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Question Number 96826 by bobhans last updated on 05/Jun/20

$$\mathrm{If}2\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(\mathrm{x}-1)={\mathrm{x}}^2.\mathrm{determine}\mathrm{f}\left(\mathrm{x}\right)$$

Answered by john santu last updated on 05/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={px}^2+qx+r$$$$\mathrm{f}(\mathrm{x}-1)={px}^2-2px+p+qx-q+r$$$$\mathrm{f}(\mathrm{x}-1)={px}^2+(q-2p)x+p-q+r$$$$\Rightarrow 2\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(\mathrm{x}-1)={\mathrm{x}}^2$$$$\Rightarrow 3{px}^2+(3q-2p)x+3r+p-q=x^2$$$$3p=1\Rightarrow p=\frac{1}{3}$$$$3q-2p=0\Rightarrow q=\frac{2}{3}p=\frac{2}{9}$$$$3r+p-q=0\Rightarrow r=\frac{q-p}{3}=\frac{\frac{2}{9}-\frac{1}{3}}{3}$$$$r=-\frac{1}{27}$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{3}{x}^2+\frac{2}{9}x-\frac{1}{27}$$$$$$

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