Mathematics - Continuity - Differentiability Question with Solution | TestHub

P. $f(x)=|x^2-1|+|x^2-4|$. The function is not differentiable where $x^2-1=0$ or $x^2-4=0$. These are $x=\pm 1$ and $x=\pm 2$. All four points are distinct. So, $f(x)$ is not differentiable at exactly four points. Q. $g(x)=\max (x,x^2)$. This function is $x^2$ for $x\le 0$ or $x\ge 1$, and $x$ for $0<x<1$. It is not differentiable at $x=0$ (LHD=0, RHD=1) and $x=1$ (LHD=1, RHD=2). So, $g(x)$ is not differentiable at exactly two points. R. $h(x)=\{\begin{array}{ll}{x}^2\sin \left(\frac{1}{x}\right)& x\ne 0\\ 0& x=0\end{array}$. $h^{\prime }(0)={\lim }_{t\to 0}\frac{t^2\sin (1/t)}{t}={\lim }_{t\to 0}t\sin (1/t)=0$. So $h(x)$ is differentiable at $x=0$. For $x\ne 0$, $h^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. ${\lim }_{x\to 0}{h}^{\prime }(x)$ does not exist due to the $\cos (1/x)$ term. Thus, $h^{\prime }(x)$ is not continuous at $x=0$. S. $k(x)=\{\begin{array}{ll}{x}^3& x\in \mathbb{Q}\\ x^2& x\notin \mathbb{Q}\end{array}$. $k(x)$ is continuous only where $x^3=x^2$, i.e., $x=0$ or $x=1$. For differentiability, it must be continuous. At $x=0$, $k^{\prime }(0)={\lim }_{h\to 0}\frac{k(h)-k(0)}{h}$. If $h\in \mathbb{Q}$, $h^2\to 0$. If $h\notin \mathbb{Q}$, $h\to 0$. So $k^{\prime }(0)=0$. At $x=1$, LHD = ${\lim }_{h\to 0}\frac{(1+h{)}^3-1}{h}=3$ (if $1+h\in \mathbb{Q}$) and RHD = ${\lim }_{h\to 0}\frac{(1+h{)}^2-1}{h}=2$ (if $1+h\notin \mathbb{Q}$). So not differentiable at $x=1$. Therefore, $k(x)$ is differentiable only at $x=0$.