Question

If $x \in R$ then range of $f(x)=\frac{x^{2}+2 x-3}{2 x^{2}+3 x-9}$ is:

• A. $(-\infty, \infty)$
• B. $R -\left\{\frac{1}{2}\right\}$
• C. $R -\left\{\frac{4}{9}, \frac{1}{2}\right\}$
• D. $R -\left\{\frac{3}{2}\right\}$

Answer 1

Answer: (C)

$f(x)=\frac{(x+3)(x-1)}{(2 x-3)(x+3)}$
$x \neq-3, \frac{3}{2}$
$f(x)=y-=\frac{x-1}{2 x-3} ; 2 x y-3 y=x-1$
$x=\frac{3 y-1}{2 y-1} $
$\therefore y \neq \frac{1}{2}$
But since $x \neq-3$ hence $y=\frac{4}{9}$.
$\therefore y \in R-\left\{\frac{4}{9}, \frac{1}{2}\right\}$