The domain of the real valued function f(x) = | vedclass

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(B) Given,$f(x) = \frac{1}{\sqrt{\log_{0.5}(2x - 3)}} + \sqrt{4 - 9x^2}$.For $f(x)$ to be defined,the following conditions must hold:$1. \log_{0.5}(2x

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Null set

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(B) Given,$f(x) = \frac{1}{\sqrt{\log_{0.5}(2x - 3)}} + \sqrt{4 - 9x^2}$.For $f(x)$ to be defined,the following conditions must hold:$1. \log_{0.5}(2x - 3) > 0$ $\Rightarrow 2x - 3 $2. 2x - 3 > 0$ $\Rightarrow 2x > 3$ $\Rightarrow x > \frac{3}{2}$.$3. 4 - 9x^2 \geq 0$ $\Rightarrow 9x^2 \leq 4$ $\Rightarrow x^2 \leq \frac{4}{9}$ $\Rightarrow -\frac{2}{3} \leq x \leq \frac{2}{3}$.Combining these conditions: $(x  \frac{3}{2}) \cap (-\frac{2}{3} \leq x \leq \frac{2}{3})$.Since there is no value of $x$ that satisfies $x > \frac{3}{2}$ and $x \leq \frac{2}{3}$ simultaneously,the domain is the null set.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

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

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