Question Number 66602 by aliesam last updated on 17/Aug/19
Commented by Rio Michael last updated on 17/Aug/19
$$\left.\mathrm{16}\right)\:{let}\:{y}\:=\:{x}^{\mathrm{3}} \\ $$$$\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\:=\:\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\:\frac{{dy}}{{dx}}\mid_{{x}\:=\:−\mathrm{1}} \:=\:\mathrm{3} \\ $$$${when}\:{x}\:=\:−\mathrm{1}\:\:,\:{y}\:=\:−\mathrm{1}\:\:{hence}\:{pt}\:\:\left(−\mathrm{1},−\mathrm{1}\right) \\ $$$$\:{equation} \\ $$$$\:{y}\:=\:{mx}\:+\:{c} \\ $$$$\:−\mathrm{1}\:=\:\mathrm{3}\left(−\mathrm{1}\right)\:+\:{c} \\ $$$$\:{c}\:=\:\mathrm{2} \\ $$$$\:{equation}\:;\:\:\:{y}\:=\:\mathrm{3}{x}\:+\:\mathrm{2}\:\:{answer}\:{C} \\ $$$$ \\ $$$$\left.\mathrm{17}\right)\:\:\frac{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{3}{x}\:+\:\mathrm{2}{y}}\:=\:\frac{\left(\mathrm{3}{x}−\:\mathrm{2}{y}\right)\left(\mathrm{3}{x}\:+\:\mathrm{2}{y}\right)}{\mathrm{3}{x}\:+\:\mathrm{2}{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3}{x}\:−\mathrm{2}{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3}\left(\mathrm{2}\right)\:−\mathrm{2}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}\:\:\:{D} \\ $$$$ \\ $$
Commented by aliesam last updated on 17/Aug/19
$${perfect} \\ $$
Commented by mathmax by abdo last updated on 17/Aug/19
$$\left.\mathrm{1}\left.\mathrm{5}\right)\:{the}\:{corrct}\:{answer}\:{is}\:{a}\right) \\ $$$$\left.\mathrm{16}\right)\:{equation}\:{of}\:{tangente}\:{at}\:{x}=−\mathrm{1}\:{is}\:{y}={f}^{'} \left(−\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{f}\left(−\mathrm{1}\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} \:\Rightarrow{f}^{'} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} \:\Rightarrow{f}^{'} \left(−\mathrm{1}\right)=\mathrm{3}\:{and}\:{f}\left(−\mathrm{1}\right)=−\mathrm{1}\:\Rightarrow \\ $$$${y}=\mathrm{3}\left({x}+\mathrm{1}\right)−\mathrm{1}\:\:\Rightarrow{y}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\left.\mathrm{18}\right)\:\begin{cases}{{x}−{y}+\mathrm{1}=\mathrm{0}\:\Rightarrow\begin{cases}{{x}−{y}=−\mathrm{1}}\\{\mathrm{7}{x}+{y}=\mathrm{17}}\end{cases}}\\{\mathrm{7}{x}+{y}=\mathrm{17}}\end{cases} \\ $$$$\Delta=\mathrm{8}\neq\mathrm{0}\:\Rightarrow{the}\:{systeme}\:{has}\:{one}\:{solution}\:\:\:\left({a}\right) \\ $$
Answered by Rio Michael last updated on 17/Aug/19
$$\left.\mathrm{15}\right)\:{A} \\ $$$$ \\ $$
Commented by aliesam last updated on 17/Aug/19
$${thank}\:{you}\:{sir} \\ $$
Answered by Tanmay chaudhury last updated on 17/Aug/19
$$\left.\mathrm{15}\right)\:{a}\:\:{edited} \\ $$$$\left.\mathrm{16}\right)\left({y}+\mathrm{1}\right)=\mathrm{3}\left(−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)\rightarrow{y}=\mathrm{3}{x}+\mathrm{2}\:\:\:{so}\:{C} \\ $$$$\left.\mathrm{17}\right)\frac{\mathrm{36}−\mathrm{36}}{\mathrm{6}+\mathrm{6}}=\mathrm{0}\:\:\:\boldsymbol{{d}} \\ $$$$\left.\mathrm{18}\right)\boldsymbol{{y}}=\boldsymbol{{x}}+\mathrm{1}\:\:\:\:\boldsymbol{{y}}+\mathrm{7}\boldsymbol{{x}}=\mathrm{17}\:\:\:\rightarrow{x}=\mathrm{2}\:\:\:{y}=\mathrm{3}\:\:\:\:\boldsymbol{{so}}\:\boldsymbol{{a}} \\ $$$$\left.\mathrm{18}\right)\boldsymbol{{sinx}}+\boldsymbol{{cosx}}+\mathrm{3} \\ $$$$\mathrm{3}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\boldsymbol{{sin}}\left(\frac{\pi}{\mathrm{4}}+{x}\right)+\mathrm{3}\geqslant\mathrm{3}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:{alwYs}\:{more}\:{than}\:\mathrm{2} \\ $$$${so}\:\:{b} \\ $$$$\left.\mathrm{19}\right){y}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1} \\ $$$${wait}\:{for}\:\mathrm{19} \\ $$