JEE MAIN 2026 Previous Year Questions (PYQs) – Page 2 of 25
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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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Solution
$\sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] \le \sum_{k=1}^{n} \frac{k^2}{3^x}$
$= \frac{n(n+1)(2n+1)}{6\cdot 3^x}$
$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] = \frac{1}{3^{x+1}}$
$\Rightarrow f(x) = \frac{1}{3^{x+1}}$
$12\sum_{j=1}^{\infty} f(j) = 12\sum_{j=1}^{\infty} \frac{1}{3^{j+1}}$
$= 12\left(\frac{1}{9} + \frac{1}{27} + \cdots\right)$
$= 12\left(\frac{1/9}{1 - 1/3}\right) = 2$
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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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 $\int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = 4I$
$= 2\left[2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5\right]$
$= 4\pi - 8\cot^{-1}(2) + \ln 5$
$\Rightarrow 2a + b = 8 + 1 = 9$
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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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 $\sin^{-1}x = -\frac{\pi}{3}$
$\Rightarrow 2\left(\frac{\pi}{3} + \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
$= 2\left(\frac{7\pi}{12}\right)^2 + \frac{\pi^2}{8}$
$= \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{29\pi^2}{36}$
$\Rightarrow m = 29,; n = 36$
$\Rightarrow m+n = 65$
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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Solution
$\prod_{r=0}^{12} \left(\frac{1}{^{15}C_r} + \frac{1}{^{15}C_{r+1}}\right)$
$= \prod_{r=0}^{12} \frac{16}{(r+1)\cdot \frac{15!}{r!(15-r)!}}$
$= \prod_{r=0}^{12} \frac{16}{^{15}C_r \cdot \frac{r+1}{15}}$
$= \prod_{r=0}^{12} \frac{16}{^{14}C_r}$
$= \frac{\left(\frac{16}{15}\right)^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$
$\Rightarrow \alpha = \frac{16}{15}$
$\Rightarrow 30\alpha = 32$
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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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)
$\Rightarrow \cos\theta + 5\sin\theta = 1,; \theta \in [0, 2\pi]$
$\Rightarrow 5\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin^2\frac{\theta}{2}$
$\therefore \sin\frac{\theta}{2} = 0 \Rightarrow \cos\theta = 1$
or
$\sin\frac{\theta}{2} = \frac{5}{2}\cos\frac{\theta}{2} \Rightarrow \cos\theta = -\frac{12}{13}$
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 $2$
$\therefore 13$ times sum of all values of abscissa of point $P$ is $2$
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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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: $9x^2 - 18x + 5 = 0 \Rightarrow x = \frac{1}{3},; \frac{5}{3}$
$\Rightarrow 2\alpha - \beta = \frac{10 - 1}{3} = 3$
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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Solution
$R = {(3,3), (7,8), (11,13)}$
To make it symmetric $(8,7), (13,11)$ must be added
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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Solution
$\frac{m}{n} = \frac{\int_{-2}^{-1} \left[(3-x) - (1+x^2)\right] dx}{\int_{-1}^{1} \left[(3-x) - (1+x^2)\right] dx} = \frac{20}{7}$
$\Rightarrow m + n = 20 + 7 = 27$
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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Solution
Required probability
$= 1 - (\text{product not divisible by } 3)$
Multiples of $3 = 16$
Not multiple of $3 = 34$
$= 1 - \frac{{}^{34}C_2}{{}^{50}C_2} = \frac{664}{1225}$
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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Solution
$f(x) = x^{2025} - x^{2000}$
$f'(x) = 0 \Rightarrow x = \left(\frac{2000}{2025}\right)^{1/25} = \alpha$
$f(0) = 0,; f(1) = 0,; f(\alpha) = \left(\frac{80}{81}\right)^{80} - \left(\frac{80}{81}\right)^{81}$
$= \left(\frac{80}{81}\right)^{80}\left(1 - \frac{80}{81}\right)$
$= \left(\frac{80}{81}\right)^{80}\cdot \frac{1}{81}$
$= \left(\frac{80}{81}\right)^{80} \cdot \left(\frac{80}{81}\right)^{-1}$
$= \left(\frac{80}{81}\right)^{-81}$
$\Rightarrow n = -81$
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