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
Let \( \vec{a} = 2\hat{i} - 3\hat{j} + \hat{k} \), \( \vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k} \) and a vector \( \vec{c} \) be such that \[ (\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k} \] and \[ \vec{a} \cdot \vec{c} = 3. \] If \( \vec{b} \times \vec{c} = \vec{d} \), then find \( |\vec{a} \cdot \vec{d}| \).
Let (a, b) be the point of intersection of the curve \(x^2 = 2y\) and the straight line \(y - 2x - 6 = 0\) in the second quadrant. Then the integral \(I = \int_{a}^{b} \frac{9x^2}{1+5^{x}} \, dx\) is equal to: