Question Number 139192 by EnterUsername last updated on 23/Apr/21 $$\mathrm{If}\:{z}_{\mathrm{1}\:} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{complex}\:{n}\mathrm{th}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{which}\:\mathrm{sub}- \\ $$$$\mathrm{tend}\:\mathrm{right}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{then}\:{n}\:\mathrm{must}\:\mathrm{be}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4K}+\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{4K}+\mathrm{2} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4K}+\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4K} \\ $$ Answered by mr W last…
Question Number 139190 by mathdanisur last updated on 23/Apr/21 $${a};{b}\in\mathbb{R}\:,\:\frac{\mid{ax}+{b}\mid}{\mathrm{1}+{x}^{\mathrm{2}} }\:\leqslant\:\mathrm{1}\:,\:\forall{x}\in\mathbb{R} \\ $$$${prove}:\:\:\mid{a}\mid\leqslant\mathrm{2}\:;\:\mid{b}\mid\leqslant\mathrm{1} \\ $$ Answered by mitica last updated on 24/Apr/21 $${x}=\mathrm{0}\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$$${x}=\mathrm{1}\Rightarrow\mid{a}+{b}\mid\leqslant\mathrm{2}…
Question Number 8115 by uchechukwu okorie favour last updated on 30/Sep/16 $${if}\:{xy}+{y}^{\mathrm{2}} =\mathrm{1}.\:{Find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:{at}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$ Answered by prakash jain last updated on 30/Sep/16 $${x}\frac{{dy}}{{dx}}+{y}+\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{0}…\left({i}\right)…
Question Number 8114 by uchechukwu okorie favour last updated on 30/Sep/16 $${Evaluate}\:\int\mathrm{cos}\:^{\mathrm{6}} {x} \\ $$ Answered by sandy_suhendra last updated on 30/Sep/16 Answered by prakash…
Question Number 8113 by uchechukwu okorie favour last updated on 30/Sep/16 $${Find}\:{an}\:{equation}\:{of}\:{the}\:{tangent} \\ $$$${line}\:{to}\:{the}\:{curve}\:{y}={tan}^{\mathrm{2}} {x}\:{at}\:{the}\: \\ $$$${point}\left(\frac{{x}}{\mathrm{3}}\:,\:\mathrm{0}\right) \\ $$ Answered by prakash jain last updated…
Question Number 73649 by aliesam last updated on 14/Nov/19 $${find}\:{the}\:{range} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{6}−\sqrt{{x}+\mathrm{2}}} \\ $$ Commented by mathmax by abdo last updated on 14/Nov/19…
Question Number 139182 by aupo14 last updated on 23/Apr/21 Commented by mr W last updated on 23/Apr/21 $${there}\:{is}\:{not}\:{much}\:{to}\:{explain}.\:{it}\:{is} \\ $$$${just}\:{the}\:{definition}\:{of}\:{a}\:{special} \\ $$$${function}. \\ $$ Commented…
Question Number 8105 by 314159 last updated on 30/Sep/16 $${Prove}\:{that}\:,{for}\:{any}\:{acute}\:{angle}\:\alpha\: \\ $$$${sec}\alpha\:{cosec}\alpha\:+{tan}\alpha\:+{cot}\alpha\:\geqslant\mathrm{4}. \\ $$ Commented by Yozzia last updated on 30/Sep/16 $${f}\left({x}\right)=\frac{\mathrm{1}}{{cosxsinx}}+\frac{{sinx}}{{cosx}}+\frac{{cosx}}{{sinx}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{cosxsinx}}+\frac{\mathrm{1}}{{sinxcosx}} \\…
Question Number 139171 by mnjuly1970 last updated on 23/Apr/21 Answered by Dwaipayan Shikari last updated on 23/Apr/21 $$\Gamma\left(\mathrm{1}+{x}\right)=\mathrm{1}−\gamma{x}+{O}\left({x}^{\mathrm{2}} \right)\Rightarrow\Gamma\left({x}\right)=\frac{\mathrm{1}}{{x}}−\gamma+{O}\left({x}\right) \\ $$$$\Gamma'\left(\mathrm{1}+{x}\right)=−\gamma+{O}\left({x}\right)\Rightarrow{x}\Gamma\left({x}\right)\psi\left({x}+\mathrm{1}\right)=−\gamma+{O}\left({x}\right) \\ $$$$\Rightarrow{x}\Gamma\left({x}\right)\psi\left({x}\right)+\Gamma\left({x}\right)=−\gamma+{O}\left({x}\right)\Rightarrow\psi\left({x}\right)\sim−\frac{\gamma}{{x}\Gamma\left({x}\right)}−\frac{\mathrm{1}}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 139165 by mathlove last updated on 23/Apr/21 Answered by mr W last updated on 23/Apr/21 $$\mathrm{13}+\sqrt{\mathrm{48}}=\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}=\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}−\sqrt{\mathrm{13}+\sqrt{\mathrm{48}}}=\mathrm{5}−\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)=\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{5}−\sqrt{\mathrm{13}+\sqrt{\mathrm{48}}}}=\mathrm{3}+\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{2}+\sqrt{\mathrm{3}} \\…