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Question Number 66602 by aliesam last updated on 17/Aug/19

Commented by Rio Michael last updated on 17/Aug/19

16) let y = x^3 ⇒ (dy/dx) = 3x^2 (dy/dx)∣_(x = −1) = 3 when x = −1 , y = −1 hence pt (−1,−1) equation y = mx + c −1 = 3(−1) + c c = 2 equation ; y = 3x + 2 answer C 17) ((9x^2 −4y^2 )/(3x + 2y)) = (((3x− 2y)(3x + 2y))/(3x + 2y)) = 3x −2y = 3(2) −2(3) =0 D

$$\left.\mathrm{16}\right)\:{let}\:{y}\:=\:{x}^{\mathrm{3}} \\ $$$$\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\:=\:\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\:\frac{{dy}}{{dx}}\mid_{{x}\:=\:−\mathrm{1}} \:=\:\mathrm{3} \\ $$$${when}\:{x}\:=\:−\mathrm{1}\:\:,\:{y}\:=\:−\mathrm{1}\:\:{hence}\:{pt}\:\:\left(−\mathrm{1},−\mathrm{1}\right) \\ $$$$\:{equation} \\ $$$$\:{y}\:=\:{mx}\:+\:{c} \\ $$$$\:−\mathrm{1}\:=\:\mathrm{3}\left(−\mathrm{1}\right)\:+\:{c} \\ $$$$\:{c}\:=\:\mathrm{2} \\ $$$$\:{equation}\:;\:\:\:{y}\:=\:\mathrm{3}{x}\:+\:\mathrm{2}\:\:{answer}\:{C} \\ $$$$ \\ $$$$\left.\mathrm{17}\right)\:\:\frac{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{3}{x}\:+\:\mathrm{2}{y}}\:=\:\frac{\left(\mathrm{3}{x}−\:\mathrm{2}{y}\right)\left(\mathrm{3}{x}\:+\:\mathrm{2}{y}\right)}{\mathrm{3}{x}\:+\:\mathrm{2}{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3}{x}\:−\mathrm{2}{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3}\left(\mathrm{2}\right)\:−\mathrm{2}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}\:\:\:{D} \\ $$$$ \\ $$

Commented by aliesam last updated on 17/Aug/19

perfect

$${perfect} \\ $$

Commented by mathmax by abdo last updated on 17/Aug/19

15) the corrct answer is a) 16) equation of tangente at x=−1 is y=f^′ (−1)(x+1)+f(−1) f(x)=x^3 ⇒f^′ (x)=3x^2 ⇒f^′ (−1)=3 and f(−1)=−1 ⇒ y=3(x+1)−1 ⇒y=3x+2 18) { ((x−y+1=0 ⇒ { ((x−y=−1)),((7x+y=17)) :})),((7x+y=17)) :} Δ=8≠0 ⇒the systeme has one solution (a)

$$\left.\mathrm{1}\left.\mathrm{5}\right)\:{the}\:{corrct}\:{answer}\:{is}\:{a}\right) \\ $$$$\left.\mathrm{16}\right)\:{equation}\:{of}\:{tangente}\:{at}\:{x}=−\mathrm{1}\:{is}\:{y}={f}^{'} \left(−\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{f}\left(−\mathrm{1}\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} \:\Rightarrow{f}^{'} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} \:\Rightarrow{f}^{'} \left(−\mathrm{1}\right)=\mathrm{3}\:{and}\:{f}\left(−\mathrm{1}\right)=−\mathrm{1}\:\Rightarrow \\ $$$${y}=\mathrm{3}\left({x}+\mathrm{1}\right)−\mathrm{1}\:\:\Rightarrow{y}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\left.\mathrm{18}\right)\:\begin{cases}{{x}−{y}+\mathrm{1}=\mathrm{0}\:\Rightarrow\begin{cases}{{x}−{y}=−\mathrm{1}}\\{\mathrm{7}{x}+{y}=\mathrm{17}}\end{cases}}\\{\mathrm{7}{x}+{y}=\mathrm{17}}\end{cases} \\ $$$$\Delta=\mathrm{8}\neq\mathrm{0}\:\Rightarrow{the}\:{systeme}\:{has}\:{one}\:{solution}\:\:\:\left({a}\right) \\ $$

Answered by Rio Michael last updated on 17/Aug/19

15) A

$$\left.\mathrm{15}\right)\:{A} \\ $$$$ \\ $$

Commented by aliesam last updated on 17/Aug/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Tanmay chaudhury last updated on 17/Aug/19

15) a edited 16)(y+1)=3(−1)^2 (x+1)→y=3x+2 so C 17)((36−36)/(6+6))=0 d 18)y=x+1 y+7x=17 →x=2 y=3 so a 18)sinx+cosx+3 3+(1/(√2))≥(1/(√2))sin((π/4)+x)+3≥3−(1/(√2)) alwYs more than 2 so b 19)y=x^3 −3x^2 +3x−1 wait for 19

$$\left.\mathrm{15}\right)\:{a}\:\:{edited} \\ $$$$\left.\mathrm{16}\right)\left({y}+\mathrm{1}\right)=\mathrm{3}\left(−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)\rightarrow{y}=\mathrm{3}{x}+\mathrm{2}\:\:\:{so}\:{C} \\ $$$$\left.\mathrm{17}\right)\frac{\mathrm{36}−\mathrm{36}}{\mathrm{6}+\mathrm{6}}=\mathrm{0}\:\:\:\boldsymbol{{d}} \\ $$$$\left.\mathrm{18}\right)\boldsymbol{{y}}=\boldsymbol{{x}}+\mathrm{1}\:\:\:\:\boldsymbol{{y}}+\mathrm{7}\boldsymbol{{x}}=\mathrm{17}\:\:\:\rightarrow{x}=\mathrm{2}\:\:\:{y}=\mathrm{3}\:\:\:\:\boldsymbol{{so}}\:\boldsymbol{{a}} \\ $$$$\left.\mathrm{18}\right)\boldsymbol{{sinx}}+\boldsymbol{{cosx}}+\mathrm{3} \\ $$$$\mathrm{3}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\geqslant\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\boldsymbol{{sin}}\left(\frac{\pi}{\mathrm{4}}+{x}\right)+\mathrm{3}\geqslant\mathrm{3}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:{alwYs}\:{more}\:{than}\:\mathrm{2} \\ $$$${so}\:\:{b} \\ $$$$\left.\mathrm{19}\right){y}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1} \\ $$$${wait}\:{for}\:\mathrm{19} \\ $$

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