Subject: Mathematics — Differential Calculus
Question: Let the functions $f:( - 1,1) \rightarrow \mathbb{R}$
and $g:( - 1,1) \rightarrow ( - 1,1)$ be defined by\
$f(x) = |2x - 1| + |2x + 1|$ and $g(x) = x - \lbrack x\rbrack$, Where
$\lbrack x\rbrack$ denotes the greatest integer less than or equal to
$x$. Let $fog:( - 1,1) \rightarrow \mathbb{R}$ be the composite function
detined 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 $f \circ g$ is NOT differentiable. Then the value of
$c + d$ is $\_\_\_\_$
Answer: 4
Solution: Given that $f(x) = |2x - 1| + |2x + 1|$ and
$g(x) = x - \lbrack x\rbrack = \{ x\}$\
$${\therefore\ (fog)(x) = |2\{ x\} - 1| + |2\{ x\} + 1|
}{\Rightarrow \ (fog)(x) = \left\{ \begin{matrix}
|2x + 1| + |2x + 3|,x \in \left( - 1, - \frac{1}{2} \right\rbrack \\
|2x + 1| + |2x + 3|,x \in \left( - \frac{1}{2},0 \right\rbrack \\
|2x - 1| + |2x + 1|,x \in \left( 0,\frac{1}{2} \right\rbrack \\
|2x - 1| + |2x + 1|,x \in \left( \frac{1}{2},1 \right\rbrack
\end{matrix} \right.\ 
}{\Rightarrow \ (fog)(x) = \left\{ \begin{matrix}
2,x \in \left\lbrack - 1, - \frac{1}{2} \right) \\
4x + 4,x \in \left( - \frac{1}{2},0 \right\rbrack \\
2,x \in \left( 0,\frac{1}{2} \right\rbrack \\
4x,x \in \left( \frac{1}{2},1 \right)
\end{matrix} \right.\ }$$
[IMAGE] $\therefore\ f(g(x))$ is discontinuous at $x = 0$.\
$$\therefore\ c = 1
$$Now, $(fog)^{'}(x) = \left\{ \begin{matrix}
0,x \in \left( - 1,\frac{1}{2} \right) \\
4,x \in \left( - \frac{1}{2},0 \right) \\
0,x \in \left( 0,\frac{1}{2} \right) \\
4,x \in \left( \frac{1}{2},1 \right)
\end{matrix} \right.\ $\
$\therefore f(g(x))$ is non-differentiable at
$x = - \frac{1}{2},0,\frac{1}{2}$\
$$\therefore\ d = 3
$$Hence, $c + d = 4$