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

$$\mathrm{How}\mathrm{to}\mathrm{find}\mathrm{maximum}\mathrm{value}\mathrm{of}k\mathrm{if}$$$$\frac{5-\cos 2\theta }{\sin \theta }⩾2k0⩽\theta ⩽\pi$$

Commented by Joel577 last updated on 28/May/17

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

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

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

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

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

Commented by Joel577 last updated on 29/May/17

$$thankyouverymuch$$

Answered by ajfour last updated on 28/May/17

$$4+(1-\cos 2\theta )⩾2k\sin \theta$$$$4+2\sin {}^2\theta -2k\sin \theta ⩾0$$$$\sin {}^2\theta -k\sin \theta +2⩾0$$$$\Rightarrow for\theta =0,\pi ;k=anyrealnumber$$$$\Rightarrow k⩽\frac{2+\sin {}^2\theta }{\sin \theta }$$$$letf\left(\theta \right)=\frac{2+\sin {}^2\theta }{\sin \theta }$$$$f{}^{\prime }\left(\theta \right)=\frac{2\sin {}^2\theta \cos \theta -\cos \theta (2+\sin {}^2\theta )}{{\sin }^2\theta }$$$$=\frac{\cos \theta (\sin {}^2\theta -2)}{{\sin }^2\theta }$$$$f{}^{\prime }\left(\theta \right)>0for\frac{\pi }{2}<\theta <\pi$$$$whilef{}^{\prime }\left(\theta \right)<0for0<\theta <\pi$$$$localminimaatx=\frac{\pi }{2}$$$$\Rightarrow f\left(\theta \right)isminimumforf{}^{\prime }\left(\theta \right)=0$$$$or\cos \theta =0,\theta =\frac{\pi }{2}.$$$$f\left(\frac{\pi }{2}\right)=\frac{2+1}{1}=3$$$$k_{max}=minimumoff\left(\frac{\pi }{2}\right)$$$$k_{max}=3.$$

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

$$\mathrm{Please}\mathrm{see}\mathrm{my}\mathrm{comment}\mathrm{I}\mathrm{think}$$$$\mathrm{minimum}\mathrm{is}6.$$

Commented by ajfour last updated on 28/May/17

$$thankyouMr.Prakash.$$

Commented by Joel577 last updated on 29/May/17

$$thankyouverymuch$$

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