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Question Stats:
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56%
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If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?
I. \(\frac{\sqrt{x+y}}{2}\)
II. \(\frac{\sqrt{x}+\sqrt{y}}{2}\)
III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)
A. I only
B. II only
C. III only
D. I and II only
E. None
I. \(\frac{\sqrt{x+y}}{2}\)
II. \(\frac{\sqrt{x}+\sqrt{y}}{2}\)
III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)
A. I only
B. II only
C. III only
D. I and II only
E. None
08 Aug 2009, 21:46
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Let's consider original statement: \(\frac{1}{\sqrt{x+y}}\)
How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \(\frac{1}{\sqrt{x+y}}\) goes to infinity.
Now, let's see what do our options at x,y ---> 0.
I) \(\frac{\sqrt{x+y}}{2}\) goes to 0 at x,y ---> 0.
II) \(\frac{\sqrt{x}+\sqrt{y}}{2}\) goes to 0 at x,y ---> 0.
III) \(\frac{\sqrt{x}-sqrt{y}}{x+y}\) hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.
So, none of the options.
How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and \(\frac{1}{\sqrt{x+y}}\) goes to infinity.
Now, let's see what do our options at x,y ---> 0.
I) \(\frac{\sqrt{x+y}}{2}\) goes to 0 at x,y ---> 0.
II) \(\frac{\sqrt{x}+\sqrt{y}}{2}\) goes to 0 at x,y ---> 0.
III) \(\frac{\sqrt{x}-sqrt{y}}{x+y}\) hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.
So, none of the options.
11 Dec 2010, 00:39
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mmcooley33
Just to add couple of words to Walker's great solution:
Note that we are asked "which of the following MUST be greater than \(\frac{1}{\sqrt{x+y}}\)?" not COULD be greater.
"MUST BE TRUE" questions:
These questions ask which of the following MUST be true (must be greater in our case), or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.
As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.
So, if we find even one set of \(x\) and \(y\) for which \(\frac{1}{\sqrt{x+y}}\) is greater than option I for example then it'll mean that option I is not ALWAYS greater then \(\frac{1}{\sqrt{x+y}}\).
How can we increase the value of \(\frac{1}{\sqrt{x+y}}\)? Testing extreme examples: if \(x\) and \(y\) are very small (when their values approach zero) then denominator approaches zero and thus the value of the fraction goes to +infinity, becomes very large. In this case:
I. \(\frac{\sqrt{x+y}}{2}\): nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction;
II. \(\frac{\sqrt{x}+\sqrt{y}}{2}\): the same here - nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction;
As for:
III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\), if \(x=y\) then this fraction equals to zero and \(\frac{1}{\sqrt{x+y}}\) has some value more than zero, so this option also is not always more than given fraction;
Answer: none of the options must be greater than the given fraction.
Hope it's clear.
