Question

Consider the rational function $f\left(x\right)=\frac{x^3-4x^2+15x-8}{x^2-2}$. Then

• A. domain of $f$ is $R-\left\{\sqrt{2}\right\}$
• B. there are three pre-images of $4$
• C. there is only one pre-image of $4$
• D. domain of $f$ is $R-\{\pm \sqrt{2}\}$

Answer 1

Answer: (B), (D)

domain of $f$ is $R-\{\pm \sqrt{2}\}$

$f$ is meaningful if
$x^2-2\ne 0$
i.e., $x\ne \pm \sqrt{2}$
$\therefore$ Domain of $f$ is $R-\{\pm \sqrt{2}\}$.

We have to find the pre-image of $4$
i.e., we have to find those values in the domain of $f$ for which $f\left(x\right)=4$
$\frac{x^3-4x^2+15x-8}{x^2-2}=4$
$\Rightarrow x^3-4x^2+15x-8=4x^2-8$
$\Rightarrow x^3-8x^2+15x=0$
$\Rightarrow x(x^2-8x+15)=0$
$\Rightarrow x(x-3)(x-5)=0$
$\Rightarrow x=0,3,5$