Question

Let S = {x $ϵ$ $R$ : x 0 and 2$\left|\sqrt{x}3\right|$ + $\sqrt{x}$($\sqrt{x}$ 6) + 6 = 0} .
Then S :

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

Answer 1

The correct option is D Contains exactly two elements

Given,

$S=\{x\in R:x⩾0$, and

$2|\sqrt{x}-3|+\sqrt{x}|\sqrt{x}-b|+b=0$

Let$\sqrt{x}=t$

Then,

$2|t-3|+t|t-6|+6=0$

Now,

taking Case I, where t<3

$-2t+6+t^{2}4b+6=0$

$t^2-8t+12=0$

$t=2,6$

$\because$ we have taken, t $<3$

$\begin{array}{cc}\therefore t& =2,\\ \sqrt{x}& =2\\ x& =4\end{array}$

Case II where

$t⩾3$

$2t-6+t^2-6t+6=0$

$t^2-4t=0$

$t=4,0$

$\because$ We have taken, $t⩾3$

$\begin{array}{cc}\therefore t& =4\\ \sqrt{x}& =4\\ x& =16\\ \therefore x& =40or16\end{array}$

Option D exactly two elements.