Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Find the range of values of x ∣((x − 1)/(2x))∣ ≥ x^2 + 1

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

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

God bless you sir

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

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

i hav nt the name but you can find this in old books...

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

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

God bless you sir. Is that all sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{Is}\:\mathrm{that}\:\mathrm{all}\:\mathrm{sir} \\ $$

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

pls see final answer

$${pls}\:{see}\:{final}\:{answer} \\ $$

Commented by Tawa1 last updated on 05/Jan/19

God bless you sir

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com