Mathematics - Continuity - Differentiability Question with Solution | TestHub

The function $f(x)$ is defined as the maximum of $\sin x$ and $\cos x$. $f(x)$ is not differentiable at points where $\sin x=\cos x$ and their derivatives are unequal. In $(0,2\pi )$, $\sin x=\cos x$ at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$. For $x\in (0,\frac{\pi }{4})$, $f(x)=\cos x$. For $x\in (\frac{\pi }{4},\frac{5\pi }{4})$, $f(x)=\sin x$. For $x\in (\frac{5\pi }{4},2\pi )$, $f(x)=\cos x$. At $x=\frac{\pi }{4}$: $f^{\prime }({\frac{\pi }{4}}^{-})=-\sin (\frac{\pi }{4})=-\frac{1}{\sqrt{2}}$ and $f^{\prime }({\frac{\pi }{4}}^{+})=\cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$. Since these are not equal, $f(x)$ is not differentiable at $x=\frac{\pi }{4}$. At $x=\frac{5\pi }{4}$: $f^{\prime }({\frac{5\pi }{4}}^{-})=\cos (\frac{5\pi }{4})=-\frac{1}{\sqrt{2}}$ and $f^{\prime }({\frac{5\pi }{4}}^{+})=-\sin (\frac{5\pi }{4})=-(-\frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}$. Since these are not equal, $f(x)$ is not differentiable at $x=\frac{5\pi }{4}$. Thus, there are 2 points in $(0,2\pi )$ where $f(x)$ is not differentiable.