Let f: R → R be defined as f(x)= {(λ|x^2-5 x+6|)/(μ(5 x-x^2-6)), x<2; e^((tan (x-2))/(x-[x])), x>2; μ, x=2} where [x] is the greatest integer less…

Let $\displaystyle f: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $$f(x)= \begin{cases}\frac{\lambda\left|x^2-5 x+6\right|}{\mu\left(5 x-x^2-6\right)} & , x<2 \\ e^{\frac{\tan (x-2)}{x-[x]}} & , x>2 \\ \mu & , x=2\end{cases} $$ where $\displaystyle [x]$ is the greatest integer less than or equal to $\displaystyle x$. If $\displaystyle f$ is continuous at $\displaystyle x=2$, then $\displaystyle \lambda+\mu$ is equal to: