Question

The domain of the definition of
f(x) $=\sqrt{{{\log }_{0.4}}\left( \dfrac{x-1}{x+5} \right)}\times \dfrac{1}{{{x}^{2}}-36}$ .
A) \left( -\infty ,0 \right)-\left\\{ -6 \right\\}
B) \left( 0,+\infty \right)-\left\\{ 1,6 \right\\}
C) \left( 1,\infty \right)-\left\\{ 6 \right\\}
D) \left( 1,\infty \right)+\left\\{ 6 \right\\}

Answer 1

Hint: - To find the domain of a definition, the most important thing to notice is that every function used in the definition should not get undefined. If any of the component functions get undefined then the whole definition gets undefined.
The most important thing to notice is that for no value of x the function should have a denominator which is turning zero or becoming zero.

Complete step-by-step answer:
As mentioned in the question, we have to find the domain of the given definition.
Now, we can see that the denominator parts of the definition become zero when x=6, -6 and -5.
So, all these points should be removed from the domain.
Now, any quantity under the square root should be positive as the square root of a negative number or quantity is undefined, so the value of the logarithmic function should not be less than zero.
For this ${{\log }_{0.4}}\left( \dfrac{x-1}{x+5} \right)$ should be greater than zero. Now, as the base of this function is less than 1, hence, the inequality can be written as follows
${{\log }_{0.4}}\left( \dfrac{x-1}{x+5} \right)>0$
Now, on taking antilog on both the side, we get

& \left( \dfrac{x-1}{x+5} \right)<{{e}^{0}} \\\ & \left( \dfrac{x-1}{x+5} \right)<1 \\\ \end{aligned}$$ (Because the base of the logarithmic function is less than 1 and hence, the anti-log will flip the inequality) And we also know that $$\left( \dfrac{x-1}{x+5} \right)>0$$ (Because the input inside the logarithmic function cannot be negative as it is only defined if a positive input is given inside the function) so, we can see that the numerator and the denominator of the expression mentioned above should be together positive, hence, we can write as follows $$\begin{aligned} & x-1>0 \\\ & x>1 \\\ \end{aligned}$$ and $$\begin{aligned} & x+5>0 \\\ & x>5 \\\ \end{aligned}$$ So, form all this information, we can get to the domain of the definition as follows $$\left( 1,\infty \right)-\left\\{ 6 \right\\}$$ Note: - The students can make an error if they don’t know about the properties of logarithmic functions and the points at which a function is termed undefined because without this information one could not get to the correct solution.