Question-69710

Question Number 69710 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19

Commented by mathmax by abdo last updated on 27/Sep/19

w e h a v e x^{2} - 6 x + 5 = x^{2} - x - 5 (x - 1) = x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

\Rightarrow {l i m}_{x \to 1} \frac{(x - 1) (x - 5)}{x^{2} + a x + b} = 2 \Rightarrow 1 + a + b = 0 \Rightarrow a + b = - 1

1 r o o t o f x^{2} + a x + b \Rightarrow x^{2} + a x + b = (x - 1) (x + α) \Rightarrow

x^{2} + α x - x + α = x^{2} + a x + b \Rightarrow α - 1 = a a n d b = α \Rightarrow

x^{2} + a x + b = (x - 1) (x + 1 + a) a f t e r s i m p l i f i c a t i o n i n l i m i t

w e g e t \frac{- 4}{2 + a} = 2 \Rightarrow 4 + 2 a = - 4 \Rightarrow 2 a = - 8 \Rightarrow a = - 4 \Rightarrow b = - 1 - a

= - 1 + 4 = 3 \Rightarrow a = - 4 a n d b = 3

Answered by $@ty@m123 last updated on 26/Sep/19

\lim_{x \to 1} \frac{x^{2} - 6 x + 5}{x^{2} + a x + b} = 2

L e t x^{2} + a x + b = (x - 1) (x - b) \dots (1)

\lim_{x \to 1} \frac{(x - 1) (x - 5)}{(x - 1) (x - b)} = 2

\frac{1 - 5}{1 - b} = 2

1 - b = - 2 \Rightarrow b = 3

∴ f r o m (1),

a = - (1 + b) = - 4