Question

If $f\left(x\right)=\left[x\right]+[x+\frac{1}{3}]+[x+\frac{2}{3}]$, then $([.]$ denotes the greatest integer function)

• A. $f\left(x\right)$ is discontinuous at $x=1,10,15$
• B. $\underset{x\to \frac{2}{3}}{lim}f\left(x\right)=2$
• C. ${\int }_0^{2/3}f\left(x\right)dx=\frac{1}{3}$
• D. $f\left(x\right)$ is continuous at $x=\frac{n}{3}$, where $n$ is any integer

Answer 1

Answer: (C)

${\int }_0^{2/3}f\left(x\right)dx=\frac{1}{3}$

$\left(A\right)$ Let $x=I$ where $I$ is an integer.

At $x=I$, $f(x=I)=\left[I\right]+[I+\frac{1}{3}]+[I+\frac{2}{3}]$

Now $[I+\frac{1}{3}]=I,[I+\frac{2}{3}]=I$ $(\because [x+h]=x$ when $hϵ[0,1))$

$\therefore f(x=I)=I+I+I=3I$

Now for small $h$ tends to $0^{+}$ ,$x=I-h\to f(x=I-h)=[I-h]+[I-h+\frac{1}{3}]+[I-h+\frac{2}{3}]$

$f(x=I-h)=(I-1)+\left(I\right)+\left(I\right)=3I-1$ $(\because [x-h]=x$ when $hϵ[0,1))$

Hence, $f\left(x\right)$ is discontinuous at all integers.

$\left(B\right)$ At $x\to \frac{2}{3}-h$, $\underset{x\to \frac{2}{3}-h}{lim}f\left(x\right)=[\frac{2}{3}-h]+[\frac{2}{3}-h+\frac{1}{3}]+[\frac{2}{3}-h+\frac{2}{3}]=0+0+1=1$

At $x\to \frac{2}{3}+h$, $\underset{x\to \frac{2}{3}+h}{lim}f\left(x\right)=[\frac{2}{3}+h]+[\frac{2}{3}+h+\frac{1}{3}]+[\frac{2}{3}+h+\frac{2}{3}]=0+1+1=2$

As Left limit and Right limit not same therefore discontinuous at $x=\frac{2}{3}$

Hence not (B)

$\left(C\right)$ Similar to $\left(B\right)$ we can see $f\left(x\right)$ is discontinuous at $x=\frac{1}{3},$

So, we need to break integral into two parts

$\Rightarrow$ ${\int }_0^{2/3}f\left(x\right)dx={\int }_0^{1/3}f\left(x\right)dx+{\int }_{1/3}^{2/3}f\left(x\right)dx$

$f\left(x\right)=0$ for $xϵ(0,\frac{1}{3})$ and $f\left(x\right)=1$ for $xϵ(\frac{1}{3},\frac{2}{3})$

${\int }_0^{2/3}f\left(x\right)dx={\int }_0^{1/3}0dx+{\int }_{1/3}^{2/3}1dx=\frac{1}{3}$

$\left(D\right)$ From $\left(B\right)$ we can see that at $x=\frac{2}{3}$, $f\left(x\right)$ was not continuous.

$\therefore$ not $\left(D\right)$

Hence, answer is $\left(C\right)$.