JEE MAIN 2026 Previous Year Questions (PYQs) – Page 11 of 25
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An equilateral triangle $OAB$ is inscribed in the parabola $y^2 = 4x$ with vertex at origin. Then the minimum distance of the circle having $AB$ as diameter from origin is:
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Solution
Let $A(at^2, 2at)$
$t = 2\sqrt{3}$
Required circle:
$(x - 12)^2 + y^2 = (4\sqrt{3})^2$
Minimum distance
$= |CP - R| = 4(3 - \sqrt{3})$
Let $\sum a_k = \alpha n^2 + \beta n$. If $a_{10} = 59$ and $a_6 = 7a_4$, then $\alpha + \beta$ is equal to:
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Solution
$a_n = S_n - S_{n-1}$
$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(n-1))$
$\Rightarrow a_n = 2\alpha n - \alpha + \beta$
$a_{10} = 59 \Rightarrow 20\alpha - \alpha + \beta = 59$
$a_6 = 7a_4$
Solving ⇒ $\alpha = 3,; \beta = 2$
$\Rightarrow \alpha + \beta = 5$
If the solution curve $y = f(x)$ of the differential equation
$(x^2 - 4)y' - 2xy + 2x(4 - x)^2 = 0,; x > 2$,
passes through the point $(3, 15)$, then the local maximum value of $f$ is ______.
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Solution
$(x^2 - 4)y' - 2xy = -2x(x^2 - 4)^2$
$\dfrac{d}{dx}\left(\dfrac{y}{x^2 - 4}\right) = -2x$
$y = (-x^2 + C)(x^2 - 4)$
For $x = 3,; y = 15$
$15 = (-9 + C)(5) \Rightarrow C = 12$
$y = (-x^2 + 12)(x^2 - 4)$
For maxima, $y' = 0 \Rightarrow x = 2\sqrt{2}$
$y_{\text{max}} = (-(2\sqrt{2})^2 + 12)\big((2\sqrt{2})^2 - 4\big)$
$= ( -8 + 12)(8 - 4) = 4 \cdot 4 = 16$
If the image of the point $P(a, 2, a)$ in the line
$\dfrac{x}{2} = \dfrac{y + a}{1} = \dfrac{z}{1}$
is $Q$ and the image of $Q$ in the line
$\dfrac{x - 2b}{2} = \dfrac{y - a}{1} = \dfrac{z + 2b}{-5}$
is $P$, then $a + b$ is equal to ______.
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Solution
Direction vector of first line: $(2,1,1)$
Using reflection property
$\dfrac{2\lambda - a + a}{2} = -5\mu - 2b$
and
$\dfrac{a + 4\lambda - a}{2} = 2\mu + 2b$
Solving
$\lambda - \mu = b$
$\lambda - \mu = 2a$
$\Rightarrow b = 2a$
Further solving
$a = 1 \Rightarrow b = 2$
$\Rightarrow a + b = 3$
The number of elements in the set
\( S = \{ x : x \in [0,100], \int_{0}^{x} t^2 \sin(x-t)\,dt = x^2 \} \)
is ______.
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Solution
$\int_0^x t^2 \sin(x - t),dt = x^2$
Use integration by parts
$-x^2\cos x + x^2 + 2x\cos x - 2x\sin x - x^2\cos x + 2x\sin x + 2\cos x - 2 = x$
$\Rightarrow \cos x = 1$
$\Rightarrow x = 0, 2\pi, 4\pi, \ldots, 30\pi$
Total elements $= 16$
Let
\(
A =
\begin{bmatrix}
0 & 2 & -3 \\
-2 & 0 & 1 \\
3 & -1 & 0
\end{bmatrix}
\)
and \( B \) be a matrix such that
\( B(I - A) = I + A \)
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Solution
$A^T = -A$
$B = (I + A)(I - A)^{-1}$
$B^T = \left((I - A)^{-1}\right)^T (I + A)^T$
$= (I - A^T)^{-1}(I + A)^T$
$= (I + A)^{-1}(I - A)$
Now
$B^T B = (I + A)^{-1}(I - A)(I + A)(I - A)^{-1}$
$= (I + A)^{-1}(I + A)(I - A)(I - A)^{-1}$
$= I$
$\Rightarrow \text{tr}(B^T B) = 3$
Let $S$ denote the set of $4$-digit numbers $abcd$ such that
$a > b > c > d$
and $P$ denote the set of $5$-digit numbers having product of its digits equal to $20$.
Then $n(S) + n(P)$ is equal to ______.
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Solution
For $n(S)$
Choose $4$ distinct digits in decreasing order
$n(S) = {}^{10}C_4 = 210$
For $n(P)$
Product of digits $= 20 = 5 \cdot 2 \cdot 2 \cdot 1 \cdot 1$
Arrangements
$= \dfrac{5!}{2!2!} = 30$
But considering permutations of positions
$n(P) = \dfrac{5!}{2!2!} = 50$
$\Rightarrow n(S) + n(P) = 210 + 50 = 260$
If the function
$f(x)=\dfrac{e^x\left(e^{\tan x}-1\right)+\ln(\sec x+\tan x)-x}{\tan x-x}$
is continuous at $x=0$, then the value of $f(0)$ is equal to:
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Solution
$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}-e^x+\ln(\sec x+\tan x)-x}{\tan x-x}$
Applying L’Hospital rule
$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}\sec^2x-e^x+\sec x-1}{\sec^2x-1}$
$=\lim_{x\to0}\dfrac{e^{\tan x}(\sec^2x-1)+(e^{\tan x}-e^x)+\sec x-1}{\tan^2x}$
$=\lim_{x\to0}\left(\dfrac{e^{\tan x}(e^{\tan x-x}-1)}{\tan^2x}+\dfrac{1}{\sec x+1}\right)$
$\Rightarrow f(0)=1+0+\frac{1}{2}=\frac{3}{2}$
Let a circle of radius $4$ pass through the origin $O$, the point $A(-\sqrt{3}a,0)$ and $B(0,-\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab\ne0$. Then the locus of centroid of $\triangle OAB$ is a circle of radius:
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Solution
$AB=8$
$3a^2+2b^2=64$
Centroid $G(h,k)$
$h=\frac{-\sqrt{3}a}{3},\quad k=\frac{-\sqrt{2}b}{3}$
$a=-\sqrt{3}h,\quad b=-\frac{3k}{\sqrt{2}}$
Substitute
$9h^2+9k^2=64$
$x^2+y^2=\frac{64}{9}$
$\Rightarrow r=\frac{8}{3}$
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