Let $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$. Let $R$ be a relation defined on $A \times B$ such that $R = \{((a_1, b_1), (a_2, b_2)) : a_1 \leq b_2 \text{ and } b_1 \leq a_2\}$. Then the number of elements in the set $R$ is

Let $f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$ be defined by $f(x) = (\log(\sec x + \tan x))^3$. Then:

If the sets $A$ and $B$ are defined as $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = x, x \in R\}$,then:

If $f:[0,2) \rightarrow R$ is defined by $f(x)=\begin{cases} 1+\frac{2x}{k} & \text{for } 0 \leq x < 1 \\ kx & \text{for } 1 \leq x < 2 \end{cases}$ where $k>0$,and $f$ is such that $\lim_{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^{+}} f(x)$,then find the value of $k^2$.

$ A $ is a set having $ 6 $ distinct elements. The number of distinct functions from $ A $ to $ A $ which are not bijections is

Consider a function $f : N \rightarrow R$,satisfying $f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x)$ for $x \geq 2$,with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to