Question-94212

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

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

(1) \lim_{x \to 1^{-}} f (x) = \underset{x \to 1^{+}}{\lim f} (x)

\lim_{x \to 1^{-}} \frac{{ax}^{2} + b}{x - 1} = \lim_{x \to 1^{+}} \frac{c}{x} + 1

(i) a {(1)}^{2} + b = 0 \Rightarrow a = - b

\lim_{x \to 1^{-}} \frac{{ax}^{2} - a}{x - 1} = c + 1

\lim_{x \to 1^{-}} \frac{a (x + 1) (x - 1)}{x - 1} = c + 1

2 a = c + 1

(2) \Rightarrow \underset{x \to 1^{-}}{\lim f}^{'} (x) = \lim_{x \to 1^{+}} f^{'} (x)

\underset{x \to 1^{-}}{\lim a} = \lim_{x \to 1^{+}} - \frac{c}{x^{2}} \Rightarrow a = - c

so c + 1 = - 2 c \Rightarrow c = - \frac{1}{3}

{\begin{cases} a = \frac{1}{3} \\ b = - \frac{1}{3} \end{cases}

∴ a + b + c = - \frac{1}{3}