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06 Feb 2020, 03:05
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C
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If \(x^2 + y^2 = 1\), what is the the maximum value of \(x^2 + 4xy − y^2\) ?
A. 1
B. √2
C. √5
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
A. 1
B. √2
C. √5
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
24 Apr 2020, 08:52
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Bunuel
Let k = x^2 + 4xy - y^2; so we need to find the maximum value of k. Since x^2 + y^2 = 1, then k(x^2 + y^2) = x^2 + 4xy - y^2. Now let’s divide this equation by y^2, obtaining:
k((x/y)^2 + 1) = (x/y)^2 + 4(x/y) - 1
k(x/y)^2 + k = (x/y)^2 + 4(x/y) - 1
(k - 1)(x/y)^2 - 4(x/y) + k + 1 = 0
Notice that this is a quadratic equation of the form au^2 + bu + c = 0 where u = x/y, a = k - 1, b = -4 and c = k + 1. Since u = x/y is real, the discriminant b^2 - 4ac must be greater than or equal to 0. That is,
(-4)^2 - 4(k - 1)(k + 1) ≥ 0
16 - 4(k^2 - 1) ≥ 0
16 - 4k^2 + 4 ≥ 0
-4k^2 ≥ -20
k^2 ≤ 5
√k^2 ≤ √5
|k| ≤ √5
Since |k| ≤ √5, the maximum value of k is √5.
Answer: C
22 Apr 2020, 13:09
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Since \(x^2+y^2 = 1\)
Put x= cos(a) and y= sin(a)
\(x^2 + 4xy − y^2\)
=\( cos^2(a) + 4cos(a) * sin(a) - sin^2(a) \)
= \([cos^2(a) - sin^2(a)] + 4cos(a) * sin(a) \)
= cos2a + 2sin2a
maximum value of \(m cosθ + n sinθ =\) \(\sqrt{m^2+n^2}\)
maximum value of cos2a + 2sin2a = \(\sqrt{1^2+2^2} = \sqrt{5}\)
Put x= cos(a) and y= sin(a)
\(x^2 + 4xy − y^2\)
=\( cos^2(a) + 4cos(a) * sin(a) - sin^2(a) \)
= \([cos^2(a) - sin^2(a)] + 4cos(a) * sin(a) \)
= cos2a + 2sin2a
maximum value of \(m cosθ + n sinθ =\) \(\sqrt{m^2+n^2}\)
maximum value of cos2a + 2sin2a = \(\sqrt{1^2+2^2} = \sqrt{5}\)
