Question
Let $$S = \left\{ {x \in R:x \geqslant 0} \right.$$ 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 element.
C.
contains exactly four element.
D.
is an empty set.
Answer :
contains exactly two element.
Solution :
Case - I : $$x \in \left[ {0,9} \right]$$
$$\eqalign{
& 2\left( {3 - \sqrt x } \right) + x - 6\sqrt x + 6 = 0 \cr
& \Rightarrow \,\,x - 8\sqrt x + 12 = 0 \cr
& \Rightarrow \,\,\sqrt x = 4,2 \cr
& \Rightarrow \,\,x = 16,4 \cr
& {\text{Since}}\,{\text{ }}x \in \left[ {0,9} \right] \cr
& \therefore \,\,x = 4 \cr} $$
Case - II : $$x \in \left[ {9,\infty } \right]$$
$$\eqalign{
& 2\left( {\sqrt x - 3} \right) + x - 6\sqrt x + 6 = 0 \cr
& \Rightarrow \,\,x - 4\sqrt x = 0 \cr
& \Rightarrow \,\,x = 16,0 \cr
& {\text{Since }}x \in \left[ {9,\infty } \right] \cr
& \therefore \,\,x = 16 \cr
& {\text{Hence, }}x = 4\,\& 16 \cr} $$