For the following questions answer them individually
Let $$S = \{1, 2, 3, \ldots, 10\}$$. Suppose $$M$$ is the set of all the subsets of $$S$$, then the relation $$R = \{(A, B) : A \cap B \neq \phi; \; A, B \in M\}$$ is :
Consider the matrix $$f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$. Given below are two statements : Statement I: $$f(-x)$$ is the inverse of the matrix $$f(x)$$. Statement II: $$f(x) f(y) = f(x + y)$$. In the light of the above statements, choose the correct answer from the options given below
The function $$f : \mathbb{N} - \{1\} \rightarrow \mathbb{N}$$; defined by $$f(n) =$$ the highest prime factor of $$n$$, is :
Consider the function $$f(x) = \begin{cases} \frac{a(7x - 12 - x^2)}{b|x^2 - 7x + 12|}, & x < 3 \\ 2^{\frac{\sin(x-3)}{x - [x]}}, & x > 3 \\ b, & x = 3 \end{cases}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. If $$S$$ denotes the set of all ordered pairs $$(a, b)$$ such that $$f(x)$$ is continuous at $$x = 3$$, then the number of elements in $$S$$ is :
If $$\int_0^1 \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx = a + b\sqrt{2} + c\sqrt{3}$$, where $$a, b, c$$ are rational numbers, then $$2a + 3b - 4c$$ is equal to :
If $$(a, b)$$ be the orthocentre 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 $$x = x(t)$$ and $$y = y(t)$$ be solutions of the differential equations $$\frac{dx}{dt} + ax = 0$$ and $$\frac{dy}{dt} + by = 0$$ respectively, $$a, b \in \mathbb{R}$$. Given that $$x(0) = 2$$; $$y(0) = 1$$ and $$3y(1) = 2x(1)$$, the value of $$t$$, for which $$x(t) = y(t)$$, is :
If $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$, $$\vec{b} = 3(\hat{i} - \hat{j} + \hat{k})$$ and $$\vec{c}$$ be the vector such that $$\vec{a} \times \vec{c} = \vec{b}$$ and $$\vec{a} \cdot \vec{c} = 3$$, then $$\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$$ is equal to
The distance, of the point $$(7, -2, 11)$$ from the line $$\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$$ along the line $$\frac{x-5}{2} = \frac{y-1}{-3} = \frac{z-5}{6}$$, is :
If the shortest distance between the lines $$\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$$ and $$\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$$ is $$\frac{6}{\sqrt{5}}$$, then the sum of all possible values of $$\lambda$$ is :
