Q. If $\lim_{x \rightarrow 1}{\frac{ax^{2} + bx + c}{(x - 1)^{2}} = 2},\$then (a, b, c) is
A) (2, -4, 2)
B) (2, 4, 2)
C) (2, 4, -2)
D) (2, -4, -2)
Explanation
(a)
Given, $\lim_{x \rightarrow 1}{\frac{ax^{2} + bx + c}{(x - 1)^{2}} = 2}$
This limit will exist, if
$ax^{2} + bx + c = 2(x - 1)^{2}$
$\Rightarrow \ \ \ \ ax^{2} + bx + c = 2x^{2} - 4x + 2$
$\Rightarrow \ \ \ \ \ \ a = 2,\ \ b = - 4,\ \ \ \ \ \ \ \ \ \ \ c = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.