Find-the-range-of-values-of-x-x-1-2x-x-2-1-

Question Number 52282 by Tawa1 last updated on 05/Jan/19

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\:\mathrm{x} \\ $$$$\:\:\:\mid\frac{\mathrm{x}\:−\:\mathrm{1}}{\mathrm{2x}}\mid\:\geqslant\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

from graph it clear ∣((x−1)/(2x))∣=x^2 +1 at x=−0.59 and x=0.313 and ∣((x−1)/(2x))∣>x^2 +1 when x (−0.59,0.313) so given ∣((x−1)/(2x))∣≥x^2 +1 [−0.59,0.313]

$${from}\:{graph}\:{it}\:{clear} \\ $$$$\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid={x}^{\mathrm{2}} +\mathrm{1}\:{at}\:{x}=−\mathrm{0}.\mathrm{59}\:\:{and}\:{x}=\mathrm{0}.\mathrm{313} \\ $$$${and}\:\:\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid>{x}^{\mathrm{2}} +\mathrm{1}\:\:{when}\:\:{x}\:\:\left(−\mathrm{0}.\mathrm{59},\mathrm{0}.\mathrm{313}\right) \\ $$$${so}\:{given}\:\:\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid\geqslant{x}^{\mathrm{2}} +\mathrm{1}\:\:\left[−\mathrm{0}.\mathrm{59},\mathrm{0}.\mathrm{313}\right] \\ $$

Commented by Tawa1 last updated on 05/Jan/19

God bless you sir. But sir. can it be solve without graph sir. if yes. please help when you are chanced.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\:\:\:\mathrm{But}\:\mathrm{sir}.\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{solve}\:\mathrm{without}\:\mathrm{graph}\:\mathrm{sir}.\:\mathrm{if}\:\mathrm{yes}.\: \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{chanced}.\: \\ $$

Commented by Tawa1 last updated on 06/Jan/19

Please sir, where can i study questions like this, on which textbook sir

$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{where}\:\mathrm{can}\:\mathrm{i}\:\mathrm{study}\:\mathrm{questions}\:\mathrm{like}\:\mathrm{this},\:\mathrm{on}\:\mathrm{which}\:\mathrm{textbook} \\ $$$$\mathrm{sir} \\ $$

Commented by Abdo msup. last updated on 05/Jan/19

for x≠0 (e) ⇔∣x−1∣≥2∣x∣(x^2 +1) ⇔ ∣x−1∣−2∣x∣(x^2 +1)≥0 let f(x)=∣x−1∣−2∣x∣(x^2 +1) so (e)⇔f(x)≥0 x −∞ 0 1 +∞ ∣x−1∣ −x+1 −x+1 0 x−1 ∣x∣ −x 0 x x f(x) 2x^3 +x+1 −2x^3 −3x+1 −2x^3 −x−1 case 1 x≤0 (e) ⇔2x^3 +x+1 ≥0 let find roots of p(x)=2x^3 +x +1 the roots are x_1 ∼−0,5898 (real) x_2 =0,2949 +0,8723i (complex) x_3 =0,2949−0,8723i (complex) ⇒ f(x)=2(x−x_1 )(x−x_2 )(x−x_3 ) =2(x−x_1 )(x−α−iβ)(x−α+iβ) =2(x−x_1 )((x−α)^2 +β^2 ) so f(x)≥0 ⇔x≥x_1 ⇒S_1 =[x_1 ,0[ case 2 0<x≤1 (e) ⇔−2x^3 −3x+1 ≥0 ⇔ 2x^3 +3x−1≤0 ...becontinued...

$${for}\:{x}\neq\mathrm{0}\:\:\left({e}\right)\:\Leftrightarrow\mid{x}−\mathrm{1}\mid\geqslant\mathrm{2}\mid{x}\mid\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\:\Leftrightarrow \\ $$$$\mid{x}−\mathrm{1}\mid−\mathrm{2}\mid{x}\mid\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\geqslant\mathrm{0}\:\:{let}\:{f}\left({x}\right)=\mid{x}−\mathrm{1}\mid−\mathrm{2}\mid{x}\mid\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${so}\:\left({e}\right)\Leftrightarrow{f}\left({x}\right)\geqslant\mathrm{0}\: \\ $$$${x}\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\mid{x}−\mathrm{1}\mid\:\:\:\:\:\:\:\:−{x}+\mathrm{1}\:\:\:\:−{x}+\mathrm{1}\:\:\:\mathrm{0}\:\:\:\:\:{x}−\mathrm{1} \\ $$$$\mid{x}\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{x}\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\:\:\: \\ $$$${f}\left({x}\right)\:\:\mathrm{2}{x}^{\mathrm{3}} \:+{x}+\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}\:\:\:−\mathrm{2}{x}^{\mathrm{3}} −{x}−\mathrm{1} \\ $$$${case}\:\mathrm{1}\:\:{x}\leqslant\mathrm{0}\:\:\left({e}\right)\:\Leftrightarrow\mathrm{2}{x}^{\mathrm{3}} \:+{x}+\mathrm{1}\:\geqslant\mathrm{0} \\ $$$${let}\:{find}\:{roots}\:{of}\:\:{p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} \:+{x}\:+\mathrm{1}\: \\ $$$${the}\:{roots}\:{are}\:\:{x}_{\mathrm{1}} \sim−\mathrm{0},\mathrm{5898}\:\left({real}\right) \\ $$$${x}_{\mathrm{2}} =\mathrm{0},\mathrm{2949}\:+\mathrm{0},\mathrm{8723}{i}\:\left({complex}\right) \\ $$$${x}_{\mathrm{3}} =\mathrm{0},\mathrm{2949}−\mathrm{0},\mathrm{8723}{i}\:\left({complex}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)=\mathrm{2}\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right) \\ $$$$=\mathrm{2}\left({x}−{x}_{\mathrm{1}} \right)\left({x}−\alpha−{i}\beta\right)\left({x}−\alpha+{i}\beta\right) \\ $$$$=\mathrm{2}\left({x}−{x}_{\mathrm{1}} \right)\left(\left({x}−\alpha\right)^{\mathrm{2}} +\beta^{\mathrm{2}} \right)\:\:{so} \\ $$$${f}\left({x}\right)\geqslant\mathrm{0}\:\Leftrightarrow{x}\geqslant{x}_{\mathrm{1}} \:\:\Rightarrow{S}_{\mathrm{1}} =\left[{x}_{\mathrm{1}} ,\mathrm{0}\left[\right.\right. \\ $$$${case}\:\mathrm{2}\:\:\mathrm{0}<{x}\leqslant\mathrm{1}\:\:\:\:\left({e}\right)\:\Leftrightarrow−\mathrm{2}{x}^{\mathrm{3}} \:−\mathrm{3}{x}+\mathrm{1}\:\geqslant\mathrm{0}\:\Leftrightarrow \\ $$$$\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{3}{x}−\mathrm{1}\leqslant\mathrm{0}\:\:…{becontinued}… \\ $$

Commented by Tawa1 last updated on 06/Jan/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by maxmathsup by imad last updated on 06/Jan/19

$${i}\:{hav}\:{nt}\:{the}\:{name}\:{but}\:{you}\:{can}\:{find}\:{this}\:{in}\:{old}\:{books}… \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

1)x≠0 2)((x−1)/(2x))=x^2 +1 2x^3 +2x−x+1=0 2x^3 +x+1=0 x=−0.59 ((−(x−1))/(2x))=x^2 +1 2x^3 +2x+x−1=0 2x^3 +3x−1=0 x=0.313 critical value of x are −0.59 and 0.313 now put condition i)1>x>0.313 ii)x<−0.59 iii)0.313 >x>−0.59 ∣((x−1)/(2x))∣>x^2 +1 when 1>x>0.313 right hand side x^2 +1 is +ve when 1>x>0.313 but ((x−1)/(2x)) is negetive so ∣((x−1)/(2x))∣≱x^2 +1 when 1> x>0.313 but for x =0.313 ((( x−1))/(2x))=x^2 +1true ii) x<−0.59 ∣((x−1)/(2x))∣is ((x−1)/(2x)) (put any value of x<−0.59 say x=−1) LHS =((−1−1)/(2×−1))=1 RHS (−1)^2 +1=2 LHS≯RHS so when x<−0.59 givencondition does not hold so x can not be less than −0.59 iii)now when 0.313> x>−0.59 [say x=0.2 LHS ∣((x−1)/(2x))∣ =∣((0.2−1)/(2×0.2))∣ =2 RHS (0.2)^2 +1=1.04 LHS>RHS so when 0.313>x>−0.59 given clnditionhold solution set for x [−0.59,0.313]

$$\left.\mathrm{1}\right){x}\neq\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}={x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}−{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=−\mathrm{0}.\mathrm{59} \\ $$$$\frac{−\left({x}−\mathrm{1}\right)}{\mathrm{2}{x}}={x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}+{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{0}.\mathrm{313} \\ $$$${critical}\:{value}\:{of}\:{x}\:\:{are}\:−\mathrm{0}.\mathrm{59}\:{and}\:\mathrm{0}.\mathrm{313} \\ $$$${now}\:{put}\:{condition} \\ $$$$\left.{i}\right)\mathrm{1}>{x}>\mathrm{0}.\mathrm{313} \\ $$$$\left.{ii}\right){x}<−\mathrm{0}.\mathrm{59} \\ $$$$\left.{iii}\right)\mathrm{0}.\mathrm{313}\:>{x}>−\mathrm{0}.\mathrm{59} \\ $$$$\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid>{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${when}\:\mathrm{1}>{x}>\mathrm{0}.\mathrm{313}\: \\ $$$${right}\:{hand}\:{side}\:{x}^{\mathrm{2}} +\mathrm{1}\:{is}\:+{ve}\:\:{when}\:\mathrm{1}>{x}>\mathrm{0}.\mathrm{313} \\ $$$${but}\:\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\:{is}\:{negetive}\:{so}\:\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid\ngeqslant{x}^{\mathrm{2}} +\mathrm{1}\:{when}\:\mathrm{1}>\:{x}>\mathrm{0}.\mathrm{313} \\ $$$$ \\ $$$${but}\:{for}\:\:\:{x}\:=\mathrm{0}.\mathrm{313}\:\:\:\frac{\left(\:\:{x}−\mathrm{1}\right)}{\mathrm{2}{x}}={x}^{\mathrm{2}} +\mathrm{1}{true} \\ $$$$\left.{ii}\right)\:{x}<−\mathrm{0}.\mathrm{59} \\ $$$$\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid{is}\:\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\:\left({put}\:{any}\:{value}\:{of}\:{x}<−\mathrm{0}.\mathrm{59}\right. \\ $$$$\left.{say}\:{x}=−\mathrm{1}\right) \\ $$$${LHS}\:=\frac{−\mathrm{1}−\mathrm{1}}{\mathrm{2}×−\mathrm{1}}=\mathrm{1} \\ $$$$\boldsymbol{{R}}{HS}\:\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{2}\:\:\: \\ $$$${LHS}\ngtr{RHS} \\ $$$${so}\:{when}\:{x}<−\mathrm{0}.\mathrm{59}\:\:{givencondition} \\ $$$${does}\:{not}\:{hold} \\ $$$${so}\:{x}\:{can}\:{not}\:{be}\:{less}\:{than}\:−\mathrm{0}.\mathrm{59} \\ $$$$\left.{iii}\right){now}\:{when}\:\mathrm{0}.\mathrm{313}>\:{x}>−\mathrm{0}.\mathrm{59} \\ $$$$\left[{say}\:{x}=\mathrm{0}.\mathrm{2}\right. \\ $$$${LHS} \\ $$$$\mid\frac{{x}−\mathrm{1}}{\mathrm{2}{x}}\mid \\ $$$$=\mid\frac{\mathrm{0}.\mathrm{2}−\mathrm{1}}{\mathrm{2}×\mathrm{0}.\mathrm{2}}\mid \\ $$$$=\mathrm{2} \\ $$$${RHS}\:\:\:\left(\mathrm{0}.\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{1}.\mathrm{04} \\ $$$${LHS}>{RHS} \\ $$$${so}\:{when}\:\mathrm{0}.\mathrm{313}>{x}>−\mathrm{0}.\mathrm{59}\: \\ $$$${given}\:{clnditionhold} \\ $$$${solution}\:{set}\:{for}\:{x}\:\left[−\mathrm{0}.\mathrm{59},\mathrm{0}.\mathrm{313}\right] \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 05/Jan/19