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Question Number 92777 by ckkim89 last updated on 09/May/20 $$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{n}} =? \\ $$$$ \\ $$ Commented by mr W…
Question Number 27204 by rish@bh last updated on 03/Jan/18 $$\mathrm{if}\:{g}\left({x}\right)={f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right) \\ $$$$\mathrm{and}\:{f}^{\left(\mathrm{2}\right)} \left({x}\right)<\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$${g}\left({x}\right)\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{in}\:\left(\mathrm{0},\mathrm{1}/\mathrm{2}\right)\:\mathrm{and} \\ $$$${g}\left({x}\right)\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{in}\:\left(\mathrm{1}/\mathrm{2},\mathrm{1}\right) \\ $$ Answered by Giannibo last…
Question Number 92712 by majid last updated on 08/May/20 $${find}\:\:\boldsymbol{{m}}\:{for}\:{fix}\:{function} \\ $$$${f}\left({x}\right)=\frac{\left(\boldsymbol{{m}}−\mathrm{1}\right){x}+\mathrm{3}}{{x}−\mathrm{1}} \\ $$ Commented by Rio Michael last updated on 08/May/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{function}? \\ $$…
Question Number 92708 by Rio Michael last updated on 08/May/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}. \\ $$$$\:\left(\mathrm{a}\right)\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{6}{y}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{b}\right)\:\left(\mathrm{2}{y}−{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:−\mathrm{2}\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$ Answered…
Question Number 27159 by RJPADHI2000@gmail.com last updated on 02/Jan/18 $$\sqrt{\mathrm{1}−{x}^{\mathrm{6}\:} \:}\:+\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }\:={k}^{\mathrm{3}} \left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)\:\:\:{then}\:{prove}\:{that}\:\:\:\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{y}^{\mathrm{2}} \sqrt{\mathrm{1}−{y}^{\mathrm{2}\Delta} }} \\ $$$$ \\ $$$$ \\ $$…
Question Number 27073 by sorour87 last updated on 01/Jan/18 $$\int\mathrm{ln}\:{x}×\mathrm{cos}\:\mathrm{2ln}\:{xdx} \\ $$ Answered by prakash jain last updated on 02/Jan/18 $${u}=\mathrm{ln}\:{x} \\ $$$${du}=\frac{\mathrm{1}}{{x}}{dx}\Rightarrow{dx}={xdu}\Rightarrow{dx}={e}^{{u}} {du} \\…
Question Number 92605 by som(math1967) last updated on 08/May/20 $${If}\:\frac{\mathrm{1}+{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}\:+\frac{\mathrm{1}−{x}}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}\:=\mathrm{1} \\ $$$${find}\:{x} \\ $$ Commented by behi83417@gmail.com last updated on 08/May/20 $$\frac{\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{x}}\right)}{−\mathrm{x}}+\frac{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{x}}\right)}{\mathrm{x}}=\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{x}\right)\sqrt{\mathrm{1}+\mathrm{x}}−\left(\mathrm{1}+\mathrm{x}\right)+\left(\mathrm{1}−\mathrm{x}\right)\sqrt{\mathrm{1}−\mathrm{x}}+\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x} \\…
Question Number 92593 by Rio Michael last updated on 08/May/20 $$\mathrm{using}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{theorem}\: \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com