Question

If $\underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{{\left(x-1\right)}^2}=2$, then $(a,b,c)=?$

• A. $2,-4,2$
• B. $2,4,2$
• C. $2,4,-2$
• D. $2,-4,-2$

Answer 1

The correct option is A

$2,-4,2$

Explanation for the correct option.

$\underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{{\left(x-1\right)}^2}=2$

If limit exits then

$\begin{array}{rcl}a{x}^2+bx+c& =& 2{\left(x-1\right)}^2\\ & \Rightarrow & a{x}^2+bx+c=2\left(x^2-2x+1\right)\\ & \Rightarrow & a{x}^2+bx+c=2x^2-4x+2\end{array}$

By comparing the coefficients, we get

$a=2,b=-4,c=2$

Therefore, $(a,b,c)=\left(2,-4,2\right)$

Hence, option A is correct.