Question Number 61241 by Tawa1 last updated on 30/May/19 Answered by meme last updated on 30/May/19 $${the}\:{larger}\:{is}\:{A}\:{because}\:\mathrm{4}^{\mathrm{2015}} −\mathrm{2}^{\mathrm{2015}} +\mathrm{1}<\mathrm{4}^{\mathrm{2015}} +\mathrm{2}^{\mathrm{2015}} +\mathrm{1} \\ $$$$ \\ $$…
Question Number 192297 by Mastermind last updated on 14/May/23 Answered by witcher3 last updated on 14/May/23 $$\mathrm{let}\:\mathrm{a}=\mathrm{supA},\mathrm{b}=\mathrm{SupB}\Rightarrow\forall\left(\mathrm{x}\in\mathrm{Aety}\in\mathrm{B}\right) \\ $$$$\mathrm{x}+\mathrm{y}\leqslant\mathrm{a}+\mathrm{b}\Rightarrow\mathrm{sup}\left(\mathrm{A}+\mathrm{B}\right)\leqslant\mathrm{a}+\mathrm{b} \\ $$$$\mathrm{let}\:\mathrm{M}=\mathrm{sup}\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\forall\epsilon>\mathrm{0}\:\exists\mathrm{x}\in\mathrm{A},\mathrm{y}\in\mathrm{B}\:\mathrm{such}\:\mathrm{a}−\epsilon<\mathrm{x}\leqslant\mathrm{a} \\ $$$$\mathrm{b}−\epsilon<\mathrm{y}\leqslant\mathrm{b}…
Question Number 192299 by Mastermind last updated on 14/May/23 Answered by witcher3 last updated on 14/May/23 $$\mathrm{R}\subseteq\mathrm{S}\Rightarrow\mathrm{x}=\mathrm{Sup}\left(\mathrm{R}\right)\leqslant\mathrm{sup}\left(\mathrm{S}\right)=\mathrm{y} \\ $$$$\forall\mathrm{r}\in\mathrm{R}\:\:\mathrm{r}\leqslant\mathrm{y},\forall\epsilon>\mathrm{0}\:\exists\mathrm{s}\in\mathrm{S}\:\mathrm{such}\:\mathrm{y}−\epsilon<\mathrm{s}\leqslant\mathrm{y} \\ $$$$\mathrm{by}\:\mathrm{definition}\:\exists\mathrm{r}\in\mathrm{R}\:\mathrm{r}\geqslant\mathrm{s} \\ $$$$\Rightarrow\forall\epsilon>\mathrm{0}\:\mathrm{y}−\epsilon\leqslant\mathrm{r} \\ $$$$\epsilon=\frac{\mathrm{1}}{\mathrm{n}}\Rightarrow\forall\mathrm{n}\in\mathbb{N}\:\:\mathrm{y}−\frac{\mathrm{1}}{\mathrm{n}}<\mathrm{r}\Rightarrow\mathrm{s}\geqslant\mathrm{y}\Rightarrow\mathrm{sup}\left(\mathrm{S}\right)\leqslant\mathrm{Sup}\left(\mathrm{R}\right)…
Question Number 192286 by pete last updated on 14/May/23 $$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{is}: \\ $$$$\mathrm{1}.\mathrm{A}\:\mathrm{function} \\ $$$$\mathrm{2}.\:\mathrm{injective} \\ $$$$\mathrm{3}.\:\mathrm{surjective} \\ $$$$\mathrm{4}.\:\mathrm{bijective} \\ $$ Answered by mehdee42 last…
Question Number 61210 by pooja24 last updated on 30/May/19 $${for}\:{what}\:{value}\:{of}\:\theta,\:\:{e}^{{i}\theta} =\mathrm{0}\:\: \\ $$ Commented by Tony Lin last updated on 30/May/19 $${e}^{{i}\theta} ={cos}\theta+{isin}\theta \\ $$$$\:\:\:\:\:{cos}\theta\in\left[−\mathrm{1},\mathrm{1}\right]\in{R}…
Question Number 126704 by Dwaipayan Shikari last updated on 23/Dec/20 $$\frac{{e}^{\pi} −\mathrm{1}}{{e}^{\pi} +\mathrm{1}}=\frac{\pi}{\mathrm{2}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}+\frac{\pi^{\mathrm{2}} }{\mathrm{10}+\frac{\pi^{\mathrm{2}} }{\mathrm{14}+….}}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61165 by Tawa1 last updated on 29/May/19 Answered by MJS last updated on 30/May/19 $$\int\frac{\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}{x}−\mathrm{5}\right)\:\mathrm{cos}\:\left(\mathrm{2}{x}−\mathrm{14}\right)}{\mathrm{cos}\:\left(\mathrm{2}{x}−\mathrm{7}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}−\mathrm{7}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cos}^{\mathrm{2}} \:\left({t}+\mathrm{2}\right)\:\mathrm{cos}\:\left({t}−\mathrm{7}\right)}{\mathrm{cos}\:\left({t}\right)}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{use}\:\mathrm{these}:\right.…
Question Number 192238 by Kallu last updated on 12/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61147 by malwaan last updated on 29/May/19 $$\boldsymbol{{prove}} \\ $$$$\int\frac{\mathrm{1}+{cos}\:{x}}{\mathrm{1}−{cos}\:{x}}{dx}=−\mathrm{2}{cot}\:\frac{{x}}{\mathrm{2}}−{x}+{c} \\ $$$$ \\ $$ Answered by tanmay last updated on 29/May/19 $$\int\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 126669 by Dwaipayan Shikari last updated on 23/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} ^{\mathrm{2}} }{{n}^{\mathrm{4}} } \\ $$ Commented by talminator2856791 last updated on 23/Dec/20…