Find the domain of each of the following real valued functions of real variable:
(i) $f(x)=\sqrt{x-2}$
(ii) $f(x)=\frac{1}{\sqrt{x^{2}-1}}$
(iii) $f(x)=\sqrt{9-x^{2}}$
(iv) $f(x)=\frac{\sqrt{x-2}}{3-x}$
(i) Given: $f(x)=\sqrt{x-2}$
Clearly, f (x) assumes real values if x
$\Rightarrow x \geq 2$
$\Rightarrow x \in[2, \infty)$
Hence, domain $(f)=[2, \infty)$.
(ii) Given: $f(x)=\frac{1}{\sqrt{x^{2}-1}}$
Clearly, $f(x)$ is defined for $x^{2}-1>0$.
$(x+1)(x-1)>0 \quad\left[\right.$ Since $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$
$x<-1$ and $x>1$
$x \in(-\infty,-1) \cup(1, \infty)$
Hence, domain $(f)=(-\infty,-1) \cup(1, \infty)$
(iii) Given: $f(x)=\sqrt{9-x^{2}}$
We observe that f (x) is defined for all satisfying
$9-x^{2} \geq 0$
$\Rightarrow x^{2}-9 \leq 0$
$\Rightarrow(x+3)(x-3) \leq 0$
$\Rightarrow-3 \leq x \leq 3$
$x \in[-3,3]$
Hence, domain $(f)=[-3,3]$
(iv) Given: $f(x)=\sqrt{\frac{x-2}{3-x}}$
Clearly, f (x) assumes real values if
$x-2 \geq 0$ and $3-x>0$
$\Rightarrow x \geq 2$ and $3>x$
$\Rightarrow x \in[2,3)$
Hence, domain ( f ) = [2, 3) .
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