If the function $f(x)=\left\{\begin{matrix}\frac{x^3-(k+4)x+2k}{x-3},&x≠3\\8,&x=3\end{matrix}\right.$ is continuous at x = 3, then k is:
Answer & explanation
Correct answer: option 4
$\underset{x→3}{\lim}\frac{x^3-(k+4)x+2k}{x-3}=\underset{x→3}{\lim}\frac{3x^2-(k+4)}{1}=23-k$
⇒ 23 - k = 8
⇒ k = 15