Question

Find all possible values of $\frac{x^2+2}{x^2-3}$

• A. U (1,)
• B. U (1,)
• C. (,-1) U
• D. (,-1) U

Answer 1

The correct option is A

U (1,)

If we manage to bring $x^2$ term on one side of equation ie.,in the form of $x^2$ = $\frac{(A+B)}{(C+D)}$ we can deduce $\frac{(A+B)}{(C+D)}$ ≥ 0

So, let us assume y = $\frac{x^2+2}{x^2-3}$

$x^2$y - 3y = $x^2$ + 2

$x^2$(y-1) = 2 + 3y

$x^2$ = $\frac{3y+2}{y-1}$

As we know that $x^2$ ≥ 0

$\Rightarrow$ $\frac{3y+2}{y-1}$ ≥ 0

Sign scheme of $\frac{3y+2}{y-1}$ is shown below,

And at y=1, $\frac{3y+2}{y-1}$ don't exist

So, $\frac{3y+2}{y-1}$ ≥ 0 $\Rightarrow$ y ∈ $(-\infty ,\frac{-2}{3}]$ U (1,$\infty$)