saikiranjadav wrote:
Is p > q ?
(1) p^2 > q^2
(2) p^3 > q^3
Condition 1) \(p^2 - q^2 > 0\)
\((p+q)(p-q) > 0\)
\(p + q > 0\) implies \(p - q > 0\) or \(p > q\).
\(p - q < 0\) implies \(p - q < 0\) or \(p < q\).
Thus we have \(p = 2\), \(q = 1\) and \(p = -2\), \(q = -1\).
Thus this condition is not sufficient.
Condition 2) \(p^3 > q^3\)
\(p^3 - q^3 = (p-q)(p^2+pq+q^2) > 0\)
Without loss of generality, we can assume \(q \ne 0\).
\(p^2 + pq + q^2\)
\(= q^2 \{ \frac{p^2}{q^2} + \frac{p}{q} + 1 \}\)
\(= q^2 \{ (\frac{p}{q})^2 + 2(\frac{1}{2})(\frac{p}{q}) + 1 \}\)
\(= q^2 \{ (\frac{p}{q})^2 + 2(\frac{1}{2})(\frac{p}{q}) + \frac{1}{4} + \frac{3}{4} \}\)
\(= q^2 \{ ( \frac{p}{q} + \frac{1}{2} )^2 + \frac{3}{4} \} \ge \frac{3}{4} q^2 > 0\)
Thus we have \(p^2 + pq + q^2 > 0\).
Since \(p^3 - q^3 > 0\), \(p^2 + pq + q^2 > 0\) and \(p^3 - q^3 = (p-q)(p^2 + pq + q^2) > 0\), we have \(p - q > 0\) or \(p > q\).
This is sufficient.
Answer: B
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