Mathematics - Continuity - Differentiability Question with Solution | TestHub

For continuity at an integer $k$: $f(k)=[k]\sin (\pi k)=k\cdot 0=0$. ${\lim }_{x\to k^{-}}f(x)={\lim }_{x\to k^{-}}[x]\sin (\pi x)=(k-1)\sin (\pi k)=(k-1)\cdot 0=0$. ${\lim }_{x\to k^{+}}f(x)={\lim }_{x\to k^{+}}[x]\sin (\pi x)=k\sin (\pi k)=k\cdot 0=0$.

Since ${\lim }_{x\to k^{-}}f(x)={\lim }_{x\to k^{+}}f(x)=f(k)=0$, $f(x)$ is continuous at all integers. For non-integers, $[x]$ is constant and $\sin (\pi x)$ is continuous, so $f(x)$ is continuous. Thus, $f(x)$ is continuous everywhere. For differentiability at an integer $k$:$f^{\prime }(k^{-})={\lim }_{h\to 0^{-}}\frac{[k+h]\sin (\pi (k+h))-0}{h}={\lim }_{h\to 0^{-}}\frac{(k-1)\sin (\pi k+\pi h)}{h}={\lim }_{h\to 0^{-}}\frac{(k-1)(-1{)}^k\sin (\pi h)}{h}=(k-1)(-1{)}^k\pi$. $f^{\prime }(k^{+})={\lim }_{h\to 0^{+}}\frac{[k+h]\sin (\pi (k+h))-0}{h}={\lim }_{h\to 0^{+}}\frac{k\sin (\pi k+\pi h)}{h}={\lim }_{h\to 0^{+}}\frac{k(-1{)}^k\sin (\pi h)}{h}=k(-1{)}^k\pi$. For differentiability, $f^{\prime }(k^{-})=f^{\prime }(k^{+})$ must hold, so $(k-1)(-1{)}^k\pi =k(-1{)}^k\pi$. This implies $k-1=k$, which is false. Thus, $f(x)$ is not differentiable at any integer.