For what range of values of x will the inequality

\(15x-\frac{2}{x}>1\) ---> \(\frac{15x^2-x-2}{x}>0\) --> \(\frac{(3x+1)(5x-2)}{x}>0\) -->3 expressions change their signs at: -1/3, 0, 2/5, so 4 ranges: \(x<-\frac{1}{3}\), \(-\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) --> in rightmost range expression is positive: so -+-+ --> \(-\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\).

Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x-\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x-\frac{2}{x}>1\) holds true than the answer could be D as well as A.

P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero.

Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486

Not a good question.