Bunuel wrote:
Official Solution:
Is x greater than 1?
(1) \(\frac{1}{x} \gt - 1\). Rewrite as \(\frac{1+x}{x} \gt 0\), two cases:
A. \(x \gt 0\) and \(1+x \gt 0\), (\(x \gt -1\)). So, \(x \gt 0\);
B. \(x \lt 0\) and \(1+x \lt 0\), (\(x \lt -1\)). So, \(x \lt -1\).
So, the given inequality holds true in two ranges: \(x \gt 0\) and \(x \lt -1\), thus \(x\) may or may not be greater than 1. Not sufficient.
(2) \(\frac{1}{x^5} \gt \frac{1}{x^3}\). Rewrite as \(\frac{1-x^2}{x^5} \gt 0\), two cases:
A. \(x \gt 0\) (it's the same as \(x^5 \gt 0\)) and \(1-x^2 \gt 0\), (\(-1 \lt x \lt 1\)). So, \(0 \lt x \lt 1\);
B. \(x \lt 0\) and \(1-x^2 \lt 0\), (\(x \lt -1\) or \(x \gt 1\)). So, \(x \lt -1\);
We got that given inequality holds true in two ranges: \(0 \lt x \lt 1\) and \(x \lt -1\), ANY \(x\) from this ranges will be less than 1. So the answer to our original question is NO. Sufficient.
Answer: B
Hi Bunuel,
I did it like this:
Stmt1: \(\frac{1}{x} >-1\)
To make left side between -1 and 0, x must be negative.
To make left side greater than 0, x must be positive.
Not sufficient.
Stmt2: \(\frac{1}{x^5} > \frac{1}{x^3}\)
Multiply both sides by \(x^2\) => \(\frac{1}{x^3} >\frac{1}{x}\)
Now: \(\frac{1}{x^3} >\frac{1}{x}\)is true only if x<1
Sufficient.
Answer B.
Tell me if I did something wrong.