Question

The domain of definition of the function $f\left(x\right)=\sqrt{{\log }_{\left(x−1\right)}\left(x^2+4x+4\right)}$ , is

• A. $\left[−3,−1\right]∪\left[1,2\right]$
• B. $\left(−2,−1\right)∪\left[2,∞\right)$
• C. $\left(−∞,−3\right]∪\left(−2,−1\right)∪\left(2,∞\right)$
• D. $\left[−2,−1\right]∪\left[2,∞\right]$

Answer 1

Answer: (C)

$f(x)=\sqrt{{\log }_{\left(x−1\right)}\left(x^2+4x+4\right)}$
$⇒{\log }_{\left(x−1\right)}\left(x^2+4x+4\right)â‰\yen 0$
Case 1: $0<âˆ\pounds xâˆ\pounds −1<1$
i.e., $1<âˆ\pounds xâˆ\pounds <2$ , then
$0<x^2+4x+4≤1$ $⇒x^2+4x+3≤0$ $\&{\left(x+2\right)}^2>0$
$⇒$ $−3≤x≤−1$ $\&xî€=−2$
So, $x∈\left(−2,−1\right)$
Case 2: $âˆ\pounds xâˆ\pounds −1>1$
i.e., $âˆ\pounds xâˆ\pounds >2$
$x^2+4x+4â‰\yen 1$ $⇒\left(x+1\right)\left(x+3\right)â‰\yen 0⇒x∈\left(−∞,−3\right]∪\left[−1,∞\right)$
$⇒x∈\left(−∞,−3\right]∪\left(2,∞\right)$
Hence, domain is $\left(−∞,−3\right]∪\left(−2,−1\right)∪\left(2,∞\right)$