Would you mind giving more of an explanation for the formal reasoning behind the answer? I can't quite understand it lol
It's probably important that we look at the original wording of the question (
VCAA 2014 Exam 2 - Question 13). Also note that in the original question, you are told that the domain is chosen so that \(h\) is one-to-one, instead of being told that the inverse of \(h\) exists.
Notice that it actually says
"Which one of the following could be the domain?" The word "could" is actually pretty significant in that there are actually infinitely many domains that could be picked to make \(h\) a one-to-one function. So, the approach I showed, although should be understood, isn't actually that important. This question is more about just checking each option, which can be done with a graph.
However, if you didn't want to draw any graphs, what I wrote above leads you to the correct answer, so I'll go through it in more detail.
For clarity, I'm defining \(g(x)=\log_a(x)\), where \(a>1\), and \(f(x)=\cos(x)\), so that \(h(x)=f(g(x))\).
The fact that \(g\) is strictly increasing means essentially: the bigger the input, the bigger the output. This would guarantee therefore, that given \(a<x<b\), we would have \(g(a)<g(x)<g(b)\).
So, if \(1\leq x\leq a^{\pi/2}\), we know then that \(0\leq g(x)\leq \pi/2\). Since \(g(x)\) is being inputted into \(f\), the only question that remains is: Is \(f(x)\) a one-to-one function (or monotone) on the interval \([0,\,\pi/2]\)? By either, imagining the graph of \(f(x)=\cos(x)\), or using the unit circle, the answer is yes.
Hence,
C is the correct answer.