Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.

Suppose $\quad f : R \rightarrow(0, \infty)$ be a differentiable function such that $5 f ( x + y )= f ( x ) \cdot f ( y ), \forall x , y \in R$. If $f(3)=320$, then $\sum \limits_{n=0}^5 f(n)$ is equal to :

If $f(x) = \cos (\log x)$, then $f(x)f(y) - \frac{1}{2}[f(x/y) + f(xy)] = $

Let $\mathrm{f}: N \rightarrow N$ be a function such that $\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in N$. If $\mathrm{f}(6)=18$ then $\mathrm{f}(2) \cdot \mathrm{f}(3)$ is equal to :

The range of values of the function $f\left( x \right) = \frac{1}{{2 - 3\sin x}}$ is

If $f(x)$ is a quadratic expression such that $f(1) + f (2)\, = 0$ , and $-1$ is a root of $f(x)\, = 0$, then the other root of $f(x)\, = 0$ is