Question

The domain of the function $y=\frac{1}{\left\{lo{g}_{10}\right(3-x\left)\right\}}+\sqrt{x+7}$

• A. [-7, 3) - {2}
• B. [-7, 3] - {1}
• C. (-7, 3) - {0}
• D. (-7, 3)

Answer 1

The correct option is A [-7, 3) - {2}

The function would be defined when the term $\frac{1}{\left\{lo{g}_{10}\right(3-x\left)\right\}}$ is real, which will occur when x < 3.
However, if x = 2, then the denominator of the term becomes 0, which should not be allowed. The other limit of the function gets defined by the constraint defined by the term $\sqrt{x+7}$. For $\sqrt{x+7}$ to be real, x $\ge$ - 7 is the requirement.
Hence, Required domain = - 7 $\le$ x < 3, x $\ne$ 2
i.e., x $ϵ$ [-7,3) - {2}
Option (a) is correct.