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Question Number 20653 by Tinkutara last updated on 30/Aug/17

$$\mathrm{Length}\mathrm{of}\mathrm{interval}\mathrm{of}\mathrm{range}\mathrm{of}\mathrm{function}$$$$f\left(\theta \right)={\cos }^2\theta -6\sin \theta \cos \theta +3{\sin }^2\theta +2$$$$\mathrm{is}$$$$\left(1\right)8$$$$\left(2\right)-8$$$$\left(3\right)\sqrt{10}$$$$\left(4\right)2\sqrt{10}$$

Answered by dioph last updated on 31/Aug/17

$$f\left(\theta \right)={\cos }^2\theta -6\sin \theta \cos \theta +{3\sin }^2\theta +2$$$$\Rightarrow f\left(\theta \right)=1+2{\sin }^2\theta -3\sin 2\theta +2$$$$\Rightarrow f\left(\theta \right)=1-\cos 2\theta -3\sin 2\theta +3$$$$\Rightarrow f\left(\theta \right)=4-\cos 2\theta -3\sin 2\theta$$$$f{}^{\prime }\left(\theta \right)=2\sin 2\theta -6\cos 2\theta$$$$f{}^{\prime }\left(\theta \right)=0\iff \sin 2\theta =3\cos 2\theta$$$$\iff \{\begin{array}{l}\sin 2\theta =3/\sqrt{10}\\ \cos 2\theta =1/\sqrt{10}\end{array}\mathrm{or}\{\begin{array}{l}\sin 2\theta =-3/\sqrt{10}\\ \cos 2\theta =-1/\sqrt{10}\end{array}$$$$f\left(\theta \right)=4-\frac{1}{\sqrt{10}}-\frac{9}{\sqrt{10}}=4-\sqrt{10}\left(\min \right)$$$$f\left(\theta \right)=4+\frac{1}{\sqrt{10}}+\frac{9}{\sqrt{10}}=4+\sqrt{10}\left(\max \right)$$$$range=2\sqrt{10}$$

Commented by dioph last updated on 31/Aug/17

$$\mathrm{I}\mathrm{am}\mathrm{sorry},\mathrm{I}\mathrm{forgot}\mathrm{to}\mathrm{add}1,\mathrm{it}\mathrm{is}$$$$\mathrm{corrected}$$

Answered by mrW1 last updated on 31/Aug/17

$$f\left(\theta \right)={\cos }^2\theta -6\sin \theta \cos \theta +3{\sin }^2\theta +2$$$$f\left(\theta \right)=3-3\sin 2\theta +2{\sin }^2\theta$$$$f\left(\theta \right)=4-(3\sin 2\theta +\cos 2\theta )$$$$f\left(\theta \right)=4-\sqrt{10}(\frac{3}{\sqrt{10}}\sin 2\theta +\frac{1}{\sqrt{10}}\cos 2\theta )$$$$f\left(\theta \right)=4-\sqrt{10}(\cos \alpha \sin 2\theta +\sin \alpha \cos 2\theta )$$$$f\left(\theta \right)=4-\sqrt{10}\sin (2\theta +\alpha )$$$$\max .\mathrm{f}\left(\theta \right)=4+\sqrt{10}$$$$\min .\mathrm{f}\left(\theta \right)=4-\sqrt{10}$$$$\max .-\min .=2\sqrt{10}$$$$\Rightarrow \mathrm{answer}\left(4\right)$$

Commented by Tinkutara last updated on 31/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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