Mathematics - Functions Question with Solution | TestHub

If $x\in (n-1,n)$ then $[x]=n-1\Rightarrow [x{]}^2=(n-1{)}^2$ and $(n-1{)}^2\le \left[n^2\right]\le n^2-1$ $\begin{array}{rlr}& 0\le \left[x^2\right]-[x{]}^2\le n^2-1-(n-1{)}^2& 0\le f(x)\le 2n-2\end{array}$

Since $f(x)$ has to be integer, range of $f(x)=\{0,1,2,3,\dots ..2n-2\}$

∴ The number of elements in range of $f$ is $(2n-1)$