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FIND THE MAXIMUM VALUE OF 1/(SIN^2X+3SINXCOSX+5COS^2X) please help me with this question.

FIND THE MAXIMUM VALUE OF 1/(SIN^2X+3SINXCOSX+5COS^2X) please help me with this question.

Grade:11

2 Answers

Nikitha
23 Points
6 years ago
So for maximum value the value in the denominator should be minimum.Let us consider only the denominator now=Sin^2x+3sinxcosx+5 cos^2x=1+4cos^2x+3sinxcosx ( as sin^2x+cos^2x=1)=1+cosx(4cosx+3sinx)By observing we can say that the minimum value will be 1 when x=90, cosx=0, sinx=1Now let us check if this is the minimum value by sign of second order differential equation method.By differentiating once we get-8cosxsinx-3sin^2x+3cos^2xAnd by differentiating again we get-8cos^2x+8sin^2x-6sinxcosx-6cosxsinxWhen we substitute x as 90,We get the equation to be equal to 8 which is positive and therefore the value is the minimum for that particular equation.Therefore the minimum value of the denominator will be 1 and therefore the maximum value of the given fraction will also be equal to 1.
shivprasad Suryawanshi
13 Points
6 years ago
Enter your answer here...for max value denominator should be min sinx has min value 0 but in denominator it would be invalid so we take min value as1/2 so1÷1/2 it will be 2

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