Question

Consider the set \[ A = \{ a \in \mathbb{R} : x^2 = a(a+1)(a+2) \text{ has a real root} \}. \] The number of connected components of \(A\) is _________.

Hint:

Connected components of sets on \(\mathbb{R}\) correspond to maximal intervals where the defining inequality holds continuously.

Answer 1

Correct Answer: 2

Step 1: Condition for real roots.
The equation is \(x^2 = a(a+1)(a+2).\) Since \(x^2 \ge 0,\) for real \(x\) to exist, we must have \(a(a+1)(a+2) \ge 0.\)
Step 2: Analyze the sign of the cubic.
The zeros are at \(a = -2, -1, 0.\) Now test sign changes in intervals: \[ \begin{array}{c|c} \text{Interval} & a(a+1)(a+2)
\hline (-\infty, -2) & -
(-2, -1) & +
(-1, 0) & -
(0, \infty) & + \end{array} \] Thus, \(a(a+1)(a+2) \ge 0\) in \([ -2, -1 ] \cup [ 0, \infty ).\)
Step 3: Conclusion.
Hence \(A = [ -2, -1 ] \cup [ 0, \infty )\), which has two connected components.