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

For $$x<2,$$ we have

$$x^2-5x+6=(x-2)(x-3)$$

Since $$x<2,$$ both $$x-2<0$$ and $$x-3<0.$$ Hence,

$$(x-2)(x-3)>0$$

Therefore,

$$|x^2-5x+6|=x^2-5x+6$$

Also,

$$5x-x^2-6=-(x^2-5x+6)$$

Thus,

$$f(x)=\frac{\lambda(x^2-5x+6)}{-\mu(x^2-5x+6)}=-\frac{\lambda}{\mu}$$

Hence,

$$\lim_{x\to2^-}f(x)=-\frac{\lambda}{\mu}$$

Now evaluate the right-hand limit.

For $$2<x<3,$$ we have

$$[x]=2$$

Therefore,

$$x-[x]=x-2$$

Hence,

$$\lim_{x\to2^+}e^{\frac{\tan(x-2)}{x-[x]}}=\lim_{x\to2^+}e^{\frac{\tan(x-2)}{x-2}}$$

Using

$$\lim_{t\to0}\frac{\tan t}{t}=1,$$

we get

$$\lim_{x\to2^+}f(x)=e$$

Since $$f$$ is continuous at $$x=2,$$

$$-\frac{\lambda}{\mu}=e=\mu$$

Hence,

$$\mu=e$$ and $$-\frac{\lambda}{e}=e$$

$$\lambda=-e^2$$

Therefore,

$$\lambda+\mu=-e^2+e=e(1-e)$$