Question Number 200984 by sonukgindia last updated on 27/Nov/23 Answered by Calculusboy last updated on 29/Nov/23 $$\mathrm{2} \\ $$ Answered by witcher3 last updated on…
Question Number 200980 by sonukgindia last updated on 27/Nov/23 Answered by MM42 last updated on 27/Nov/23 $$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}={y}\Rightarrow\:{x}={y}^{\mathrm{2}} −{y}\Rightarrow{dx}=\left(\mathrm{2}{y}−\mathrm{1}\right){dy} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\:\:\:\:\:\Rightarrow\:\:\:{x}=\mathrm{0}\rightarrow{y}=\mathrm{1}\:\:\:\:;\:\:{x}=\mathrm{1}\rightarrow{y}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left.\Rightarrow\int_{\mathrm{1}} ^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \:\:\frac{\mathrm{2}{y}−\mathrm{1}}{{y}}\:{dy}\:=\left(\mathrm{2}{y}−{lny}\right)\right]_{\mathrm{1}} ^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}…
Question Number 201008 by Mingma last updated on 27/Nov/23 Answered by mr W last updated on 28/Nov/23 $$\frac{{r}_{{n}} −{r}_{{n}+\mathrm{1}} }{{r}_{{n}} +{r}_{{n}+\mathrm{1}} }=\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}={k} \\ $$$$\Rightarrow{r}_{{n}+\mathrm{1}} =\frac{\mathrm{1}−{k}}{\mathrm{1}+{k}}{r}_{{n}}…
Question Number 200976 by Blackpanther last updated on 27/Nov/23 Answered by Mathspace last updated on 28/Nov/23 $$\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{cos}\left({n}\pi\right)}{{ln}\mathrm{3}}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{ln}\mathrm{3}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}}…
Question Number 200978 by sonukgindia last updated on 27/Nov/23 Answered by BaliramKumar last updated on 27/Nov/23 $$\mathrm{put}\:\:\:\mathrm{x}\:=\:\mathrm{tan}\theta\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\mathrm{dx}\:=\:\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \theta+\mathrm{1}}}\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\int\:\mathrm{sec}\theta\mathrm{d}\theta\:=\:\mathrm{ln}\left(\mathrm{sec}\theta\:+\:\mathrm{tan}\theta\right) \\…
Question Number 200979 by sonukgindia last updated on 27/Nov/23 Answered by Frix last updated on 27/Nov/23 $$\mathrm{Qu}\:\mathrm{200934} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 201004 by Mingma last updated on 27/Nov/23 Answered by ajfour last updated on 27/Nov/23 Commented by Mingma last updated on 27/Nov/23 Very elegant, sir! Answered…
Question Number 200972 by Mingma last updated on 27/Nov/23 Answered by AST last updated on 27/Nov/23 $${a}^{\mathrm{2}} ={b}^{\mathrm{2}} \left({b}+\mathrm{2}\right);{a}={b}\sqrt{{b}+\mathrm{2}} \\ $$$${b}=\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}\Rightarrow\left({a},{b}\right)=\left(\frac{{x}^{\mathrm{3}} −\mathrm{2}{xy}^{\mathrm{2}} }{{y}^{\mathrm{3}} },\frac{{x}^{\mathrm{2}}…
Question Number 200973 by Mingma last updated on 27/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201000 by mokys last updated on 27/Nov/23 $${find}\: \\ $$$$ \\ $$$$\left.\:\mathrm{1}\right)\left(\mathrm{0},\mathrm{2}\right)\:\cup\:\left\{\mathrm{3}\right\}\:\: \\ $$$$ \\ $$$$\left.\:\:\mathrm{2}\right)\left[\mathrm{0},\mathrm{2}\right]\:\cup\:\left\{\mathrm{3}\right\} \\ $$$$ \\ $$$$\left.\:\:\mathrm{3}\right)\:\left(−\mathrm{5},\mathrm{5}\right)\:\cup\:\left\{\mathrm{6}\right\} \\ $$$$\:\: \\…