vrajesh wrote:
If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}?\)
I. \(\frac{\sqrt{x+y}}{2x}\)
II.\(\frac{\sqrt{x} + \sqrt{y}}{\sqrt{x+y}}\)
III. \(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x+y}}\)
A. None
B. I only
C. II only
D. I and III
E. II and III
Can someone please explain what is the best way to solve this problem in under 2 minutes?
I would really like to understand how to solve these type of problems.
What makes this problem especially difficult, is the condition x and y are positive
i.e. x > 0 and y > 0
hence x and y can be any of the following 1/2, 3/4, 1, 2, 3, 4, ....., and more
Okay, so the main fraction can be simplified to the following by means of rationalization.
\(\frac{1}{\sqrt{x+y}}\) = \(\frac{\sqrt{x+y}}{x+y}\)
Now the question asks for something that MUST be true for all the values of x and y, which are positive.
So look at the options:
I: Here, your denominator is 2x. If x+y > 2x, then this option is greater than the given number. If not, it's smaller. So this can't be the answer.
II: This always has to be greater. In our simplified form, our numerator is 1, and here the numerator is \(\sqrt{x} + \sqrt{y}\) which has to be greater than 1, since the smallest possible value that x can take is 1 (Remember 0 is not a positive integer) - So this option is good.
III. \(\sqrt{x} - \sqrt{y}\). This can be greater than or lesser than one depending on the values that x and y takes, so this need not ALWAYS be greater than what's given to us. Hence incorrect.
Thus the final answer is C, just option II. Hope this helps.