Question

The domain of the function f(x) defined by
$f\left(x\right)=\frac{1}{{\log }_{10}\left(2-x\right)}+\sqrt{x+2}$is

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

Answer 1

The correct option is D $[-2,1)\cup \left(1,2\right)$

$\mathrm{We}\mathrm{have}f\left(x\right)=\frac{1}{{\log }_{10}\left(2-x\right)}+\sqrt{x+2}\mathrm{Let}g\left(x\right)=\frac{1}{{\log }_{10}\left(2-x\right)}\mathrm{and}h\left(x\right)=\sqrt{x+2}\mathrm{Thus},f\left(x\right)=g\left(x\right)+h\left(x\right)\mathrm{Domain}\left(f\right)=\mathrm{Domain}\left(g\left(x\right)\right)\cap \mathrm{Domain}\left(h\left(x\right)\right)\mathrm{Now},g\left(x\right)=\frac{1}{{\log }_{10}\left(2-x\right)}\mathrm{is}\mathrm{defined}\mathrm{for}\mathrm{all}x\mathrm{for}\mathrm{which}{\log }_{10}\left(2-x\right)\mathrm{is}\mathrm{defined}\mathrm{and}{\log }_{10}\left(2-x\right)\ne 0\Rightarrow 2-x>0\mathrm{and}2-x\ne 1\Rightarrow x<2\mathrm{and}x\ne 1\Rightarrow x\in \left(-\infty ,1\right)\cup \left(1,2\right)\mathrm{Domain}\left(g\right)=\left(-\infty ,1\right)\cup \left(1,2\right)\mathrm{And},h\left(x\right)=\sqrt{x+2}\mathrm{is}\mathrm{defined}\mathrm{for}\mathrm{all}x\mathrm{satisfying},x+2\ge 0\Rightarrow x\ge -2\Rightarrow x\in [-2,\infty ).\mathrm{Domain}\left(h\right)=[-2,\infty )\mathrm{Hence},\mathrm{Domain}\left(f\right)=\left\{\right(-\infty ,1)\cup (1,2\left)\right\}\cap [-2,\infty )\Rightarrow \mathrm{Domain}\left(f\right)=[-2,1)\cup \left(1,2\right)$