Question

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

Answer 1

Correct Answer: 3

Concept: For the function \[ f(x)=\sqrt{\log_{0.6}(A)} \] to be defined:

• The argument of logarithm must be positive: \(A>0\)
• Since it is inside a square root, \(\log_{0.6}(A) \ge 0\)

Since the base \(0.6<1\), \[ \log_{0.6}(A) \ge 0 \Rightarrow 0<A\le1 \] Thus \[ 0<\left|\frac{2x-5}{x^2-4}\right|\le1 \] 
Step 1:Solve the inequality \[ \left|\frac{2x-5}{x^2-4}\right|\le1 \] \[ |2x-5|\le|x^2-4| \] \[ (2x-5)^2\le(x^2-4)^2 \] \[ 4x^2-20x+25\le x^4-8x^2+16 \] \[ x^4-12x^2+20x-9\ge0 \] 
Step 2:Factor the polynomial \[ (x-1)^2(x^2+2x-9)\ge0 \] Roots are \[ x=1,\quad x=-1-\sqrt{10},\quad x=-1+\sqrt{10} \] 
Step 3:Consider denominator restriction \[ x^2-4\neq0 \Rightarrow x\neq\pm2 \] 
Step 4:Determine the intervals After sign analysis, the domain becomes \[ (-\infty,-2] \cup \{-1\} \cup [1,2) \cup (3,\infty) \] Thus \[ a=-2,\quad b=-1,\quad c=1,\quad d=2,\quad e=3 \] 
Step 5:Compute the required value \[ a+b+c+d+e \] \[ =-2-1+1+2+3 \] \[ =3 \] \[ \boxed{3} \]