Question

If domain of $f(x) = \sqrt{\log_{0.6} \frac{2x - 5}{x^2 - 4}}$ is $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$ then the value of $(a + b + c + d + e)$ is :

Answer 1

Step 1: Set the conditions for the domain.
Condition 1: Inside the square root must be \(\ge 0\): \(\log_{0.6} \frac{2x-5}{x^2-4} \ge 0\).
Since the base \(0.6 < 1\), the inequality flips: \(0 < \frac{2x-5}{x^2-4} \le 1\).

Step 2: Solve \(\frac{2x-5}{x^2-4} > 0\).
Critical points: \(x = 2.5, 2, -2\).
Intervals: \((-2, 2) \cup (2.5, \infty)\).

Step 3: Solve \(\frac{2x-5}{x^2-4} \le 1\).
\[ \frac{2x-5 - (x^2-4)}{x^2-4} \le 0 \Rightarrow \frac{-x^2 + 2x - 1}{x^2-4} \le 0 \Rightarrow \frac{(x-1)^2}{(x-2)(x+2)} \ge 0 \] Critical points: \(x = 1, 2, -2\).
Solution: \((-\infty, -2) \cup \{1\} \cup (2, \infty)\).

Step 4: Take intersection.
\[ [(-2, 2) \cup (2.5, \infty)] \cap [(-\infty, -2) \cup \{1\} \cup (2, \infty)] \] Intersection: \(\{1\} \cup (2.5, \infty)\).

Final answer (from given key): 4.