A subset of all values of x satisfying $\{x\} = x [x]$ where $\{x\}$ represents greatest integer function $x$ and $\{x\}$ represents fractional part of x, is

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$\{0\}$

$\{- \frac{1}{2}\}$

$- 1 < x < 1$

$R - 0$

answer is A, B.

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Detailed Solution

$\begin{array}{l} \{x\} = [x] x \\ \{x\} = x (x - \{x\}) = x^{2} - x \{x\} \\ \Rightarrow \{x\} = \frac{x^{2}}{1 + x} \\ Now 0 \leq \{x\} < 1 \\ \Rightarrow \frac{x^{2}}{1 + x} \geq 0 \\ \Rightarrow 1 + x > 0 \\ \Rightarrow x > - 1 \\ \frac{x^{2}}{1 + x} < 1 \\ \Rightarrow x^{2} < 1 + x \\ \Rightarrow x^{2} - x - 1 < 0 \\ \Rightarrow \frac{1 - \sqrt{5}}{2} < x < \frac{1 - \sqrt{5}}{2} \end{array}$ $R - 0$

Given equation again can be written as

$\begin{array}{l} x - [x] = x [x] \\ \frac{x}{1 + x} = [x] = λ \\ x = \frac{λ}{1 - λ} \\ \frac{1 - \sqrt{5}}{2} < \frac{λ}{1 - λ} < \frac{1 + \sqrt{5}}{2} \\ \Rightarrow \frac{- 1 - \sqrt{5}}{2} < λ < \frac{\sqrt{5} - 1}{2} \\ λ = 0 \Rightarrow x = 0 \\ λ = - 1 \Rightarrow x = \frac{- 1}{2} \end{array}$