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Category: Arithmetic

Question-211979

Question Number 211979 by Spillover last updated on 25/Sep/24 Answered by BHOOPENDRA last updated on 25/Sep/24 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} {x}+\mathrm{tan}^{−\mathrm{1}} {x}\:+{x}^{\mathrm{2}} \pi+\pi\right)} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{tan}^{−\mathrm{1}} {x}+\pi\right)}…

Question-211895

Question Number 211895 by Spillover last updated on 23/Sep/24 Answered by mehdee7396 last updated on 23/Sep/24 $${lim}_{{x}\rightarrow\infty} \:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)\mathrm{2}{x} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\left(\frac{{ax}−\mathrm{4}}{{x}^{\mathrm{2}} }\right)\mathrm{2}{x}=\mathrm{2}{a} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\infty}…

Question-211871

Question Number 211871 by Spillover last updated on 22/Sep/24 Answered by Ghisom last updated on 23/Sep/24 $$=−\underset{\mathrm{0}} {\overset{\mathrm{arccos}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}} {\int}}\frac{\mathrm{2}+\mathrm{tan}\:{x}}{\mathrm{3}−\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{2}}\mathrm{tan}\:{x}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}}…

Question-211873

Question Number 211873 by Spillover last updated on 22/Sep/24 Answered by IbtisamAdnan last updated on 23/Sep/24 $$\:\:\:\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{4}} \frac{\left(\boldsymbol{\mathrm{cos}\alpha}\right)^{\mathrm{x}} −\left(\mathrm{sin}\alpha\right)^{\mathrm{x}} −\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{x}−\mathrm{4}} \\ $$$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{4}} \:\frac{\left(\boldsymbol{\mathrm{cos}\alpha}\right)^{\mathrm{x}} .\:\mathrm{ln}\:\mathrm{cos}\alpha\:\:−\:\left(\mathrm{sin}\:\alpha\right)^{\mathrm{x}} .\mathrm{ln}\:\mathrm{sin}\alpha}{\mathrm{1}}\left[\mathrm{L}\:\mathrm{hospital}\:\mathrm{rule}\right]…

Question-211636

Question Number 211636 by BaliramKumar last updated on 15/Sep/24 Answered by Rasheed.Sindhi last updated on 15/Sep/24 $$\blacktriangleright{A}={x}^{\mathrm{8}{k}+\mathrm{3}} +{x}^{\mathrm{8}{k}+\mathrm{6}} +{x}^{\mathrm{8}{k}+\mathrm{9}} +{x}^{\mathrm{8}{k}+\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:={x}^{\mathrm{8}{k}+\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +{x}^{\mathrm{9}}…

Question-211105

Question Number 211105 by peter frank last updated on 28/Aug/24 Answered by mm1342 last updated on 28/Aug/24 $${z}={z}_{\mathrm{1}} {z}_{\mathrm{2}} ={cos}\frac{\mathrm{12}\pi}{\mathrm{5}}+{isin}\frac{\mathrm{12}\pi}{\mathrm{5}} \\ $$$$={cos}\frac{\mathrm{2}\pi}{\mathrm{5}}+{isin}\frac{\mathrm{2}\pi}{\mathrm{5}}={e}^{\frac{\mathrm{2}\pi}{\mathrm{5}}{i}\:} \Rightarrow{z}^{\mathrm{5}} ={e}^{\mathrm{2}\pi{i}} =\mathrm{1}…