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by vk_vinayak » Sun Sep 02, 2012 4:11 am
shrey2287 wrote:x2 + y2 = 100. All of the following could
be true EXCEPT
A. |x| + |y| = 10
B. |x| > |y|
C. |x| > |y| +10
D. |x| = |y|
E. |x| - |y| = 5
If C is true, it means that x^2 > 100. Since y^2 can't be negative, C can't be possible. Ans C.
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by neelgandham » Sun Sep 02, 2012 9:35 am
x^2 + y^2 = 100. All of the following could be true EXCEPT
A. |x| + |y| = 10

If x = 0 and y = 10, then x2 + y2 = 100 and |x| + |y| = 10.
B. |x| > |y|
If x = 8 and y = 6, then x2 + y2 = 100 and |x| > |y|.
C. |x| > |y| +10
Squaring on both sides - |x|^2 > (|y| +10)^2
Since |x| and |y| +10 are positive integers and integers greater than 1, the inequality doesn't change when squared.
|x|^2 > |y|^2 + 20|y| + 100
Add |y|^2 on both sides,
|x|^2 + |y|^2 > |y|^2 + |y|^2 + 20|y| + 100.
|x|^2 + |y|^2 > 2*|y|^2 + 20|y| + 100
Replace |x|^2 + |y|^2 = 100, replace |x|^2 + |y|^2 by 100.
100 > 2*|y|^2 + 20|y| + 100
0 > 2*|y|^2 + 20|y|. Which is not possible because the least value of 2*|y|^2 + 20|y| is 0 when y = 0. So, the value of 2*|y|^2 + 20|y| can never be less than 0.
D. |x| = |y|
if x = y = 5√2, then |x| = |y| and x^2 + y^2 = 100.
E. |x| - |y| = 5
|x| = |y| + 5
We already know that, x^2 + y^2 = 100. Replace the value of x with |y| + 5.
(|y| + 5)^2 + y^2 = 100.
y^2 + 10|y| + 25 + y^2 = 100.
2*y^2 + 10|y| = 75
|y| = 4.p and |x|=5.k.

IMO C
Anil Gandham
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