Question

If ${\int }_0^x\left[x\right]dx={\int }_0^{\left[x\right]}xdx$, then which of the following is true?

(where $[.]$ and $\{.\}$ denote the greatest integer and fractional parts respectively.)

• A. $x$ is an integer
• B. $x$ is purely fractional
• C. $\left\{x\right\}=\frac{1}{2}$
• D. $\left[x\right]>1$

Answer 1

Answer: (B), (C)

$x$ is purely fractional
$\left\{x\right\}=\frac{1}{2}$

Let $x=I+f⟹[x=I]$ (i)
Now, ${\int }_0^{\left[x\right]}xdx={\int }_0^{I}xdx=\frac{I^2}{2}$
and ${\int }_0^x\left[x\right]dx={\int }_0^{I+f}\left[x\right]dx={\int }_0^{1}0dx+{\int }_1^{2}1dx+\cdots +{\int }_{I-1}^I(I-1)dx+{\int }_I^{I+f}Idx$
$=\{1+2+3+\cdots +(I-1\left)\right\}+I(I+f-I)$
$=\frac{I(I-1)}{2}+I\left(f\right)=\frac{I(I-1)}{2}+I(x-I)$ [using equation (i)]
Given, ${\int }_0^x\left[x\right]dx={\int }_0^{\left[x\right]}xdx$
$⟹\frac{I^2}{2}=\frac{I(I-1)}{2}+I(x-I)$
$⟹I(I+1-2x+2I-1)=0$
$⟹I(2I-2x+1)=0$
$⟹I=0$, $2I-2x+1=0$
ie, $\left[x\right]=0$ or $x=\frac{2I+1}{2}=I+\frac{1}{2}$
$⟹0\le x<1$ or $x=\left[x\right]+\frac{1}{2}$
$⟹0\le x<1$ or $\left\{x\right\}=\frac{1}{2}$
Ans: B,C