Question

Which of the following intervals belong(s) to the domain of the function $f\left(x\right)=\frac{\sqrt{-{\log }_{0.4}(x^4-1)}}{x^2+2x+8}$

• A. $(1,\infty )$
• B. $[-\sqrt[4]{42},\sqrt[4]{42}]$
• C. $[2,\infty )$
• D. $[-\sqrt[3]{32},-\sqrt[4]{42}]$

Answer 1

Answer: (C), (D)

The correct options are
C $[2,\infty )$
D $[-\sqrt[3]{32},-\sqrt[4]{42}]$

$\left(i\right)x^2+2x+8=(x+1{)}^2+7\ne 0\forall x\in R$

$\left(ii\right)x^4-1>0$
$\Rightarrow (x^2+1)(x+1)(x-1)>0$
$\Rightarrow x\in (-\infty ,-1)\cup (1,\infty )$

$\left(iii\right)-{\log }_{0.4}(x^4-1)\ge 0$
$\Rightarrow {\log }_{0.4}(x^4-1)\le 0$
$\Rightarrow x^4-1\ge 1$
$\Rightarrow x^4-2\ge 0$
$\Rightarrow (x^2+\sqrt{2})(x+\sqrt[4]{42})(x-\sqrt[4]{42})\ge 0$
$\Rightarrow x\in (-\infty ,-\sqrt[4]{42}]\cup [\sqrt[4]{42},\infty )$

$\therefore$ Domain is $(-\infty ,-\sqrt[4]{42}]\cup [\sqrt[4]{42},\infty )$