Question

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

Answer 1

\Step 1: Finding the value of $g\left(x\right)$

$$\begin{gathered} f\left(x\right)=|2x-1|+|2x+1| \\ f\left(x\right)=-4xforx\le -\frac{1}{2} \\ f\left(x\right)=2for-\frac{1}{2}<x<\frac{1}{2} \\ f\left(x\right)=4xforx\ge \frac{1}{2} \end{gathered}$$

Also,

$\begin{array}{rcl}g\left(x\right)& =& x-\left[x\right]\\ & =& \left\{x\right\}\end{array}$

Step 2: Finding $f\circ g\left(x\right)$

$\begin{array}{rcl}f\circ g\left(x\right)& =& \left\{\begin{array}{cc}-4g\left(x\right)& g\left(x\right)\le \frac{-1}{2}\\ 2& \frac{-1}{2}<g\left(x\right)<\frac{1}{2}\\ 4g\left(x\right)& g\left(x\right)\ge \frac{1}{2}\end{array}\right.\\ & =& \left\{\begin{array}{cc}2& -1<x<\frac{-1}{2}\\ 4\left\{x\right\}& \frac{-1}{2}\le x<0\\ 2& 0\le x<\frac{1}{2}\\ 4\left\{x\right\}& \frac{1}{2}\le x<1\end{array}\right.\\ & =& \left\{\begin{array}{cc}2& -1<x<\frac{-1}{2}\\ 4\left(x+1\right)& \frac{-1}{2}\le x<0\\ 2& 0\le x<\frac{1}{2}\\ 4x& \frac{1}{2}\le x<1\end{array}\right.\end{array}$

Step 3: Checking the solution

$f\circ g\left(x\right)$ is not continuous at$x=0$ only

$\Rightarrow c=1$

$f\circ g\left(x\right)$ is not differentiable at $x=(-\frac{1}{2}),0,\left(\frac{1}{2}\right),\Rightarrow d=3$

Substituting the value of $c\text{and}d$, we get;

So, $c+d=4$

Therefore, The value of $c+d$ is $4$.