Question

If $f(x)=\left\{\begin{array}{l}\frac{x^2-9}{x-3} \text { if } x \neq 3 \\ 2 x+k \end{array}\right.$, otherwise is continuous at $x=3$, then $k$ is equal to

(A) 3

(B) 0

(C) -6

(D) $1 / 6$

Answer 1

$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3}(x+3)=6$

and $f(3)=2(3)+k=6+k$

$\because f$ is continuous at $x=3$

$\begin{aligned}\therefore & f(3) & =\lim _{x \rightarrow 3} f(x) \\\Rightarrow & 6+k & =6 \Rightarrow k=0\end{aligned}$

So, The correct option will be (B).