Question

The domain of $f\left(x\right)={\left(\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}\right)}^{1/2022}$ is

• A. $R$
• B. $R-\{\frac{1}{2},1\}$
• C. $[0,\infty )$
• D. $R-\left\{\frac{1}{2}\right\}$

Answer 1

The correct option is A $R$

We know that, domain of $\left(g\right(x){)}^{1/\text{even}}=$ domain of $\sqrt{g\left(x\right)}$

Given denominator
$2x^2-2x+1$
$=2[x^2-x]+1$
$=2[x^2-x+\frac{1}{4}-\frac{1}{4}]+1$
$=2\left[{(x-\frac{1}{2})}^2\right]+\frac{1}{2}\ge 0\forall x\in R$

Now for $f\left(x\right)$ to be defined
$x^4-2x^3+3x^2-2x+2\ge 0$
$\Rightarrow (x^4+3x^2+2)-(2x^3+2x)\ge 0$
$\Rightarrow (x^2+1)(x^2+2)-2x(x^2+1)\ge 0$
$\Rightarrow (x^2+1)(x^2-2x+2)\ge 0$
$\Rightarrow (x^2+1)\left(\right(x-1{)}^2+1)\ge 0,\Rightarrow x\in R$

Hence, Domain is $R$