Is x < x^2 < x^3? (1) x < x^3 (2) x^3 > x^2

Is \(x < x^2 < x^3\)?

Observe that the question basically asks whether x > 1. That's because x cannot be negative since x^2 < x^3, and x cannot be a proper positive fraction since x < x^2 (or since x^2 < x^2). However, if x > 1, then x < x^2 < x^3 holds true.

(1) \(x < x^3\)

x - x^3 < 0
x(1 - x^2) < 0
x(1 - x)(1 + x) < 0

Here is how to solve the above inequality easily. The "roots", in ascending order, are -1, 0, and 1, which gives us 4 ranges:

\(x < -1\);

\(-1 < x < 0\);

\(0 < x < 1\);

\(x > 1\).

Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 10, then two of the multiples will be positive and one, 1 - x, will be negative, giving a negative result for the whole expression. So when \(x > 1\), the expression is negative. Now the trick: as in the 4th range, the expression is negative, then in the 3rd it'll be positive, in the 2nd it'll be negative, and finally in the 1st, it'll be positive: \(\text{(+ - + -)}\). So, the ranges when the expression is negative are: \(-1 < x < 0\) and \(x > 1\).

Hence, x > 1 may or may not be true . Not sufficient.

P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative are: \(-2 < x < 4\).

(2) \(x^3 > x^2\)

The above is only true when x > 1. Neither negative numbers nor a proper positive fraction satisfy x^3 > x^2. However, if x > 1, then x^3 > x^2 holds true. Hence, we have a YES answer to the question.

Answer: B.