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Differential Calculus

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30 Nov 2011 18:55:23 IST

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determine extreme value of a^2cos^x+b^2sin^x.show further that if a^2>b^2,then a^2 is maximum and b^

Mathematics

determine extreme value of a^2cos^x+b^2sin^x.show further that if a^2>b^2,then a^2 is maximum and b^2 is minimum

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Soumik Halder

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6 Dec 2011 13:38:41 IST

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The max value of psinx+qcosx is [p^2+q^2]^(1/2)

Thus, here the max value is [a^4+b^4]^1/2

Bipin Dubey

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29 Jan 2012 17:40:39 IST

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$a^{2}cos^{2}x+b^{2}sin^{2}x=a^{2}(1-sin^{2}x)+ b^{2}sin^{2}x$

$=a^{2}+(b^{2}-a^{2})sin^{2}x$

If a^2 > b^2 then maximum value occurs at sinx=0 which is a^2 and minimum value occurs at sinx=1 which is b^2.

Similarly for b^2 > a^2, max. value is b^2 and min. value is a^2

Therefore maximum value = max(a^2,b^2) and minimum value = min(a^2,b^2)

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