Is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? (1) x > 0 (2) x < 1

Official Solution:


Is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}?\)

Let's simplify the question:

Is \(1+x+x^2+x^3+x^4+x^5-\frac{1}{1-x} < 0?\)

Is \(\frac{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}{1-x} < 0?\)

Is \(\frac{(1+x+x^2+x^3+x^4+x^5-x-x^2-x^3-x^4-x^5-x^6)-1}{1-x} < 0?\)

Is \(\frac{-x^6}{1-x} < 0?\)

Is \(\frac{x^6}{x-1} < 0?\)

For \(\frac{x^6}{x-1}\) to be negative \(x\) must not be 0 (because if \(x=0\), then \(\frac{x^6}{x-1}\) would be 0) AND \(x-1\) must be negative. In this case we'd have \(\frac{x^6}{x-1}=\frac{positive}{negative}=negative\). So, the question basically asks whether \(x < 1\) (\(x -1 < 0\) means \(x < 1 \)), bearing in mind that \(x\) cannot be 0.

(1) \(x > 0\). Not sufficient.

(2) \(x < 1\). Not sufficient because if \(x = 0\), then \(\frac{x^6}{x-1}=0\), not \( < 0 \).

(1)+(2) When combining the statements we get \(0 < x < 1\), so \(\frac{x^6}{x-1} < 0 \). Sufficient.


Answer: C