Let \( \alpha, \beta; \, \alpha > \beta \), be the roots of the equation $$ x^2 - \sqrt{2}x - \sqrt{3} = 0. $$ Let \( P_n = \alpha^n - \beta^n, \, n \in \mathbb{N} \). Then $$ \left( 11\sqrt{3} - 10\sqrt{2} \right) P_{10} + \left( 11\sqrt{2} + 10 \right) P_{11} - 11P_{12} $$ is equal to:
Let $$ B = \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix} $$ and $A$ be a $2 \times 2$ matrix such that $$ AB^{-1} = A^{-1}. $$ If $BCB^{-1} = A$ and $$ C^4 + \alpha C^2 + \beta I = O, $$ then $2\beta - \alpha$ is equal to:
Let the foci of a hyperbola $ H $ coincide with the foci of the ellipse $ E : \frac{(x - 1)^2}{100} + \frac{(y - 1)^2}{75} = 1 $ and the eccentricity of the hyperbola $ H $ be the reciprocal of the eccentricity of the ellipse $ E $. If the length of the transverse axis of $ H $ is $ \alpha $ and the length of its conjugate axis is $ \beta $, then $ 3\alpha^2 + 2\beta^2 $ is equal to:
\(\lim_{x \to 0} \frac{e - (1 + 2x)^{\frac{1}{2x}}}{x} \quad \text{is equal to:}\)
It is known that 20% of the students in a school have above 90% attendance and 80% of the students are irregular. Past year results show that 80% of students who have above 90% attendance and 20% of irregular students get “A” grade in their annual examination. At the end of a year, a student is chosen at random from the school and is found to have an “A” grade. What is the probability that the student is irregular?