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:
If \( (a, b) \) be the orthocenter of the triangle whose vertices are \( (1, 2) \), \( (2, 3) \), and \( (3, 1) \), and \( I_1 = \int_a^b x \sin(4x - x^2) dx \), \( I_2 = \int_a^b \sin(4x - x^2) dx \), then \( 36 \frac{I_1}{I_2} \) is equal to:
Let
\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix} \)
and \(|2A|^3 = 2^{21}\) where \(\alpha, \beta \in \mathbb{Z}\). Then a value of \(\alpha\) is:
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:
The value of $\frac{1 \times 2^2 + 2 \times 3^2 + \dots + 100 \times (101)^2}{1^2 \times 2 + 2^2 \times 3 + \dots + 100^2 \times 101}$ is:
Let the line $L$ intersect the lines$x - 2 = -y = z - 1$, $\quad 2(x + 1) = 2(y - 1) = z + 1$and be parallel to the line$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$.Then which of the following points lies on $L$?