^{2}x + cos

^{4}x

A= sin^{2}x + cos^{4}x

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It may be worthwhile in general to confirm via the second derivative test as such:

Re-writing dA/dx as -sin(2x)cos(2x)= -sin(4x)/2, we can easily compute the second derivative: -2cos(4x)

At x=0, the second derivative is <0, and thus this is indeed a maximum-generating point

At x= pi/4, the second derivative is >0 and thus this is indeed a minimum-generating point

(no boundary value conditions stated)

A=sin^{2}x + (1-sin^{2}x)^{2}

A= sin^{2}x + 1 -2sin^{2}x + sin^{4}x

A = sin^{4}x - sin^{2}x + 1

Now to find the max and min vale of A we will find dA/dx and equal it to 0

dA / dx = 4sin^{3}xcosx - 2sinxcosx = 0

2sinxcox (2sin^{2}x - 1) = 0

Sin2x (2sin^{2}x-1 ) = 0

So Either Sin2x=0 or 2sin^{2}x-1 = 0

if sin2x=0, we have x=0,90,180 degree and so on

if 2sin^{2}x-1 = 0, we have x=45, 135, 225 degree and so on^{
}

So when x = 0 degree , A = 1

when x = 45 degree, A = 3/4

Max and min value of A is 1 and 3/4 respectively

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