Mathematics - Continuity - Differentiability Question with Solution | TestHub

P. $f(x)=\{\begin{array}{ll}\frac{e^{1/x}-1}{e^{1/x}+1}& x\ne 0\\ 0& x=0\end{array}$. ${\lim }_{x\to 0^{+}}f(x)={\lim }_{y\to \infty }\frac{e^y-1}{e^y+1}={\lim }_{y\to \infty }\frac{1-e^{-y}}{1+e^{-y}}=1$. ${\lim }_{x\to 0^{-}}f(x)={\lim }_{y\to -\infty }\frac{e^y-1}{e^y+1}=\frac{0-1}{0+1}=-1$. Since the left and right limits at $x=0$ are not equal, $f(x)$ is not continuous at $x=0$. Q. $g(x)=x|x-1|$. This can be written as $g(x)=\{\begin{array}{ll}{x}^2-x& x\ge 1\\ -x^2+x& x<1\end{array}$. It is continuous everywhere. For differentiability at $x=1$, LHD = ${\lim }_{x\to 1^{-}}(-2x+1)=-1$ and RHD = ${\lim }_{x\to 1^{+}}(2x-1)=1$. Since LHD $\ne$ RHD, $g(x)$ is not differentiable at $x=1$. It is differentiable for $x\ne 1$. Thus, $g(x)$ is continuous but not differentiable at exactly one point. R. $h(x)={\int }_0^x(t^2-1{)}^{2}dt$. The integrand $(t^2-1{)}^2$ is a polynomial, and thus continuous for all $t\in \mathbb{R}$. By the Fundamental Theorem of Calculus, $h(x)$ is differentiable for all $x\in \mathbb{R}$, and $h^{\prime }(x)=(x^2-1{)}^2$, which is also a polynomial and hence differentiable for all $x\in \mathbb{R}$. Thus, $h(x)$ is differentiable for all $x\in \mathbb{R}$. S. $k(x)=|x^2-4|$. This function is not differentiable where $x^2-4=0$, i.e., at $x=\pm 2$. It is differentiable for all other $x$. Thus, $k(x)$ is differentiable for all $x\in \mathbb{R}$ except at exactly two points.