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If $[x]$ denotes the greatest integer not exceeding the number $x$, then $f(x)$ defined by
$f(x)=\left\{\begin{array}{ll}{[x],} & \text { if } x < 2 \\ {[x]-1,} & \text { if } x \geq 2\end{array}\right.$ is continuous in the interval.
Options:
$f(x)=\left\{\begin{array}{ll}{[x],} & \text { if } x < 2 \\ {[x]-1,} & \text { if } x \geq 2\end{array}\right.$ is continuous in the interval.
Solution:
1849 Upvotes
Verified Answer
The correct answer is:
$[1,3)$
Given function,
$$
\begin{aligned}
& \qquad f(x)=\left\{\begin{aligned}
{[x], } & \text { if } x < 2 \\
{[x]-1, } & \text { if } x \geq 2
\end{aligned}\right. \\
& \text { At } \quad x=2, f(2)=1 \\
& \text { LHL } \\
& \qquad(\text { at } x=2)=\lim _{h \rightarrow 0}[2-h]=1 \\
& \text { and RHL (at } x=2)=\lim _{h \rightarrow 0}([2+h]-1)=2-1=1
\end{aligned}
$$
So, $f(x)$ is continuous at $x=2$
And $[x]$ is continuous in $[n, n+1), \forall n \in$ integer.
So, given function continuous in the interval $[1,3)$.
$$
\begin{aligned}
& \qquad f(x)=\left\{\begin{aligned}
{[x], } & \text { if } x < 2 \\
{[x]-1, } & \text { if } x \geq 2
\end{aligned}\right. \\
& \text { At } \quad x=2, f(2)=1 \\
& \text { LHL } \\
& \qquad(\text { at } x=2)=\lim _{h \rightarrow 0}[2-h]=1 \\
& \text { and RHL (at } x=2)=\lim _{h \rightarrow 0}([2+h]-1)=2-1=1
\end{aligned}
$$
So, $f(x)$ is continuous at $x=2$
And $[x]$ is continuous in $[n, n+1), \forall n \in$ integer.
So, given function continuous in the interval $[1,3)$.
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