If $A=\{x \in R: \sqrt{x^2-8x+15} \in R\}$ and $B=\{x \in R: \frac{x-3}{2x-5} < \frac{x-6}{2x-11}\}$, then $A \cap B=$

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

Let $A$ denote the set of all real numbers $x$ such that $x^3-[x]^3=(x-[x])^3$,where $[x]$ is the greatest integer less than or equal to $x$. Then,

For a real number $x$,$[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $\left[\frac{1}{2}\right] + \left[\frac{1}{2} + \frac{1}{100}\right] + \left[\frac{1}{2} + \frac{2}{100}\right] + \left[\frac{1}{2} + \frac{3}{100}\right] + \ldots + \left[\frac{1}{2} + \frac{99}{100}\right] = $

In a regular graph of $15$ vertices,the sum of the degrees of the vertices is $60$. Then,the degree of each vertex is:

If $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$ is a polynomial such that $P(0) = 1, P(1) = 2, P(2) = 5, P(3) = 10$ and $P(4) = 17$,then $P(5) =$