Question

Find the domain of the function:\[f\left( x \right)=\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)-\dfrac{3}{{{\log }_{2}}\left| \sin x \right|} \right\}\]?

Answer 1

Hint: In the given question, we have been asked to find the domain of the given function. In order to find the domain of any given function, we will use the definition of Domain. To find the domain of any given function, we will need to find the set of all the possible values which will satisfy the given function in the question. Domain is basically the set of all the values that the \[x\] in the given function can take, and those values of \[x\] will satisfy the given function.

Complete step by step answer:
We have given that,
\[f\left( x \right)=\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)-\dfrac{3}{{{\log }_{2}}\left| \sin x \right|} \right\}\]
Simplifying the function, we get
\[\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)-\dfrac{3}{{{\log }_{2}}\left| \sin x \right|} \right\}=\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)+{{\log }_{\left| \sin x \right|}}\dfrac{1}{8} \right\}\]
\[\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)-\dfrac{3}{{{\log }_{2}}\left| \sin x \right|} \right\}=\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)+{{\log }_{\left| \sin x \right|}}\dfrac{1}{8} \right\}\]
Now, simplifying further
We will get,
\[\log \left\{ {{\log }_{\left| \sin x \right|}}\left( {{x}^{2}}-8x+23 \right)+{{\log }_{\left| \sin x \right|}}\dfrac{1}{8} \right\}=\log \left\{ {{\log }_{\left| \sin x \right|}}\dfrac{\left( {{x}^{2}}-8x+23 \right)}{8} \right\}\]
Simplifying further,
\[\log \left\{ {{\log }_{\left| \sin x \right|}}\dfrac{\left( {{x}^{2}}-8x+23 \right)}{8} \right\}=\log \left( \dfrac{\dfrac{\left( {{x}^{2}}-8x+23 \right)}{8}}{{{\log }_{\left| \sin x \right|}}} \right)\]
For the domain of the given function;
\[f\left( x \right)\]is defined if \[\log \left( \dfrac{\dfrac{\left( {{x}^{2}}-8x+23 \right)}{8}}{{{\log }_{\left| \sin x \right|}}} \right)>0\]
As we know that,
\[{{\log }_{\left| \sin x \right|}}\]is always negative.
Thus, the numerator must also be negative.
\[\log \left( \dfrac{\left( {{x}^{2}}-8x+23 \right)}{8}>0 \right)\]
Now,
We have,
\[\dfrac{{{x}^{2}}-8x+23}{8}<1\]
Thus,
\[\Rightarrow {{x}^{2}}-8x+23<8\]
\[\Rightarrow {{x}^{2}}-8x+15<0\]
\[\Rightarrow {{x}^{2}}-5x-3x+15<0\]
\[\Rightarrow \left( x-3 \right)\left( x-5 \right)<0\]
Therefore,
\[\Rightarrow x\in \left( 3,5 \right)\]
The above condition is true only if,
\[\left| \sin x \right|\ne 0,1\ and\ \dfrac{{{x}^{2}}-8x+23}{8}<1\]
This is because, as \[\left| \sin x \right|<1\Rightarrow {{\log }_{\left| \sin x \right|}}a>0\Rightarrow a<1\]
Therefore,
\[\Rightarrow x\in \left( 3,5 \right)-\left\{ \pi ,\dfrac{3\pi }{2} \right\}\]
Hence, this is the required answer.

Note: While solving these types of questions where we would have to find the domain of any given function, you need to know that or you have to keep in mind that all the values that go into a given function are called the domain of function. You can also identify the domain from the relation set, which will show the collection of all possible ordered pairs for the given function. The domain is always the set of all the first element i.e. x coordinates of ordered pairs.