There are two functions, $f$ and $g$ such that $f:( - 1,1) \to R$ and $g:( - 1,1) \to ( - 1,1)$ . They are defined by $f(x) = \left| {2x - 1} \right| + \left| {2x + 1} \right|$ and $g(x) = x - \left[ x \right]$ . Here, $\left[ x \right]$ denotes the greatest integer function. Now, consider the composite function $fog$ defined by $f(g(x))$ . Let’s say $C$ denotes the number of points in (-1,1) at which $fog$ is not continuous and $D$ denotes the number of points at which $fog$ is not differentiable. Find out the value of $C + D$.

Hint: For a function $f(x)$ to be continuous at a point it is necessary for it to be continuous. It can also be said that every differentiable function is continuous, but it is not necessary for every continuous function to be differentiable as the function can take a sharp turn at ant any point, hence, making it non-differentiable.

Complete step by step solution: 
Consider the function $f(x) = \left| {2x - 1} \right| + \left| {2x + 1} \right|$ .
For $2x - 1 = 0$ , we get $x = \dfrac{1}{2}$ .
Similarly, for $2x + 1 = 0$ , we get $x = - \dfrac{1}{2}$ .
Now, consider the function $g(x) = x - \left[ x \right]$ , where $\left[ x \right]$ is the greatest integer function.
Also $x - \left[ x \right] = \left\{ x \right\}$ , where $\left\{ x \right\}$ is the fractional part function.
Therefore, $g(x) = \left\{ x \right\}$ . … (2)
Now, consider $fog$ which is defined as $f(g(x))$ .
For the interval $( - 1, - \dfrac{1}{2})$ ,
$f(g(x)) = \left| {2\left\{ x \right\} - 1} \right| + \left| {2\left\{ x \right\} + 1} \right|$
Removing the moduli, we get:
$f(g(x)) = - \left( {2\left\{ x \right\} - 1} \right) + \left( {2\left\{ x \right\} + 1} \right)$ as $\{ x\} \leqslant \dfrac{1}{2}$ which makes $2\{ x\} \leqslant 1$ .
On simplifying,
$f(g(x)) = 2,\forall x \in \left( { - 1, - \dfrac{1}{2}} \right)$
Similarly, for the interval $\left[ { - \dfrac{1}{2},0} \right)$ ,
$f(g(x)) = \left| {2\left\{ x \right\} - 1} \right| + \left| {2\left\{ x \right\} + 1} \right|$
Removing the moduli, we get:
$f(g(x)) = \left( {2\left\{ x \right\} - 1} \right) + \left( {2\left\{ x \right\} + 1} \right)$ as $\{ x\} \geqslant \dfrac{1}{2}$ which makes $2\{ x\} \geqslant 1$ .
On simplifying,
$f(g(x)) = 4x,\forall x \in \left[ { - \dfrac{1}{2},0} \right)$
Similarly, for the interval $\left[ {0,\dfrac{1}{2}} \right)$ ,
$f(g(x)) = \left| {2\left\{ x \right\} - 1} \right| + \left| {2\left\{ x \right\} + 1} \right|$
Removing the moduli, we get:
$f(g(x)) = - \left( {2\left\{ x \right\} - 1} \right) + \left( {2\left\{ x \right\} + 1} \right)$ as $\{ x\} \leqslant \dfrac{1}{2}$ which makes $2\{ x\} \leqslant 1$ .
On simplifying,
$f(g(x)) = 2,\forall x \in \left[ {0,\dfrac{1}{2}} \right)$
Similarly, for the interval $\left[ {\dfrac{1}{2},1} \right)$ ,
$f(g(x)) = \left| {2\left\{ x \right\} - 1} \right| + \left| {2\left\{ x \right\} + 1} \right|$
Removing the moduli, we get:
$f(g(x)) = \left( {2\left\{ x \right\} - 1} \right) + \left( {2\left\{ x \right\} + 1} \right)$ as $\{ x\} \geqslant \dfrac{1}{2}$ which makes $2\{ x\} \geqslant 1$ .
On simplifying,
$f(g(x)) = 4x,x \in \left[ {\dfrac{1}{2},1} \right)$
Hence, the function $f(g(x))$ can be defined as:
$f(x) = \left\{
  2,x \in \left( { - 1, - \dfrac{1}{2}} \right) \\
  4\{ x\} ,x \in \left[ { - \dfrac{1}{2},0} \right) \\
  2,x \in \left( {0,\dfrac{1}{2}} \right) \\
  4\{ x\} ,x \in \left[ {\dfrac{1}{2},1} \right) \\
   \right.$ … (1)
Now, we’ll check the continuity of the composite function at $x = - \dfrac{1}{2}$ .
For $x = - \dfrac{1}{2}$ ,
$\mathop {\lim }\limits_{x \to - {{\dfrac{1}{2}}^ - }} fog = \mathop {\lim }\limits_{x \to - {{\dfrac{1}{2}}^ - }} 2 = 2$
And
$\mathop {\lim }\limits_{x \to - {{\dfrac{1}{2}}^ + }} fog = \mathop {\lim }\limits_{x \to - {{\dfrac{1}{2}}^ + }} 4\{ x\} = 4 \times \dfrac{1}{2} = 2$
As both the left-hand and right-hand limits are equal, $f(g(x))$ is continuous at $x = - \dfrac{1}{2}$ .
Similarly, it was found that the composite function is continuous at $x = \dfrac{1}{2}$ .
Now, we’ll check the continuity of the composite function at $x = 0$ .
For $x = 0$ ,
$\mathop {\lim }\limits_{x \to {0^ - }} fog = \mathop {\lim }\limits_{x \to {0^ - }} 4\{ x) = 0$
And
$\mathop {\lim }\limits_{x \to {0^ + }} fog = \mathop {\lim }\limits_{x \to {0^ + }} 2 = 2$
As the left-hand limit is not equal to the right-hand limit, $f(g(x))$ is not continuous at $x = 0$ . … (1)
Thus, the composite function $fog$ is not continuous at one point, that is at $x = 0$ respectively.
Therefore, $C = 1$ . … (2)
Now, we’ll check the differentiability of the composite function at $x = - \dfrac{1}{2}$ .
For checking the differentiability of a function at a point $x = a$ , we check the existence of the following limit:
$f'(a) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$
For $a = - \dfrac{1}{2}$ , checking the differentiability of the composite function $fog(x)$ ,
$fog'\left( { - \dfrac{1}{2}} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h}$
Calculating the left-hand limit of the above derivative,
$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{4\left\{ { - \dfrac{1}{2} + h} \right\} - 4\left\{ { - \dfrac{1}{2}} \right\}}}{h}$
Using the property of the fractional part function,
$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{2 - 4\left( {\dfrac{1}{2}} \right)}}{h}$
On simplifying further,
$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{2 - 2}}{h} = 0$
Calculating the right-hand limit of the above derivative,
$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{4\left\{ { - \dfrac{1}{2} + h} \right\} - 4\left\{ { - \dfrac{1}{2}} \right\}}}{h}$
Using the property of the fractional part function,
$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{4\left( { - \dfrac{1}{2} + h - ( - 1)} \right) - 4\left( {\dfrac{1}{2}} \right)}}{h}$
On simplifying further,
$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{fog\left( { - \dfrac{1}{2} + h} \right) - fog\left( { - \dfrac{1}{2}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{2 + 4h - 2}}{h} = 4$
As the left-hand limit is not equal to the right-hand limit, $f(g(x))$ is not differentiable at $x = - \dfrac{1}{2}$ .
Similarly, the composite function is not differentiable at $x = \dfrac{1}{2}$ .
Using equation (1), as the composite function is not continuous $x = 0$ , therefore, the function is also not differentiable at that point.
Thus, the function $fog$ is not differentiable at three points, that is, at $x = \pm \dfrac{1}{2},0$ respectively.
Therefore, $D = 3$ . … (3)
Using equations (2) and (3), $C + D = 1 + 3 = 4$ .
Therefore, $C + D = 4$ .

Note: The above question can be easily solved on drawing the composite function $fog(x)$ after defining it. It is observed that the composite function breaks at the point, $x = 0$ which makes it both non-continuous and non-differentiable at that point. Also, the function is not differentiable at a point at which it takes a sharp turn. It is also observed that the composite function takes a sharp turn at two points, $x = \pm \dfrac{1}{2}$ respectively, making it non-differentiable at that point.