JEE MAIN 2026 Previous Year Questions (PYQs) – Page 4 of 25
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The coefficient of $x^{48}$ in $(1 + x)^2(1 + x^2 + 3(1 + x)^3 + \cdots + 100(1 + x)^{100})$ is equal to
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Solution
Let $1 + x = r$
$S = 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \cdots + 100\cdot r^{100}$ ……(1)
$rS = 1\cdot r^2 + 2\cdot r^3 + \cdots + 99r^{100} + 100r^{101}$ ……(2)
(1) – (2) gives
$S = \frac{(1+x)^{101}}{x^2} + \frac{100(1+x)^{101}}{x} + 100$
Coefficient of $x^{48}$ in $S$
$= $ coefficient of $x^{48}$ in $\frac{(1+x)^{101}}{x^2} + \frac{100(1+x)^{101}}{x}$
Coefficient of $x^{48}$ in $\frac{(1+x)^{101}}{x}$
$= {}^{100}C_{49} - {}^{100}C_{50}$
$\Rightarrow$ answer $= 100\cdot {}^{100}C_{49} - {}^{100}C_{50}$
If the chord joining the points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ on the parabola $y^2 = 12x$ subtends a right angle at the vertex of the parabola, then $x_1x_2 - y_1y_2$ is equal to
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Solution
$(x_1,y_1) = (3t_1^2, 6t_1),; (x_2,y_2) = (3t_2^2, 6t_2)$
$t_1t_2 = -4$
$x_1x_2 = 9(t_1t_2)^2,; y_1y_2 = 36t_1t_2$
$x_1x_2 - y_1y_2 = 9(16) - 36(-4)$
$= 144 + 144 = 288$
The number of solutions of $\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to
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Solution
$\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$
$\Rightarrow \tan^{-1}\left(\frac{4x + 6x}{1 - 24x^2}\right) = \frac{\pi}{6}$
$\Rightarrow \frac{10x}{1 - 24x^2} = \frac{1}{\sqrt{3}}$
$\Rightarrow 24x^2 + 10\sqrt{3}x - 1 = 0$
$x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48}$
$x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}$
Only one solution lies in interval $\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$
$\Rightarrow$ number of solutions $= 1$
Let the solution curve of the differential equation
$xdy - ydx = \sqrt{x^2 + y^2},dx,; x>0,; y(1)=0$
be $y = y(x)$. Then $y(3)$ is equal to
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Solution
$xdy - ydx = \sqrt{x^2 + y^2},dx$
$\Rightarrow \frac{xdy - ydx}{x^2} = \frac{\sqrt{x^2 + y^2}}{x^2}dx$
$\Rightarrow d\left(\frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^2}\cdot \frac{dx}{x}$
$\Rightarrow \int \frac{d(y/x)}{\sqrt{1 + (y/x)^2}} = \int \frac{dx}{x}$
$\Rightarrow \ln\left(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}\right) = \ln x + C$
$\Rightarrow y + \sqrt{x^2 + y^2} = kx^2$
At $x=1,; y=0 \Rightarrow k=1$
$\Rightarrow y + \sqrt{x^2 + y^2} = x^2$
At $x=3$
$y + \sqrt{9 + y^2} = 9$
$\Rightarrow y = 4$
If the sum of the first four terms of an A.P. is $6$ and the sum of its first six terms is $4$, then the sum of its first twelve terms is
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Solution
$\frac{4}{2}(2a + 3d) = 6 \Rightarrow 2a + 3d = 3 \quad ...(1)$
Sum of first $6$ terms $S_6 = 4$
$\frac{6}{2}(2a + 5d) = 4 \Rightarrow 2a + 5d = \frac{4}{3} \quad ...(2)$
(2) – (1)
$(2a + 5d) - (2a + 3d) = \frac{4}{3} - 3$
$\Rightarrow 2d = -\frac{5}{3} \Rightarrow d = -\frac{5}{6}$
$2a + 3\left(-\frac{5}{6}\right) = 3 \Rightarrow 2a = \frac{11}{2} \Rightarrow a = \frac{11}{4}$
$S_{12} = \frac{12}{2}\left(2a + 11d\right)$
$= 6\left(\frac{11}{2} + 11\left(-\frac{5}{6}\right)\right)$
$= 6\left(\frac{33 - 55}{6}\right) = -22$
Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If
$(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m$, then $m$ is ______.
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Solution
$(9 + 7\alpha - 7\beta)^{20} + \alpha^{40}(9 + 7\alpha - 7\beta)^{20}$
$+ \omega^{40}(9 + 7\alpha - 7\beta)^{20} + (14 + 7(\alpha + \beta))^{20}$
$(9 + 7\alpha - 7\beta)^{20}(1 + \omega + \omega^2) + (14 - 7)^{20}$
$= 7^{20} = 49$
22. Let A be a 3 × 3 matrix such that A + A^T = O. If
$$
A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}
= \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}, \quad
A^2 \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}
= \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}
$$
and
$$
\det(\text{adj}(2\text{adj}(A + I))) = (2)^\alpha (3)^\beta (11)^\gamma
$$
, α, β, γ are non-negative integers, then α + β + γ is equal to ______.
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Solution
If $\displaystyle \int (\sin x)^{-\frac{11}{2}} (\cos x)^{\frac{5}{2}} dx = -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C$,
where $p_i, q_i$ are positive integers with $\gcd(p_i,q_i)=1$ for $i = 1,2,3,4$ and $C$ is the constant of integration, then
$\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4}$ is equal to ______.
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Solution
$\displaystyle \int (\tan x)^{-\frac{11}{2}} \sec^8 x , dx$
$= \int (\tan x)^{-\frac{11}{2}} (1+\tan^2 x)^3 \sec^2 x , dx$
Put $\tan x = t$
$\Rightarrow \int t^{-\frac{11}{2}} (1+t^2)^3 , dt$
$= \int \left(t^{-\frac{11}{2}} + 3t^{-\frac{7}{2}} + 3t^{-\frac{3}{2}} + t^{\frac{1}{2}}\right) dt$
$= -\frac{2}{9}t^{-\frac{9}{2}} - \frac{6}{5}t^{-\frac{5}{2}} - 2t^{-\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + C$
Back substitute $t = \cot x$
$\Rightarrow p_1 = 2,; p_2 = 6,; p_3 = 2,; p_4 = 2$
$q_1 = 9,; q_2 = 5,; q_3 = 1,; q_4 = 3$
$\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} = \frac{15 \cdot 2 \cdot 6 \cdot 2 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$
If $\displaystyle \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}$, where $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to ______.
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Solution
Use identities
$\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$
$\cos(A+B)\cos(A-B) = \cos^2 A - \cos^2 B$
$\displaystyle \frac{\cos 60^\circ \cos 36^\circ}{\sin 30^\circ \sin 18^\circ}
= \frac{\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}-1}{4}}
= \frac{\sqrt{5}+1}{\sqrt{5}-1}$
$= \frac{(\sqrt{5}+1)^2}{5-1}
= \frac{6+2\sqrt{5}}{4}
= \frac{3+\sqrt{5}}{2}$
$\Rightarrow \alpha = 3,; \beta = 1$
$\Rightarrow \alpha + \beta = 4$
Let $ABC$ be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side $AB$, five points $p_5, p_6, p_7, p_8, p_9$ on the side $BC$ and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side $AC$. None of these points is a vertex of the triangle $ABC$. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is ______.
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Solution
Case 1:
$2$ from $AB$, $2$ from $BC$, $1$ from $AC$
$\binom{4}{2}\binom{5}{2}\binom{4}{1} = 6 \cdot 10 \cdot 4 = 240$
Case 2:
$2$ from $AB$, $1$ from $BC$, $2$ from $AC$
$\binom{4}{2}\binom{5}{1}\binom{4}{2} = 6 \cdot 5 \cdot 6 = 180$
Case 3:
$1$ from $AB$, $2$ from $BC$, $2$ from $AC$
$\binom{4}{1}\binom{5}{2}\binom{4}{2} = 4 \cdot 10 \cdot 6 = 240$
Total $= 240 + 180 + 240 = 660$
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