The domain of the function f(x)=log(|x|−1)(x2+4x+4) is

f(x)=log|x|−1(x2+4x+4) is definediff x2+4x+4≥0,|x|−1>0 and |x|−1≠1⇒ (x+2)2≥0,|x|>1,|x|≠2∀x∈R,x 1 ,x≠±2(−∞, −2)∪(−2, −1)∪(1, 2)∪(2, ∞)

The domain of the function f(x)=log(|x|−1)(x2+4x+4) is

f(x)=log|x|−1(x2+4x+4) is definediff x2+4x+4≥0,|x|−1>0 and |x|−1≠1⇒ (x+2)2≥0,|x|>1,|x|≠2∀x∈R,x 1 ,x≠±2(−∞, −2)∪(−2, −1)∪(1, 2)∪(2, ∞)

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$T h e d o m a i n o f t h e f u n c t i o n f (x) = \sqrt{\log_{(| x | - 1)} (x^{2} + 4 x + 4)} i s$

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$[- 3, - 1] \cup [1, 2]$

$(- 2, - 1) \cup [2, \infty)$

$[- \infty, - 3) \cup (- 2, - 1) \cup (2, \infty)$

$[- 3, - 2] \cup [0, 2]$

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\log_{| x | - 1} (x^{2} + 4 x + 4)} i s d e f i n e d \\ i f f x^{2} + 4 x + 4 \geq 0, | x | - 1 > 0 a n d | x | - 1 \neq 1 \\ \Rightarrow {(x + 2)}^{2} \geq 0, | x | > 1, | x | \neq 2 \\ \forall x \in R, x < - 1 o r x > 1, x \neq \pm 2 \\ (- \infty, - 2) \cup (- 2, - 1) \cup (1, 2) \cup (2, \infty) \end{array}$