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Question Number 94212 by abony1303 last updated on 17/May/20

Answered by john santu last updated on 17/May/20

$$\left(1\right)\underset{x\to 1^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{x\to 1^{+}}{\lim f}\left(\mathrm{x}\right)$$$$\underset{x\to 1^{-}}{\lim }\frac{{\mathrm{ax}}^2+\mathrm{b}}{\mathrm{x}-1}=\underset{x\to 1^{+}}{\lim }\frac{\mathrm{c}}{\mathrm{x}}+1$$$$\left(\mathrm{i}\right)\mathrm{a}{\left(1\right)}^2+\mathrm{b}=0\Rightarrow \mathrm{a}=-\mathrm{b}$$$$\underset{x\to 1^{-}}{\lim }\frac{{\mathrm{ax}}^2-\mathrm{a}}{\mathrm{x}-1}=\mathrm{c}+1$$$$\underset{x\to 1^{-}}{\lim }\frac{\mathrm{a}(\mathrm{x}+1)(\mathrm{x}-1)}{\mathrm{x}-1}=\mathrm{c}+1$$$$2\mathrm{a}=\mathrm{c}+1$$$$\left(2\right)\Rightarrow \underset{x\to 1^{-}}{\lim f}{}^{\prime }\left(\mathrm{x}\right)=\underset{x\to 1^{+}}{\lim }\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)$$$$\underset{x\to 1^{-}}{\lim a}=\underset{x\to 1^{+}}{\lim }-\frac{\mathrm{c}}{{\mathrm{x}}^2}\Rightarrow \mathrm{a}=-\mathrm{c}$$$$\mathrm{so}\mathrm{c}+1=-2\mathrm{c}\Rightarrow \mathrm{c}=-\frac{1}{3}$$$$\{\begin{array}{l}\mathrm{a}=\frac{1}{3}\\ \mathrm{b}=-\frac{1}{3}\end{array}$$$$\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\frac{1}{3}$$

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