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(B) Given the function $f(x) = \begin{cases} [x], & 	ext{if } x < 2 \ [x]-1, & 	ext{if } x \geq 2 \end{cases}$.First,check continuity at $x = 2$:$f

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$[1,3)$

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(B) Given the function $f(x) = \begin{cases} [x], & 	ext{if } x First,check continuity at $x = 2$:$f(2) = [2] - 1 = 2 - 1 = 1$.Left-hand limit $(LHL)$ at $x = 2$:$\lim_{x 	o 2^-} f(x) = \lim_{h 	o 0} [2-h] = 1$.Right-hand limit $(RHL)$ at $x = 2$:$\lim_{x 	o 2^+} f(x) = \lim_{h 	o 0} ([2+h] - 1) = 2 - 1 = 1$.Since $LHL = RHL = f(2)$,the function is continuous at $x = 2$.For $x \in [1, 2)$,$f(x) = [x]$. The greatest integer function $[x]$ is continuous on any interval $[n, n+1)$ where $n \in \mathbb{Z}$. Thus,$f(x)$ is continuous on $[1, 2)$.For $x \in [2, 3)$,$f(x) = [x] - 1$. Similarly,this is continuous on $[2, 3)$.Since the function is also continuous at the junction $x = 2$,$f(x)$ is continuous on the interval $[1, 3)$.

If $[x]$ denotes the greatest integer not exceeding the number $x$,then $f(x)$ defined by $f(x) = \begin{cases} [x], & \text{if } x < 2 \\ [x]-1, & \text{if } x \geq 2 \end{cases}$ is continuous in the interval.

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