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Author: Tinku Tara

2x-2-1-2x-x-2-1-x-2-x-x-1-x-2-1-dx-dx-x-x-1-1-x-3-

Question Number 73202 by MJS last updated on 08/Nov/19 $$\int\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} −{x}+\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}=? \\ $$$$\int\frac{{dx}}{{x}\sqrt{{x}+\mathrm{1}}\sqrt{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }}=? \\ $$ Commented by mathmax by abdo last…

lim-x-x-x-2-log-1-1-x-

Question Number 7667 by upendrakishor99@gmail.com last updated on 08/Sep/16 $$\underset{{x}\rightarrow\alpha} {\mathrm{lim}}\left[{x}−{x}^{\mathrm{2}} \mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right] \\ $$ Answered by FilupSmith last updated on 08/Sep/16 $$=\alpha−\alpha^{\mathrm{2}} \mathrm{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\alpha}\right) \\ $$…

Question-73203

Question Number 73203 by Tanmay chaudhury last updated on 08/Nov/19 Commented by kaivan.ahmadi last updated on 08/Nov/19 $${xsin}\frac{\mathrm{1}}{{x}}=\mathrm{1}\Rightarrow{sin}\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}}\Rightarrow\frac{\mathrm{1}}{{x}}=\mathrm{0}\Rightarrow{there}\:{is}\:{no}\:{root} \\ $$$${so}\:{A}\:{has}\:{exactly}\:{one}\:{element},\:{A}=\left\{\mathrm{0}\right\} \\ $$ Terms of Service…

d-dx-x-2-4-

Question Number 7665 by upendrakishor99@gmail.com last updated on 08/Sep/16 $$\frac{{d}}{{dx}}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}} \\ $$ Answered by Rasheed Soomro last updated on 08/Sep/16 $$\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{1}/\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\left(\mathrm{1}/\mathrm{2}\right)−\mathrm{1}}…

mathematical-analysis-suppose-f-a-b-R-is-a-function-and-a-b-R-is-an-increasing-function-on-a-b-meanwhile-is-continuous-at-y-0-whe

Question Number 138733 by mnjuly1970 last updated on 17/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{mathematical}\:….{analysis}….. \\ $$$$\:\:{suppose}\:\:\:\:{f}\::\left[{a}\:,\:{b}\right]\rightarrow\mathbb{R}\:{is}\:{a}\:{function} \\ $$$$\:\:\:{and}\:\:\:\alpha:\left[{a}\:,\:{b}\right]\overset{\alpha\nearrow} {\rightarrow}\mathbb{R}\:\left(\alpha\:{is}\:{an}\:{increasing}\:{function}\right. \\ $$$$\left.\:{on}\:\left[{a}\:,\:{b}\right]\right)\:\:{meanwhile}\:\alpha\:{is}\:{continuous}\:{at}\:{y}_{\mathrm{0}} \: \\ $$$$\:\:{where}\:\:\:{a}\leqslant{y}_{\mathrm{0}} \leqslant{b}\:\:.\:{defining}\: \\ $$$$\:\:\:{f}\left({x}\right)=\begin{cases}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:{x}={y}_{\mathrm{0}} }\\{\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:{x}\neq{y}_{\mathrm{0}} }\end{cases}…