Question

The domain of the definition of the function $y=\frac{1}{{\log }_{10}(1-x)}+\sqrt{(x+2)}$ is .

• A. $-2\le x<1$
• B. $x\ge -2$
• C. $-3<x\le -2$
• D. $-2\le x<0$

Answer 1

Answer: (D)

Given, $y=\frac{1}{{\log }_{10}(1-x)}+\sqrt{x+2}$
$y$ is defined for
${\log }_{10}(1-x)\ne 0$ and $x+2\ge 0$
$\Rightarrow 1-x\ne 1x\ge -2$
$x\ne 0$
and $1-x>0$ and $1-x\ne 0$
$x<1$ and $x\ne 1$
$\therefore y$ is defined for values $x$ for which both the terms of the function is defined
$\therefore$ y is defined, when
$-2\le x<0$
And $0<x<1$