Question Number 80027 by jagoll last updated on 30/Jan/20 $${find}\:{minimum} \\ $$$${value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}} \\ $$$${for}\:{x}\geqslant\mathrm{0}\:{in}\:\mathbb{R} \\ $$ Commented by john santu last updated on…
Question Number 79964 by ubaydulla last updated on 29/Jan/20 $$\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=\left[\mathrm{tan}\:^{\mathrm{2}} {x}={t}\right]=\int\left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dx}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$ Commented by $@ty@m123 last updated on…
Question Number 145449 by solihin last updated on 05/Jul/21 Answered by Olaf_Thorendsen last updated on 05/Jul/21 $${p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\pi{a}+\mathrm{2}{a}+\mathrm{2}{b}\:=\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}+\mathrm{2}{b}\:=\:\mathrm{5} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{ab} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{a}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}\right) \\ $$$${A}\:=\:−\left(\frac{\pi}{\mathrm{8}}+\mathrm{1}\right){a}^{\mathrm{2}}…
Question Number 145450 by solihin last updated on 05/Jul/21 Answered by Olaf_Thorendsen last updated on 05/Jul/21 $$\left.{a}\right) \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}+\mathrm{1}} },\:{n}\in\mathbb{N}^{\ast} \\ $$$${a}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}}…
Question Number 145399 by Willson last updated on 04/Jul/21 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow+\infty} {\boldsymbol{\mathrm{lim}}}\:\:\underset{\:\mathrm{0}} {\int}^{\:\boldsymbol{\mathrm{n}}} \:\frac{\boldsymbol{\mathrm{t}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!}\:\boldsymbol{{e}}^{−\boldsymbol{\mathrm{t}}} \:\boldsymbol{\mathrm{dt}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by ArielVyny last updated on…
Question Number 79689 by ~blr237~ last updated on 27/Jan/20 $${Find}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:−{x}\:\:\:\:\:{without}\:\:{DL}\:{and}\:{if}\:{you}\:{have}\:{to}\:{use}\:{the}\: \\ $$$${hospital}\:{rule}\:{please}\:\:{justify}\:{that}\:{your}\:{function}\:{is}\:{C}^{\mathrm{1}} \: \\ $$ Commented by mr W last updated on 27/Jan/20…
Question Number 79679 by ~blr237~ last updated on 27/Jan/20 $${Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x} \\ $$ Commented by mathmax by abdo last updated on 27/Jan/20 $${let}\:{f}\left({x}\right)={x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x}\:\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}}…
Question Number 79644 by loveineq. last updated on 26/Jan/20 $${f}\left({x}\right)\:=\:\frac{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:{f}\left({x}\right)\:\geqslant\:\mathrm{2}\:. \\ $$ Commented by john santu last updated on 27/Jan/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)<\mathrm{2}…
Question Number 145164 by mim24 last updated on 02/Jul/21 Answered by mathmax by abdo last updated on 02/Jul/21 $$\mathrm{this}\:\mathrm{question}\:\mathrm{is}\:\mathrm{solved}\:\mathrm{see}\:\mathrm{the}\:\mathrm{platform} \\ $$ Commented by mim24 last…
Question Number 145163 by mim24 last updated on 02/Jul/21 Answered by mathmax by abdo last updated on 02/Jul/21 $$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…+\infty}}}\:\Rightarrow\mathrm{y}=\sqrt{\mathrm{x}+\mathrm{y}}\:\Rightarrow\mathrm{y}^{\mathrm{2}} \:=\mathrm{x}+\mathrm{y}\:\Rightarrow \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{x}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}+\mathrm{4x}\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{y}=\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{y}\geqslant\mathrm{0}\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}+\sqrt{\mathrm{4x}+\mathrm{1}}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{4x}+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}}\:\Rightarrow…