List of top Questions asked in JEE Main

The number of solutions of the equation \[ \left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0 \] is:

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is:

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word "MATHS" such that any letter that appears in the word must appear at least twice is:

If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440^th position in this arrangement is:

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.

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.

If the number of seven-digit numbers, such that the sum of their digits is even, is $ m \cdot n \cdot 10^a $; $ m, n \in \{1, 2, 3, ..., 9\} $, then $ m + n $ is equal to

The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1’s and exactly three 2’s, is equal to:

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.

Let \[ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b \quad \text{and} \quad H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1. \] Let the distance between the foci of $ E $ and the foci of $ H $ be $ 2\sqrt{3} $. If $ a - A = 2 $, and the ratio of the eccentricities of $ E $ and $ H $ is $ \frac{1}{3} $, then the sum of the lengths of their latus rectums is equal to:

Let $$ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b \quad \text{and} \quad H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1. $$ Let the distance between the foci of $E$ and the foci of $H$ be $2\sqrt{3}$. If $a - A = 2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$, then the sum of the lengths of their latus rectums 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

Consider the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, $ having one of its foci at $ P(-3, 0) $. If the latus rectum through its other focus subtends a right angle at $ P $, and $ a^2b^2 = \alpha\sqrt{2} - \beta, \quad \alpha, \beta \in \mathbb{N}, $ then find $ \alpha $ and $ \beta $.

Let one focus of the hyperbola $ H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $ and the corresponding directrix be $ x = \dfrac{9}{\sqrt{10}} $. If $ e $ and $ l $ respectively are the eccentricity and the length of the latus rectum of $ H $, then $ 9 \left(e^2 + l \right) $ is equal to:

If $ S $ and $ S' $ are the foci of the ellipse \[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \] and $ P $ is a point on the ellipse, then \[ \min (SP \cdot S'P) + \max (SP \cdot S'P) \] is equal to:

Let O be the origin, the point A be $ z_1 = \sqrt{3} + 2\sqrt{2}i $, the point B $ z_2 $ be such that $ \sqrt{3}|z_2| = |z_1| $ and $ \arg(z_2) = \arg(z_1) + \frac{\pi}{6} $. Then:

Let the ellipse $ E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, $ a>b $ and $ E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 $, $ A<B $, have the same eccentricity $ \frac{1}{\sqrt{3}} $. Let the product of their lengths of latus rectums be $ \frac{32}{\sqrt{3}} $ and the distance between the foci of $ E_1 $ be 4. If $ E_1 $ and $ E_2 $ meet at A, B, C, and D, then the area of the quadrilateral ABCD equals:

Let $ A = \left\{ x \in (0, \pi) \mid - \log\left(\frac{2}{\pi}\right)\sin x + \log\left(\frac{2}{\pi}\right)\cos x = 2 \right\} $ and

\[ B = \left\{ x \geq 0 : \sqrt{x}(\sqrt{x - 4}) - 3\sqrt{x - 2} + 6 = 0 \right\}. \]

Then $ n(A \cup B) $ is equal to:

Let the curve $($$ z(1 + i) + \overline{z(1 - i)} = 4$, $\, z \in \mathbb{C}$ $)$, divide the region $|z - 3| \leq 1$ into two parts of areas $ \alpha $ and $ \beta $. Then $ |\alpha - \beta| $ equals:

If $ \alpha $ and $ \beta $ are the roots of the equation $ 2z^2 - 3z - 2i = 0 $, where $ i = \sqrt{-1} $, then \[ 16 \cdot {Re} \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \cdot {Im} \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \] is equal to:

Let $ z_1, z_2, z_3 $ be three complex numbers on the circle $ |z| = 1 $ with $ \arg(z_1) = -\frac{\pi}{4}, \arg(z_2) = 0 $ and $ \arg(z_3) = \frac{\pi}{4} $. If $ |z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2} $, where $ \alpha, \beta \in \mathbb{Z} $, then the value of $ \alpha^2 + \beta^2 $ is :

The number of complex numbers $ z $, satisfying $ |z| = 1 $ and \[ \left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1, \] is:

Let O be the origin, the point A be $ z_1 = \sqrt{3} + 2\sqrt{2}i $, the point B $ z_2 $ be such that $ \sqrt{3}|z_2| = |z_1| $ and $ \arg(z_2) = \arg(z_1) + \frac{\pi}{6} $. Then: