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Question Number 150698 by saly last updated on 14/Aug/21

Commented by saly last updated on 14/Aug/21

  Do you help me?

$$\:\:{Do}\:{you}\:{help}\:{me}? \\ $$

Answered by Olaf_Thorendsen last updated on 14/Aug/21

f(x) = 11cos^2 x+10sinxcosx+13sin^2 x  f(x) = 11+10sinxcosx+2sin^2 x  f(x) = 11+5sin2x+(1−cos2x)  f(x) = 12+5sin2x−cos2x  f′(x) = 10cos2x+2sin2x  f′(x) = 0 ⇔ tan2x = −5  cos^2 2x = (1/(1+tan^2 2x)) = (1/(26))  ⇒ sin^2 2x = ((25)/(26))  f_(max)  = 12+5.(5/( (√(26))))+(1/( (√(26)))) = 12+(√(26))

$${f}\left({x}\right)\:=\:\mathrm{11cos}^{\mathrm{2}} {x}+\mathrm{10sin}{x}\mathrm{cos}{x}+\mathrm{13sin}^{\mathrm{2}} {x} \\ $$$${f}\left({x}\right)\:=\:\mathrm{11}+\mathrm{10sin}{x}\mathrm{cos}{x}+\mathrm{2sin}^{\mathrm{2}} {x} \\ $$$${f}\left({x}\right)\:=\:\mathrm{11}+\mathrm{5sin2}{x}+\left(\mathrm{1}−\mathrm{cos2}{x}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{12}+\mathrm{5sin2}{x}−\mathrm{cos2}{x} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{10cos2}{x}+\mathrm{2sin2}{x} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:\mathrm{tan2}{x}\:=\:−\mathrm{5} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{x}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{2}{x}}\:=\:\frac{\mathrm{1}}{\mathrm{26}} \\ $$$$\Rightarrow\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:=\:\frac{\mathrm{25}}{\mathrm{26}} \\ $$$${f}_{{max}} \:=\:\mathrm{12}+\mathrm{5}.\frac{\mathrm{5}}{\:\sqrt{\mathrm{26}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{26}}}\:=\:\mathrm{12}+\sqrt{\mathrm{26}} \\ $$

Commented by saly last updated on 16/Aug/21

 Thank you

$$\:{Thank}\:{you} \\ $$

Answered by liberty last updated on 15/Aug/21

y=11(((1+cos 2x)/2))+5sin 2x+13(((1−cos 2x)/2))  y=12+5sin 2x−cos 2x  y=12+(√(26)) sin (2x+β) ; β=tan^(−1) (−(1/5))  y_(max) =12+(√(26))  y_(min) =12−(√(26))

$$\mathrm{y}=\mathrm{11}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right)+\mathrm{5sin}\:\mathrm{2x}+\mathrm{13}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right) \\ $$$$\mathrm{y}=\mathrm{12}+\mathrm{5sin}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{2x} \\ $$$$\mathrm{y}=\mathrm{12}+\sqrt{\mathrm{26}}\:\mathrm{sin}\:\left(\mathrm{2x}+\beta\right)\:;\:\beta=\mathrm{tan}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\mathrm{y}_{\mathrm{max}} =\mathrm{12}+\sqrt{\mathrm{26}} \\ $$$$\mathrm{y}_{\mathrm{min}} =\mathrm{12}−\sqrt{\mathrm{26}}\: \\ $$

Commented by saly last updated on 16/Aug/21

Thank you

$${Thank}\:{you} \\ $$

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