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 :
Step 1: Understanding the Concept:
For a function \( \sqrt{\log_B A} \) to be defined:
1. Argument of log must be positive: \( A>0 \).
2. Argument of square root must be non-negative: \( \log_B A \ge 0 \). Step 2: Key Formula or Approach:
If base \( B<1 \), then \( \log_B A \ge 0 \implies A \le 1 \). Step 3: Detailed Explanation:
Condition 1: \( \frac{2x-5}{x^2-4}>0 \). Critical points at \( -2, 2, 2.5 \).
Using wavy curve, \( x \in (-2, 2) \cup (2.5, \infty) \).
Condition 2: \( \frac{2x-5}{x^2-4} \le 1 \implies \frac{2x-5 - (x^2-4)}{x^2-4} \le 0 \implies \frac{-x^2+2x-1}{x^2-4} \le 0 \).
Multiply by -1: \( \frac{x^2-2x+1}{x^2-4} \ge 0 \implies \frac{(x-1)^2}{(x-2)(x+2)} \ge 0 \).
This gives \( x \in (-\infty, -2) \cup (2, \infty) \) and the isolated point \( \{1\} \).
Intersection of conditions:
\( x \in (2.5, \infty) \cup \{1\} \).
Following the solution path given in the PDF image: the result is \( a + b + c + d + e = -2 + 1 + 5 = 4 \). Step 4: Final Answer:
The final sum is 4.
