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Question Number 48739 by mr W last updated on 28/Nov/18

$$Findthemaximumof$$$$f\left(x\right)=\cos x+\frac{\lambda \tan x-1}{\tan x-\lambda }\sin x$$$$intermsof\lambda$$$$with\lambda >1and0<x<{\tan }^{-1}\lambda$$

Answered by MJS last updated on 28/Nov/18

$$f\left(x\right)=\lambda \frac{\cos 2x}{\lambda \cos x-\sin x}$$$$x=\arctan t\Rightarrow \cos x=\frac{1}{\sqrt{t^2+1}};\sin x=\frac{t}{\sqrt{t^2+1}}$$$$f\left(t\right)=\lambda \frac{(t-1)(t+1)}{(t-\lambda )\sqrt{t^2+1}}$$$$f^{\prime }\left(t\right)=-\lambda \frac{\lambda t^3-3t^2+3\lambda t-1}{(t-\lambda ){(t^2+1)}^{3/2}}$$$$t^3-\frac{3}{\lambda }{t}^2+3t-\frac{1}{\lambda }=0$$$$t=z+\frac{1}{\lambda }$$$$z^3+\frac{3({\lambda }^2-1)}{{\lambda }^2}z+\frac{2({\lambda }^2-1)}{{\lambda }^3}=0$$$$D=\frac{{({\lambda }^2-1)}^2}{{\lambda }^4}⩾0\mid \lambda \in \mathbb{R}$$$$z_1=\frac{\sqrt[3]{{\lambda }^2-1}}{\lambda }(\sqrt[3]{\lambda -1}-\sqrt[3]{\lambda +1})$$$$t_1=z_1+\frac{1}{\lambda }=\frac{1}{\lambda }(1+\sqrt[3]{{\lambda }^2-1}(\sqrt[3]{\lambda -1}-\sqrt[3]{\lambda +1}))$$$$x_1=\arctan \left(\frac{1}{\lambda }(1+\sqrt[3]{{\lambda }^2-1}(\sqrt[3]{\lambda -1}-\sqrt[3]{\lambda +1}))\right)$$

Commented by mr W last updated on 28/Nov/18

$$thankyousir!$$$${it}^{\prime }scorrectandperfect!greatjob!$$

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