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Length-of-interval-of-range-of-function-f-cos-2-6-sin-cos-3-sin-2-2-is-1-8-2-8-3-10-4-2-10-

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Question Number 20653 by Tinkutara last updated on 30/Aug/17
Length of interval of range of function f(θ) = cos^2 θ − 6 sin θ cos θ + 3 sin^2 θ + 2 is (1) 8 (2) −8 (3) (√(10)) (4) 2(√(10))
$$\mathrm{Length}\:\mathrm{of}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function} \\ $$$${f}\left(\theta\right)\:=\:\mathrm{cos}^{\mathrm{2}} \:\theta\:−\:\mathrm{6}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:+\:\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{2} \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{8} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{8} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{10}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\sqrt{\mathrm{10}} \\ $$
Answered by dioph last updated on 31/Aug/17
f(θ) = cos^2 θ−6 sin θ cos θ+3sin^2 θ+2 ⇒ f(θ) = 1+2 sin^2 θ − 3 sin 2θ + 2 ⇒ f(θ) = 1−cos 2θ − 3 sin 2θ + 3 ⇒ f(θ) = 4 − cos 2θ − 3 sin 2θ f ′(θ) = 2sin 2θ − 6 cos 2θ f ′(θ) = 0 ⇔ sin 2θ = 3 cos 2θ ⇔ { ((sin 2θ = 3/(√(10)))),((cos 2θ = 1/(√(10)))) :} or { ((sin 2θ = −3/(√(10)))),((cos 2θ = −1/(√(10)))) :} f(θ) = 4 − (1/( (√(10)))) − (9/( (√(10)))) = 4−(√(10)) (min) f(θ) = 4 + (1/( (√(10)))) + (9/( (√(10)))) = 4+(√(10)) (max) range = 2(√(10))
$${f}\left(\theta\right)\:=\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{6}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta+\mathrm{3sin}^{\mathrm{2}} \:\theta+\mathrm{2} \\ $$$$\Rightarrow\:{f}\left(\theta\right)\:=\:\mathrm{1}+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:−\:\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:+\:\mathrm{2} \\ $$$$\Rightarrow\:{f}\left(\theta\right)\:=\:\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\:−\:\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:+\:\mathrm{3} \\ $$$$\Rightarrow\:{f}\left(\theta\right)\:=\:\mathrm{4}\:−\:\mathrm{cos}\:\mathrm{2}\theta\:−\:\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$${f}\:'\left(\theta\right)\:=\:\mathrm{2sin}\:\mathrm{2}\theta\:−\:\mathrm{6}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$${f}\:'\left(\theta\right)\:=\:\mathrm{0}\:\Leftrightarrow\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{3}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Leftrightarrow\:\begin{cases}{\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{3}/\sqrt{\mathrm{10}}}\\{\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{1}/\sqrt{\mathrm{10}}}\end{cases}\:\mathrm{or}\:\begin{cases}{\mathrm{sin}\:\mathrm{2}\theta\:=\:−\mathrm{3}/\sqrt{\mathrm{10}}}\\{\mathrm{cos}\:\mathrm{2}\theta\:=\:−\mathrm{1}/\sqrt{\mathrm{10}}}\end{cases} \\ $$$${f}\left(\theta\right)\:=\:\mathrm{4}\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}\:−\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{10}}}\:=\:\mathrm{4}−\sqrt{\mathrm{10}}\:\left(\mathrm{min}\right) \\ $$$${f}\left(\theta\right)\:=\:\mathrm{4}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}\:+\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{10}}}\:=\:\mathrm{4}+\sqrt{\mathrm{10}}\:\left(\mathrm{max}\right) \\ $$$${range}\:=\:\mathrm{2}\sqrt{\mathrm{10}} \\ $$
Commented by dioph last updated on 31/Aug/17
I am sorry, I forgot to add 1, it is corrected
$$\mathrm{I}\:\mathrm{am}\:\mathrm{sorry},\:\mathrm{I}\:\mathrm{forgot}\:\mathrm{to}\:\mathrm{add}\:\mathrm{1},\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{corrected} \\ $$
Answered by mrW1 last updated on 31/Aug/17
f(θ) = cos^2 θ − 6 sin θ cos θ + 3 sin^2 θ + 2 f(θ) = 3 − 3 sin 2θ + 2 sin^2 θ f(θ) = 4 − (3 sin 2θ +cos 2θ) f(θ) = 4 − (√(10))((3/( (√(10)))) sin 2θ +(1/( (√(10))))cos 2θ) f(θ) = 4 − (√(10))(cos α sin 2θ +sin α cos 2θ) f(θ) = 4 − (√(10)) sin (2θ+α) max. f(θ)=4+(√(10)) min. f(θ)=4−(√(10)) max.−min.=2(√(10)) ⇒answer (4)
$${f}\left(\theta\right)\:=\:\mathrm{cos}^{\mathrm{2}} \:\theta\:−\:\mathrm{6}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:+\:\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{2} \\ $$$${f}\left(\theta\right)\:=\:\mathrm{3}\:−\:\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:+\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta\: \\ $$$${f}\left(\theta\right)\:=\:\mathrm{4}\:−\:\left(\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:+\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$${f}\left(\theta\right)\:=\:\mathrm{4}\:−\:\sqrt{\mathrm{10}}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}\:\mathrm{sin}\:\mathrm{2}\theta\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$${f}\left(\theta\right)\:=\:\mathrm{4}\:−\:\sqrt{\mathrm{10}}\left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{2}\theta\:+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$${f}\left(\theta\right)\:=\:\mathrm{4}\:−\:\sqrt{\mathrm{10}}\:\mathrm{sin}\:\left(\mathrm{2}\theta+\alpha\right) \\ $$$$\mathrm{max}.\:\mathrm{f}\left(\theta\right)=\mathrm{4}+\sqrt{\mathrm{10}} \\ $$$$\mathrm{min}.\:\mathrm{f}\left(\theta\right)=\mathrm{4}−\sqrt{\mathrm{10}} \\ $$$$\mathrm{max}.−\mathrm{min}.=\mathrm{2}\sqrt{\mathrm{10}} \\ $$$$\Rightarrow\mathrm{answer}\:\left(\mathrm{4}\right) \\ $$
Commented by Tinkutara last updated on 31/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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