For the following questions answer them individually
Let $$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$$ and $$B = I + \text{adj}(A) + (\text{adj } A)^2 + \ldots + (\text{adj } A)^{10}$$. Then, the sum of all the elements of the matrix $$B$$ is:
Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1, 1]$$ such that $$\cos^{-1} x - \sin^{-1} y = \alpha$$, $$\frac{-\pi}{2} \leq \alpha \leq \pi$$. Then, the minimum value of $$x^2 + y^2 + 2xy \sin \alpha$$ is
If the function $$f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ a \log_e 2 \log_e 3, & x = 0 \end{cases}$$ is continuous at $$x = 0$$, then the value of $$a^2$$ is equal to
Let $$f(x) = 3\sqrt{x - 2} + \sqrt{4 - x}$$ be a real valued function. If $$\alpha$$ and $$\beta$$ are respectively the minimum and the maximum values of $$f$$, then $$\alpha^2 + 2\beta^2$$ is equal to
If the value of the integral $$\int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} dx$$ is $$\frac{2}{\pi}$$. Then, a value of $$\alpha$$ is
The area (in sq. units) of the region described by $$\{(x, y) : y^2 \leq 2x, y \geq 4x - 1\}$$ is
Let $$y = y(x)$$ be the solution of the differential equation $$(x^2 + 4)^2 dy + (2x^3 y + 8xy - 2)dx = 0$$. If $$y(0) = 0$$, then $$y(2)$$ is equal to
Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$$ and $$\vec{c} = x\hat{i} + 2\hat{j} + 3\hat{k}$$, $$x \in \mathbb{R}$$. If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b} + \vec{c}$$ such that $$\vec{a} \cdot \vec{d} = 1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to
For $$\lambda > 0$$, let $$\theta$$ be the angle between the vectors $$\vec{a} = \hat{i} + \lambda\hat{j} - 3\hat{k}$$ and $$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$$. If the vectors $$\vec{a} + \vec{b}$$ and $$\vec{a} - \vec{b}$$ are mutually perpendicular, then the value of $$(14 \cos \theta)^2$$ is equal to
Let $$P$$ be the point of intersection of the lines $$\frac{x-2}{1} = \frac{y-4}{5} = \frac{z-2}{1}$$ and $$\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-3}{2}$$. Then, the shortest distance of $$P$$ from the line $$4x = 2y = z$$ is
