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Question Number 100130 by bemath last updated on 25/Jun/20

$$\mathrm{how}\mathrm{many}\mathrm{integer}\mathrm{number}$$$$\mathrm{satisfy}\mathrm{the}\mathrm{equation}$$$$\frac{\sin \mathrm{x}-\mid \mathrm{x}+2\mid }{{\mathrm{x}}^2-4\mathrm{x}-5}⩾0$$

Answered by mr W last updated on 25/Jun/20

$$letf\left(x\right)=\sin x-\mid x+2\mid$$$$forx⩽-2:$$$$f\left(x\right)=\sin x+x+2$$$$f^{\prime }\left(x\right)=\cos x+1⩾0\Rightarrow increasing$$$$f\left(x\right)⩽f(-2)=\sin (-2)\approx -0.909$$$$$$$$forx⩾-2:$$$$f\left(x\right)=\sin x-x-2$$$$f^{\prime }\left(x\right)=\cos x-1⩽0\Rightarrow decreasing$$$$f\left(x\right)⩽f(-2)=\sin (-2)\approx -0.909$$$$$$$$\Rightarrow f\left(x\right)=\sin x-\mid x+2\mid ⩽\sin (-2)\approx -0.909<0$$$$$$$$\frac{\sin \mathrm{x}-\mid \mathrm{x}+2\mid }{{\mathrm{x}}^2-4\mathrm{x}-5}⩾0$$$$sincealways\sin x-\mid x+2\mid <0$$$$\Rightarrow x^2-4x-5<0$$$$i.e.(x+1)(x-5)<0$$$$\Rightarrow -1<x<5$$$$\Rightarrow x=0,1,2,3,4\Rightarrow 5integers$$

Commented by Rasheed.Sindhi last updated on 25/Jun/20

$${\mathcal{T}}_{{\mathcal{H}}_{\mathcal{A}}\mathcal{N}}{k}_{\mathcal{S}}$$$${\mathcal{S}}_{\mathcal{I}}\mathcal{R}$$$$!$$

Commented by bobhans last updated on 26/Jun/20

$$\mathrm{cooll}\mathrm{great}$$

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