JEE Main 2025 Jan 28 Shift 2 — Mathematics

[Stem missing: page 1 of the source PDF is corrupted — the stem and first three options of Question 1 could not be recovered in either English or Hindi. Only the fourth option survives.]

Let $f : [0,\ 3] \to \mathrm{A}$ be defined by $f(x) = 2x^3 - 15x^2 + 36x + 7$ and $g : [0,\ \infty) \to \mathrm{B}$ be defined by $g(x) = \frac{x^{2025}}{x^{2025}+1}$. If both the functions are onto and $S = \{x \in \mathbf{Z} : x \in \mathrm{A} \text{ or } x \in \mathrm{B}\}$, then $n(S)$ is equal to :

Let $f : \mathbf{R} - \{0\} \to (-\infty,\ 1)$ be a polynomial of degree 2, satisfying $f(x)\,f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$. If $f(\mathrm{K}) = -2\mathrm{K}$, then the sum of squares of all possible values of $\mathrm{K}$ is :

If $\alpha + i\beta$ and $\gamma + i\delta$ are the roots of $x^2 - (3-2i)x - (2i-2) = 0$, $i = \sqrt{-1}$, then $\alpha\gamma + \beta\delta$ is equal to :

Let $\mathrm{A} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$ and $\mathrm{P} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, $\theta > 0$. If $\mathrm{B} = \mathrm{PAP^T}$, $\mathrm{C} = \mathrm{P^T B^{10} P}$ and the sum of the diagonal elements of $\mathrm{C}$ is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $m + n$ is :

For positive integers $n$, if $4a_n = (n^2 + 5n + 6)$ and $S_n = \sum_{k=1}^{n} \left(\frac{1}{a_k}\right)$, then the value of $507\, S_{2025}$ is :

Let the coefficients of three consecutive terms $\mathrm{T}_r$, $\mathrm{T}_{r+1}$ and $\mathrm{T}_{r+2}$ in the binomial expansion of $(a+b)^{12}$ be in a G.P. and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $\left(\sqrt[4]{3} + \sqrt[3]{4}\right)^{12}$. Then $p + q$ is equal to :

Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the select… [remainder of stem corrupted in the source PDF; the Hindi duplicate is also lost]

Bag $\mathrm{B}_1$ contains 6 white and 4 blue balls, Bag $\mathrm{B}_2$ contains 4 white and 6 blue balls, and Bag $\mathrm{B}_3$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $\mathrm{B}_2$, is :

Two equal sides of an isosceles triangle are along $-x + 2y = 4$ and $x + y = 4$. If $m$ is the slope of its third side, then the sum, of all possible distinct values of $m$, is :

If the midpoint of a chord of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is $\left(\sqrt{2},\ \frac{4}{3}\right)$, and the length of the chord is $\frac{2\sqrt{\alpha}}{3}$, then $\alpha$ is :

If A and B are the points of intersection of the circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ and a point P moves on the line $2x - 3y + 4 = 0$, then the centroid of $\Delta \mathrm{PAB}$ lies on the line :

If $\sum_{r=1}^{13} \left\{\frac{1}{\sin\left(\frac{\pi}{4} + (r-1)\frac{\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{r\pi}{6}\right)}\right\} = a\sqrt{3} + b$, $a,\ b \in \mathbf{Z}$, then $a^2 + b^2$ is equal to :

Let A, B, C be three points in $xy$-plane, whose position vector are given by $\sqrt{3}\hat{i} + \hat{j}$, $\hat{i} + \sqrt{3}\hat{j}$ and $a\hat{i} + (1-a)\hat{j}$ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ is $\frac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is :

If the components of $\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ along and perpendicular to $\vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ respectively, are $\frac{16}{11}\left(3\hat{i} + \hat{j} - \hat{k}\right)$ and $\frac{1}{11}\left(-4\hat{i} - 5\hat{j} - 17\hat{k}\right)$, then $\alpha^2 + \beta^2 + \gamma^2$ is equal to :

The square of the distance of the point $\left(\frac{15}{7},\ \frac{32}{7},\ 7\right)$ from the line $\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ in the direction of the vector $\hat{i} + 4\hat{j} + 7\hat{k}$ is :

Let $f : \mathbf{R} \to \mathbf{R}$ be a twice differentiable function such that $f(2) = 1$. If $\mathrm{F}(x) = x f(x)$ for all $x \in \mathbf{R}$, $\int_0^2 x\, \mathrm{F}'(x)\, \mathrm{d}x = 6$ and $\int_0^2 x^2\, \mathrm{F}''(x)\, \mathrm{d}x = 40$, then $\mathrm{F}'(2) + \int_0^2 \mathrm{F}(x)\, \mathrm{d}x$ is equal to :

Let $f$ be a real valued continuous function defined on the positive real axis such that $g(x) = \int_0^x \mathrm{t}\, f(\mathrm{t})\, \mathrm{dt}$. If $g(x^3) = x^6 + x^7$, then value of $\sum_{r=1}^{15} f(\mathrm{r}^3)$ is :

If $f(x) = \int \frac{1}{x^{1/4}\left(1 + x^{1/4}\right)} \mathrm{d}x$, $f(0) = -6$, then $f(1)$ is equal to :

The area of the region bounded by the curves $x(1 + y^2) = 1$ and $y^2 = 2x$ is:

The interior angles of a polygon with $n$ sides, are in an A.P. with common difference $6°$. If the largest interior angle of the polygon is $219°$, then $n$ is equal to __________.

The number of n… [remainder of stem corrupted in the source PDF; the Hindi duplicate is also lost]

Let A and B be the points of intersection of the mirror image of the parabola $y^2 = 4x$ in the line $x + y + 4 = 0$, and the line $y + 5 = 0$. If the distance between A and B is $d$ and the area of $\Delta \mathrm{SAB}$ is $a$, where S is the focus of the parabola $y^2 = 4x$, then the value of $(a + d)$ is __________. [English block corrupted in the source PDF; transcribed from the Hindi duplicate of the same question]

Let $f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left(\frac{\tan\left(x/2^{r+1}\right) + \tan^3\left(x/2^{r+1}\right)}{1 - \tan^2\left(x/2^{r+1}\right)}\right)$. Then $\lim_{x \to 0} \frac{\mathrm{e}^x - \mathrm{e}^{f(x)}}{(x - f(x))}$ is equal to __________.

If $y = y(x)$ is the solution of the differential equation, $\sqrt{4 - x^2}\, \frac{\mathrm{d}y}{\mathrm{d}x} = \left(\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2 - y\right) \sin^{-1}\left(\frac{x}{2}\right)$, $-2 \le x \le 2$, $y(2) = \frac{\pi^2 - 8}{4}$, then $y^2(0)$ is equal to __________.