First slide

Functions (XII)

Question

If x is real, then the values of the expression $\frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3}$ are

Easy

A

$\begin{array}{l} [4, 5] \end{array}$
B

$\begin{array}{l} [- 4, 5] \end{array}$
C

$[- 5, 4]$
D

None of these

Solution

$\begin{array}{l} \frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3} = y \\ or x^{2} + 14 x + 9 = x^{2} y + 2 x y + 3 y \\ or x^{2} (y - 1) + 2 x (y - 7) + (3 y - 9) = 0 \end{array}$
since x is real, we have
$\begin{array}{l} 4 {(y - 7)}^{2} - 4 (3 y - 9) (y - 1) \geq 0 \\ or 4 (y^{2} + 49 - 14 y) - 4 (3 y^{2} + 9 - 12 y) \geq 0 \\ or (y + 5) (y - 4) \leq 0 \\ ∴ y \in [- 5, 4] \end{array}$

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