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Author: Tinku Tara

Question-211419

Question Number 211419 by maryaaaa last updated on 08/Sep/24 Answered by mahdipoor last updated on 08/Sep/24 $$\left.{salam}\:\::\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}\left[{x}\right]\:\:\:,\:\:\:{f}\left(−\frac{{f}\left(\sqrt{\mathrm{3}}\right)}{\mathrm{2}}\right)=? \\ $$$${f}\left(\sqrt{\mathrm{3}}\right)=\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\left[\sqrt{\mathrm{3}}\right]=\mathrm{3}−\mathrm{2}×\mathrm{1}=\mathrm{1} \\ $$$${f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(−\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}}…

Question-211414

Question Number 211414 by AlagaIbile last updated on 08/Sep/24 Answered by Frix last updated on 08/Sep/24 $$\mathrm{Let}\:{x}_{{j}} \leqslant{x}_{{j}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }=\mathrm{1}\:\Rightarrow\:\mathrm{max}\:\left({x}_{\mathrm{1}} \right)\:=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }+\frac{\mathrm{1}}{{x}_{\mathrm{2}}…

Question-211415

Question Number 211415 by AlagaIbile last updated on 08/Sep/24 Commented by loredcs2357 last updated on 11/Oct/24 $$\mathrm{Let}\:{S}\left({n}\right)=\underset{{k}\leqslant{n}} {\sum}\mathrm{lcm}\left({k},{n}\right). \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:{S}\left({n}\right)=\frac{{n}}{\mathrm{2}}\left[\underset{{d}\mid{n}} {\sum}\varphi\left({d}^{\mathrm{2}} \right)+\mathrm{1}\right].\:\mathrm{Moreover}\:\mathrm{we}\:\mathrm{have}: \\ $$$${f}\left({n}\right)=\underset{{d}\mid{n}} {\sum}\varphi\left({d}^{\mathrm{2}}…

Question-211402

Question Number 211402 by efronzo1 last updated on 08/Sep/24 Answered by A5T last updated on 08/Sep/24 $${Q}\mathrm{211029},\:{we}\:{can}\:{find}\:{a}\:{polynomial}\:{f}\left({x}\right) \\ $$$${f}\left({x}\right)={x}^{{n}} \underset{−} {+}\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{28}\Rightarrow{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{1} \\…

Question-211398

Question Number 211398 by Spillover last updated on 08/Sep/24 Answered by Frix last updated on 08/Sep/24 $$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}\:\overset{{t}=\frac{{x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}{\:\sqrt{\mathrm{2}}}} {=} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2tan}^{−\mathrm{1}} \:\sqrt{\mathrm{2}}{t}\:= \\…