Answer:
Option A.
Step-by-step explanation:
The given piecewise function is
[tex]f(x)=\begin{cases}-\left(x+1\right)^{2}+2 & \text{ if } x<-1 \\ -x+2 & \text{ if } -1\le x<2 \\ \sqrt{x-1\ } & \text{ if } x\ge2 \end{cases}[/tex]
From the given function it is clear that function is divided at x=-1 and x=2. It means we check the discontinuity at x=-1 and x=2.
For x=-1,
[tex]f(-1)=-(-1)+2=3[/tex]
LHL: [tex]lim_{x\rightarrow -1^-}f(-1)=-\left(-1+1\right)^{2}+2=2[/tex]
Since LHL ≠ f(-1), therefore the given function is discontinuous at x=-1.
For x=2,
[tex]f(2)=\sqrt{2-1\ }=1[/tex]
LHL: [tex]lim_{x\rightarrow 2^-}f(2)=-(2)+2=0[/tex]
Since LHL ≠ f(2), therefore the given function is discontinuous at x=2.
Therefore, the correct option is A.
