Question

Let $f(x)=\{\begin{array}{ll}\frac{\sin (x-[x])}{x-[x]},& x\in (-2,-1)\\ \max \{2x,3[|x|]\}& |x|<1\\ 1& \text{ otherwise }\end{array}$
where [ $t$ ] denotes greatest integer $\le t$. If $m$ is the number of points where $f$ is not continuous and $n$ is the number of points where $f$ is not differentiable, then the ordered pair $(m,n)$ is :

• A. $(3,3)$
• B. $(2,4)$
• C. $(2,3)$
• D. $(3,4)$

Answer 1

Answer: (C)

$f(x)=\{\begin{array}{ll}\frac{\sin (x+2)}{x+2},& x\in (-2,-1)\\ \max \{2x,0\}& x\in (-1,1)\\ 1& \text{ otherwise }\end{array}$
$f\left(-2^{+}\right)=\underset{h\to 0}{\lim }f(-2+h)=\underset{h\to 0}{\lim }\frac{\sinh }{h}=1$
$f$ is continuous at $x=-2$
$f\left(-1^{-}\right)=\underset{h\to 0}{\lim }\frac{\sin (-1-h+2)}{(-1-h+2)}=\sin 1$
$f(-1)=f\left(-1^{+}\right)=0$
$f\left(1^{+}\right)=1\&f\left(1^{-}\right)=0\Rightarrow f$ is not continuous at $x=1$
$f$ is continuous but not diff. at $x=0$