Question Number 83956 by jagoll last updated on 08/Mar/20 $$\mathrm{for}\:\mathrm{x}\:\in\:\mathbb{R}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{3x}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{2019}\right)\:.\: \\ $$ Commented by mr W last updated on 08/Mar/20 $$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{3x}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)\:\:\:…\left({i}\right)…
Question Number 18416 by Tinkutara last updated on 20/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{values}\:\mathrm{of}\:{x} \\ $$$$\mathrm{which}\:\mathrm{satisfies} \\ $$$$\frac{\left({x}\:−\:\mathrm{5}\right)^{\mathrm{10}} \left({x}\:−\:\mathrm{13}\right)^{\mathrm{20}} \left({x}\:−\:\mathrm{19}\right)^{\mathrm{13}} }{\left({x}\:−\:\mathrm{10}\right)^{\mathrm{18}} \left({x}\:−\:\mathrm{25}\right)^{\mathrm{19}} }\:\geqslant\:\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{2}\:\leqslant\:{x}\:\leqslant\:\mathrm{30}\:\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{23} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{24}…
Question Number 18333 by ajfour last updated on 19/Jul/17 Commented by ajfour last updated on 19/Jul/17 $$\mathrm{In}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{Q}.\mathrm{18327} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 83662 by jagoll last updated on 05/Mar/20 $$\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{3}}} \\ $$$$\mathrm{find}\:\mathrm{domain}\:\mathrm{function}\: \\ $$$$\left(\mathrm{g}\:\bullet\:\mathrm{f}\right)\left(\mathrm{x}\right) \\ $$ Commented by MJS last updated on 05/Mar/20…
Question Number 83610 by john santu last updated on 04/Mar/20 $$\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:+\:\frac{\mathrm{5}}{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{6}+\mathrm{x}−\mathrm{x}^{\mathrm{2}} }}\:>\:\frac{\mathrm{1}}{\mathrm{1}+\mid\mathrm{x}−\mathrm{1}\mid} \\ $$ Commented by john santu last updated on 04/Mar/20 $$\mathrm{ans}\::\:\left[−\mathrm{2},\:−\mathrm{1}\right)\:\cup\:\left(\mathrm{1},\:\frac{\mathrm{6}}{\mathrm{5}}\right)\:\cup\:\left(\mathrm{2},\mathrm{3}\:\right] \\ $$…
Question Number 83539 by jagoll last updated on 03/Mar/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}? \\ $$ Commented by john santu last updated on 03/Mar/20 $$\mathrm{domain}\:\mathrm{x}^{\mathrm{2}} −\mathrm{1}>\mathrm{0}\:\Rightarrow\:\mathrm{x}<−\mathrm{1}\:\vee\mathrm{x}\:>\mathrm{1}…
Question Number 17895 by ajfour last updated on 12/Jul/17 Commented by ajfour last updated on 12/Jul/17 $$\mathrm{solution}\:\mathrm{to}\:\mathrm{Q}.\mathrm{17872}\: \\ $$$$\mathrm{which}\:\mathrm{said}:\:\:\mathrm{Solve} \\ $$$$\:\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{x}\mid+\mid\mathrm{x}+\mathrm{1}\mid=\mathrm{x}+\mathrm{2} \\ $$ Commented by…
Question Number 83420 by jagoll last updated on 02/Mar/20 $$\mathrm{find}\:\mathrm{range}\:\mathrm{x}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{x}−\mathrm{4}\sqrt{\mathrm{y}}\:=\:\mathrm{2}\sqrt{\mathrm{x}−\mathrm{y}} \\ $$ Commented by mathmax by abdo last updated on 02/Mar/20 $${we}\:{have}\:\mathrm{0}\leqslant{y}\leqslant{x}\:\:\:\left({e}\right)\Rightarrow\left({x}−\mathrm{4}\sqrt{{y}}\right)^{\mathrm{2}} =\mathrm{4}\left({x}−{y}\right)\:\Rightarrow…
Question Number 17872 by Tinkutara last updated on 11/Jul/17 $$\mathrm{Solve}\:: \\ $$$$\mid{x}\:−\:\mathrm{1}\mid\:+\:\mid{x}\mid\:+\:\mid{x}\:+\:\mathrm{1}\mid\:=\:{x}\:+\:\mathrm{2} \\ $$ Answered by ajfour last updated on 11/Jul/17 $$\:\:\:\mathrm{x}\in\left[\mathrm{0},\:\mathrm{1}\right] \\ $$ Answered…
Question Number 83406 by jagoll last updated on 02/Mar/20 $$\mathrm{If}\:\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} +\mathrm{u}_{\mathrm{3}} +…+\mathrm{u}_{\mathrm{n}} \:=\:\mathrm{2n}^{\mathrm{2}} +\mathrm{n}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\:\mathrm{AP}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} +\mathrm{u}_{\mathrm{3}} +…+\mathrm{u}_{\mathrm{2n}−\mathrm{2}} +\mathrm{u}_{\mathrm{2n}−\mathrm{1}} \:. \\…