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Question Number 107728 by mathdave last updated on 12/Aug/20

Answered by john santu last updated on 12/Aug/20

            ((♣JS♣)/…)  let f(x)=x^(2020) +ax^2 +bx+c   (1)f(2)= 2^(2020) +4a+2b+c = 0  (2) f ′(2)=2020.2^(2019) +4a+b=0  (3) f′′(2) = 6  2020×2019.2^(2018) +2a = 6   a =(3/(1010×2019.2^(2018) ))  b=−2020.2^(2019) −((12)/(1010×2019.2^(2018) ))  now you can get the value of c

$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\clubsuit\mathcal{JS}\clubsuit}{\ldots} \\ $$$${let}\:{f}\left({x}\right)={x}^{\mathrm{2020}} +{ax}^{\mathrm{2}} +{bx}+{c}\: \\ $$$$\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)=\:\mathrm{2}^{\mathrm{2020}} +\mathrm{4}{a}+\mathrm{2}{b}+{c}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{f}\:'\left(\mathrm{2}\right)=\mathrm{2020}.\mathrm{2}^{\mathrm{2019}} +\mathrm{4}{a}+{b}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:{f}''\left(\mathrm{2}\right)\:=\:\mathrm{6} \\ $$$$\mathrm{2020}×\mathrm{2019}.\mathrm{2}^{\mathrm{2018}} +\mathrm{2}{a}\:=\:\mathrm{6}\: \\ $$$${a}\:=\frac{\mathrm{3}}{\mathrm{1010}×\mathrm{2019}.\mathrm{2}^{\mathrm{2018}} } \\ $$$${b}=−\mathrm{2020}.\mathrm{2}^{\mathrm{2019}} −\frac{\mathrm{12}}{\mathrm{1010}×\mathrm{2019}.\mathrm{2}^{\mathrm{2018}} } \\ $$$${now}\:{you}\:{can}\:{get}\:{the}\:{value}\:{of}\:{c}\: \\ $$

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