Q.
The domain of definition of the function $f\left(x\right)= \, \sqrt{\log_{\left(x - 1\right)} \left(x^{2} + 4 x + 4\right)}$ , is
NTA AbhyasNTA Abhyas 2022
Solution:
$f(x)= \, \sqrt{\log_{\left(x - 1\right)} \left(x^{2} + 4 x + 4\right)}$
$\Rightarrow \log_{\left( x - 1\right)} \left( \, x^{2} + 4 x + 4\right)\geq 0$
Case 1: $0 < \left|x\right|-1 < 1$
i.e., $1 < \left|x\right| < 2$ , then
$0 < x^{2}+4x+4 \, \leq 1$ $\Rightarrow \, x^{2}+4x+3 \, \leq 0$ $\& \, \left(x + 2\right)^{2}>0$
$\Rightarrow $ $-3 \, \leq x \, \leq -1$ $\& \, x\neq -2$
So, $x\in \left(- 2 , \, - 1\right)$
Case 2: $\left|x\right|-1>1$
i.e., $\left|x\right|>2$
$x^{2}+4x+4 \, \geq 1$ $\Rightarrow \left(x + 1\right)\left(x + 3\right)\geq 0\Rightarrow x\in \left( \, - \infty , - 3 \, \right] \, \cup\left[ \, - 1 \, , \infty\right)$
$\Rightarrow x\in \left(- \infty , \, - 3\right]\cup\left(2 , \, \infty\right)$
Hence, domain is $\left(- \infty , - 3\right] \, \cup\left(-2,-1\right)\cup\left(2 , \infty\right)$