Mathematics - Continuity - Differentiability Question with Solution | TestHub

We analyze the function $f(x)$ based on the magnitudes of $x$ and $x-1$:

Case 1: $|x|>|x-1|$. This implies $x^2>(x-1{)}^2⟹x^2>x^2-2x+1⟹2x>1⟹x>1/2$. In this case, divide numerator and denominator by $x^n$. As $n\to \infty$, ${\left(\frac{x-1}{x}\right)}^n\to 0$. So, $f(x)=\frac{\sin (\pi x)+0\cdot \cos (\pi x)}{1+0}=\sin (\pi x)$ for $x>1/2$.

Case 2: $|x-1|>|x|$. This implies $(x-1{)}^2>x^2⟹x^2-2x+1>x^2⟹-2x+1>0⟹x<1/2$. In this case, divide numerator and denominator by $(x-1{)}^n$. As $n\to \infty$, ${\left(\frac{x}{x-1}\right)}^n\to 0$. So, $f(x)=\frac{0\cdot \sin (\pi x)+\cos (\pi x)}{0+1}=\cos (\pi x)$ for $x<1/2$. Case 3: $|x|=|x-1|$. This implies $x=1/2$. Substituting $x=1/2$ into the expression for $f(x)$:$f(1/2)={\lim }_{n\to \infty }\frac{(1/2{)}^n\sin (\pi /2)+(-1/2{)}^n\cos (\pi /2)}{(1/2{)}^n+(-1/2{)}^n}={\lim }_{n\to \infty }\frac{(1/2{)}^n\cdot 1+(-1/2{)}^n\cdot 0}{(1/2{)}^n+(-1/2{)}^n}={\lim }_{n\to \infty }\frac{(1/2{)}^n}{(1/2{)}^n+(-1/2{)}^n}$. If $n$ is even, $f(1/2)={\lim }_{n\to \infty }\frac{(1/2{)}^n}{2(1/2{)}^n}=1/2$. If $n$ is odd, $f(1/2)={\lim }_{n\to \infty }\frac{(1/2{)}^n}{0}$. The limit does not exist. Thus, $f(1/2)$ is undefined. The function $f(x)$ is defined as:$$f(x)=\{\begin{array}{ll}\cos (\pi x)& x<1/2\\ \sin (\pi x)& x>1/2\\ \text{undefined}& x=1/2\end{array}$$ We check for continuity at $x=1/2$: LHL at $x=1/2$: ${\lim }_{x\to (1/2{)}^{-}}f(x)={\lim }_{x\to (1/2{)}^{-}}\cos (\pi x)=\cos (\pi /2)=0$. RHL at $x=1/2$: ${\lim }_{x\to (1/2{)}^{+}}f(x)={\lim }_{x\to (1/2{)}^{+}}\sin (\pi x)=\sin (\pi /2)=1$.

Since LHL $\ne$ RHL, $f(x)$ is discontinuous at $x=1/2$. The function is continuous for $x<1/2$ and $x>1/2$ as $\cos (\pi x)$ and $\sin (\pi x)$ are continuous functions. Therefore, there is only 1 point of discontinuity.