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Author: Tinku Tara

Question-194913

Question Number 194913 by cortano12 last updated on 19/Jul/23 Answered by witcher3 last updated on 19/Jul/23 $$\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}''\left(\mathrm{x}\right)\right)\mathrm{cos}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\left.\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} \mathrm{f}''\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{f}'\mathrm{cos}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int\mathrm{f}'\mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\left.=\left.−\mathrm{f}'\left(\mathrm{0}\right)+\mathrm{fsin}\left(\mathrm{x}\right)\right]_{\mathrm{0}}…

lim-x-x-2-5x-1-x-2-2x-1-x-2-3-x-2-4x-9-16x-2-8-

Question Number 194928 by cortano12 last updated on 19/Jul/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{9}}−\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{8}}\:=?\right. \\ $$ Answered by horsebrand11 last updated on 20/Jul/23…

Question-194899

Question Number 194899 by kapoorshah last updated on 19/Jul/23 Answered by Frix last updated on 19/Jul/23 $${y}={px}\wedge{z}={qx} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{1}−{pq}}=\frac{\mathrm{8}}{{p}^{\mathrm{2}} −{q}}=\frac{\mathrm{10}}{{q}^{\mathrm{2}} −{p}} \\…

Question-194914

Question Number 194914 by Mingma last updated on 19/Jul/23 Answered by a.lgnaoui last updated on 20/Jul/23 $$\mathrm{distance}\:\mathrm{betwen}\:\mathrm{Dand}\:\mathrm{E}\:=\mathrm{distance}\:\mathrm{betwen}\: \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{D};\:\:\:\mathrm{so}\:\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{point} \\ $$$$\:\mathrm{4}^{\mathrm{9}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\boldsymbol{\mathrm{A}} \\ $$$$ \\…

Question-194879

Question Number 194879 by tri26112004 last updated on 18/Jul/23 Answered by cortano12 last updated on 18/Jul/23 $$\:\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2}{u}\right)^{{u}} −\mathrm{1}}{\left(\mathrm{1}+\mathrm{3}{u}\right)^{{u}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\left[\:{u}\:=\frac{\mathrm{1}}{{x}}\:\right] \\ $$$$\:{then}\:{apply}\:{L}'{Hopital}\: \\…