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

The domain of the real valued function $f(x)=\frac{1}{\sqrt{\log _{0.5}(2 x-3)}}+\sqrt{4-9 x^2}$
  1. $\left[\frac{2}{3}, \frac{3}{2}\right)$
  2. Null set
  3. $\left[\frac{2}{3}, 2\right)$
  4. $\left[-\frac{2}{3}, \frac{2}{3}\right]$

Solution

Given, $f(x)=\frac{1}{\sqrt{\log _{0.5}(2 x-3)}}+\sqrt{4-9 x^2}$ For defined to be $f(x)$, $\begin{aligned} & \log _{0.5}(2 x-3)\gt0,(2 x-3)\gt0,\left(4-9 x^2\right) \geq 0 \\ & \Rightarrow(2 x-3) \lt 1, x\gt\frac{3}{2} \text { and }\left(9 x^2-4\right) \leq 0 \\ & \Rightarrow 2 x \lt 4, x\gt\frac{3}{2} \text { and }(3 x-2)(3 x+2) \leq 0 \\ & \Rightarrow x \lt 2, x\gt\frac{3}{2} \text { and } \frac{-2}{3} \lt x \lt \frac{2}{3} \Rightarrow x \in \phi \end{aligned}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

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