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What is max and min value of A

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2 Answers

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=sin2x + (1-sin2x)2
A= sin2x + 1 -2sin2x + sin4x
A = sin4x - sin2x + 1
 
Now to find the max and min vale of A we will find dA/dx and equal it to 0
 
dA / dx = 4sin3xcosx - 2sinxcosx = 0
 
2sinxcox (2sin2x - 1) = 0
Sin2x (2sin2x-1 ) = 0
 
So Either Sin2x=0 or 2sin2x-1 = 0
 
if sin2x=0, we have x=0,90,180 degree and so on
if 2sin2x-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