Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

(A) Consider the function $f(x) = |x| + |x - 1|$.
Since the modulus function is continuous everywhere and the sum of two continuous functions is also continuous,$f(x)$ is continuous for all $x \in \mathbb{R}$.
To check the differentiability of $f(x)$,we examine the points where the expression inside the modulus changes sign,which are $x = 0$ and $x = 1$.
At $x = 0$:
Left-hand derivative $(LHD)$ $= \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{(|x| + |x - 1|) - 1}{x} = \lim_{x \to 0^-} \frac{(-x - x + 1) - 1}{x} = \lim_{x \to 0^-} \frac{-2x}{x} = -2$.
Right-hand derivative $(RHD)$ $= \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{(|x| + |x - 1|) - 1}{x} = \lim_{x \to 0^+} \frac{(x - x + 1) - 1}{x} = \lim_{x \to 0^+} \frac{0}{x} = 0$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = 0$.
At $x = 1$:
$LHD$ $= \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^-} \frac{(|x| + |x - 1|) - 1}{x - 1} = \lim_{x \to 1^-} \frac{(x - x + 1) - 1}{x - 1} = \lim_{x \to 1^-} \frac{0}{x - 1} = 0$.
$RHD$ $= \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{(|x| + |x - 1|) - 1}{x - 1} = \lim_{x \to 1^+} \frac{(x + x - 1) - 1}{x - 1} = \lim_{x \to 1^+} \frac{2x - 2}{x - 1} = 2$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = 1$.
Thus,$f(x) = |x| + |x - 1|$ is continuous everywhere but not differentiable at exactly two points,$x = 0$ and $x = 1$.