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Question Number 118546 by putriamalia last updated on 18/Oct/20

$$nilaimaksimum?fungsiy=1+sin2x+cos2x$$

Commented by mr W last updated on 18/Oct/20

$$y=1+\sin 2x+\cos 2x$$$$=1+\sqrt{2}\sin (2x+\frac{\pi }{4})$$$$⩽1+\sqrt{2}$$$$\Rightarrow maximum=1+\sqrt{2}$$

Commented by benjo_mathlover last updated on 18/Oct/20

$$hellomrW,howareyousir?$$

Commented by mr W last updated on 18/Oct/20

$$thingsareok.thankssir!$$

Commented by benjo_mathlover last updated on 18/Oct/20

$$doyourememberme?$$

Commented by benjo_mathlover last updated on 18/Oct/20

I am a long time member of this forum, but have been inactive for a while. I forgot my old account password. we've argued about math problems. haha .. at that time sir said my results were just a coincidence

Commented by mr W last updated on 18/Oct/20

$$irememberIDcontainingbenjo.$$$$whichIDwereyouusing?$$

Commented by benjo_mathlover last updated on 18/Oct/20

$$yes.youareright$$

Answered by TANMAY PANACEA last updated on 18/Oct/20

$$y=1+sin2x+cos2x$$$$y=1+\sqrt{2}(\frac{1}{\sqrt{2}}sin2x+\frac{1}{\sqrt{2}}cos2x)$$$$y=1+\sqrt{2}sin(2x+\frac{\pi }{4})$$$$maxvalueofsin(2x+\frac{\pi }{4})=1$$$$minvalueofsin(2x+\frac{\pi }{4})=-1$$$$y_{max}=1+\sqrt{2}\times 1=1+\sqrt{2}$$$$y_{min}=1-\sqrt{2}$$

Answered by mathmax by abdo last updated on 18/Oct/20

$$\mathrm{y}\left(\mathrm{x}\right)=1+\sin \left(2\mathrm{x}\right)+\cos \left(2\mathrm{x}\right)\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=1+\sqrt{2}\cos (2\mathrm{x}-\frac{\pi }{4})\Rightarrow$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=-2\sqrt{2}\sin (2\mathrm{x}-\frac{\pi }{4})$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=0\Rightarrow 2\mathrm{x}-\frac{\pi }{4}=\mathrm{k}\pi \Rightarrow 2\mathrm{x}=\mathrm{k}\pi +\frac{\pi }{4}\Rightarrow \mathrm{x}=\frac{\pi }{8}+\mathrm{k}\frac{\pi }{2}$$$$\mathrm{the}\mathrm{period}\mathrm{of}\mathrm{y}\mathrm{is}\pi \mathrm{so}\mathrm{we}\mathrm{search}\mathrm{the}\mathrm{solution}\mathrm{on}[0,\pi ]$$$$0⩽\frac{\pi }{4}+\frac{\mathrm{k}\pi }{2}⩽\pi \Rightarrow \mathrm{o}⩽\frac{1}{4}+\frac{\mathrm{k}}{2}⩽1\Rightarrow -\frac{1}{4}⩽\frac{\mathrm{k}}{2}⩽\frac{3}{4}\Rightarrow$$$$-\frac{1}{2}⩽\mathrm{k}⩽\frac{3}{2}\Rightarrow \mathrm{k}=0\mathrm{or}1\Rightarrow \mathrm{x}=\frac{\pi }{8}\mathrm{or}\mathrm{x}=\frac{5\pi }{8}$$$$\mathrm{y}\left(\frac{\pi }{8}\right)=1+\sqrt{2}\cos \left(0\right)=1+\sqrt{2}\mathrm{after}\mathrm{dressing}\mathrm{the}\mathrm{variation}\mathrm{we}$$$$\mathrm{get}\max \mathrm{f}\left(\mathrm{x}\right)=1+\sqrt{2}$$

Answered by 1549442205PVT last updated on 18/Oct/20

$${(\sin 2\mathrm{x}+\cos 2\mathrm{x})}^2=2({\sin }^{2}2\mathrm{x}+{\cos }^{2}2\mathrm{x})$$$$-{(\sin 2\mathrm{x}-\cos 2\mathrm{x})}^2⩽2({\sin }^{2}2\mathrm{x}+{\cos }^{2}2\mathrm{x})=2$$$$\iff \mid \sin 2\mathrm{x}+\cos 2\mathrm{x}\mid ⩽\sqrt{2}$$$$\Rightarrow -\sqrt{2}⩽\sin 2\mathrm{x}+\cos 2\mathrm{x}⩽\sqrt{2}$$$$\Rightarrow 1-\sqrt{2}⩽\mathrm{y}=1+\sin 2\mathrm{x}+\cos 2\mathrm{x}⩽1+\sqrt{2}$$$$\Rightarrow {\mathrm{y}}_{\min }=1-\sqrt{2}\mathrm{when}\sin 2\mathrm{x}=\cos 2\mathrm{x}=-\sqrt{2}/2$$$$\iff 2\mathrm{x}=-\frac{\pi }{4}+2\mathrm{k}\pi \iff \mathrm{x}=-\frac{\pi }{8}+\mathrm{k}\pi$$$${\mathrm{y}}_{\max }=1+\sqrt{2}\mathrm{when}\sin 2\mathrm{x}=\cos 2\mathrm{x}=\sqrt{2}/4$$$$\iff 2\mathrm{x}=\frac{\pi }{4}+2\mathrm{m}\pi \iff \mathrm{x}=\frac{\pi }{8}+\mathrm{m}\pi$$

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