Question

The domain of the function $f\left(x\right)=\sqrt{10-\sqrt{x^4-21x^2}}$ is

• A. $[5,\infty )$
• B. $[-\sqrt{21},\sqrt{21}]$
• C. $[-5,-\sqrt{21}]\cup [\sqrt{21},5]\cup \left\{0\right\}$
• D. $(-\infty ,-5)$

Answer 1

The correct option is C $[-5,-\sqrt{21}]\cup [\sqrt{21},5]\cup \left\{0\right\}$

We have,
$f\left(x\right)=\sqrt{10-\sqrt{x^4-21x^2}}$
For $f$ to be defined,
$x^4-21x^2\ge 0$
$\Rightarrow x^2(x^2-21)\ge 0$
$\Rightarrow x^2(x-\sqrt{21})(x+\sqrt{21})\ge 0$
$\therefore x\in (-\infty ,-\sqrt{21}]\cup [\sqrt{21},\infty )\cup \left\{0\right\}\cdots \left(1\right)$
Also,
$10-\sqrt{x^4-21x^2}\ge 0$
$\Rightarrow \sqrt{x^4-21x^2}\le 10$
$\Rightarrow x^4-21x^2\le 100$
$\Rightarrow x^4-21x^2-100\le 0$
$\Rightarrow (x^2-25)(x^2+4)\le 0$
$\Rightarrow (x^2-25)\le 0$
$\therefore x\in [-5,5]\cdots \left(2\right)$
Therefore, domain of $f\left(x\right)$ is
$[-5,-\sqrt{21}]\cup [\sqrt{21},5]\cup \left\{0\right\}$