Question

Solution set of $x-\sqrt{1-\left|x\right|}<0$ is

• A. $\left[-1,\left(\frac{-1+\sqrt{5}}{2}\right)\right]$
• B. $\left[-1,1\right]$
• C. $\left[-1,\left(\frac{1+\sqrt{5}}{2}\right)\right]$
• D. $\left[1,-1\right]$

Answer 1

The correct option is A

$\left[-1,\left(\frac{-1+\sqrt{5}}{2}\right)\right]$

Explanation for the correct answer:

Solving the given expression for $x\ge 0$ and $x<0$:

Given $x-\sqrt{1-\left|x\right|}<0$

Case 1: If $x\ge 0$ then

$x-\sqrt{1-\left|x\right|}<0\left[\left|x\right|=x\text{since}1–x\ge 0\text{and}1–x\ge x^2\right]$

Since $0\le x\le 1\text{and}\left(\frac{-1–\sqrt{5}}{2}\right)<x<\left(\frac{-1+\sqrt{5}}{2}\right)$

Hence $x\in \left[0,\frac{-1+\sqrt{5}}{2}\right]..\left(1\right)$

Case 2: If $x<0$ then

$$\begin{gathered} x-\sqrt{1-\left|x\right|}<0 \\ \Rightarrow \sqrt{1-x}>0 \\ \Rightarrow x+1\ge 0 \\ \Rightarrow x\ge -1 \end{gathered}$$

Hence $x\in \left(-1,0\right)...\left(2\right)$

Hence $x\in \left[-1,\left(\frac{-1+\sqrt{5}}{2}\right)\right]...\left[\text{From}\left(1\right)\text{and}\left(2\right)\right]$

Therefore, the correct answer is option (A).