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Author: Tinku Tara

Solve-for-x-x-1-x-2-x-1-gt-x-2-x-3-

Question Number 17354 by ajfour last updated on 04/Jul/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:: \\ $$$$\mid\mathrm{x}−\mathrm{1}\mid−\mid\mathrm{x}−\mathrm{2}\mid+\mid\mathrm{x}+\mathrm{1}\mid>\mid\mathrm{x}+\mathrm{2}\mid+\mid\mathrm{x}\mid−\mathrm{3}\:. \\ $$ Commented by ajfour last updated on 04/Jul/17 $$\mathrm{my}\:\mathrm{answer}:\:\mathrm{x}\in\:\left(−\mathrm{3},\:\mathrm{3}\right)−\left\{−\mathrm{1},\:\mathrm{1}\:\right\} \\ $$$$\mathrm{answer}\:\mathrm{in}\:\mathrm{book}:\:\:\mathrm{x}\in\:\left(\mathrm{1},\:\mathrm{3}\right)\:. \\…

Question-148427

Question Number 148427 by Rustambek last updated on 27/Jul/21 Answered by Ar Brandon last updated on 27/Jul/21 $$\mathrm{S}=\mathrm{2}+\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{2020}} {\sum}}\frac{\left(\mathrm{k}+\mathrm{1}\right)\mathrm{k}}{\frac{\mathrm{1}}{\mathrm{k}!}+\frac{\mathrm{1}}{\left(\mathrm{k}−\mathrm{1}\right)!}} \\ $$$$\:\:\:=\mathrm{2}+\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{2020}} {\sum}}\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)!}{\mathrm{1}+\mathrm{k}}=\mathrm{2}+\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{2020}}…

The-number-of-points-in-for-which-x-2-x-sin-x-cos-x-0-is-1-6-2-4-3-2-4-0-

Question Number 17348 by Tinkutara last updated on 04/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{points}\:\mathrm{in}\:\left(−\infty,\:\infty\right)\:\mathrm{for} \\ $$$$\mathrm{which}\:{x}^{\mathrm{2}} \:−\:{x}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{6} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{0} \\ $$ Answered by…

lim-x-x-cos-2x-3x-1-

Question Number 148416 by mathdanisur last updated on 27/Jul/21 $$\underset{\boldsymbol{{x}}\rightarrow\infty} {{lim}}\frac{{x}!\:-\:{cos}\left(\mathrm{2}{x}\right)}{\mathrm{3}{x}\:+\:\mathrm{1}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 28/Jul/21 $$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{x}!−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{3x}+\mathrm{1}}=\frac{\mathrm{x}!}{\mathrm{3x}+\mathrm{1}}−\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{3x}+\mathrm{1}} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty}…

e-2-x-x-y-x-dx-dy-1-x-0-

Question Number 82883 by jagoll last updated on 26/Feb/20 $$\left[\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{y}}{\:\sqrt{\mathrm{x}}}\:\right]\:.\frac{\mathrm{dx}}{\mathrm{dy}}\:=\:\mathrm{1}\:,\:\mathrm{x}\:\neq\:\mathrm{0} \\ $$ Answered by john santu last updated on 26/Feb/20 $$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}\:−\:\frac{\mathrm{y}}{\:\sqrt{\mathrm{x}}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)\:\mathrm{y}\:=\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}}…