Question

Considering only the principal values of the inverse trigonometric functions, the domain of the function $f\left(x\right)={\cos }^{-1}\left(\frac{x^2-4x+2}{x^2+3}\right)$ is

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

Answer 1

The correct option is A $[-\frac{1}{4},\infty )$

$-1\le \frac{x^2-4x+2}{x^2+3}\le 1$

$\Rightarrow -x^2-3\le x^2-4x+2\le x^2+3$

$\begin{array}{ccc}\Rightarrow 2x^2-4x+5\ge 0& \&& -4x\le 1\\ x\in R& \&& x\ge -\frac{1}{4}\end{array}$

So domain is $[-\frac{1}{4},\infty )$