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(B) Given the condition $|f(x)-f(y)|=|x-y|$ for all $x, y \in [0,1]$.This implies that the slope of the function $f$ must be $\pm 1$,i.e.,$f'(x) = 1$

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exactly $2$

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(B) Given the condition $|f(x)-f(y)|=|x-y|$ for all $x, y \in [0,1]$.This implies that the slope of the function $f$ must be $\pm 1$,i.e.,$f'(x) = 1$ or $f'(x) = -1$.Case $1$: If $f'(x) = 1$,then $f(x) = x + c$. Since the codomain is $[0,1]$,for $f:[0,1] \rightarrow [0,1]$,we must have $f(0) \ge 0$ and $f(1) \le 1$. Thus,$0+c \ge 0$ and $1+c \le 1$,which gives $c=0$. So,$f(x) = x$.Case $2$: If $f'(x) = -1$,then $f(x) = -x + c$. For $f:[0,1] \rightarrow [0,1]$,we must have $f(0) \le 1$ and $f(1) \ge 0$. Thus,$0+c \le 1$ and $-1+c \ge 0$,which gives $c=1$. So,$f(x) = 1-x$.Therefore,there are exactly $2$ such functions: $f(x) = x$ and $f(x) = 1-x$.

The number of functions $f:[0,1] \rightarrow [0,1]$ satisfying $|f(x)-f(y)|=|x-y|$ for all $x, y \in [0,1]$ is

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