Re: x > 0
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23 Oct 2018, 20:13
While you can skip a lot of steps on an easy question like this, its good to engrain what you would be doing on ANY problem like this, regardless of the difficulty.
The first step should therefore be to read the "variable modifier" (in this case x>0) and determine if it disallows +/-/0/Fractions. x>0 disallows 0 and negatives, but not positives and fractions, and so you know you need to check if the answer is the same when x is a positive fraction and when it is a positive integer. When checking if something is a positive, its good to check both 1 and another number above 1 since sometimes 1 can give you an answer that contradicts, telling you know the answer is actually D. Even without getting to checking if fractions affect the answer, you can see that plugging in 1 gives: Quantity A = 2 and Quantity B = 0, while plugging in 2 gives you: Quantity A= 5 Quantity B=7. Since "when x = 1" gives a contradictory answer to "when x=2" we know that the answer must be D.