Question

The domain of the real valued function \( f(x) = \frac{1}{\sqrt{log_{0.5}(2x-3)}} + \sqrt{4 - 9x^2} \) is:

• A. \( \left[ \frac{2}{3}, \frac{3}{2} \right] \)
• B. Null Set
• C. \( \left[ \frac{2}{3}, 2 \right) \)
• D. \( \left( -\frac{2}{3}, \frac{3}{2} \right) \)

Hint:

When dealing with square roots, ensure the terms inside are non-negative and satisfy all given inequalities to find the domain of the function.

Answer 1

The Correct Option is B

We need to find the domain of the given function: \[ f(x) = \frac{1}{\sqrt{\log_{0.5}(2x-3)}} + \sqrt{4 - 9x^2} \] Step 1: Condition for the first term to be defined and real
The first term is defined if and only if: \[ \log_{0.5}(2x - 3)<0 \] Recall that for the logarithm base \(0.5<1\), the logarithm function is decreasing. So, \[ \log_{0.5} y<0 \quad \text{implies} \quad y<1 \] Thus, \[ 2x - 3<1 \implies 2x<4 \implies x<2 \] Additionally, since the logarithm is defined only for positive values, \[ 2x - 3<0 \implies 2x<3 \implies x<\frac{3}{2} \] Combining these inequalities: \[ x<\frac{3}{2} \quad \text{and} \quad x<2 \] This gives the condition: \[ \frac{3}{2}<x<2 \] Step 2: Condition for the second term to be defined and real
The second term is defined if and only if: \[ 4 - 9x^2 \geq 0 \] \[ 9x^2 \leq 4 \] \[ x^2 \leq \frac{4}{9} \] \[ -\frac{2}{3} \leq x \leq \frac{2}{3} \] Step 3: Intersection of Both Conditions
- From the first condition: \( x \in \left( \frac{3}{2}, 2 \right) \) - From the second condition: \( x \in \left[ -\frac{2}{3}, \frac{2}{3} \right] \) Since these two intervals have no overlap, the function is not defined for any real value of \(x\). Conclusion: The domain is the Null Set. Final Answer: (2) Null Set