Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ is defined by $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right]-\frac{1}{2}$ for $\mathrm{x}\in \mathrm{R}$, where $\left[\mathrm{x}\right]$ is the greatest integer not exceeding x, then $\left\{\mathrm{x}\in \mathrm{R}:\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}\right\}$ is equal to

• A. Z, the set of all integers
• B. N, the set of all natural numbers
• C. $\mathrm{\varphi }$
• D. R, the set of all rational numbers

Answer 1

The correct option is C

$\mathrm{\varphi }$

Explanation for the correct answer:

Given function $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right]-\frac{1}{2}$

$$\begin{gathered} \therefore \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right]-\frac{1}{2} \\ \Rightarrow \mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\} \\ \Rightarrow \left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ \therefore \mathrm{f}\left(\mathrm{x}\right)=\left\{\mathrm{x}\right\}-\frac{1}{2} \end{gathered}$$

For, $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}$

$\begin{array}{rcl}& \Rightarrow & \left\{\mathrm{x}\right\}-\frac{1}{2}=\frac{1}{2}\\ \therefore \left\{\mathrm{x}\right\}& =& 1\end{array}$

Therefore, the fractional part of $\left\{\mathrm{x}\right\}$ lies between $0\le \left\{\mathrm{x}\right\}\le 1$

This is not possible that is $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}$ in the interval $\mathrm{\varphi }$.

Hence, the correct answer is an option (C).