Question

The domain of the function $f\left(x\right)=lo{g}_{3+x}(x^2-1)$ is

• A. $\left(-3,-1\right)\cup \left(1,\infty \right)$
• B. $[-3,-1)\cup [1,\infty )$
• C. $\left(-3,-2\right)\cup \left(-2,-1\right)\cup \left(1,\infty \right)$
• D. $[-3,-1)\cup [-2,-1)\cup [1,\infty )$

Answer 1

Answer: (C)

$\left(-3,-2\right)\cup \left(-2,-1\right)\cup \left(1,\infty \right)$

${\log }_{x+3}(x^2-1)$ is defined when

$x+3\ne 1$ and $x+3>0$ and $x^2-1>0$

$\Rightarrow x\ne 1-3$ and $x>-3$ and $(x-1)(x+1)>0$

$\Rightarrow x\ne -2$ and $x>-3$ and $(x-1)(x+1)>0$

$\Rightarrow x\ne -2$ and $x\in (-3,\infty )$ and $x\in (-\infty ,-1)\cup (1,\infty )$

$\Rightarrow x\ne -2$ and $x\in (-3,\infty )\cap [(-\infty ,-1)\cup (1,\infty )]$

$\Rightarrow x\ne -2$ and $x\in (-3,-1)\cup (1,\infty )$

$\Rightarrow x\in (-3,-1)\cup (1,\infty )-\{-2\}$

$\therefore x\in (-3,-2)\cup (-2,-1)\cup (1,\infty )$

Hence Domain is $x\in (-3,-2)\cup (-2,-1)\cup (1,\infty )$