gaurav1418z wrote:
Bunuel, thanks for the wonderful set, i had a question
. Is x^4 + y^4 > z^4 ?
(1) x^2 + y^2 > z^2
(2) x+y > z
x^4 + y^4 = ((x^2) + (y^2))^2 - 2*x^2*y^2
So basically we want to find out if ((x^2) + (y^2))^2 - 2*x^2*y^2 > z^4
OR if
((x^2) + (y^2))^2 > z^4 + 2*x^2*y^2
Now by (1), we know that x^2 + y^2 > z^2
Or
((x^2) + (y^2))^2 > z^4 ( squaring both sides, both sides are absolutely positive)...... (a)
We also know that 2*x^2*y^2 will be less than ((x^2) + (y^2))^2......(b)
Combining (a) and (b), can we not conclude that ((x^2) + (y^2))^2 > z^4 + 2*x^2*y^2 and hence making (1) sufficient
Is \(x^4+y^4>z^4\)? The best way to deal with this problem is plugging numbers. Remember on DS questions when plugging numbers,
goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.
(1) \(x^2+y^2>z^2\)
It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).
For NO answer let's try numbers from Pythagorean triples:
\(x^2=3\), \(y^2=4\) and \(z^2=5\) (\(x^2+y^2=7>5=z^2\)) --> \(x^4+y^4=9+16=25=z^4\), so we have answer NO (\(x^4+y^4\) is NOT more than \(z^4\), it's equal to it).
Not sufficient.
(2) \(x+y>z\). This one is even easier: again we can get YES answer with big x and y, and small z.
As for NO try to make z some big enough negative number: so if \(x=y=1\) and \(z=-5\), then \(x^4+y^4=1+1=2<25=z^4\).
Not sufficient.
(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of \(x^2=3\), \(y^2=4\) and \(z^2=5\) again: \(x+y=\sqrt{3}+\sqrt{4}>\sqrt{5}\) (\(\sqrt{3}+2\) is more than 3 and \(\sqrt{5}\) is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied (\(x^2+y^2=7>5=z^2\)) and \(x^4+y^4=9+16=25=z^4\). Not sufficient.
Answer: E.
OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-4-y-4-z-101358.html _________________