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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

$$\frac{\mathrm{♣}\mathcal{JS}\mathrm{♣}}{\dots }$$$$letf\left(x\right)=x^{2020}+{ax}^2+bx+c$$$$\left(1\right)f\left(2\right)=2^{2020}+4a+2b+c=0$$$$\left(2\right)f{}^{\prime }\left(2\right)=2020.2^{2019}+4a+b=0$$$$\left(3\right)f^{″}\left(2\right)=6$$$$2020\times 2019.2^{2018}+2a=6$$$$a=\frac{3}{1010\times 2019.2^{2018}}$$$$b=-2020.2^{2019}-\frac{12}{1010\times 2019.2^{2018}}$$$$nowyoucangetthevalueofc$$

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