Mathematics - Functions Question with Solution | TestHub

Given,

$f\left(x\right)={\cos }^{-1}\left(\frac{x^2-4x+2}{x^2+3}\right)$

We know that domain of ${\cos }^{-1} \text{is }\left[-1,1\right]$

So, $\left|\frac{x^2+4x+2}{x^2+3}\right|\le 1$

$\Rightarrow {\left(x^2-4x+2\right)}^2\le {\left(x^2+3\right)}^2$

$\Rightarrow {\left(x^2-4x+2\right)}^2-{\left(x^2+3\right)}^2\le 0$

$\Rightarrow \left(2x^2-4x+5\right)\left(-4x-1\right)\le 0$

Since $\left(2x^2-4x+5\right)>0$ as its discriminant $D<0$

So, $-4x-1\le 0\Rightarrow x\ge -\frac{1}{4}$

$\Rightarrow x\in [-\frac{1}{4},\infty )$