Question

Range of $\frac{x^4-81}{x-3}$ is

• A. $R-\left\{3\right\}$
• B. $R-\left\{81\right\}$
• C. $R-\left\{108\right\}$
• D. $R$

Answer 1

The correct option is C $R-\left\{108\right\}$

$f\left(x\right)=\frac{x^4-81}{x-3}=\frac{(x+3)(x-3)(x^2+9)}{(x-3)},x\ne 3$
$=(x+3)(x^2+9),x\ne 3$
Hence, domain of $f$ is $R-\left\{3\right\}$

$f\left(x\right)$ is a cubic polynomial for all $x\in R-\left\{3\right\}$
We know that a cubic polynomial has range $R$ for domain $R$.
$f\left(x\right)=(x+3)(x^2+9)$
$f\left(3\right)=108$
$\therefore$ Range is $R-\left\{108\right\}$