First slide

Functions (XII)

Question

$If ∣ f (x) + 6 - x^{2}] = | f (x) | + |4 - x^{2}| + 2, then f (x) is necessarily non-negaive for :$

Moderate

A

$x \in [- 2, 2]$
B

$x \in (- \infty, - 2) \cup (2, \infty)$
C

$x \in [- \sqrt{6}, \sqrt{6}]$
D

$x \in [- 5, - 2] \cup [2, 5]$

Solution

$\begin{array}{ll} Gives |f (x) + 6 - x^{2}| = | f (x) | + |4 - x^{2}| + 2 \\ \Rightarrow |f (x) + 2 + (4 - x^{2})| = | f (x) | + |4 - x^{2}| + 2 \end{array}$
$| a + b + c | = | a | + | b | + | c |$
$If a \geq 0, b \geq 0, c \geq 0 or a \leq 0, b \leq 0, c \leq 0$
$\Rightarrow f (x) \geq 0 and 4 - x^{2} \geq 0 \Rightarrow - 2 \leq x \leq 2 and f (x) \geq 0$

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