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1.Set Theory
hard
Let $S = \{ x \in R:x \ge 0$ and $2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0\} $ then $S:$ . . .
A
contains exactly one element.
B
contains exactly two elements
C
contains exactly four element.
D
is an empty set
(JEE MAIN-2018)
Solution
Case – $I$ : $x\,\in \,[0,\,9]$
$2(3 – \sqrt x )\, + \,x\, – \,6\sqrt x \, + \,6\, = \,0$
$ \Rightarrow \,x\, – \,8\sqrt x \, + \,12\, = \,0\, \Rightarrow \,\sqrt x \, = \,4,2\, \Rightarrow \,x\, = \,16,4$
Since $x\, \in \,[0,\,9]$
$\therefore \,\,x\,=\,4$
Case – $II$ : $x\, \in \,[9,\,\infty ]$
$2(\sqrt x – 3)\, + \,x\, – \,6\sqrt x \, + \,6\, = \,0$
$ \Rightarrow \,x\, – \,4\sqrt x \,\, = \,0\, \Rightarrow \,x\, = \,16,0$
Since $x\, \in \,[9,\,\infty ]$
$\therefore \,\,x\,=\,16$
Hence , $x\,=\,4$ and $16$
Standard 11
Mathematics
