Question Number 107728 by mathdave last updated on 12/Aug/20
Answered by john santu last updated on 12/Aug/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\clubsuit\mathcal{JS}\clubsuit}{\ldots} \\ $$$${let}\:{f}\left({x}\right)={x}^{\mathrm{2020}} +{ax}^{\mathrm{2}} +{bx}+{c}\: \\ $$$$\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)=\:\mathrm{2}^{\mathrm{2020}} +\mathrm{4}{a}+\mathrm{2}{b}+{c}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{f}\:'\left(\mathrm{2}\right)=\mathrm{2020}.\mathrm{2}^{\mathrm{2019}} +\mathrm{4}{a}+{b}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:{f}''\left(\mathrm{2}\right)\:=\:\mathrm{6} \\ $$$$\mathrm{2020}×\mathrm{2019}.\mathrm{2}^{\mathrm{2018}} +\mathrm{2}{a}\:=\:\mathrm{6}\: \\ $$$${a}\:=\frac{\mathrm{3}}{\mathrm{1010}×\mathrm{2019}.\mathrm{2}^{\mathrm{2018}} } \\ $$$${b}=−\mathrm{2020}.\mathrm{2}^{\mathrm{2019}} −\frac{\mathrm{12}}{\mathrm{1010}×\mathrm{2019}.\mathrm{2}^{\mathrm{2018}} } \\ $$$${now}\:{you}\:{can}\:{get}\:{the}\:{value}\:{of}\:{c}\: \\ $$