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Question Number 171762 by pete last updated on 20/Jun/22
The curve for which (dy/dx)=a(x−p)(x−q),  where a, p and q are constants, has turning  points at (2,0) and (1,1).  i) state the value of p and q.  ii) using these values, determine the value  of a
$$\mathrm{The}\:\mathrm{curve}\:\mathrm{for}\:\mathrm{which}\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}\left(\mathrm{x}−\mathrm{p}\right)\left(\mathrm{x}−\mathrm{q}\right), \\ $$$$\mathrm{where}\:\mathrm{a},\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{constants},\:\mathrm{has}\:\mathrm{turning} \\ $$$$\mathrm{points}\:\mathrm{at}\:\left(\mathrm{2},\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{1},\mathrm{1}\right). \\ $$$$\left.\mathrm{i}\right)\:\mathrm{state}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}. \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{using}\:\mathrm{these}\:\mathrm{values},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{a} \\ $$
Answered by mr W last updated on 21/Jun/22
a(x−p)(x−q)=0  ⇒x=p, q  ⇒p=1, q=2  (dy/dx)=a(x−1)(x−2)=a(x^2 −3x+2)  y=a((x^3 /3)−((3x^2 )/2)+2x)+C  0=a((8/3)−((3×4)/2)+2×2)+C  1=a((1/3)−(3/2)+2)+C  ⇒a=6, C=−4
$${a}\left({x}−{p}\right)\left({x}−{q}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}={p},\:{q} \\ $$$$\Rightarrow{p}=\mathrm{1},\:{q}=\mathrm{2} \\ $$$$\frac{{dy}}{{dx}}={a}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)={a}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\right) \\ $$$${y}={a}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{x}\right)+{C} \\ $$$$\mathrm{0}={a}\left(\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{3}×\mathrm{4}}{\mathrm{2}}+\mathrm{2}×\mathrm{2}\right)+{C} \\ $$$$\mathrm{1}={a}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\right)+{C} \\ $$$$\Rightarrow{a}=\mathrm{6},\:{C}=−\mathrm{4} \\ $$
Commented by pete last updated on 21/Jun/22
Thank you very much sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

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