Question

Let \(S=\{x\in \mathbb{R}\mid x\ge 0 \text{ and } 2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}\), then \(S=\)

• A. Contains exactly one element
• B. Contains exactly two elements
• C. Contains exactly four elements
• D. \(\phi\)

Hint:

When equations involve \(\sqrt{x}\):
Substitute \(t=\sqrt{x}\)
Always check domain restrictions
Handle absolute values using cases

Answer 1

The Correct Option is B

Step 1: Let \(t=\sqrt{x}\), where \(t\ge 0\). The given equation becomes: \[ 2|t-3|+t(t-6)+6=0 \]
Step 2: Simplify the polynomial part: \[ t(t-6)+6=t^2-6t+6 \] So, \[ 2|t-3|+t^2-6t+6=0 \] 
Step 3: Consider cases due to the absolute value. Case I: \(t\ge 3 \Rightarrow |t-3|=t-3\) \[ 2(t-3)+t^2-6t+6=0 \] \[ t^2-4t=0 \] \[ t(t-4)=0 \] Valid solution (since \(t\ge 3\)): \[ t=4 \] Case II: \(0\le t<3 \Rightarrow |t-3|=3-t\) \[ 2(3-t)+t^2-6t+6=0 \] \[ t^2-8t+12=0 \] \[ (t-2)(t-6)=0 \] Valid solution (since \(t<3\)): \[ t=2 \] 
Step 4: Convert back to \(x\): \[ x=t^2 \Rightarrow x=16,\;4 \] 
Step 5: Hence, \[ S=\{4,16\} \] which contains exactly two elements.