Question

Let $S=\{xϵR:x\ge 0and2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}$. Then $S$

• A. Contains exactly two elements
• B. Contains exactly four elements
• C. Is an empty set
• D. Contains exactly one element

Answer 1

The correct option is A Contains exactly two elements

$x\ge 0$ (given ) ------ (1)

$2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$ ------(2)

Consider $x<9,\sqrt{x}-3<0$.

For $x<9$, eq $\left(2\right)$ becomes

$-2(\sqrt{x}-3)+\sqrt{x}.\sqrt{x}-6\sqrt{x}+6=0$

$x-8\sqrt{x}+12=0$

$x-6\sqrt{x}-2\sqrt{x}+12=0$

$(\sqrt{x}-6)(\sqrt{x}-2)=0$

$\Rightarrow \sqrt{x}=6,2$

$x=36,4$

Since $x<9$, so $x=4$ is valid

For $x\ge 9$

$2\sqrt{x}-6+x-6\sqrt{x}+6=0$

$x=4\sqrt{x}$

$\Rightarrow \sqrt{x}=0or\sqrt{x}=4$

Since $x\ge 9,x=16$ is valid

Thus, $S$ has only $2$ valid elements $4,16$.

Hence, option A is correct.