Mathematics - Continuity - Differentiability Question with Solution | TestHub

We analyze $f(x)$ based on the value of $|x|$. If $|x|<1$, $f(x)=x^2$. If $|x|>1$, $f(x)=\sin (\pi x)$. If $x=1$, $f(1)=\frac{1^{2n}\sin (\pi )+1^2}{1^{2n}+1}=\frac{1}{2}$. If $x=-1$, $f(-1)=\frac{(-1{)}^{2n}\sin (-\pi )+(-1{)}^2}{(-1{)}^{2n}+1}=\frac{1}{2}$. At $x=1$: ${\lim }_{x\to 1^{-}}f(x)=1$, ${\lim }_{x\to 1^{+}}f(x)=0$. Since $f(1)=1/2$ and the limits are different, $f(x)$ is discontinuous at $x=1$. At $x=-1$: ${\lim }_{x\to -1^{-}}f(x)=0$, ${\lim }_{x\to -1^{+}}f(x)=1$. Since $f(-1)=1/2$ and the limits are different, $f(x)$ is discontinuous at $x=-1$. Thus, there are 2 points of discontinuity in $[-2,2]$, namely $x=1$ and $x=-1$.