Question

The domain of which of the following functions is : $[2,\infty )$ ?

• A. $f\left(x\right)=\sqrt{x-2}$
• B. $f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$
• C. $f\left(x\right)=\sqrt{9-x^2}$
• D. $f\left(x\right)=\frac{\sqrt{x-2}}{3-x}$

Answer 1

The correct option is A $f\left(x\right)=\sqrt{x-2}$

$f\left(x\right)=\sqrt{x-2}x-2\ge 0x\ge 2\Rightarrow x\in [2,\infty )$

$f\left(x\right)=\frac{1}{\sqrt{x^2-1}}{x}^2-1>0(x-1)(x+1)<0\Rightarrow x\Rightarrow \in (-1,1)$

$f\left(x\right)=\sqrt{9-x^2}9-x^2\ge 0x^2-9\le 0(x-3)(x+3)\le 0\Rightarrow x\Rightarrow \in [-3,3]$

$f\left(x\right)=\frac{\sqrt{x-2}}{3-x}x-2\ge 0x\ge 23-x\ne 0x\ne 3x\in [2,\infty )-\left\{3\right\}$

Thus the correct answer is $A$.