{f'(x+2)}{2(x-3)}}$ ; $0/0$ form apply L’Hospital

$\Rightarrow T = e^{\lim \frac{f''(x+2)}{2}} = e^2$

$\Rightarrow \log_e (T) = 2$

Qus : 218 JEE MAIN PYQ 2026

Let the foci of hyperbola coincide with the foci of the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Let $e_1$ be eccentricity of ellipse

$\Rightarrow e_1 = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{4}{9}} = \frac{\sqrt{5}}{3}$

So $ae_1 = 6 \cdot \frac{\sqrt{5}}{3} = 2\sqrt{5}$

Now $H : \frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$

$p.e = ae_1$

$p \cdot 5 = 2\sqrt{5}$

$p = \frac{2}{\sqrt{5}}$

$e^2 = 1 + \frac{q^2}{p^2}$

$25 = 1 + \frac{q^2}{p^2} \Rightarrow 25 = 1 + \frac{5q^2}{4}$

$q^2 = \frac{96}{5}$

So length of LR $= \frac{2q^2}{p} = \frac{96}{\sqrt{5}}$

Qus : 219 JEE MAIN PYQ 2026

The value of $\int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x}{1 - \sin(x + \pi/6)} \right) dx$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$= 2\pi \int_0^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$ let $x + \frac{\pi}{6} = t \Rightarrow dx = dt$

$= 2\pi \int_{\pi/6}^{\pi/3} \frac{dt}{1 - \sin t}$

$= 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$

$= 2\pi \left[ \int_{\pi/6}^{\pi/3} \sec^2 t , dt + \int_{\pi/6}^{\pi/3} \sec t \tan t , dt \right]$

$= 2\pi \left[ (\tan t){\pi/6}^{\pi/3} + (\sec t){\pi/6}^{\pi/3} \right]$

$= 2\pi \left[ \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) + \left(2 - \frac{2}{\sqrt{3}}\right) \right]$

$= 2\pi \left[ \sqrt{3} + 2 - \sqrt{3} \right] = 4\pi$

Qus : 220 JEE MAIN PYQ 2026

Let the mean and variance of $7$ observations $2, 4, 10, x, 12, 14, y$, $x > y$, be $8$ and $16$ respectively. Two numbers are chosen from ${1, 2, 3, x-4, y, 5}$ one after another without replacement, then the probability that the smaller number among the two chosen numbers is less than $4$, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Mean $(\bar{x}) = 8$ (Given)

$\Rightarrow \frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$

$\Rightarrow x + y = 14 \quad ...(1)$

Variance $(\sigma^2) = 16$ (Given)

$\Rightarrow 16 = \frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2$

$\Rightarrow x^2 + y^2 = 100 \quad ...(2)$

$(x + y)^2 = x^2 + y^2 + 2xy$

$\Rightarrow 14^2 = 100 + 2xy \Rightarrow xy = 48$

Since $x > y$

$\Rightarrow x = 8,; y = 6$

Now set $X = {1,2,3,4,6,5}$

Now we choose two numbers one after another without replacement

Total outcomes $= 6 \times 5 = 30$

We want the probability that smaller number $< 4$

$P(\text{smaller} < 4) = 1 - P(\text{smaller} \ge 4)$

$= 1 - \frac{6}{30} = \frac{4}{5}$

Qus : 221 JEE MAIN PYQ 2026

Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. The length of the projection of the vector $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$A : (5,4,2)$

$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$

$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$

Any general point $P$ on the line is $(2\lambda - 1, 3\lambda + 3, -\lambda + 1)$

Let the given point is $A(5,4,2)$

$\overrightarrow{AP} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k}$

$\because \overrightarrow{AP} \perp \text{Line } (L)$

$\overrightarrow{AP} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$

$2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0$

$\Rightarrow \lambda = 1$

$\alpha = 1,; \beta = 6,; \gamma = 0$

Let the vector $\vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$

$\vec{u} = \hat{i} + 6\hat{j} + 0\hat{k}$

and $\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$

So projection $= \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|} = \frac{18}{7}$

Qus : 222 JEE MAIN PYQ 2026

Let PQ and MN be two straight lines touching the circle $x^2 + y^2 - 4x - 6y - 3 = 0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB = \pi/3$. Then the locus of the point of intersection of the lines PQ and MN is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Given circle

$x^2 + y^2 - 4x - 6y - 3 = 0$

$C(2,3)$ & $r = 4$

$\cos 30^\circ = \frac{r}{OR} = \frac{4}{OR}$

$\Rightarrow OR = \frac{8}{\sqrt{3}}$

Now

$OR^2 = (h - 2)^2 + (k - 3)^2$

$\Rightarrow 3(x^2 + y^2) - 12x - 18y - 25 = 0$

Qus : 223 JEE MAIN PYQ 2026

If the coefficient of $x$ in the expansion of $(ax^2 + bx + c)(1 - 2x)^{26}$ is $-56$ and the coefficients of $x^2$ and $x^3$ are both zero, then $a + b + c$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$(ax^2 + bx + c)\sum_{r=0}^{6} {}^6C_r(-2x)^r$

Coeff. of $x^2$:
$a \cdot {}^6C_0(-2)^0 + b \cdot {}^6C_1(-2)^1 + c \cdot {}^6C_2(-2)^2 = 0$

$\Rightarrow a - 12b + 60c = 0 \quad ...(1)$

Coeff. of $x^3$:
$a \cdot {}^6C_1(-2)^1 + b \cdot {}^6C_2(-2)^2 + c \cdot {}^6C_3(-2)^3 = 0$

$\Rightarrow -12a + 60b - 160c = 0 \quad ...(2)$

Coeff. of $x = -56$:
$b \cdot {}^6C_0(-2)^0 + c \cdot {}^6C_1(-2)^1 = -56$

$\Rightarrow b - 12c = -56 \quad ...(3)$

After solving (1), (2) & (3):

$a = 1300,; b = 100,; c = 3$

$\Rightarrow a + b + c = 1403$

Qus : 224 JEE MAIN PYQ 2026

If $x^2 + x + 1 = 0$, then the value of $\left(x + \frac{1}{x}\right)^4 + \left(x^2 + \frac{1}{x^2}\right)^4 + \left(x^3 + \frac{1}{x^3}\right)^4 + \cdots + \left(x^{25} + \frac{1}{x^{25}}\right)^4$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$x^2 + x + 1 = 0$

$\Rightarrow x = \omega$ or $\omega^2$

$\therefore \alpha = \omega,; \beta = \omega^2$

$= (\omega + \omega^2)^4 + (\omega^2 + \omega)^4 + (\omega^4 + \omega^2)^4 + \cdots$

$= [1 + 1 + \cdots + 1] + [(1+1)^4 + (-1+1)^4 + \cdots]$

$= 17 + 128 = 145$

Qus : 225 JEE MAIN PYQ 2026

The value of $\csc 10^\circ - \sqrt{3} \sec 10^\circ$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$= \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$

$= \frac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$

$= \frac{2\sin(30^\circ - 10^\circ)}{\sin 20^\circ}$

$= \frac{2\sin 20^\circ}{\sin 20^\circ} = 2 \cdot 2 = 4$

Qus : 226 JEE MAIN PYQ 2026

The sum of all the roots of the equation $(x - 1)^2 - 5|x - 1| + 6 = 0$, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Let $|x - 1| = t$

$t^2 - 5t + 6 = 0$

$t = 2$ & $t = 3$

$|x - 1| = 2$ & $|x - 1| = 3$

$x - 1 = \pm 2,; x - 1 = \pm 3$

$x = 3, -1, 4, -2$

Sum of roots $= 3 + (-1) + 4 + (-2) = 4$

Qus : 227 JEE MAIN PYQ 2026

Let $O$ be the vertex of the parabola $x^2 = 4y$ and $Q$ be any point on it. Let the locus of the point $P$, which divides the segment $OQ$ internally in the ratio $2 : 3$, be the conic $C$. Then equation of the chord of $C$, which is bisected at the point $(1, 2)$, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$h = \frac{4t}{5}$

$k = \frac{2t^2}{5} = \frac{2}{5}\left(\frac{5h}{4}\right)^2$

$8k = 5h^2$

$\Rightarrow 5x^2 = 8y$

$T = S_1$

$5(xx_1) - 4(y + y_1) = 5x_1^2 - 8y_1$

$5(xx_1) - 4(y + 2) = 5 - 8 \cdot 2$

$5x - 4y + 3 = 0$

Qus : 228 JEE MAIN PYQ 2026

Let $f : R \to R$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f(x)m + f'(x) = 0$ in $m$, has two equal roots for every $x \in R$. If $f(0) = 1$, $f'(0) = 2$ and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

Given quadratic equation has equal roots, thus

$D = 0 \Rightarrow (r(x))^2 = r'(x)\cdot f(x)$

$\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}$

Integrate

$\ln(f'(x)) = \ln(f(x)) + \ln C \Rightarrow f'(x) = C f(x)$

Put $x = 0$

$1 = C \cdot 2 \Rightarrow C = \frac{1}{2}$

Now $2f'(x) = f(x)$

$\Rightarrow \frac{f'(x)}{f(x)} = 2$

Integrate

$\ln(f(x)) = 2x + d$

$\Rightarrow d = 0$

$\Rightarrow \ln(f(x)) = 2x \Rightarrow f(x) = e^{2x}$

Now let $g(x) = f(\ln x - x) = e^{2(\ln x - x)}$

$\therefore g'(x) = 2e^{2(\ln x - x)}\left(\frac{1}{x} - 1\right) \ge 0$

$\Rightarrow \frac{1 - x}{x} \ge 0$

$\Rightarrow x \in (0,1]$

$\Rightarrow \alpha = 0,; \beta = 1$

$\alpha + \beta = 1$

Qus : 229 JEE MAIN PYQ 2026

Let $a_1 = 1$ and for $n \ge 1$, $a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$. Then $\sum_{n=1}^{\infty} \left(a_n - \frac{2}{n^2}\right)$ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2} = \frac{2n^2 - (n+1)^3}{n^2(n+1)^2}$

$\Rightarrow a_{n+1} - \frac{1}{2}a_n = \frac{2}{(n+1)^2} - \frac{1}{n^2}$

$n = 1 \Rightarrow a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$

$2\left[a_3 - \frac{1}{2}a_2\right] = \frac{2}{3^2} - \frac{1}{2^2}$

$2^2\left[a_4 - \frac{1}{2}a_3\right] = \frac{2}{4^2} - \frac{1}{3^2}$

$\vdots$

$2^{n-1}\left[a_n - \frac{1}{2}a_{n-1}\right] = \frac{2}{n^2} - \frac{1}{(n-1)^2}$

Adding

$a_n = \frac{2}{n^2} - \frac{1}{2^{n-1}}$

$\Rightarrow \sum_{n=1}^{\infty} \left(a_n - \frac{2}{n^2}\right) = \sum_{n=1}^{\infty} \left(-\frac{1}{2^{n-1}}\right) = 2$

Qus : 230 JEE MAIN PYQ 2026

Let $S = {(m,n) : m,n \in {1,2,3,\ldots,50}}$. If the number of elements $(m,n)$ in $S$ such that $6^m + 9^n$ is a multiple of $5$ is $p$ and the number of elements $(m,n)$ in $S$ such that $m + n$ is a square of a prime number is $q$, then $p + q$ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$S = {1,2,3,\ldots,50}$

$p = (6^m + 9^n)$ is divisible by $5$

No. of ways

$6^m = (5\lambda + 1)^m = 5k + 1$

$9^n = (10 - 1)^n = 10\mu - 1$ if $n$ is odd

$\Rightarrow n$ must be odd

$10\mu + 1$ if $n$ is even

$\Rightarrow$ No. of ways $= 50 \times 25 = 1250$

$q \Rightarrow (m+n)$ is square of a prime

$m+n = 4, 9, 25, 49$

No. of ways: $3, 8, 24, 48$

$q = 3 + 8 + 24 + 48 = 83$

$p + q = 1250 + 83 = 1333$

Qus : 231 JEE MAIN PYQ 2026

For some $\alpha, \beta \in R$, let $A = \begin{bmatrix} \alpha & 2 \ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \ 1 & \beta \end{bmatrix}$ be such that $A^2 - 4A + 2I = B^2 - 3B + I = O$. Then $(\det(\text{adj}(A^3 - B^3)))$ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$\text{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$

$\text{Tr}(B) = 3 \Rightarrow \beta + 1 = 3 \Rightarrow \beta = 2$

$A^2 - 4A + 2I = 0$

$A^3 = 4A^2 - 2A = 16A - 8I - 2A = 14A - 8I$

$= \begin{bmatrix} 28 & 28 \ 14 & 20 \end{bmatrix}$

$B^2 - 3B + I = 0 \Rightarrow B^2 = 3B - I$

$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$

$= \begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix}$

$A^3 - B^3 = \begin{bmatrix} 15 & 20 \ 6 & 7 \end{bmatrix}$

$\Rightarrow |A^3 - B^3| = 105 - 120 = -15$

$\Rightarrow \det(\text{adj}(A^3 - B^3)) = |A^3 - B^3|^2 = 225$

Qus : 232 JEE MAIN PYQ 2026

$6\int_{0}^{\pi/2} (\sin 3x + \sin 2x + \sin x) dx$ is equal to ______

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Morning Shift) PYQ

Solution

$6\int_{0}^{\pi/2} 2\sin 2x \cos x + \sin 2x , dx$

$= 6\int_{0}^{\pi/2} (4\sin x \cos^2 x + 2\sin x \cos x) dx$

$I = 12\int_{0}^{\pi/2} \sin x (2\cos^2 x + \cos x) dx$

Put $\cos x = t \Rightarrow -\sin x dx = dt$

$I = -12\int_{1}^{0} (2t^2 + t) dt$

$I = 12\int_{0}^{1} (2t^2 + t) dt$

$I = 12\left[\frac{2t^3}{3} + \frac{t^2}{2}\right]_0^1 = 12\left(\frac{2}{3} + \frac{1}{2}\right) = 17$

Qus : 233 JEE MAIN PYQ 2026

The positive integer $n$, for which the solutions of the equation $x(x+2) + (x+2)(x+4) + \cdots + (x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$x(x+2) + (x+2)(x+4) + \cdots + (x+2n-2)(x+2n) = \frac{8n}{3}$

$\Rightarrow \sum_{r=1}^{n} (x+2r-2)(x+2r) = \frac{8n}{3}$

$nx^2 + 2x\sum_{r=1}^{n}(2r-1) + 4\sum_{r=1}^{n} r(r-1) = \frac{8n}{3}$

$nx^2 + 2xn^2 + \frac{4n(n^2-1)}{3} - \frac{8n}{3} = 0$

$x^2 + 2nx + \frac{4(n^2-1)}{3} - \frac{8}{3} = 0$

Let roots be $\alpha, \beta$ $\because |\alpha - \beta| = 2$

$\Rightarrow \frac{\sqrt{D}}{|a|} = 2 \Rightarrow D = 4$

$\Rightarrow 4n^2 - 4\left(\frac{4(n^2-1)}{3} - \frac{8}{3}\right) = 4$

$\Rightarrow n^2 - \frac{4n^2}{3} = -3$

$\Rightarrow n^2 = 9$ $\Rightarrow n = 3$

Qus : 234 JEE MAIN PYQ 2026

Let $f : R \to R$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in R$ and $f'(a-1) = 0$, where $a$ is real number. Let

$g(x) = f(\tan x - 2\tan x + a), \quad 0 < x < \frac{\pi}{2}$

Consider the following two statements:

(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$

(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

Then,

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$g(x) = f((\tan x - 1)^2 + a - 1)$

$g'(x) = f'((\tan x - 1)^2 + a - 1)\cdot 2(\tan x - 1)\sec^2 x$

$\because f'(a-1) = 0$ and $f''(x) > 0$

$\Rightarrow f'((\tan x - 1)^2 + a - 1) > 0$

$\Rightarrow g'(x) > 0$ if $(\tan x - 1) > 0$

$g$ is increasing in $x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

$g'(x) < 0$ if $\tan x - 1 < 0$

$g$ is decreasing in $x \in \left(0, \frac{\pi}{4}\right)$

Qus : 235 JEE MAIN PYQ 2026

Let $f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3)$, $x \in R$. Then the value of $f'(5)$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$f'(x) = 3x^2 + 2x f'(1) + 2f''(2)$

$f''(x) = 6x + 2f'(1)$ $f'''(x) = 6$

$f(2) = 12 + 2f'(1)$

$f(x) = 3x^2 + 2x f'(1) + (12 + 2f'(1))$

Putting $x = 1$

$f(1) = 3 + 6f'(1) + 24$ $-5f'(1) = 27 \Rightarrow f'(1) = -\frac{27}{5}$

$\therefore f''(2) = 12 + 2\left(-\frac{27}{5}\right) = \frac{6}{5}$

$\therefore f'(x) = 3x^2 - \frac{54}{5}x + \frac{12}{5}$

$\therefore f'(5) = 75 - 54 + \frac{12}{5} = \frac{117}{5}$

Qus : 236 JEE MAIN PYQ 2026

In the line $ax + 4y = \sqrt{7}$, where $a \in R$, touches the ellipse $3x^2 + 4y^2 = 1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$ax + 4y - \sqrt{7} = 0$ touches $3x^2 + 4y^2 = 1$

$\Rightarrow \frac{c^2}{a^2} = \frac{7}{16} = \frac{1}{3}\cdot \frac{a^2}{16} + \frac{1}{4}$

$\Rightarrow a = 3, -3$

Tangent is $3x + 4y - \sqrt{7} = 0$

Let the point of contact is $P(x_1, y_1)$

$\Rightarrow 3x_1 + 4y_1 = \sqrt{7}$

$\Rightarrow \frac{3x_1}{3} + \frac{4y_1}{4} = \frac{\sqrt{7}}{1}$

$\Rightarrow P\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$

$e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$

$PS = e \cdot PM$

$\Rightarrow PS = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$

Qus : 237 JEE MAIN PYQ 2026

Let $y^2 = 12x$ be the parabola with its vertex at $O$. Let $P$ be a point on the parabola and $A$ be a point on the y-axis such that $OPA = 90^\circ$. Then the locus of the centroid of triangle $OPA$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

Let $P(3t^2, 6t)$ $m_{AP} = \frac{t}{2}$ Equation of $AP$ is $y - 6t = \frac{t}{2}(x - 3t^2)$ Put $y = 0 \Rightarrow x = 12 + 3t^2$ $\Rightarrow A(12 + 3t^2, 0)$ Let centroid of $\triangle OPA$ be $(h,k)$ $\Rightarrow 3h = 0 + 3t^2 + 12 + 3t^2$ $\Rightarrow 3k = 0 + 6t + 0$ $\Rightarrow t = \frac{k}{2},; h = 2t^2 + 4$ $\Rightarrow h = \frac{k^2}{2} + 4$ $\Rightarrow$ locus of $(h,k)$ is $y^2 = 2x - 8$

Qus : 238 JEE MAIN PYQ 2026

Let one end of a focal chord of the parabola $y^2 = 16x$ be $(16, 16)$. If $P(\alpha, \beta)$ divides this focal chord internally in the ratio $5 : 2$, then the minimum value of $\alpha + \beta$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$y^2 = 16x$

$\therefore$ parameter of point $A$ is $t = 2$

$\Rightarrow$ Parameter of point $B$ is $t = -\frac{1}{2}$

$\Rightarrow$ Coordinates of $B$ is $(1, -4)$

Case 1:

$A(16,16),; P(\alpha,\beta),; B(1,-4)$

$\alpha = \frac{5\cdot1 + 2\cdot16}{7} = \frac{37}{7}$

$\beta = \frac{5(-4) + 2\cdot16}{7} = \frac{12}{7}$

$\Rightarrow \alpha + \beta = 7$

Case 2:

$A(16,16),; P(\alpha,\beta),; B(1,-4)$

$\alpha = \frac{2\cdot1 + 5\cdot16}{7}$

$\beta = \frac{2(-4) + 5\cdot16}{7}$

$\Rightarrow \alpha + \beta = 22$

So minimum value of $\alpha + \beta = 7$

Qus : 239 JEE MAIN PYQ 2026

Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

Equation of line is

$\frac{x+3}{1} = \frac{y-5}{1} = \frac{z-2}{1} = \lambda$

$\therefore$ General point $R$ on line is $R(\lambda - 3, \lambda + 5, \lambda + 2)$

$P(-2, r, 1)$

$\vec{PR} = (\lambda - 1,; \lambda + 5 - r,; \lambda + 1)$

Now $\vec{PR} \cdot \vec{d} = 0$

$\Rightarrow (\lambda - 1) + (\lambda + 5 - r) + (\lambda + 1) = 0$

$\Rightarrow 3\lambda - r + 5 = 0$

$\Rightarrow \lambda = \frac{r - 5}{3}$

$\therefore R\left(\frac{r-5}{3} - 3,; \frac{r-5}{3} + 5,; \frac{r-5}{3} + 2\right)$

$= \left(\frac{r-14}{3},; \frac{r+10}{3},; \frac{r+1}{3}\right)$

Now

$PR = \sqrt{\frac{14}{3}}$

$\Rightarrow (PR)^2 = \frac{14}{3}$

$\Rightarrow \left(\frac{r-8}{3}\right)^2 + \left(\frac{10-2r}{3}\right)^2 + \left(\frac{r-2}{3}\right)^2 = \frac{14}{3}$

$\Rightarrow r^2 - 10r + 21 = 0$

$\Rightarrow r = 3, 7$

Sum of possible values of $r = 10$

Qus : 240 JEE MAIN PYQ 2026

Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $|CD|^2$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$L_1 : \frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4}$

Point $C$ on $L_1$: $(-3\lambda_1 + 2,; 2\lambda_1 + 6,; 4\lambda_1 + 7)$

$L_2 : \frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3}$

Point $D$ on $L_2$: $(2\lambda_2 + 4,; \lambda_2 + 3,; 3\lambda_2 + 5)$

D.R’s of line $L_3$:

$\frac{2\lambda_1 + 3\lambda_1 + 2}{-3} = \frac{\lambda_2 - 2\lambda_1 - 3}{5} = \frac{3\lambda_2 - 4\lambda_1 - 2}{16}$

$\lambda_1 = -3,; \lambda_2 = 2$

$C(11, 0, -5)$

$D(8, 5, 11)$

$|CD|^2 = 3^2 + 5^2 + 16^2 = 290$

Qus : 241 JEE MAIN PYQ 2026

Let $\alpha$ and $\beta$ be the roots of equation $x^2 + 2ax + (3a + 10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:

1
2
3
4
Go to Discussion

JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$\alpha < 1 < \beta$

$f(1) < 0$

$\Rightarrow 1 + 2a + (3a + 10) < 0$

$\Rightarrow 5a + 11 < 0$

$\Rightarrow a < -\frac{11}{5}$

$\therefore a \in (-\infty, -\frac{11}{5})$

Qus : 242 JEE MAIN PYQ 2026

A random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30},b$ respectively, where $a, b \in R$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2 + \mu^2 = 2$. Then $\frac{1}{b}$ is equal to:

1
2
3
4
Go to Discussion

JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (21 January Evening Shift) PYQ

Solution

$x \quad 0 \quad 1 \quad 2 \quad 3$

$p(x) \quad \frac{2a+1}{30} \quad \frac{8a-1}{30} \quad \frac{4a+1}{30} \quad b$

$\sigma^2 = \sum x^2 p(x) - \mu^2$

$\sigma^2 + \mu^2 = \sum x^2 p(x)$

$= 0 + 1\cdot \frac{8a-1}{30} + 4\cdot \frac{4a+1}{30} + 9b$

$= \frac{24a + 270b + 3}{30} = 2$

$\Rightarrow 24a + 270b = 57$

$\Rightarrow 8a + 90b = 19 \quad ...(1)$

Also $\sum p(x) = 1$

$\Rightarrow \frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$

$\Rightarrow 14a + 30b = 29 \quad ...(2)$

Solving (1) & (2):

$a = 2,; b = \frac{1}{60}$

$\therefore \frac{1}{b} = 60$

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