JEE MAIN 2026 Previous Year Questions (PYQs) – Page 7 of 25
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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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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 $=6$.
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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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 $\dfrac{1}{\pi}\cdot4\cdot\dfrac{\pi}{2}=2$.
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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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 $r+s=5$.
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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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 $9\times\frac{2}{3}=6$.
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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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 $1984$.
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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Solution
By solving line & equation of ellipse we get $x = 0$ and $x = -\frac{4}{3}$
$\therefore B\left(-\frac{4}{3}, -\frac{1}{3}\right)$
$m_{OB} = \tan\theta = \frac{1}{4}$
$\therefore \angle AOB = \frac{\pi}{2} + \theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$
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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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
$n = 6$
$\therefore \ell + m + n = 23 + 4 + 6 = 33$
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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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, $c = 0$
$\left(\frac{\pi}{3}\right)^4 f\left(\frac{\pi}{3}\right) = 2\cos \frac{\pi}{3}$
$\Rightarrow \pi^4 f\left(\frac{\pi}{3}\right) = 81$
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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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 $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 10$
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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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 $= 2 + 3 + 4 = 9$
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