First slide

Functions (XII)

Question

$The domain of f (x) = x^{2} + \sqrt{\frac{| x + 3 | + x}{x + 2} - 1} is$

Moderate

A

$(- 5, - 3) \cup (1, \infty)$
B

$(- 5, - 2) \cup (1, \infty)$
C

$(- 5, - 2) \cup (- 1, \infty)$
D

$R - {- 2}$

Solution

$\begin{array}{l} We must have \frac{| x + 3 | + x}{x + 2} \geq 1 \\ \Rightarrow \frac{| x + 3 | - 2}{x + 2} \geq 0 \\ For | x + 3 | - 2 = 0 \Rightarrow x + 3 = \pm 2 \Rightarrow x = - 5, - 1 \end{array}$
$From the sign scheme we have x \in (- 5, - 2) \cup (- 1, \infty) .$

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