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Question-200980

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-200976

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-200978

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) \\…

find-1-0-2-3-2-0-2-3-3-5-5-6-4-2-2-3-5-

Question Number 201000 by mokys last updated on 27/Nov/23 $${find}\: \\ $$$$ \\ $$$$\left.\:\mathrm{1}\right)\left(\mathrm{0},\mathrm{2}\right)\:\cup\:\left\{\mathrm{3}\right\}\:\: \\ $$$$ \\ $$$$\left.\:\:\mathrm{2}\right)\left[\mathrm{0},\mathrm{2}\right]\:\cup\:\left\{\mathrm{3}\right\} \\ $$$$ \\ $$$$\left.\:\:\mathrm{3}\right)\:\left(−\mathrm{5},\mathrm{5}\right)\:\cup\:\left\{\mathrm{6}\right\} \\ $$$$\:\: \\…

Question-200960

Question Number 200960 by sonukgindia last updated on 27/Nov/23 Answered by Frix last updated on 27/Nov/23 $${t}=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\int…=\int\left(\mathrm{1}−\frac{\mathrm{1}}{{t}}\right){dt}={t}−\mathrm{ln}\:{t}\:= \\ $$$$=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}}…