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Author: Tinku Tara

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Question Number 131144 by SWPlaysMC last updated on 16/Dec/21 $${delete}\:{this}\:{question}\:{please}\:{because}\:{I}\:{can}'{t} \\ $$$${was}\:{just}\:{trying}\:{it}\:{out} \\ $$$${if}\:{x}=\mathrm{10}{y}^{\mathrm{2}} \\ $$$${then}\:{x}>{x}−\mathrm{14} \\ $$$${else}\:{if}\:{x}=\mathrm{11}{y}^{\mathrm{2}} \\ $$$${then}\:{x}<{x}+\mathrm{14} \\ $$$${end} \\ $$ Terms…

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Question Number 131147 by EDWIN88 last updated on 02/Feb/21 $$\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\sqrt{{x}−{x}^{\mathrm{3}} }}\:{dx}\:=?\: \\ $$ Commented by EDWIN88 last updated on 02/Feb/21 $$\:\sqrt{{x}−{x}^{\mathrm{3}} }\:=\:{w}−\mathrm{1}\:\Rightarrow{x}−{x}^{\mathrm{3}}…

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Question Number 131146 by EDWIN88 last updated on 02/Feb/21 $${Given}\:{f}\left({x}\right)={f}\left({x}+\mathrm{6}\right)\:\forall{x}\in\mathbb{R} \\ $$$${If}\:\int_{−\mathrm{2}} ^{\:\mathrm{2}} {f}\left({x}\right){dx}=−\mathrm{2}\:{and}\:\int_{−\mathrm{2}} ^{\:\mathrm{4}} {f}\left({x}\right){dx}=\mathrm{2} \\ $$$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{4}} ^{\:\mathrm{10}} {f}\left({x}\right){dx}. \\ $$ Commented by EDWIN88…

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Question Number 71 by mike last updated on 25/Jan/15 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}=? \\ $$ Answered by jayant last updated on 17/Nov/14 $$\mathrm{1} \\ $$ Commented by 123456…

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Question Number 131138 by liberty last updated on 02/Feb/21 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{arctan}\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2}}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2}}}\:\mathrm{dx}\:=? \\ $$ Answered by EDWIN88 last updated on 02/Feb/21 $$=\frac{\mathrm{5}\pi^{\mathrm{2}}…

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Question Number 65601 by AnjanDey last updated on 31/Jul/19 $$\mathrm{1}.\mathrm{If}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\boldsymbol{{n}}−\mathrm{1}} \mathrm{log}\:\boldsymbol{{x}},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\boldsymbol{{x}}^{\mathrm{2}} \frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }+\left(\mathrm{3}−\mathrm{2}\boldsymbol{{n}}\right)\boldsymbol{{x}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{y}}=\mathrm{0} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{I}}\mathrm{f}\:\frac{\mathrm{mtan}\:\left(\alpha−\theta\right)}{\mathrm{cos}\:^{\mathrm{2}} \theta}=\frac{{n}\mathrm{tan}\:\theta}{\mathrm{cos}\:^{\mathrm{2}} \left(\alpha−\theta\right)},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{n}−{m}}{{n}+{m}}\mathrm{tan}\:\alpha\right)\right] \\ $$ Commented by Prithwish…

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Question Number 131132 by 676597498 last updated on 01/Feb/21 $$\mathrm{u}\left(\mathrm{x}\right)=\mathrm{cosh}\left(\mathrm{x}\right)−\mathrm{x} \\ $$$$\mathrm{v}_{\mathrm{n}+\mathrm{1}} =\mathrm{u}\left(\mathrm{v}_{\mathrm{n}} \right) \\ $$$$\mathrm{m}=\sqrt{\mathrm{2}}−\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{v}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{A}.\:\mathrm{V}_{\mathrm{n}} −\mathrm{V}_{\mathrm{0}} =\mathrm{nm} \\ $$$$\mathrm{B}.\:\mathrm{v}_{\mathrm{n}}…