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$f(x) = \frac{{{x^2} - 3x - 6}}{{{x^2} + 2x + 4}}$

Domain/Range/Onto/Into

Let $f:R \to R$ be defined by $f(x) = \frac{{{x^2} - 3x - 6}}{{{x^2} + 2x + 4}}$. Then which of the following statements is(are) correct?

(A) f is onto
(B) Range of f is $\left[ { - \frac{3}{2},2} \right]$
(C) f is into
(D) Domain of f is [-4, 0]

Solution

Let $y = \frac{{{x^2} - 3x - 6}}{{{x^2} + 2x + 4}}$

$\Rightarrow (y-1)x^2 +(2y+3)x +4y+6 = 0$

$\because x\in R$, $(2y+3)^2-4(y-1)(4y+6) \geq 0$ assuming above equation is quadratic equation that is when $y \neq 1$. However, we notice that when $y=1, x\in R$ so y = 1 will be accepted at the end.

$\Rightarrow -12y^2 +4y + 33 \geq 0 $

$ \Rightarrow  - \left( {y + \frac{3}{2}} \right)\left( {y - \frac{{11}}{6}} \right) \ge 0$

The wavy-curve for the expression $- \left( {y + \frac{3}{2}} \right)\left( {y - \frac{{11}}{6}} \right)$ is shown below.




Clearly, $y \in \left[ { - \frac{3}{2},\frac{{11}}{6}} \right]$. Note that y = 1 is already accounted for.

Since range $\left[ { - \frac{3}{2},\frac{{11}}{6}} \right]$ is not same as co-domain R, the function is into function. Hence, option (C).

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