Question

Let $S=\{x\in R:x\ge 0\text{and}2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}$. Then $S$

• A. contains exactly four elements
• B. is an empty set
• C. contains exactly one element
• D. contains exactly two elements

Answer 1

The correct option is D contains exactly two elements

Given: $2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$
Case $\left(1\right):$ when $0\le \sqrt{x}<3$
$\therefore -2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}-6)+6=0$
$\Rightarrow 6-2\sqrt{x}+(\sqrt{x}{)}^2-6\sqrt{x}+6=0$
$\Rightarrow (\sqrt{x}{)}^2-8\sqrt{x}+12=0$
$\Rightarrow (\sqrt{x}-6)(\sqrt{x}-2)=0$
$\Rightarrow \sqrt{x}=2,\sqrt{x}=6$ (rejected)
$\therefore x=4$

Case $\left(2\right):$ when $\sqrt{x}\ge 3$
$\therefore 2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}-6)+6=0$
$\Rightarrow 2\sqrt{x}-6+(\sqrt{x}{)}^2-6\sqrt{x}+6=0$
$\Rightarrow (\sqrt{x}{)}^2-4\sqrt{x}=0$
$\Rightarrow \sqrt{x}(\sqrt{x}-4)=0$
$\Rightarrow \sqrt{x}=4,\sqrt{x}=0$ (rejected)
$\therefore x=16$

Hence, $S$ contains exactly two elements.