The missing term in the following table is
$\begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y=f(x) & 1 & 3 & 9 & ? & 81 \\ \hline \end{array}$

For a real number $x$,$[x]$ denotes the greatest integer less than or equal to $x$. The value of $\left[ \frac{1}{2} \right] + \left[ \frac{1}{2} + \frac{1}{100} \right] + \left[ \frac{1}{2} + \frac{2}{100} \right] + \dots + \left[ \frac{1}{2} + \frac{99}{100} \right]$ is

Let $A = \{\theta \in R \mid \cos^2(\sin \theta) + \sin^2(\cos \theta) = 1\}$ and $B = \{\theta \in R \mid \cos(\sin \theta) \sin(\cos \theta) = 0\}$. Then,$A \cap B$ is:

Let $A$ be the set of first $101$ terms of an $A$.$P$.,whose first term is $1$ and the common difference is $5$, and let $B$ be the set of first $71$ terms of an $A$.$P$.,whose first term is $9$ and the common difference is $7$. Then, the number of elements in $A \cap B$ which are divisible by $3$ is:

Let $A$ and $B$ be finite sets and $P_A$ and $P_B$ respectively denote their power sets. If $P_B$ has $112$ elements more than those in $P_A$,then the number of injective functions from $A$ to $B$ is

If $A, B, C$ are three sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$,then