Let f(x) be defined in the interval [0,4] such that $\mathrm{f}\left(\mathrm{x}\right)$ $=\left\{\begin{array}{l}1-\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le \mathrm{x}\le 1\\ \mathrm{x}+2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1<\mathrm{x}<2\\ 4-\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le \mathrm{x}\le 4\end{array}\right.$ Then the number of points where f(f (x)) is discontinues is

f (x) is discontinuous at x = 1 and x = 2. therefore, f ( f (x) may be discontinuous when
f (x) = 1 or Now,
1 -x = 1$\Rightarrow$x = 0, where f (x) is continuous
x + 2 = 1$\Rightarrow$ x = -1$\notin$(1,2)
4 - x = 1$\Rightarrow$ x = 3 $\notin$[2,4]
Now, 1 - x = 2 $\Rightarrow$ x = -1$\notin$[0,1]
x + 2 = 2$\Rightarrow$ x = 0 $\notin$(0,1]
4 - x = 2$\Rightarrow$ x = 2 $\notin$[2,4]
Hence, f(f(x) is discontinuous at x = 2,3