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Question Number 37108 by rahul 19 last updated on 09/Jun/18

$$\mathrm{If}4x+8\cos x+\tan x-2\sec x-4\log \left\{\cos x(1+\sin x)\right\}⩾6$$$$\mathrm{\forall }xϵ[0,\psi )\mathrm{then}\mathrm{largest}\mathrm{value}\mathrm{of}\psi \mathrm{is}?$$

Answered by ajfour last updated on 09/Jun/18

$$f\left(x\right)=4x+8\cos x+\tan x-4\ln \left\{\cos x(1+\sin x)\right\}-6$$$$f\left(0\right)=8$$$$f{}^{\prime }\left(x\right)=4-8\sin x+\sec {}^{2}x-\frac{4(-\sin x+\cos {}^{2}x-\sin {}^{2}x)}{\cos x(1+\sin x)}$$$$f{}^{\prime }\left(x\right)=4-8\sin x+\sec {}^{2}x-\frac{4(1-2\sin x)}{\cos x}$$$$=4-8\sin x+\sec {}^{2}x-4\sec x+8\tan x$$$$={(\sec x-2)}^2+8(\tan x-\sin x)$$$$\Rightarrow f{}^{\prime }\left(x\right)>0forallx>0.$$$$sof\left(x\right)>f\left(0\right)$$$$\Rightarrow f\left(x\right)>8forallx⩾0$$$$Hence\psi \to +\mathrm{\infty }.$$

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