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Given-the-function-f-defined-by-f-x-2e-x-e-x-1-x-0-0-x-0-i-study-the-differentiability-of-f-at-x-0-ii-Show-that-the-point-0-1-is-the-centre-of-symetry-to-the-cur

Question Number 141340 by physicstutes last updated on 17/May/21 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:{f}\left({x}\right)\:=\:\begin{cases}{\frac{\mathrm{2}{e}^{{x}} }{{e}^{{x}} −\mathrm{1}},{x}\neq\:\mathrm{0}}\\{\mathrm{0},\:{x}\:=\:\mathrm{0}}\end{cases} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{study}\:\mathrm{the}\:\mathrm{differentiability}\:\mathrm{of}\:{f}\:\mathrm{at}\:{x}\:=\:\mathrm{0}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{0},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symetry}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\mathrm{of}\:{f}. \\ $$ Terms of Service…

Write-the-next-three-terms-1-3-7-8-3-

Question Number 141197 by paulpadas last updated on 16/May/21 $${Write}\:{the}\:{next}\:{three}\:{terms}\:\mathrm{1},\:\frac{\mathrm{3}}{\mathrm{7}},\:\frac{\mathrm{8}}{\mathrm{3}},\:\_\_\_,\:\_\_\_,\:\_\_\_,\:… \\ $$ Answered by MJS_new last updated on 16/May/21 $$\mathrm{zillions}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{answers}. \\ $$$$\mathrm{I}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one} \\ $$$${a}_{{n}} =−\frac{\mathrm{3}}{\mathrm{10}}−\frac{\mathrm{38}{n}−\mathrm{25}}{\mathrm{10}\left(\mathrm{5}{n}^{\mathrm{2}}…

Question-10108

Question Number 10108 by ridwan balatif last updated on 24/Jan/17 Answered by nume1114 last updated on 24/Jan/17 $${from}\:{Vieta}'{s}\:{fomula}: \\ $$$$\begin{cases}{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{2}}\\{{x}_{\mathrm{1}} {x}_{\mathrm{2}} =−\mathrm{5}}\end{cases} \\…

Question-75623

Question Number 75623 by liki last updated on 14/Dec/19 Commented by liki last updated on 14/Dec/19 $$…\boldsymbol{{Please}}\:\boldsymbol{{anyone}}\:\boldsymbol{{to}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{this}}\:\boldsymbol{{Qn}}\:! \\ $$ Answered by $@ty@m123 last updated on…

Question-141163

Question Number 141163 by Niiicooooo last updated on 16/May/21 Answered by mindispower last updated on 16/May/21 $$\zeta\left({n}\right)=\mathrm{1}+\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{k}^{{n}} } \\ $$$${k}^{{n}} \geqslant{nk}^{\mathrm{2}} \:{easy}\:{to}\:{see}\:{that}\:,{k}\geqslant\mathrm{2} \\ $$$${for}\:{n}>>\mathrm{2}…