Answer:
Option A
Explanation:
We have ,
$f(x)=\begin{cases}\cos 2x & for -\infty<x<0\\e^{3x} & for 0\leq x<3\\ x^{2}-4x+3& ,for 3\leq x\leq6\\\frac{\log(15x-89)}{x-6}&, for x>6\end{cases}$
$\because \lim_{x \rightarrow 3-}e^{3x}= e^{9}$ and $\lim_{x \rightarrow 3+} x^{2}-4x+3=9-12+3=0$
Clearly , $\lim_{x \rightarrow 3-}f(x)\neq \lim_{x \rightarrow 3+}f(x) $
$\therefore$ f(x) is discontinuous at x=3
$\therefore$ a=3,
Now, $\lim_{x \rightarrow 3}\frac{x^{2}-9}{x^{3}-5x^{2}+9x-9}=\lim_{x \rightarrow 3}\frac{(x-3)(x+3)}{(x-3)(x^{2}-2x+3)}$
= $\frac{3+3}{(3)^{2}-2(3)+3}= \frac{2}{2}=1$
