JEE MAIN 2026 Previous Year Questions (PYQs) – Page 3 of 25
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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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Solution
If a random variable $x$ has the probability distribution
then $P(3 < x \le 6)$ is equal to:
then $P(3 < x \le 6)$ is equal to:
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Solution
$\sum P(x) = 1$
$9k + 10k^2 = 1$
$\Rightarrow k = \frac{1}{10}$
$P(3 < x \le 6) = 3k + 3k^2$
$= \frac{3}{10} + \frac{3}{100} = 0.33$
The number of distinct real solutions of the equation x|x+4| + 3|x+2| + 10 = 0 is
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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.
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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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 $x = 1,; y = 1 \Rightarrow C = 2$
$\Rightarrow f(x) = -x^2 + 2x$
$f(2) - f(3) = ( -4 + 4 ) - ( -9 + 6 ) = 3$
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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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)
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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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):
$a = 3,; c = 15$
$\Rightarrow a^2 + b^2 + c^2 = 9 + 64 + 225 = 298$
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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Solution
$(x-2)^2 + (y-1)^2 = 9$ & $(x+1)^2 + (y+4)^2 = r^2$
$r_1 < c < r_1 + r_2$
$|r-3| < \sqrt{(2+1)^2 + (1+4)^2} < r + 3$
$|r-3| < \sqrt{34} < r + 3$
$-\sqrt{34} < r-3 < \sqrt{34}$
$\Rightarrow r \in (3-\sqrt{34}, 3+\sqrt{34})$
$\therefore \alpha\beta = (3-\sqrt{34})(3+\sqrt{34}) = 9 - 34 = -25$
$\Rightarrow 25$
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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Solution
$A = \begin{bmatrix} 2 & 3 \ 3 & 5 \end{bmatrix}, \quad A^2 = \begin{bmatrix} 13 & 21 \ 21 & 34 \end{bmatrix}$
$|A^{2025} - 3A^{2024} + A^{2023}|$
$= |A^{2023}(A^2 - 3A + I)|$
$= |A^{2023}| \cdot |A^2 - 3A + I|$
$= |A|^{2023} \cdot \left|\begin{bmatrix} 8 & 12 \ 12 & 20 \end{bmatrix}\right|$
$= 1^{2023}(160 - 144) = 16$
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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Solution
$-1 \le \frac{5-x}{2x+3} \le 1,; 10-x > 0,; 10-x \ne 1$
$\left|\frac{5-x}{2x+3}\right| \le 1$ & $x < 10$ & $x \ne 9$
$(5-x)^2 - (2x+3)^2 \le 0$
$(x+8)(3x-2) \ge 0$ & $x < 10$ & $x \ne 9$
$\Rightarrow (-\infty,-8] \cup \left[\frac{2}{3},10\right) - {9}$
$\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left(-8 + \frac{2}{3} + 10 + 9\right)$
$= 70$
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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Solution
$I = \int_{-\pi/2}^{\pi/2} \frac{1}{[x] + 4},dx$
$I = \int_{-\pi/2}^{-1} \frac{dx}{2} + \int_{-1}^{0} \frac{dx}{3} + \int_{0}^{1} \frac{dx}{4} + \int_{1}^{\pi/2} \frac{dx}{5}$
$= \frac{1}{2}\left(-1 + \frac{\pi}{2}\right) + \frac{1}{3}(1) + \frac{1}{4}(1) + \frac{1}{5}\left(\frac{\pi}{2} - 1\right)$
$= \frac{7\pi}{20} - \frac{7}{60} = \frac{7}{60}(3\pi - 1)$
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