How-to-find-maximum-value-of-k-if-5-cos-2-sin-2k-0-pi-

Question Number 14130 by Joel577 last updated on 28/May/17

How to find maximum value of k if ((5 − cos 2θ)/(sin θ)) ≥ 2k 0 ≤ θ ≤ π

$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{if} \\ $$$$\frac{\mathrm{5}\:−\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{sin}\:\theta}\:\:\geqslant\:\mathrm{2}{k}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\pi \\ $$

Commented by Joel577 last updated on 28/May/17

((5 − cos 2θ)/(sin θ)) ≥ 2k ((4 + 2sin^2 θ)/(sin θ)) ≥ 2k sin θ is minimum when θ = (π/2) so k = 3 Am I right?

$$\frac{\mathrm{5}\:−\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{sin}\:\theta}\:\geqslant\:\mathrm{2}{k} \\ $$$$\frac{\mathrm{4}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}\:\theta}\:\geqslant\:\mathrm{2}{k} \\ $$$$\mathrm{sin}\:\theta\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when}\:\theta\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{so}\:{k}\:=\:\mathrm{3} \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}? \\ $$

Commented by prakash jain last updated on 28/May/17

$$\mathrm{sin}\:\theta\:\mathrm{is}\:\mathrm{actually}\:\mathrm{maximum}\:\mathrm{when} \\ $$$$\theta=\frac{\pi}{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 28/May/17

f(θ)=((4 + 2sin^2 θ)/(sin θ)) f′(θ)=cos θ(2−4cosec^2 θ) f′(θ)=0 at θ=(π/2) f′′(θ)=−2sin θ+8cosec^3 θ−4cosec θ f′′((π/2))=−2+8−4>0 f((π/2)) is local minima for f(θ) f((π/2))=((4+2)/1)=6⇒k=3

$${f}\left(\theta\right)=\frac{\mathrm{4}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}\:\theta}\: \\ $$$${f}'\left(\theta\right)=\mathrm{cos}\:\theta\left(\mathrm{2}−\mathrm{4cosec}^{\mathrm{2}} \theta\right) \\ $$$${f}'\left(\theta\right)=\mathrm{0} \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${f}''\left(\theta\right)=−\mathrm{2sin}\:\theta+\mathrm{8cosec}^{\mathrm{3}} \theta−\mathrm{4cosec}\:\theta \\ $$$${f}''\left(\frac{\pi}{\mathrm{2}}\right)=−\mathrm{2}+\mathrm{8}−\mathrm{4}>\mathrm{0} \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{local}\:\mathrm{minima}\:\mathrm{for}\:{f}\left(\theta\right) \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{4}+\mathrm{2}}{\mathrm{1}}=\mathrm{6}\Rightarrow{k}=\mathrm{3} \\ $$

Commented by Joel577 last updated on 29/May/17

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by ajfour last updated on 28/May/17

4+(1−cos 2θ)≥2ksin θ 4+2sin^2 θ−2ksin θ≥0 sin^2 θ−ksin θ+2≥0 ⇒for θ=0,π ; k=any real number ⇒ k≤((2+sin^2 θ)/(sin θ)) let f(θ)=((2+sin^2 θ)/(sin θ)) f ′(θ)=((2sin^2 θcos θ−cos θ(2+sin^2 θ))/(sin^2 θ)) =((cos θ(sin^2 θ−2))/(sin^2 θ)) f ′(θ)>0 for (π/2)<θ<π while f ′(θ)<0 for 0<θ<π local minima at x=(π/2) ⇒ f(θ) is minimum for f ′(θ)=0 or cos θ=0, θ=(π/2). f((π/2))=((2+1)/1)=3 k_(max) = minimum of f((π/2)) k_(max) =3 .

$$\mathrm{4}+\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)\geqslant\mathrm{2}{k}\mathrm{sin}\:\theta \\ $$$$\mathrm{4}+\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{2}{k}\mathrm{sin}\:\theta\geqslant\mathrm{0} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \theta−{k}\mathrm{sin}\:\theta+\mathrm{2}\geqslant\mathrm{0} \\ $$$$\Rightarrow{for}\:\theta=\mathrm{0},\pi\:\:\:;\:\:{k}={any}\:{real}\:{number} \\ $$$$\Rightarrow\:{k}\leqslant\frac{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta} \\ $$$${let}\:{f}\left(\theta\right)=\frac{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:{f}\:'\left(\theta\right)=\frac{\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta−\mathrm{cos}\:\theta\left(\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\:\:\:\:=\frac{\mathrm{cos}\:\theta\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{2}\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta}\: \\ $$$${f}\:'\left(\theta\right)>\mathrm{0}\:{for}\:\frac{\pi}{\mathrm{2}}<\theta<\pi \\ $$$${while}\:{f}\:'\left(\theta\right)<\mathrm{0}\:{for}\:\mathrm{0}<\theta<\pi \\ $$$${local}\:{minima}\:{at}\:{x}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{f}\left(\theta\right)\:{is}\:{minimum}\:{for}\:{f}\:'\left(\theta\right)=\mathrm{0} \\ $$$${or}\:\:\mathrm{cos}\:\theta=\mathrm{0},\:\:\theta=\frac{\pi}{\mathrm{2}}. \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{1}}=\mathrm{3} \\ $$$${k}_{{max}} =\:{minimum}\:{of}\:{f}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$${k}_{{max}} =\mathrm{3}\:. \\ $$

Commented by prakash jain last updated on 28/May/17