Mathematics - Continuity - Differentiability Question with Solution | TestHub

Doubtful points : $-1,0,1,\sqrt{2},\sqrt{3},2$ at $x=\sqrt{2},\sqrt{3}$ $\begin{array}{c}f(x)=\left(2x^2+x-[x]\right)+\left[x^2\right]=\text{Discount}\\ \downarrow \downarrow \\ \text{Cont.}\text{Cont.}\end{array}$
at $x=-1$ : $\begin{array}{rl}\text{ RHL }\Rightarrow & f(x)=(2-1-(-1))+0=2\\ & f(-1)=2-1-(-1)+1=3\end{array}\}$ Dis. at $x=2$ : $\begin{array}{rl}\text{LHL }\Rightarrow f(x)& =8+2-1+3=12\\ f(2)& =8+2-2+4=12\end{array}\}\text{ Cont. }$ at $\mathrm{x}=0$ : $\begin{array}{rl}\text{LHL }\Rightarrow & 0+0-(-1)+0=1\\ & \mathrm{f}(0)=0\end{array}\}\text{ Dis. }$ at $\mathrm{x}=1$ $\begin{array}{rl}\text{LHL }\Rightarrow & 2+1-0+0=3\\ & f(1)=3-1+1=3\\ \text{RHL }\Rightarrow & 2+1-1+1=3\end{array}\}\text{ Cont. }$