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Question Number 77218 by john santu last updated on 04/Jan/20
what is solution in R   ((sin (x)−∣x+2∣)/(x^2 −4x−5))≥0 ?
$${what}\:{is}\:{solution}\:{in}\:\mathbb{R}\: \\ $$$$\frac{\mathrm{sin}\:\left({x}\right)−\mid{x}+\mathrm{2}\mid}{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}}\geqslant\mathrm{0}\:?\: \\ $$
Answered by MJS last updated on 04/Jan/20
sin x −∣x+2∣≥0∧(x^2 −4x−5)>0  ∨  sin x −∣x+2∣<0∧(x^2 −4x−5)<0    sin x −∣x+2∣<0 ∀x∈R  x^2 −4x−5<0 ⇔ −5<x<1  ⇒  solution is −5<x<1
$$\mathrm{sin}\:{x}\:−\mid{x}+\mathrm{2}\mid\geqslant\mathrm{0}\wedge\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}\right)>\mathrm{0} \\ $$$$\vee \\ $$$$\mathrm{sin}\:{x}\:−\mid{x}+\mathrm{2}\mid<\mathrm{0}\wedge\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}\right)<\mathrm{0} \\ $$$$ \\ $$$$\mathrm{sin}\:{x}\:−\mid{x}+\mathrm{2}\mid<\mathrm{0}\:\forall{x}\in\mathbb{R} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}<\mathrm{0}\:\Leftrightarrow\:−\mathrm{5}<{x}<\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{solution}\:\mathrm{is}\:−\mathrm{5}<{x}<\mathrm{1} \\ $$
Commented by john santu last updated on 05/Jan/20
thanks sir
$${thanks}\:{sir} \\ $$

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