List of top Mathematics Questions asked in JEE Main

Let $ A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\4 & 6 + 2p & 8 + 3p + 2q \\6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix} $ If $ \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n, \, m, n \in \mathbb{N}, $ then $ m + n $ is equal to:

Let the ellipse $ 3x^2 + py^2 = 4 $ pass through the centre $ C $ of the circle $ x^2 + y^2 - 2x - 4y - 11 = 0 $ of radius $ r $. Let $ f_1, f_2 $ be the focal distances of the point $ C $ on the ellipse. Then $ 6f_1 f_2 - r $ is equal to

If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... \infty = \frac{\pi^4}{90}, $ $ \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... \infty = \alpha, $ $ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + ... \infty = \beta, $ then $ \frac{\alpha}{\beta} $ is equal to:

The number of rational terms in the binomial expansion of $ \left( 5^{1/2} + 7^{1/8} \right)^{1016} $ is:
The product of the last 2 digits of $ (1919)^{1919} $ is:

Consider two statements: 
Statement 1: $ \lim_{x \to 0} \frac{\tan^{-1} x + \ln \left( \frac{1+x}{1-x} \right) - 2x}{x^5} = \frac{2}{5} $ 
Statement 2: $ \lim_{x \to 1} x \left( \frac{2}{1-x} \right) = e^2 \; \text{and can be solved by the method}  \lim_{x \to 1} \frac{f(x)}{g(x) - 1} $

Find the area of the region defined by the conditions: $ \left\{ (x, y): 0 \leq y \leq \sqrt{9x}, y^2 \geq 3 - 6x \right\} \text{(in square units)} $

Let $ f $ be defined as $ R \setminus \{0\} \to R $ such that $ f(x) = \frac{x}{3} + \frac{3}{x} + 3 $ If $ f(x) $ is strictly increasing in $ (-\infty, \alpha_1) \cup (\alpha_2, \infty) $ and strictly decreasing in $ (\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5) $, then $ \sum_{i=1}^5 (\alpha_i)^2 $ is equal to:

Let $ A = \{0, 1, 2, 3, 4, 5, 6\} $ and $ R_1 = \{(x, y): \max(x, y) \in \{3, 4 \}$. Consider the two statements:
Statement 1: Total number of elements in $ R_1 $ is 18.
Statement 2: $ R $ is symmetric but not reflexive and transitive.

Let $ I_1 = \int_{\frac{1}{2}}^{1} 2x \cdot f(2x(1 - 2x)) \, dx $ 
and $ I_2 = \int_{-1}^{1} f(x(1 - x)) \, dx \; \text{then}  \frac{I_2}{I_1} \text{ equals to:} $

The sum of squares of roots of $ |x - 2|^2 - |x - 2| - 2 = 0 $ and $ x^2 - 2|x - 3| - 5 = 0 $ equals to:
A circle touches the parabola $ y^2 = 9x $ at $ (4, 6) $ and the positive x-axis. Find the radius of the circle.

If $ A = \begin{pmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{pmatrix} $, then the value of $ \det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n $, then $ m + n $ is equal to:

Let $ \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k} $ and $ \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} $, and $ \vec{c} $ is a vector perpendicular to $ \vec{a} $ and lies in the plane of $ \vec{a} $, $ \vec{b} $. Then $ \vec{c} $ is equal to:
If $ f(x) = x - 1 $ and $ g(x) = e^x $, and the differential equation is: $ \frac{dy}{dx} = \left( e^{-2\sqrt{x}} g(f(f(f(x)))) - \frac{y}{\sqrt{x}} \right) $ with the initial condition $ y(0) = 0 $, then $ y(1) $ equals:
The value of the following expression: $ \cot^{-1} \left( \frac{\sqrt{1 + \tan^2 2} + 1}{\tan 2} \right) - \cot^{-1} \left( \frac{\sqrt{1 + \tan^2 2} - 1}{\tan 2} \right) $
Given: $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{\pi^4}{90}, $ $ \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = \alpha, $ $ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots = \beta $ Find the value of $ \frac{\alpha}{\beta} $.
Evaluate the integral: $ \int_{-1}^{3/2} | \pi^2 x \sin(\pi x) | \, dx $
Probability of event A is 0.7 and event B is 0.4, and $ P(A \cap B^c) = 0.5 $, then the value of $ P(B | A \cup B^c) $ is equal to:
There are 12 points in a plane in which 5 are collinear such that no three of them are in a straight line. Then the number of triangles that can be formed from any 3 vertices from 12 points.
Let $ A = [a_{ij}] $ be a $3 \times 3 $ matrix such that: $$ A = \begin{bmatrix} 0 & 0 & 4 \\ 1 & 0 & 0 \\ 0 & 1 & 3 \\ \end{bmatrix}, A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 0 \end{bmatrix}. $$ Then $ a_{23} $ equals:
Let $f: [0, 3] \to A$ be defined by $f(x) = 2x^3 - 15x^2 + 36x + 7$ and $g: [0, \infty) \to B$ be defined by $g(x) = \frac{x}{x^{2025} + 1}$. If both functions are onto and $S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \}$, then $n(S)$ is equal to:
Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$.
Let $$ L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \quad \lambda \in \mathbb{R} $$ and $$ L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \quad \mu \in \mathbb{R} $$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$, and is parallel to $\vec{a} + \vec{b}$, then $L_3$ passes through the point:
If the equation of the parabola with vertex $\left( \frac{3}{2}, 3 \right)$ and the directrix $x + 2y = 0$ is $$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, \text{ then } \alpha + \beta + \gamma \text{ is equal to:} $$
If the square of the shortest distance between the lines $$ \frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3} \quad \text{and} \quad \frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}, $$ is $ \frac{m}{n} $, where $m, n$ are coprime numbers, then $m+n$ is equal to: