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The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x

You did all the work

Yes, answer is D

With the work I did above, I thought the answer would be B? I found 2 negative integers for statement 1, 1 which satisfies the condition and one which does not, meaning that statement 1 should be insufficient?
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The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x

You did all the work

Yes, answer is D

With the work I did above, I thought the answer would be B? I found 2 negative integers for statement 1, 1 which satisfies the condition and one which does not, meaning that statement 1 should be insufficient?

Yes, maybe I should read your entire post. Sorry about that. Yes, the answer should be B Where did you find this problem?
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Factorials were someone's attempt to make math look exciting!!!

Hmm. What is the explanation in the back of the book for how they arrived at D? Because even moreover, if you try -1 the inequality is > and if you try -3 the inequality is < and if you try -2, it is an equality --> Its all over the charts!!
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Factorials were someone's attempt to make math look exciting!!!

Is this a proof (1) is sufficient ?? You just test it for 2 values (why -1 and -2 by the way ?) and therefore it is sufficient ? Are you sure it works for x=-1/8 ?

I would say something like: if x<0, then x |x| <0

And since 2^x is positive, then we have x |x| < 2^x (this is no longer that plugging numbers )

The x is small, and half way up the 2, meaning it is an exponent. I hope that helps. If it were 2x, then the x would be on same line as the 2.

My understanding is that this is printed incorrectly in earlier printings of the Guide, and was later corrected. In some books, it is actually printed as 2x, and not as 2^x, in the question (though not in the solution section of the book).
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