List of practice Questions

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 \_\_\_\_\_$.

A physical quantity P is related to four observations a, b, c, and d as follows: P = a³ b² (c / √d) The percentage errors of measurement in a, b, c, and d are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity P is:

Which of the following statements is/are correct?

If $ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 $, $ |\overrightarrow{a}| = \sqrt{37} $, $ |\overrightarrow{b}| = 3 $, and $ |\overrightarrow{c}| = 4 $, then the angle between $ \overrightarrow{b} $ and $ \overrightarrow{c} $ is:

“बिसनाथ को अपने गाँव बिस्कोहर से जो लगाव है वह मूलतः मनुष्य की अपनी स्मृतियों के प्रति लगाव का ही एक रूप है।” ‘बिस्कोहर की माटी’ पाठ के आधार पर सोदाहरण टिप्पणी कीजिए।

Solve the following linear programming problem graphically: Maximise $ Z = x + 2y $ Subject to the constraints: \[ x - y \geq 0 \] \[ x - 2y \geq -2 \] \[ x \geq 0, \, y \geq 0 \]

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:

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

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