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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

(1) lim_(x→1^− ) f(x)=lim_(x→1^+ ) f(x) lim_(x→1^− ) ((ax^2 +b)/(x−1)) = lim_(x→1^+ ) (c/x)+1 (i) a(1)^2 +b = 0 ⇒ a = −b lim_(x→1^− ) ((ax^2 −a)/(x−1)) = c+1 lim_(x→1^− ) ((a(x+1)(x−1))/(x−1)) = c+1 2a = c+1 (2)⇒lim_(x→1^− ) f ′(x)=lim_(x→1^+ ) f ′ (x) lim_(x→1^− ) a = lim_(x→1^+ ) −(c/x^2 ) ⇒ a = −c so c + 1 = −2c ⇒ c = −(1/3) { ((a = (1/3))),((b = −(1/3))) :} ∴ a+b+c = −(1/3)

$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}f}\left(\mathrm{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{b}}{\mathrm{x}−\mathrm{1}}\:=\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\mathrm{c}}{\mathrm{x}}+\mathrm{1} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{a}\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{b}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{a}\:=\:−\mathrm{b}\: \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\frac{\mathrm{ax}^{\mathrm{2}} −\mathrm{a}}{\mathrm{x}−\mathrm{1}}\:=\:\mathrm{c}+\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\frac{\mathrm{a}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)}{\mathrm{x}−\mathrm{1}}\:=\:\mathrm{c}+\mathrm{1} \\ $$$$\mathrm{2a}\:=\:\mathrm{c}+\mathrm{1}\: \\ $$$$\left(\mathrm{2}\right)\Rightarrow\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}f}\:'\left(\mathrm{x}\right)=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\mathrm{f}\:'\:\left(\mathrm{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}a}\:=\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}−\frac{\mathrm{c}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\:\mathrm{a}\:=\:−\mathrm{c}\: \\ $$$$\mathrm{so}\:\mathrm{c}\:+\:\mathrm{1}\:=\:−\mathrm{2c}\:\Rightarrow\:\mathrm{c}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\begin{cases}{\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{b}\:=\:−\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$\therefore\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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