Question

Consider the function $f\left(x\right)=2\sqrt{6-\left[x\right]}$ at $2<x\le 3$ , where $\left[x\right]$ denotes step up function, then at $x=2$ the function

• A. has missing point removable discontinuity
• B. has isolates missing point removable discontinuity
• C. has non removable discontinuity finite type
• D. is continuous

Answer 1

Answer: (D)

is continuous

For functions defined on open intervals $(a,b)$, continuity can be checked by

$1.$ RHL $=f\left(a\right)$

$2.$ LHL$=f\left(b\right)$

there we need to show

RHL$=\underset{x\to 2^{+}}{lim}f\left(x\right)=f\left(2\right)$

$f\left(2\right)=2\sqrt{6-\left[2\right]}=2\sqrt{4}=4$

$\underset{x\to 2^{+}}{lim}f\left(x\right)=2\sqrt{6-2}=4$

as $x\to 2^{+},\left[x\right]=2$

Hence, it is continuous.