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Question Number 195208 Answers: 4 Comments: 0
Question Number 195202 Answers: 0 Comments: 0
Question Number 195203 Answers: 1 Comments: 0
$$\mathrm{Calculer}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{dt}}{\left(\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} } \\ $$
Question Number 195195 Answers: 1 Comments: 1
$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{limit}: \\ $$$$\:\underset{\mathrm{x}\rightarrow\mathrm{a}\:} {\overset{\mathrm{lim}} {\:}}\:\:\:\frac{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\frac{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$
Question Number 195194 Answers: 2 Comments: 0
Question Number 195192 Answers: 1 Comments: 0
$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{sinx}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{x}} =? \\ $$
Question Number 195206 Answers: 1 Comments: 0
Question Number 195185 Answers: 0 Comments: 0
$$\mathrm{1}/\:\:\mathrm{Montrer}\:\mathrm{que}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx}=\left(\frac{\mathrm{2}^{\mathrm{2}{p}−\mathrm{3}} }{{p}}\right)\pi\left[\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\sum}}{cos}^{\mathrm{2}{p}−\mathrm{1}} \left(\frac{{k}\pi}{\mathrm{2}{p}}\right)\right] \\ $$$$\mathrm{2}/\:\:\:\:\mathrm{En}\:\mathrm{d}\acute {\mathrm{e}duire}\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx} \\ $$
Question Number 195180 Answers: 1 Comments: 0
$$ \\ $$$$\mathrm{1}.\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{sinx}\:\:\:\:\:\:,\:\:\:\:\frac{\pi}{\mathrm{2}}<\mathrm{x}\leqslant\mathrm{2}\pi}\\{\mathrm{cosx}\:\:\:\:\:\:,\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{f}^{'} \left(\frac{\pi}{\mathrm{2}}\right)\:=? \\ $$$$\mathrm{2}.\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{sinx}\:\:\:\:\:\:,\:\:\:\:\frac{\pi}{\mathrm{2}}<\mathrm{x}\leqslant\mathrm{2}\pi}\\{\mathrm{cosx}\:\:\:\:\:\:,\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{f}'\left(\mathrm{2}\pi\right)\:=? \\ $$$$ \\ $$
Question Number 195178 Answers: 2 Comments: 0
Question Number 195175 Answers: 1 Comments: 0
Question Number 195171 Answers: 0 Comments: 0
Question Number 195170 Answers: 2 Comments: 0
$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{7}} +\mathrm{2}{x}+\mathrm{1}\:\:\:\:\:\:\:;{x}\geqslant\mathrm{2}}\\{{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{4}\:\:\:\:\:\:\:\:;{x}<\mathrm{1}}\end{cases} \\ $$$${f}^{'} \left(\mathrm{1}\right)=? \\ $$
Question Number 195165 Answers: 0 Comments: 1
$$\mathrm{If}\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{1}} }=\left(\mathrm{sin}\theta,\mathrm{cos}\theta,\theta\right),\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{2}} }=\left(\mathrm{cos}\theta,−\mathrm{sin}\theta,−\mathrm{3}\right)\:\mathrm{and} \\ $$$$\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{3}} }=\left(\mathrm{2},\mathrm{3},−\mathrm{1}\right),\:\mathrm{find}\:\frac{\mathrm{d}}{\mathrm{d}\theta}\left\{\overset{\rightarrow} {\mathrm{r}_{\mathrm{1}} }×\left(\overset{\rightarrow} {\mathrm{r}_{\mathrm{2}} }×\overset{\rightarrow} {\mathrm{r}_{\mathrm{3}} }\right)\right\}\:\mathrm{at}\:\theta=\mathrm{0} \\ $$
Question Number 195157 Answers: 1 Comments: 0
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\:\frac{{x}}{{y}}\right)}+\frac{{y}^{\mathrm{3}} }{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\:\frac{{y}}{{x}}\right)}=\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$
Question Number 195154 Answers: 1 Comments: 0
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{2}\pi} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:\left(\pi\:\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{2}} \left({x}−\mathrm{5}\pi\right)+\mathrm{4}\pi^{\mathrm{2}} \left(\mathrm{2}{x}−\pi\right)}\right)=? \\ $$$$ \\ $$
Question Number 195148 Answers: 3 Comments: 0
$$\:\:\begin{array}{|c|}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{x}\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}=?}\\\hline\end{array} \\ $$
Question Number 195137 Answers: 1 Comments: 0
$${f}\left({x}\right)={arctan}\left(\frac{\mathrm{4}{sinx}}{\mathrm{3}+\mathrm{5}{cosx}}\right)\:\:\:{then}\:{f}^{'} \left(\frac{\pi}{\mathrm{3}}\right)=? \\ $$
Question Number 195136 Answers: 1 Comments: 0
$${f}\left({x}\right)={arctan}\left({sinx}\right) \\ $$$${and}\:\:{cosa}=\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:\:\:\:\:{faind}\:\:\:{f}^{'} \left({a}\right)=? \\ $$
Question Number 195135 Answers: 4 Comments: 0
Question Number 195129 Answers: 3 Comments: 0
$$\mathrm{Calculer}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\mathrm{la}\:\mathrm{serie}\:\mathrm{suivante}: \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{32}}+\frac{\mathrm{9}}{\mathrm{128}}+..... \\ $$
Question Number 195126 Answers: 1 Comments: 0
$$\mathrm{Soit}\:{f}_{{n}} \left({x}\right)=\mathrm{2}^{{n}+\mathrm{1}} \left[\frac{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{cotan}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)−{cotanx}}{{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\right] \\ $$$${Calculer}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}f}_{{n}} \left({x}\right)\:{et}\:\underset{{n}\rightarrow+\infty} {{lim}}\:\frac{{f}_{{n}} \left({x}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} } \\ $$
Question Number 195124 Answers: 1 Comments: 0
Question Number 195120 Answers: 0 Comments: 0
Question Number 195118 Answers: 0 Comments: 1
$${a},{b},{c}>\mathrm{0}\:\&\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{{a}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{b}}{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{c}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{a}+{b}+{c}}{{ab}+{bc}+{ac}}\right)^{\mathrm{2}} \\ $$
Question Number 195122 Answers: 0 Comments: 0
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