List of top Mathematics Questions asked in JEE Main

Let \( \alpha, \beta \) be the roots of the equation \( x^2 - \sqrt{6}x + 3 = 0 \) such that \( \operatorname{Im}(\alpha) > \operatorname{Im}(\beta) \). Let \( a, b \) be integers not divisible by 3 and \( n \) be a natural number such that\[\frac{\alpha^{99}}{\beta} + \alpha^{98} = 3^n (a + ib), i = \sqrt{-1}.\]Then \( n + a + b \) is equal to ______.

Let \( f : \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow \mathbb{R} \) be a differentiable function such that \( f(0) = \frac{1}{2} \). If the \( \lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha \), then \( 8\alpha^2 \) is equal to:

If \( 2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0 \) has exactly 3 solutions in the interval \( \left[ 0, \frac{n \pi}{2} \right] \), \( n \in \mathbb{N} \), then the roots of the equation \( x^2 + nx + (n - 3) = 0 \) belong to:

Let \( A (2, 3, 5) \) and \( C(-3, 4, -2) \) be opposite vertices of a parallelogram \( ABCD \). If the diagonal \( \overrightarrow{BD} = \hat{i} + 2 \hat{j} + 3 \hat{k} \), then the area of the parallelogram is equal to

Let a unit vector \( \hat{u} = x\hat{i} + y\hat{j} + z\hat{k} \) make angles \( \frac{\pi}{2}, \frac{\pi}{3} \), and \( \frac{2\pi}{3} \) with the vectors \( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k} \), \( \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \), and \( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \) respectively. If \( \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \), then \( |\hat{u} - \vec{v}|^2 \) is equal to

Let $y = y(x)$ be the solution of the differential equation $(1 + y^2)e^{\tan x} dx + \cos^2 x (1 + e^{2\tan x}) dy = 0$, $y(0) = 1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to:

If \[ f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix} \] then \( \frac{1}{5} f'(0) \) is equal to

Let H: $\frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$, $\alpha>0$ lies on H. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to:

Let M denote the median of the following frequency distribution.

\(x_i\)	\(f_i\)
0 - 4	2
4 - 8	4
8 - 12	7
12 - 16	8
16 - 20	6

Then 20M is equal to:

An integer is chosen at random from the integers 1,2, 3, ..., 50. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:

If \( R \) is the smallest equivalence relation on the set \( \{1, 2, 3, 4\} \) such that \( \{(1,2), (1,3)\} \subseteq R \), then the number of elements in \( R \) is ______.

Let \( y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right), -1 < x < 1 \). Then at \( x = \frac{1}{2} \), the value of \( 225(y' - y'') \) is equal to

The function \( f(x) = \frac{x}{x^2 - 6x - 16} \), \( x \in \mathbb{R} - \{-2, 8\} \)

The set of all α, for which the vectors $\vec{a} = \alpha \hat{ti} + 6\hat{j} - 3\hat{k}$ and $\vec{b} = \hat{ti} - 2\hat{j} - 2\alpha t\hat{k}$ are inclined at an obtuse angle for all $t \in \mathbb{R}$ is:

For the function $f(x) = (\cos x) - x + 1, x \in \mathbb{R}$, find the correct relationship between the following two statements
(S1) $f(x) = 0$ for only one value of x is $[0, \pi]$.
(S2) $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right]$.

If the circles \( (x+1)^2 + (y+2)^2 = r^2 \) and \( x^2 + y^2 - 4x - 4y + 4 = 0 \) intersect at exactly two distinct points, then

Consider the system of linear equations \( x + y + z = 4\mu \), \( x + 2y + 2\lambda z = 10\mu \), \( x + 3y + 4\lambda^2 z = \mu^2 + 15 \), where \( \lambda, \mu \in \mathbb{R} \). Which one of the following statements is NOT correct?

Let \( x = \frac{m}{n} \) ( \( m, n \) are co-prime natural numbers) be a solution of the equation \( \cos \left( 2 \sin^{-1} x \right) = \frac{1}{9} \) and let \( \alpha, \beta (\alpha > \beta) \) be the roots of the equation \( mx^2 - nx - m + n = 0 \). Then the point \( (\alpha, \beta) \) lies on the line

Let $z$ be a complex number such that $|z + 2| = 1$ and $\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$. Then the value of $|\text{Re}(z+2)|$ is:

If the set $R = {(a, b) : a + 5b = 42, a, b \in \mathbb{N}}$ has $m$ elements and $\sum_{n=1}^m (1 + i^n) = x + iy$, where $i = \sqrt{-1}$, then the value of $m + x + y$ is:

If the domain of the function \[ f(x) = \cos^{-1} \left( \frac{2 - |x|}{4} \right) + \left( \log_e (3 - x) \right)^{-1} \] is \([-\alpha, \beta) - \{ \gamma \}\), then \( \alpha + \beta + \gamma \) is equal to:

If each term of a geometric progression \( a_1, a_2, a_3, \dots \) with \( a_1 = \frac{1}{8} \) and \( a_2 \neq a_1 \), is the arithmetic mean of the next two terms and \( S_n = a_1 + a_2 + \dots + a_n \), then \( S_{20} - S_{18} \) is equal to

Let \( A \) be the point of intersection of the lines \( 3x + 2y = 14 \), \( 5x - y = 6 \) and \( B \) be the point of intersection of the lines \( 4x + 3y = 8 \), \( 6x + y = 5 \). The distance of the point \( P(5, -2) \) from the line \( AB \) is

Two integers \( x \) and \( y \) are chosen with replacement from the set \( \{0, 1, 2, 3, \ldots, 10\} \). Then the probability that \( |x - y| > 5 \) is: