Question

The possible values of expression $\frac{1}{x^2-x-3}$ is

• A.
• B. U (0,)
• C. U [0,)
• D.

Answer 1

The correct option is B

U (0,)

Take the denominator $x^2$ - x - 3

$x^2$ - x - 3 = $x^2$ - 2($\frac{1}{2}$).x + $(\frac{1}{2}{)}^2$ - $(\frac{1}{2}{)}^2$ - 3

= $(x-\frac{1}{2}{)}^2$ - $\frac{1}{4}$ - 3

= $(x-\frac{1}{2}{)}^2$ - $\frac{13}{4}$

Now, we know that $(x-\frac{1}{2}{)}^2$ ≥ 0

$\Rightarrow$ $(x-\frac{1}{2}{)}^2$ - $\frac{13}{4}$ ≥ $\frac{-13}{4}$, ∀ x ∈ R

(or)

$\infty$ > $(x-\frac{1}{2}{)}^2$ - $\frac{13}{4}$ ≥ $\frac{-13}{4}$
$\Rightarrow$ $\frac{1}{x^2-x-3}$ is in $(-\infty ,\frac{-4}{13}]$ U (0,$\infty$)