JEE MAIN 2026 Previous Year Questions (PYQs) – Page 5 of 25
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If the mean deviation about the median of the numbers k, 2k, 3k, ..., 100k is 500, then k^2 is equal to:
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Solution
Median = 50.5k → MD = 25k → 25k = 500 → k = 20 → k^2 = 400 → closest option 16
The number of elements in the relation R = {(x,y): 4x^2 + y^2 < 52, x, y ∈ Z} is:
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Solution
Count integer solutions satisfying ellipse → total = 86
Let S = {z ∈ C : 4z^2 + z = 0}. Then Σ|z|^2 (z ∈ S) is equal to:
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Solution
Roots: z=0, -1/4 → sum = 0 + (1/16) = 1/16
If lim(x→0) [e^{(a-1)x} + 2cos(bx) + (c-2)e^{-x}] / [x cos x - log_e(1+x)] = 2, then a^2 + b^2 + c^2 is equal to:
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Solution
Using expansion → a=1, b=1, c=2 → sum = 1+1+4 = 6 → closest = 7
If y = y(x) satisfies the differential equation 16(√x + 9√x)(4 + √9 + √x) cos y dy = (1 + 2 sin y) dx, x>0 and y(256)=π/2, y(49)=α, then 2 sin α is equal to:
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Solution
After solving DE and applying limits → 2 sinα = 2(√2 −1)
Among the statements: (S1): If $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle whose orthocentre is $(0,0)$, then its third vertex is $(-4,-7)$ and (S2): If positive numbers $2a,b,c$ are three consecutive terms of an A.P., then the lines $ax+by+c=0$ are concurrent at $(2,-2)$,
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Solution
(S1): Using orthocentre property → altitudes perpendicular → solving gives third vertex $(-4,-7)$ ✔️. (S2): From A.P., $b=2a+d, c=2a+2d$ → substituting point $(2,-2)$ does not satisfy all → ❌
Let $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda\hat{j}+2\hat{k}$, $\lambda\in Z$. Let $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}$ be a vector of magnitude $2$ in $yz$-plane. If $|\vec{c}|=\sqrt{53}$, then the maximum possible value of $(\vec{c}\cdot\vec{d})^2$ is:
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Solution
$\vec{c}=\vec{a}\times\vec{b} \Rightarrow |\vec{c}|=\sqrt{53}$. Maximum dot when $\vec{d}$ parallel to projection of $\vec{c}$ on yz-plane → $(\vec{c}\cdot\vec{d})_{max}=|\vec{c}|\cdot|\vec{d}|=\sqrt{53}\cdot2$. Square = $4\times53=104$
If $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ is a solution of $AX=B$, where $\text{adj }A=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ and $B=\begin{bmatrix}4\\0\\2\end{bmatrix}$, then $x+y+z$ is equal to:
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Solution
Using $A^{-1}=\dfrac{adjA}{|A|}$ and $AX=B$ → $X=\dfrac{adjA\cdot B}{|A|}$. Compute gives sum $=2$
Let $L$ be the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$ and let $S$ be the set of all points $(a,b,c)$ on $L$, whose distance from the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$ along the line $L$ is $7$. Then $\sum_{(a,b,c)\in S}(a+b+c)$ is equal to:
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Solution
Parametrize line $L$: $x=-1+2t,y=-1+3t,z=-3+6t$. Distance condition gives two values of $t$. Sum $(a+b+c)$ for both points gives $28$
Let $P(10,2\sqrt{15})$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\Delta PSS'$ is equal to :
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Solution
For hyperbola, latus rectum length $=\dfrac{2b^2}{a}=8\Rightarrow b^2=4a$. Since $P(10,2\sqrt{15})$ lies on it, $\dfrac{100}{a^2}-\dfrac{60}{b^2}=1$. Using $b^2=4a$, we get $a=5$, so $b^2=20$ and $c^2=a^2+b^2=45$. Hence $SS'=2c=6\sqrt5$. Area of $\Delta PSS' = \dfrac12\cdot 6\sqrt5\cdot 2\sqrt{15}=30\sqrt3$. Therefore square of area $=(30\sqrt3)^2=2700$.
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