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

1. x < x^2
=> x(x-1)>0
=> x<0 and x>1

Now, when x is < 0,
The answer to the question will be NO; as in that case x^2 will always be positive and x^3 will always be negative.

And when x>1, the answer will be YES. As in that case x^3 will always be greater than x^2.

Therefore, (1) alone is not sufficient.

2. x<1
We can check this for 3 ranges of x -

A. If x is a +ve fraction, the answer is NO.

B. If x is 0, the answer again will be NO. As both LHS and RHS are equal.

C. If x is -ve, the answer again will be NO.

Therefore, (2) alone is sufficient.

(B)

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Official Solution:


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

We can simplify the question by dividing the inequality by \(x^2\) to get: is \(1 < x\)? We can safely do this because \(x^2\) cannot be negative. Also, note that \(x = 0\) does not satisfy either \(x^2 < x^3\) or the rephrased version \(1 < x\). Therefore, by dividing by \(x^2\), we are not introducing the possibility of \(x\) being 0 for the rephrased version, which ensures the validity of this operation.

(1) \(x < x^2\)

Rewrite the inequality as \(x^2 - x > 0\), and then as \(x(x - 1) > 0\). The roots are \(x=0\) and \(x = 1\). The "\(>\)" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root; therefore, the given inequality holds true for: \(x < 0\) and \(x > 1\). Hence, \(x\) may or may not be greater than 1. Not sufficient.

(2) \(x < 1\)

This statement directly provides a NO answer to the question. Sufficient.


Answer: B