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Author: Tinku Tara

let-p-be-a-prime-number-amp-let-a-1-a-2-a-3-a-p-be-integers-show-that-there-exists-an-integer-k-such-that-the-numbers-a-1-k-a-2-k-a-3-k-a-p-k-produce-at-least-1-2-p-distinct-

Question Number 194710 by York12 last updated on 14/Jul/23 $${let}\:{p}\:{be}\:{a}\:{prime}\:{number} \\ $$$$\&\:{let}\:{a}_{\mathrm{1}} \:,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,…,{a}_{{p}\:} {be}\:{integers} \\ $$$${show}\:{that}\:,\:{there}\:{exists}\:{an}\:{integer}\:{k}\:{such}\:{that}\:{the}\:{numbers} \\ $$$${a}_{\mathrm{1}} +{k},\:{a}_{\mathrm{2}} +{k},{a}_{\mathrm{3}} +{k},….,{a}_{{p}} +{k} \\…

Question-194735

Question Number 194735 by sonukgindia last updated on 14/Jul/23 Answered by TheHoneyCat last updated on 14/Jul/23 $$\mathrm{If}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\:\left({i}.{e}.\:{x}\left(\mathrm{2}\right)={x}×\mathrm{2}\:\right) \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}: \\ $$$${x}\left(\mathrm{2}\right)=\frac{\mathrm{4}\sqrt{\mathrm{7}}−\mathrm{2}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{I}'\mathrm{m}\:\mathrm{guessing}\:{x}\:\mathrm{is}\:\mathrm{here}\:\mathrm{a}\:\mathrm{function}……

Question-194685

Question Number 194685 by cortano12 last updated on 13/Jul/23 Answered by qaz last updated on 14/Jul/23 $$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)} ={e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…} \\ $$$$={e}\left(\mathrm{1}+\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} +…\right.…

Question-194697

Question Number 194697 by cortano12 last updated on 13/Jul/23 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by MM42 last updated on 13/Jul/23 $$\frac{{tanx}−{tan}\mathrm{3}{x}}{{tanx}}=\mathrm{3}\Rightarrow\frac{{tan}\mathrm{3}{x}}{{tanx}}=−\mathrm{2} \\ $$$$\frac{{cotx}}{{cotx}+{cot}\mathrm{3}{x}}=\frac{{tan}\mathrm{3}{x}}{{tanx}+{tan}\mathrm{3}{x}} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{1}−\mathrm{2}}=\mathrm{2}\:\checkmark \\…