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Author: Tinku Tara

if-p-x-2-x-2-x-1-2-x-4-1-4-x-2-and-x-4-x-2-then-show-that-32-p-2-80-

Question Number 203066 by mathlove last updated on 09/Jan/24 $${if}\:{p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}\:\:\:{and}\:{x}−\frac{\mathrm{4}}{{x}}=\mathrm{2} \\ $$$${then}\:{show}\:{that}\:\:\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\mathrm{80} \\ $$ Answered by Rasheed.Sindhi last updated on 09/Jan/24 $${if}\:{p}=\frac{{x}^{\mathrm{2}}…

Question-203062

Question Number 203062 by ajfour last updated on 09/Jan/24 Answered by ajfour last updated on 09/Jan/24 $$\mathrm{tan}\:\theta=\frac{{R}−\left(\mathrm{1}−{R}\right)}{{x}−{c}}=\frac{\mathrm{1}}{{x}} \\ $$$$\&\:\:{x}^{\mathrm{2}} ={R}^{\mathrm{2}} −\left(\mathrm{1}−{R}\right)^{\mathrm{2}} =\mathrm{2}{R}−\mathrm{1} \\ $$$$\Rightarrow\:\frac{{x}−{c}}{{x}}=\frac{\mathrm{2}{R}−\mathrm{1}}{\mathrm{1}}={x}^{\mathrm{2}} \\…

Question-203063

Question Number 203063 by LowLevelLump last updated on 09/Jan/24 Answered by MM42 last updated on 09/Jan/24 $${f}'={e}^{{x}} −{a}=\mathrm{0}\Rightarrow\alpha={lna} \\ $$$$\Rightarrow{minf}={a}−{alna} \\ $$$$\:{g}'={a}−\frac{\mathrm{1}}{{x}}=\mathrm{0}\Rightarrow\beta=\frac{\mathrm{1}}{{a}} \\ $$$$\Rightarrow{ming}=\mathrm{1}+{lna} \\…

14-1-x-ln-x-3-d-x-

Question Number 203061 by York12 last updated on 08/Jan/24 $$\mathrm{14}\int\frac{\mathrm{1}}{\mathrm{x}\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{3}} \:}\:\:\mathrm{d}\left(\mathrm{x}\right) \\ $$ Answered by MM42 last updated on 09/Jan/24 $${ans} \\ $$$$\:\frac{−\mathrm{7}}{\left({lnx}\right)^{\mathrm{2}} } \\…

Question-203059

Question Number 203059 by hassanmpsy last updated on 08/Jan/24 Commented by witcher3 last updated on 11/Jan/24 $$\mathrm{U}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{2}\frac{\mathrm{k}}{\mathrm{n}}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{f}\left(\frac{\mathrm{k}}{\mathrm{n}}\right) \\…