Question

Let $f\left(x\right)=\left[x\right]+\sqrt{x-\left[x\right]}$, where [x] denotes the greatest integer function. Then

• A. f(x) is continuous on $R^{+}$
• B. f(x) is continuous on R
• C. f(x) is continuous on R - l
• D. f(x) is continuous on $(R-l)\cup \left\{0\right\}$

Answer 1

The correct option is B

f(x) is continuous on R

$f\left(x\right)=\left[x\right]+\sqrt{x-\left[x\right]}$

$\because \left[x\right]\le x\Rightarrow x-\left[x\right]\ge o$

$\therefore \text{Domain of f(x) is R}$

For continuity, crtical points are the integer values with greatest integer function.

$\therefore f\left(x\right)=\left[x\right]+\sqrt{I-\left[I\right]}$

$=I+\sqrt{I-I}$

$=I$

$\left(ii\right)Letx=I+f;wheref>0\&f\to o$

$\therefore f\left(x\right)=[I+f]+\sqrt{I+f-[I+f]}$

$=I+\sqrt{f}$

$\because f\left(x\right)\to I$

$\left(iii\right)Letx=I-f;wheref>o\&f\to o$

$f\left(x\right)=[I-f]+\sqrt{I-f-[I+1]}$

$=I-1+\sqrt{\text{/}I-f-\text{/}I+1}$

$=I-1+\sqrt{1-f}$

$\because f\to o;1-f\to 1\Rightarrow \sqrt{1-f}\to 1$

$\therefore f\left(x\right)\to I-1+1=I$

$\therefore$ f(x) leads to I for the neighbour hood value of I always

Hence f(x) is continuous $\forall xϵR$