Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{x^3+x^2-16x+20}{(x-2{)}^2},& ifx\ne 2\\ & k,ifx=2\end{array}$ f f(x) be continuous for all x, then k =

• A. 7
• B. -7
• C. $\pm$7
• D. None of these

Answer 1

The correct option is A 7

For continuous ${lim}_{x\to 2}f\left(x\right)=f\left(2\right)=k$
$\Rightarrow k={lim}_{x\to 2}\frac{x^3+x^2-16x+20}{(x-2{)}^2}$
$={lim}_{x\to 2}\frac{(x^2-4x+4)(x+5)}{(x-2{)}^2}=7$