Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that \[\sum_{i=1}^{10} (x_i - 2) = 30, \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 98, \quad \beta > 2\]and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of \[ 2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, \ldots, 2(x_{10} - 1) + 4\beta\] then $\frac{\beta \mu}{\sigma^2}$ is equal to:
Let R = {(1, 2), (2, 3), (3, 3)}} be a relation defined on the set \( \{1, 2, 3, 4\} \). Then the minimum number of elements needed to be added in \( R \) so that \( R \) becomes an equivalence relation, is:
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to:
If \(\int e^x \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2} \right) dx = g(x) + C\), where C is the constant of integration, then \(g\left( \frac{1}{2} \right)\)equals:
The value of $\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^3 + 6k^2 + 11k + 5}{(k+3)!}$ is:
Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ be a function defined by $ f(x) = ||x+2| - 2|x|| $. If m is the number of points of local maxima of f and n is the number of points of local minima of f, then m + n is