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Question-135838

Question Number 135838 by abdurehime last updated on 16/Mar/21 Answered by liberty last updated on 16/Mar/21 $${f}\left({x}\right)=\:\frac{{x}}{{x}^{\mathrm{2}} +{k}}\:\Rightarrow\mathrm{ln}\:{y}\:=\:\mathrm{ln}\:{x}−\mathrm{ln}\:\left({x}^{\mathrm{2}} +{k}\right) \\ $$$$\:\frac{{y}'}{{y}}\:=\:\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +{k}} \\ $$$$\frac{{y}'}{{y}}\:=\:\frac{{k}−{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{2}}…

Question-135840

Question Number 135840 by abdurehime last updated on 16/Mar/21 Answered by liberty last updated on 16/Mar/21 $${from}\:{p}\Rightarrow\:\backsim{q}\:{False}\:{we}\:{get}\:\begin{cases}{{p}\::\:{T}}\\{{q}\::\:{T}}\end{cases} \\ $$$${so}\:{the}\:{following}\:{statements}\:{is} \\ $$$${True}\::\:\left({p} \sim{q}\right)\Leftrightarrow\left(\sim{r} {r}\right) \\ $$$${answer}\::\:{C}…