Question Number 39695 by maxmathsup by imad last updated on 09/Jul/18
$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{2}{xarctan}\sqrt{\left.\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}\right)}\right. \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{'} \left({x}\right)\:{and}\:\:{determine}\:{its}\:{sign}. \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{the}\:{equation}\:{of}\:{assymptote}\:{at}\:{pont}\:{A}\left(\mathrm{1},{f}\left(\mathrm{1}\right)\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{a}\:{and}\:{b}\:{from}\:{R}\:/\:\:{f}\left({x}\right)\sim\:{a}\left({x}−\mathrm{1}\right)\:+{b}\:\:\left({x}\rightarrow\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{5}\right)\:{calculate}\:{f}^{''} \left({x}\right) \\ $$
Commented by maxmathsup by imad last updated on 14/Jul/18
![1) we have f(x)=ln(2x) +ln(arctan((√(2x^2 −1)))) x ∈ D_f ⇔ x>0 and 2x^2 −1>0 and arctan((√(2x^2 −1)))>0 but 2x^2 −1>0 ⇔x^2 >(1/2) ⇒∣x∣>(1/( (√2))) ⇒ x∈]−∞,−(1/( (√2)))[∪](1/( (√2))),+∞[ ⇒ D_f =](1/( (√2))),+∞[ 2) f^, (x) = (1/x) +(((arctan((√(2x^2 −1))))^′ )/(arctan((√(2x^2 −1))))) =(1/x) +(1/(arctan((√(2x^2 −1))))) { (((4x)/(2(√(2x^2 −1))))/(1+2x^2 −1))} =(1/x) + (1/(x arctan((√(2x^2 −1)))(√(2x^2 −1)))) .](https://www.tinkutara.com/question/Q39989.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)={ln}\left(\mathrm{2}{x}\right)\:+{ln}\left({arctan}\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}}\right)\right) \\ $$$${x}\:\in\:{D}_{{f}} \:\:\Leftrightarrow\:{x}>\mathrm{0}\:{and}\:\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}>\mathrm{0}\:{and}\:{arctan}\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}}\right)>\mathrm{0}\:{but} \\ $$$$\left.\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}>\mathrm{0}\:\Leftrightarrow{x}^{\mathrm{2}} >\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow\mid{x}\mid>\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow\:{x}\in\right]−\infty,−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\cup\right]\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},+\infty\left[\:\Rightarrow\right. \\ $$$$\left.{D}_{{f}} =\right]\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},+\infty\left[\right. \\ $$$$\left.\mathrm{2}\right)\:{f}^{,} \left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\:+\frac{\left({arctan}\left(\sqrt{\left.\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)}\right)^{'} \right.}{{arctan}\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}}\right)}\:\:=\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{arctan}\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}}\right)}\:\left\{\:\frac{\frac{\mathrm{4}{x}}{\mathrm{2}\sqrt{\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}}}}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{{x}}\:+\:\:\frac{\mathrm{1}}{{x}\:{arctan}\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}\right)\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}}\:. \\ $$
Commented by maxmathsup by imad last updated on 14/Jul/18
$${we}\:{have}\:{x}>\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow\:{f}^{'} \left({x}\right)>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{equation}\:{of}\:{tangent}\:{to}\:{C}_{{f}} \:{at}\:{point}\:{A}\left(\mathrm{1},{f}\left(\mathrm{1}\right)\right){is}\:{y}={f}^{'} \left(\mathrm{1}\right)\left({x}−\mathrm{1}\right)\:+{f}\left(\mathrm{1}\right) \\ $$$${but}\:{f}\left(\mathrm{1}\right)={ln}\left(\mathrm{2}\right)\:+{ln}\left(\frac{\pi}{\mathrm{4}}\right)\:\:{and}\:{f}^{'} \left(\mathrm{1}\right)\:=\:\mathrm{1}+\:\frac{\mathrm{4}}{\pi}\:\Rightarrow \\ $$$${y}\:=\left(\mathrm{1}+\frac{\mathrm{4}}{\pi}\right)\left({x}−\mathrm{1}\right)\:+{ln}\left(\mathrm{2}\right)\:+{ln}\left(\frac{\pi}{\mathrm{4}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 14/Jul/18
$$\left.\mathrm{3}\right)\:{in}\:{this}\:{Q}\:{change}\:{assymptote}\:{by}\:{tangent}\:. \\ $$
Commented by maxmathsup by imad last updated on 14/Jul/18
$$\left.\mathrm{4}\right)\:{a}=\left(\mathrm{1}+\frac{\mathrm{4}}{\pi}\right)\:\:{and}\:{b}={ln}\left(\mathrm{2}\right)+{ln}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$