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-

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Question Number 171762 by pete last updated on 20/Jun/22

$$\mathrm{The}\mathrm{curve}\mathrm{for}\mathrm{which}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}(\mathrm{x}-\mathrm{p})(\mathrm{x}-\mathrm{q}),$$$$\mathrm{where}\mathrm{a},\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{are}\mathrm{constants},\mathrm{has}\mathrm{turning}$$$$\mathrm{points}\mathrm{at}(2,0)\mathrm{and}(1,1).$$$$\mathrm{i})\mathrm{state}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{p}\mathrm{and}\mathrm{q}.$$$$\mathrm{ii})\mathrm{using}\mathrm{these}\mathrm{values},\mathrm{determine}\mathrm{the}\mathrm{value}$$$$\mathrm{of}a$$

Answered by mr W last updated on 21/Jun/22

$$a(x-p)(x-q)=0$$$$\Rightarrow x=p,q$$$$\Rightarrow p=1,q=2$$$$\frac{dy}{dx}=a(x-1)(x-2)=a(x^2-3x+2)$$$$y=a(\frac{x^3}{3}-\frac{3x^2}{2}+2x)+C$$$$0=a(\frac{8}{3}-\frac{3\times 4}{2}+2\times 2)+C$$$$1=a(\frac{1}{3}-\frac{3}{2}+2)+C$$$$\Rightarrow a=6,C=-4$$

Commented by pete last updated on 21/Jun/22

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$

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