fluke wrote:
It's not a good question because it is not formed correctly or because it is subpar or both?
What if the question read, "Which of the following represents the complete range of x over which 15x - (2/x) > 1"?
Then; D would be the only choice, right?
Right.
fluke wrote:
More important question is:
How x<-1/3, -1/3<x<0, 0<x<2/5 and x>2/5; get reduced to -1/3<x<0, x>2/5.
Is there any comprehensive material available for these types of inequalities and ranges apart from the shortcut method suggested by gurpreetsingh?
This is actually quite simple. We want to solve \(\frac{(3x+1)(5x-2)}{x}>0\):
1. Find zeros of the multiples (the values of x for which x, 3x+1 and 5x-2 equal to zero): -1/3, 2/5 and 0;
2. Arrange them in ascending order to get 4 ranges: \(x<-\frac{1}{3}\), \(-\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\);
3. Check extreme value: if \(x\) is some very large number (so some number from the 4th range) then \(\frac{(3x+1)(5x-2)}{x}\) will obviously be positive;
4. Trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative:
- + - +, so expression is positive for 2nd range: \(-\frac{1}{3}<x<0\), and for fourth range: \(x>\frac{2}{5}\).
This works for all such kind of inequalities.
Links in from previous post:
everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here:
inequalities-trick-91482.html xy-plane-71492.html?hilit=solving%20quadratic#p841486One more:
data-suff-inequalities-109078.html