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Author: Tinku Tara

Question-137016

Question Number 137016 by PGeeman last updated on 28/Mar/21 Answered by Dwaipayan Shikari last updated on 28/Mar/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{a}^{{x}} \sim\mathrm{1}+{xlog}\left({a}\right)\:\:\:{sin}\left({x}\right)\sim{x}\:;\:\:{log}\left(\mathrm{1}+{x}\right)\sim{x} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{4}^{{x}} −\mathrm{1}\right)^{\mathrm{3}} }{{sin}\left(\frac{{x}}{{p}}\right){log}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}}…

Question-5940

Question Number 5940 by sanusihammed last updated on 05/Jun/16 Answered by FilupSmith last updated on 06/Jun/16 $${Are}\:{of}\:{non}\:{shaded}: \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{r}\left(\theta−\mathrm{sin}\:\theta\right) \\ $$$$\therefore\mathrm{Total}\:\mathrm{A}{rea}: \\ $$$${A}=\pi{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{r}\left(\theta−\mathrm{sin}\:\theta\right)…

Hi-guyz-For-R-1-2-3-4-5-6-223-224-and-S-2-3-4-5-6-7-224-225-Prove-that-R-lt-1-15-lt-S-

Question Number 137005 by greg_ed last updated on 28/Mar/21 $$\boldsymbol{\mathrm{Hi}},\:\boldsymbol{\mathrm{guyz}}\:! \\ $$$$\boldsymbol{\mathrm{For}}\:\boldsymbol{\mathrm{R}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×…×\frac{\mathrm{223}}{\mathrm{224}}\:\:\:\boldsymbol{\mathrm{and}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{6}}{\mathrm{7}}×…×\frac{\mathrm{224}}{\mathrm{225}}\:. \\ $$$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\::\:\:\boldsymbol{\mathrm{R}}\:<\:\frac{\mathrm{1}}{\mathrm{15}}\:<\:\boldsymbol{\mathrm{S}}. \\ $$ Commented by greg_ed last updated on 01/May/21…

find-x-x-1-x-1-dx-

Question Number 137004 by Mathspace last updated on 28/Mar/21 $${find}\:\int\:\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}{dx} \\ $$ Answered by aleks041103 last updated on 28/Mar/21 $$\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}= \\ $$$$=\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}= \\ $$$$=\sqrt{{x}}\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}\right) \\…