Question

If the range of the function, $f(x)=\frac{x^3-3 x^2+2 x}{2 x^3+2 x^2-4 x}$ is $R-\{\alpha, \beta, \gamma\}$, then $(\alpha+\beta+\gamma)$ equals (where $R$ is the set of all real numbers)

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{-1}{2}$
• D. $\frac{-1}{6}$

Answer 1

Answer: (D)

$f(x)=\frac{x(x-1)(x-2)}{2 x(x-1)(x+2)}, D_f: R-\{0,1,-2\}$
$f ( x )=\frac{ x -2}{2( x +2)}, x \neq 0,1 $
$\therefore R _{ f }: R -\left\{\frac{1}{2}, \frac{-1}{2}, \frac{-1}{6}\right\}$
$\therefore \alpha+\beta+\gamma=\frac{1}{2}-\frac{1}{2}-\frac{1}{6}=\frac{-1}{6} $