JEE MAIN 2026 Previous Year Questions (PYQs) – Page 10 of 25
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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:
1
2
3
4
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
$= 2^6 - 2 = 62$
The system of linear equations
$x + y + z = 6$
$2x + 5y + az = 36$
$x + 2y + 3z = b$
has
1
2
3
4
Solution
\(
\begin{vmatrix}
1 & 1 & 1 \\
2 & 5 & a \\
1 & 2 & 3
\end{vmatrix}
= 0 \Rightarrow a = 8
\)
---
Now
\(
D_1 =
\begin{vmatrix}
6 & 1 & 1 \\
36 & 5 & a \\
b & 2 & 3
\end{vmatrix}
= 0
\Rightarrow ab - 5b - 12a + 54 = 0
\)
---
\(
D_2 =
\begin{vmatrix}
1 & 6 & 1 \\
2 & 36 & a \\
1 & b & 3
\end{vmatrix}
= 0
\Rightarrow ab - 6a - 2b - 36 = 0
\)
---
\(
D_3 =
\begin{vmatrix}
1 & 1 & 6 \\
2 & 5 & 36 \\
1 & 2 & b
\end{vmatrix}
= 0
\Rightarrow b = 14
\)
---
\(
\Rightarrow \text{infinitely many solutions}
\)
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:
1
2
3
4
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 $= 0$
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:
1
2
3
4
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
$= 2 \times \frac{1}{2} \times OM \times MP$
$= 2\sec\theta \cdot 2\sqrt{2}\tan\theta$
$= 4\sqrt{2} \cdot \frac{\sin\theta}{\cos^2\theta}$
$= 4\sqrt{2} \cdot \frac{1}{\sqrt{6}\left(1 - \frac{1}{6}\right)}$
$= \frac{8\sqrt{3}}{5}$
The area of the region enclosed between the circles
$x^2 + y^2 = 4$ and $x^2 + (y-2)^2 = 4$ is:
1
2
3
4
Solution
$A = 2\int_{0}^{\sqrt{3}} \left[\sqrt{4-x^2} - (2 - \sqrt{4-x^2})\right] dx$
$= 2\int_{0}^{\sqrt{3}} (2\sqrt{4-x^2} - 2),dx$
$= 4\int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1),dx$
$= 4\left[\frac{1}{2}\left(x\sqrt{4-x^2} + 4\sin^{-1}\frac{x}{2}\right) - x\right]_{0}^{\sqrt{3}}$
$= \frac{8\pi}{3} - 2\sqrt{3}$
The least value of
$\cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$
is:
1
2
3
4
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 $a\sin x + b\cos x = \sqrt{a^2 + b^2}$
$\Rightarrow \min f = 4 - \sqrt{(-3)^2 + (-1)^2} = 4 - \sqrt{10}$
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:
1
2
3
4
Solution
Let $4x+6 = \dfrac{1}{t}$
$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + c$
$I(0) = \frac{\sqrt{3}}{4} + c \Rightarrow c = 20$
$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + 20$
$I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{8}{8}} + 20 = \frac{3\sqrt{2}}{8} + 20$
$\Rightarrow a + b + c = 3 + 8 + 20 = 31$
The number of ways, in which $16$ oranges can be distributed to four children such that each child gets at least one orange, is:
1
2
3
4
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
$= {}^{12+4-1}C_{4-1} = {}^{15}C_3 = 455$
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:
1
2
3
4
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
If $z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2},; i = \sqrt{-1}$, then $(z^{201} - i)^5$ is equal to:
1
2
3
4
Solution
$z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$
$z^{201} = \cos\left(\frac{201\pi}{6}\right) + i\sin\left(\frac{201\pi}{6}\right) = -i$
$(z^{201} - i) = -2i$
$(-2i)^8 = 256$
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