Let the functions $f:(-1,1) \rightarrow R$ and $g:(-1,1)$ $\rightarrow(-1,1)$ be defined by $f(x)=|2

Question

Let the functions $f:(-1,1) \rightarrow R$ and $g:(-1,1)$ $\rightarrow(-1,1)$ be defined by $f(x)=|2 x-1|+|2 x+1|$ and $g(x)=x-[x]$, where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ g:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f_{\circ} g$ is NOT continuous, and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f o g$ is not differentiable. Then the value of $c+d$ is

Professor PC Thomas Classes · Answer

(4) : We have, $f(x)=|2 x-1|+|2 x+1|$ and $g(x)=x-[x]=\{x\}$\ $	herefore f(g(x))=f(\{x\})=|2\{x\}-1|+|2\{x\}+1|$\ $\Rightarrow f(g(x))=2\{x\}-1+2\{x\}+1=4\{x\}$,\ if $\{x\}>\frac{1}{2}$\ and $f(g(x))=1-2\{x\}+2\{x\}+1=2$, if $\{x\} \leq \frac{1}{2}$\ Now as $x \in(-1,1)$, we have\ $\{x\} \leq \frac{1}{2} \Rightarrow x \in\left(-1,-\frac{1}{2}ight] \cup\left[0, \frac{1}{2}ight]$\ $\{x\}>\frac{1}{2} \Rightarrow x \in\left(-\frac{1}{2}, 0ight) \cup\left(\frac{1}{2}, 1ight)$\  Note that -1 and 1 are not in the domain. So the point where $f(g(x))$ is discontinuous is $x=0$.\ $	herefore \quad c=1$\ $f(g(x))$ is not differentiable at 3 points :\ $-\frac{1}{2}, 0, \frac{1}{2}$.\ $	herefore \quad d=3$. Thus $c+d=4$\