Question

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

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

Answer 1

Answer: (C)

$-2\le x<0$

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