Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

how many integer number satisfy the equation ((sin x−∣x+2∣)/(x^2 −4x−5)) ≥ 0

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

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

let f(x)=sin x −∣x+2∣ for x≤−2: f(x)=sin x+x+2 f′(x)=cos x+1≥0 ⇒increasing f(x)≤f(−2)=sin (−2)≈−0.909 for x≥−2: f(x)=sin x−x−2 f′(x)=cos x−1≤0 ⇒decreasing f(x)≤f(−2)=sin (−2)≈−0.909 ⇒f(x)=sin x−∣x+2∣≤sin (−2)≈−0.909<0 ((sin x−∣x+2∣)/(x^2 −4x−5)) ≥ 0 since always sin x−∣x+2∣<0 ⇒x^2 −4x−5<0 i.e. (x+1)(x−5)<0 ⇒−1<x<5 ⇒x=0,1,2,3,4 ⇒5 integers

$${let}\:{f}\left({x}\right)=\mathrm{sin}\:{x}\:−\mid{x}+\mathrm{2}\mid \\ $$$${for}\:{x}\leqslant−\mathrm{2}: \\ $$$${f}\left({x}\right)=\mathrm{sin}\:{x}+{x}+\mathrm{2} \\ $$$${f}'\left({x}\right)=\mathrm{cos}\:{x}+\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow{increasing} \\ $$$${f}\left({x}\right)\leqslant{f}\left(−\mathrm{2}\right)=\mathrm{sin}\:\left(−\mathrm{2}\right)\approx−\mathrm{0}.\mathrm{909} \\ $$$$ \\ $$$${for}\:{x}\geqslant−\mathrm{2}: \\ $$$${f}\left({x}\right)=\mathrm{sin}\:{x}−{x}−\mathrm{2} \\ $$$${f}'\left({x}\right)=\mathrm{cos}\:{x}−\mathrm{1}\leqslant\mathrm{0}\:\Rightarrow{decreasing} \\ $$$${f}\left({x}\right)\leqslant{f}\left(−\mathrm{2}\right)=\mathrm{sin}\:\left(−\mathrm{2}\right)\approx−\mathrm{0}.\mathrm{909} \\ $$$$ \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{sin}\:{x}−\mid{x}+\mathrm{2}\mid\leqslant\mathrm{sin}\:\left(−\mathrm{2}\right)\approx−\mathrm{0}.\mathrm{909}<\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}−\mid\mathrm{x}+\mathrm{2}\mid}{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}−\mathrm{5}}\:\geqslant\:\mathrm{0}\: \\ $$$${since}\:{always}\:\mathrm{sin}\:{x}−\mid{x}+\mathrm{2}\mid<\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}<\mathrm{0} \\ $$$${i}.{e}.\:\left({x}+\mathrm{1}\right)\left({x}−\mathrm{5}\right)<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}<{x}<\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\Rightarrow\mathrm{5}\:{integers} \\ $$

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

T_(H_A N) k_(S ) S_I R !

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

Commented by bobhans last updated on 26/Jun/20

cooll great

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com