Question

The domain of the function $f(x)=\dfrac{1}{\log {}_{10}(1-x)}+\sqrt{x+2}$
A.[-3,-2.5] $\cap$[-2.5,-2]
B.[-2, 0) $\cup$(0, 1)
C.[0, 1]
D.None of the above

Answer 1

Hint: the domain of a function is a set of its possible values or set of all values for which the function is defined. The logarithms only take positive values and negative values of logarithms are not defined. The square root also takes positive values and it cannot take negative values and negative values in square root are not defined and they are not in the domain of the function.

Complete step-by-step answer:
Given that $f(x)=\dfrac{1}{\log {}_{10}(1-x)}+\sqrt{x+2}$
We know that when denominator is zero then the function is not defined because function goes to infinity so, $\dfrac{1}{\log {}_{10}(1-x)}$is defined only when
$\log {}_{10}(1-x)\ne 0$
$(1-x)\ne 1$
$x\ne 0$. . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. (1)
The logarithm takes only positive values and for negative values of logarithm it is undefined
So, $\log {}_{10}(1-x)$is defined only when
$(1-x)>0$
$(x-1)<0$
$x<1$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..(2)
$\sqrt{x+2}$ is defined only when
$x+2\ge 0$(because square root takes only positive values and for negative values of square root it is undefined.
$x\ge -2$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(3)
So taking all these values into consideration,
The first condition we got is $x\ne 0$
The second condition we got is $x<1$
The third condition we got is $x\ge -2$
So, the set of numbers satisfying all these values of x is [-2, 0) $\cup$(0, 1)
So, the correct option is option (B)

Note: While finding the numerator carefully note that the denominator of fraction cannot be zero because if it happens means function tends to infinity. If they asked to find range also then note that range is the complete set of all resultant values we get by substituting values of the domain in the given function.