Prove y=4sin x/(2+cos x) - x that y is an increasing funciton
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SOLUTION;-
y=4sinθ2+cosθ−θHence for the function to be increasing, y'>0Hence on differentiation we gety' =4cosθ(2+cosθ)−4sinθ(−sinθ)(2+cosθ)2−1y' =8cosθ+ 4cos2θ+4sin2θ(2+cosθ)2−1Or y' =8cosθ+ 4(2+cosθ)2−1Or y' = 8cosθ+ 4−(4 +cos2θ+4cosθ)(2+cosθ)2 y' =4cosθ−cos2θ(2+cosθ)2So y' =cosθ(4−cosθ)(2+cosθ)2So cosθ is positive in (0,π/2), (4−cosθ) >0, as cosθ max value can be 1 only, and the denominator is always positive.Hence y' is always positive in (0,π/2)So y=4sinθ2+cosθ−θ is increasing in (0,π/2
SOLUTION;-
y=4sinθ2+cosθ−θHence for the function to be increasing, y'>0Hence on differentiation we gety' =4cosθ(2+cosθ)−4sinθ(−sinθ)(2+cosθ)2−1y' =8cosθ+ 4cos2θ+4sin2θ(2+cosθ)2−1Or y' =8cosθ+ 4(2+cosθ)2−1Or y' = 8cosθ+ 4−(4 +cos2θ+4cosθ)(2+cosθ)2 y' =4cosθ−cos2θ(2+cosθ)2So y' =cosθ(4−cosθ)(2+cosθ)2So cosθ is positive in (0,π/2), (4−cosθ) >0, as cosθ max value can be 1 only, and the denominator is always positive.Hence y' is always positive in (0,π/2)So y=4sinθ2+cosθ−θ is increasing in (0,π/2
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y=4sinθ2+cosθ−θ
Hence for the function to be increasing, y'>0
Hence on differentiation we
gety' =4cosθ(2+cosθ)−4sinθ(−sinθ)(2+cosθ)2−1y' =8cosθ+ 4cos2θ+4sin2θ(2+cosθ)2−1
Or
y' =8cosθ+ 4(2+cosθ)2−1Or y' = 8cosθ+ 4−(4 +cos2θ+4cosθ)(2+cosθ)2 y' =4cosθ−cos2θ(2+cosθ)2So y' =cosθ(4−cosθ)(2+cosθ)2So cosθ is positive in (0,π/2), (4−cosθ) >0, as cosθ max value can be 1 only, and the denominator is always positive.Hence y' is always positive in (0,π/2)So y=4sinθ2+cosθ−θ is increasing in (0,π/2)
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Hence for the function to be increasing, y'>0
Hence on differentiation we
gety' =4cosθ(2+cosθ)−4sinθ(−sinθ)(2+cosθ)2−1y' =8cosθ+ 4cos2θ+4sin2θ(2+cosθ)2−1
Or
y' =8cosθ+ 4(2+cosθ)2−1Or y' = 8cosθ+ 4−(4 +cos2θ+4cosθ)(2+cosθ)2 y' =4cosθ−cos2θ(2+cosθ)2So y' =cosθ(4−cosθ)(2+cosθ)2So cosθ is positive in (0,π/2), (4−cosθ) >0, as cosθ max value can be 1 only, and the denominator is always positive.Hence y' is always positive in (0,π/2)So y=4sinθ2+cosθ−θ is increasing in (0,π/2)
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