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16 Mar 2023, 13:55
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Dipanjan005
If x = y, say, then the third option will be equal to zero, so will always be less than 1/√(x + y), since that's positive. The third option can even be negative if y > x.
08 Nov 2024, 05:00
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In comparison, we always need to look for something common. To compare fractions, either numerator or denominator must be the same.
I. \(\frac{\sqrt{x+y}}{2x}\) Let's compare it with our given expression.
\(\frac{1}{\sqrt{x+y}} = \frac{\sqrt{x+y}}{x+y}\) (Dividing and multiply by sqrt(x+y))
Their numerators are same so now we just compare denominators. Is 2x ( = x + x) smaller or greater than (x+y)? That simply depends on whether y is smaller or greater than x. Hence this expression can be greater or smaller than the given expression.
II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) Let's compare it with our given expression.
\(\frac{1}{\sqrt{x+y}} = \frac{\sqrt{x+y}}{x+y}\) (Dividing and multiply by sqrt(x+y))
Here denominator is same so we compare numerators. If you recall that \(\sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\) then great. Equality holds when x or y is 0. Since x and y are positive, this expression will always be greater.
Else try some values. It seems likely. But we still need to convince ourselves that it will work in all cases.
\((\sqrt{x}+\sqrt{y})^2 = x + y + 2\sqrt{xy}\)
\((\sqrt{x+y})^2 = x + y\)
So we see that \(\sqrt{x}+\sqrt{y}\) is greater.
Hence II is greater in all cases.
III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)
Since there is a negative in the numerator, it is easy to eliminate this option. It can be negative. Say if x = 1 and y = 4. Our given expression cannot be negative.
Answer (C)
Varun291
Joined: 27 Jun 2021
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11 Jan 2026, 21:10
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Looks like I am doing something wrong which I cant understand. I took x=6 and y =3. That way in the second statement I get root 3 + root 6= root 9/ root 9 which gives me 1/3 which is similar to the initial equation which is 1/3. Hence I get that statement must always not be greater. What am I doing wrong?
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