if lim x➡0 [x(1+alpha cos2x) + bita sin2x] then the value of [ 5 alpha + 2 bita ]
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Calculating sin2(x)=y and cos2(x)=1−y, then you want d(2y+21−y)/dy=0 you should find y=1/2 which implies x=π/4.
consider z=2sin2x, thus 1⩽z⩽2 and we want to minimize z+2/ zon [1,2] and Derivative 1−2/z2, thus the function decreases on z < 2–√ and increases on z>2–√, the minimal value is at z = 2–√, which happens to be in [1,2], hence min=22–√z .
minx∈R{2sin2x+2cos2x} = mint∈[0,1]{2t+21−t}=minr∈[1,2]{r+2r}=22–√
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