Prove that the Greatest Integer Function $f: R \rightarrow R ,$ given by $f(x)=[x]$, is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.

Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R$. Then $2 f(0)+3 f(1)$ is equal to

If $f\left( x \right) = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} – 1$ , $x \in R$ , then the equation $f(x) = 0$ has

The graph of function $f$ contains the point $P (1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( {\frac{{{s^2} + 2s – 3}}{{s – 1}}} \right)$ $x – 1 – s$. The value of $f ‘ (1)$, is

Let $x$ be a non-zero rational number and $y$ be an irrational number. Then $xy$ is