JEE MAIN 2026 Previous Year Questions (PYQs) – Page 8 of 25
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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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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 $(\triangle ABC) = \frac{1}{2} \times \sqrt{13} \times 10 = 5\sqrt{13}$
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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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 $[-3\pi, 2\pi]$
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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Solution
$\frac{3}{2} \le \left|z - \frac{3}{2}(1+i)\right| \le \frac{7}{2}$
$\min_{z \in S} \left|z - \left(-\frac{5}{2} - \frac{3}{2}i\right)\right| = PB$
$PB = PC - \frac{7}{2} = 5 - \frac{7}{2} = \frac{3}{2}$
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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Solution
$f(\theta) = 4(\cos^4\theta + \sin^4\theta) - 2(\cos^6\theta + \sin^6\theta)$
$= 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta)$
$= 2 - \sin^2(2\theta)/2$
$\alpha = 2,; \beta = \frac{3}{2}$
$\Rightarrow \alpha + 2\beta = 2 + 3 = 5$
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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Solution
Let time taken by $B = x$ days
Then time by $A = x - 24$
$\frac{1}{x} + \frac{1}{x-24} = \frac{1}{22.5}$
$\frac{2x - 24}{x(x-24)} = \frac{2}{45}$
$45(2x - 24) = 2x(x - 24)$
$90x - 1080 = 2x^2 - 48x$
$2x^2 - 138x + 1080 = 0$
$x = 60$
$\Rightarrow A = 60 - 24 = 36$
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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Solution
The value of the integral
$\displaystyle \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt{3}\tan 2x}$
is:
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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)
$2I = \int_{\pi/24}^{5\pi/24} dx$
$I = \frac{1}{2}\left(\frac{5\pi}{24} - \frac{\pi}{24}\right) = \frac{\pi}{12}$
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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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 \( x = 1 \)
\(
p + q =
\begin{vmatrix}
2 & 2 & -1 \\
4 & 3 & 3 \\
6 & 1 & 1
\end{vmatrix}
= 42
\)
\( p = 54 \)
---
\( p^2 = 54^2,\quad 196q^2 = 196 \cdot (-12)^2 \)
\( p^2 \ne 196q^2 \)
The value of $\dfrac{{}^{100}C_{50}}{51} + \dfrac{{}^{100}C_{51}}{52} + \cdots + \dfrac{{}^{100}C_{100}}{101}$ is:
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Solution
$S = \sum_{r=50}^{100} \dfrac{{}^{100}C_r}{r+1}$
$= \sum_{r=50}^{100} \dfrac{1}{101} \cdot \dfrac{r+1}{r+1} \cdot {}^{100}C_r$
$= \dfrac{1}{101} \sum_{r=50}^{100} {}^{101}C_{r+1}$
$= \dfrac{1}{101} \sum_{r=51}^{101} {}^{101}C_r$
$= \dfrac{1}{101} \cdot \dfrac{2^{101}}{2}$
$= \dfrac{2^{100}}{101}$
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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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
$\dfrac{x + y + (x+y+1) + |x-y|}{4}$
$= \dfrac{12 + 7 + 20 + 5}{4} = \dfrac{44}{4} = 11$
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