Recently I have been studying high school mathematics in Russia. Here they have classification tests that allow you to get into the top universities. In these tests, there are some interesting problems, and it is not hard to find russian websites dedicated to explaining how such problems may be solved. I have been looking for sites in English with similar problems that may explain them, but I have not been lucky.
The following is a problem involving a parameter value 'a' and you have to find the values of 'a' that make the domain of a function fulfill certain criteria.
For the following function, find all the values of the parameter 'a', for which the domain of the function contains only one integer. y, x and a, may only be real numbers.
$y = \sqrt{a^{x+0.5}+a^{4}\sqrt{x}-x^{0.5+x\log_x a}-a^{4.5}}$
I am mostly interested in finding english sources where I may find more problems like this and maybe even their discussion. If you fancy a solution to the above problem, please let me know the sources that you feel helped you better understand the problem.
I found the problem in this file: (all in Russian)
http://gimnazia5bryansk.narod.ru/romashko/z2.doc
You can find other problems like this in these sites: (also in Russian)
http://postupivuz.ru/vopros/8521.htm
http://xn--i1abbnckbmcl9fb.xn--p1ai/%D1%81%D1%82%D0%B0%D1%82%D1%8C%D0%B8/410368/
NOTE: There are some similar problems in the exchange, but they usually involve making the roots into something specific. I believe this problem is different enough to merit at least a small discussion.
Best Answer
Note that we just want the thing under the sqrt to be non-negative.
$$y = \sqrt{a^{x+0.5}+a^{4}\sqrt{x}-x^{0.5+x\log_x a}-a^{4.5}}$$
So make the thing under the square root non-negative.
$$a^x\sqrt{a}+a^4\sqrt{x}-a^x\sqrt{x}-a^4\sqrt{a}\ge0$$
Thus
$$(a^x-a^4)(\sqrt{a}-\sqrt{x})\ge0$$
Note that this means $x=4$ is always a solution.
If $a\ge4$, then if $x\ge4$, $x\le a$. Therefore if $a\ge 5$ the function is undefined as $x=5$ is a solution. If $x<4$ then the first bracket is negative, the second is positive so we die immediately.
If $a\le4$, then if $x\le4$, $x\ge a$. Therefore if $a\le3$ the function is undefined as $x=3$ is a solution. If $x>4$ then the first term is positive whilst the second is negative so we die immediately.
Therefore the only solution is $a\in(3,5)$, the open interval from $3$ to $5$.
Note: negative $a,x$ die even faster because the term under the sqrt becomes imaginary, which is just bad.
As for where you can find more fun questions like this, basically I got good at maths by solving ISL problems, from here. These questions are designed to require ingenuity and creativity to solve, rather than formula memorisation and the like. Give them a shot! https://artofproblemsolving.com/community/c3223_imo_shortlist