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y=cos^2(45+x)+(sin x-cos x)^2.
We kn ow that:-
2.cos ^2A-1=cos2A
or cos^2.A=(1+cos2A)/2
or cos^2.(45+x) =[1+cos(90+2x)]^2=(1-sin2x)/2.
Now y=(1-sin2x)/2+sin^2x+cos^2x-2.sin x.cos x.
y=(1-sin2x)/2+(1-sin2x)=3/2.(1-sin2x)
y= 3/2 -3/2.sin2x.
dy/dx= (-3/2).cos2x.(2) = -3cos2x
For maxima or minima put dy/dx=0
-3cos2x =0
cos2x =0 => 2x =90
x=45°
d2y/dx^2= 6 sin2x
d2y/dx^2 at x=45° is +ve , therefore there exist minima at x= 45°.
Minimum value y=3/2(1-sin90°) =3(1–1) =0
while maximum value = 3
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