Question Number 133180 by mathlove last updated on 19/Feb/21 Commented by mr W last updated on 19/Feb/21 $${no}\:{real}\:{solution}! \\ $$$${x}^{{x}} \geqslant\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}\approx\mathrm{0}.\mathrm{692} \\ $$$$\frac{\mathrm{1}}{\mathrm{256}}<<\mathrm{0}.\mathrm{692} \\ $$$$\Rightarrow{x}^{{x}}…
Question Number 67574 by pete last updated on 28/Aug/19 $$\mathrm{Solve}\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}<−\mathrm{5} \\ $$ Commented by mathmax by abdo last updated on 28/Aug/19 $$\left({e}\right)\Rightarrow{x}^{\mathrm{2}} +\mathrm{6}<\mathrm{0}\:\:\:{impossible}\:{equation}\:\Rightarrow{no}\:{solution}\: \\…
Question Number 133072 by bemath last updated on 18/Feb/21 $$\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{5}}{\mathrm{2}−\sqrt[{\mathrm{4}}]{\mathrm{3}}}\:\mathrm{is}\:\mathrm{in}\:\mathrm{F}_{\mathrm{2}} \:\mathrm{by}\:\mathrm{expressing} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{in}\:\mathrm{form}\:{a}_{\mathrm{1}} +{b}_{\mathrm{1}} \sqrt{{k}_{\mathrm{1}} }\:\mathrm{where} \\ $$$${a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,\:{k}_{\mathrm{1}} \:{are}\:{in}\:{F}_{\mathrm{1}} \\ $$ Answered by…
Question Number 2000 by Yozzi last updated on 29/Oct/15 $${Find}\:{a}\:{non}−{constant}\:{function}\:{f}\: \\ $$$${satisfying}\:{f}\left(\mathrm{0}\right)=\mathrm{1},{f}\left(−\mathrm{2}\right)=\mathrm{0}\:{and} \\ $$$${f}\left({x}−{y}\right)={f}\left({x}\right){f}\left({y}\right)−{f}\left(−\mathrm{2}−{x}\right){f}\left({y}−\mathrm{2}\right). \\ $$ Commented by prakash jain last updated on 29/Oct/15 $${x}=\mathrm{0}…
Question Number 1988 by Rasheed Soomro last updated on 28/Oct/15 $${x}^{\mathrm{2}} =\:\frac{{f}\left({x}\right)+{f}\left(−{x}\right)}{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$ Answered by 123456 last updated on 28/Oct/15 $$\mathrm{supossing}\:\mathrm{that}\:{f}\:\mathrm{is}\:\mathrm{poly} \\…
Question Number 133053 by pete last updated on 18/Feb/21 $$\mathrm{When}\:\mathrm{the}\:\mathrm{polynomial}\:\mathrm{f}\left({x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{4}\:\mathrm{and}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{divided} \\ $$$$\left(\mathrm{x}−\mathrm{3}\right)\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{7}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{f}\left({x}\right) \\ $$$$\mathrm{may}\:\mathrm{be}\:\mathrm{written}\:\mathrm{in}\:\mathrm{the}\:\mathrm{formf}\left({x}\right)=\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)\mathrm{Q}\left(\mathrm{x}\right)+\mathrm{ax}+\mathrm{b}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right).\:\mathrm{If}\:\mathrm{also}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{cubic}\:\mathrm{function} \\ $$$$\mathrm{in}\:\mathrm{which}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{unity}\:\mathrm{and} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1},\:\mathrm{determine}\:\mathrm{Q}\left(\mathrm{x}\right).…
Question Number 67501 by TawaTawa last updated on 28/Aug/19 $$\mathrm{Show}\:\mathrm{that}\:\:\mathrm{1n}^{\mathrm{3}} \:+\:\mathrm{2n}\:+\:\mathrm{3n}^{\mathrm{2}} \:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}. \\ $$ Commented by Prithwish sen last updated on 28/Aug/19 $$\mathrm{Another}\:\mathrm{approch} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}…
Question Number 67492 by MJS last updated on 28/Aug/19 $$\mathrm{please}\:\mathrm{check}\:\mathrm{my}\:\mathrm{comment}\:\mathrm{to}\:\mathrm{qu}.\:\mathrm{67471} \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{confusing}\:\mathrm{myself}… \\ $$ Commented by MJS last updated on 28/Aug/19 �� Terms of Service…
Question Number 1952 by prakash jain last updated on 25/Oct/15 $$\mathrm{Inequality}\:\mathrm{relation}\:\mathrm{starting}\:\mathrm{a}\:\mathrm{new}\:\mathrm{thread} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\geqslant\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{q}} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{1},\:{x}=\mathrm{1} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\frac{{x}^{{p}}…
Question Number 1937 by Rasheed Soomro last updated on 25/Oct/15 $$\bullet{Is}\:\:\:'\Leftrightarrow'\:\:{necessary}\:{and}\:{suficient}\:{for}\:{two} \\ $$$${inequalities}\:{to}\:{be}\:{equivalent}? \\ $$$$\bullet{If}\:\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}\:: \\ $$$${Are}\:\:\boldsymbol{\mathrm{A}}>\boldsymbol{\mathrm{B}}\:{and}\:\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{a}}\:>\:\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{b}}\:{equivalent}? \\ $$ Answered by 123456 last updated on…