Question Number 21313 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{k}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\sqrt{{x}\:−\:\mathrm{3}}\:+\:\sqrt{\mathrm{6}\:−\:{x}}\:\geqslant\:{k}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{solution}\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k} \\ $$$$\mathrm{is} \\ $$ Commented by mrW1 last updated on 20/Sep/17…
Question Number 21311 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{a}\:\mathrm{and}\:{b}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{with}\:{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:=\:{a}\:−\:{b},\:\mathrm{and}\:{k}\:=\:{a}^{\mathrm{2}} \:+\:\mathrm{4}{b}^{\mathrm{2}} , \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{k}\:<\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{k}\:>\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{k}\:=\:\mathrm{1} \\…
Question Number 21308 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{complex}\:\mathrm{numbers}\:{z}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid{z}\:−\:\mid{z}\:+\:\mathrm{1}\mid\mid\:=\:\mid{z}\:+\:\mid{z}\:−\:\mathrm{1}\mid\mid \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21309 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Suppose}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{with}\:\mathrm{complex} \\ $$$$\mathrm{coefficients}\:\mathrm{and}\:\mathrm{an}\:\mathrm{even}\:\mathrm{degree}.\:\mathrm{If}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{p}\:\mathrm{are}\:\mathrm{complex}\:\mathrm{non}-\mathrm{real} \\ $$$$\mathrm{numbers}\:\mathrm{with}\:\mathrm{modulus}\:\mathrm{1},\:\mathrm{prove}\:\mathrm{that} \\ $$$${p}\left(\mathrm{1}\right)\:\in\:{R}\:\mathrm{iff}\:{p}\left(−\mathrm{1}\right)\:\in\:{R}. \\ $$ Terms of Service Privacy Policy…
Question Number 21307 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{ii}\right)\:{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:\neq\:\mathrm{0}…
Question Number 152365 by mathdanisur last updated on 27/Aug/21 Answered by ghimisi last updated on 28/Aug/21 $${a}={x}+{y};{b}={y}+{z};{c}={x}+{z} \\ $$$${p}={x}+{y}+{z};{q}={ab}+{bc}+{ac};{r}={abc} \\ $$$$\Leftrightarrow….\Leftrightarrow{p}^{\mathrm{3}} +\mathrm{9}{r}\geqslant\mathrm{4}{pq}\Leftrightarrow{schur} \\ $$ Commented…
Question Number 21294 by Tinkutara last updated on 19/Sep/17 $$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{not} \\ $$$$\mathrm{all}\:\mathrm{real},\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{2}\left({z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \right)\:−\:\mathrm{3}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}}…
Question Number 152364 by mathdanisur last updated on 27/Aug/21 Answered by Kamel last updated on 28/Aug/21 $$ \\ $$$$\Omega\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{Ln}\left({tan}\left({ax}\right)\right)}{\mathrm{1}−\mathrm{2}{bcos}\left({x}\right)+{b}^{\mathrm{2}} }{dx}\:,\:\mid{b}\mid<\mathrm{1},\:\mathrm{0}<{a}\leqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$${We}\:{have}:\:{Ln}\left({tan}\left({ax}\right)\right)=−\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{+\infty}…
Question Number 21293 by Tinkutara last updated on 19/Sep/17 $$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{even}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{such} \\ $$$$\mathrm{that}\:\frac{{n}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{let}\:\alpha_{\mathrm{0}} ,\:\alpha_{\mathrm{1}} ,\:….,\:\alpha_{{n}−\mathrm{1}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{of}\:\mathrm{order}\:{n}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}\:+\:{b}\alpha_{{k}} ^{\mathrm{2}} \right)\:=\:\left({a}^{\frac{{n}}{\mathrm{2}}} \:+\:{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}}…
Question Number 152358 by Jonathanwaweh last updated on 27/Aug/21 Commented by JDamian last updated on 27/Aug/21 $$\mathrm{5} \\ $$ Answered by Olaf_Thorendsen last updated on…