Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f:[0,\infty )\to [0,\infty )$, $f\left(x\right)={\int }_0^x\left[y\right]dy$

Let $x=n+f,f\in (0,1)$

$\Rightarrow \left[x\right]=n, \left\{x\right\}=f$.

So

$$\begin{gathered} f\left(x\right)={\int }_0^1\left[y\right]dy+{\int }_1^2\left[y\right]dy+{\int }_2^3\left[y\right]dy+ \\ \dots +{\int }_{n-1}^n\left[y\right]dy+{\int }_n^{n+f}\left[y\right]dy \end{gathered}$$

$$\begin{gathered} f\left(x\right)={\int }_0^{1}0dy+{\int }_1^{2}1dy+{\int }_2^{3}2dy+ \\ \dots +{\int }_{n-1}^n\left(n-1\right)dy+{\int }_n^{n+f}ndy \end{gathered}$$

$$\begin{gathered} f\left(x\right)=0\left[1-0\right]+1\left[2-1\right]+2\left[3-2\right]+ \\ \dots +\left(n-1\right)\left[n-\left(n-1\right)\right]+n\left[n+f-n\right] \end{gathered}$$

$f\left(x\right)=0+1+2+\dots +\left(n-1\right)+nf$

$\Rightarrow f\left(x\right)=\frac{n(n-1)}{2}+nf$

$=\frac{\left[x\right]\left(\right[x]-1)}{2}+\left[x\right]\left\{x\right\}$

Note $\underset{x\to n^{+}}{\lim }f\left(x\right)=\frac{n\left(n-1\right)}{2}$

$\underset{x\to n^{-}}{\lim }f\left(x\right)=\frac{(n-1)(n-2)}{2}+\left(n-1\right)$

$=\frac{n(n-1)}{2}$

$f\left(x\right)=\frac{n\left(n-1\right)}{2}\left(n\in N_0\right)$

so $f\left(x\right)$ is continuous $\forall x\ge 0$ and differentiable except at integer.'