Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if and only if $\mathop {\lim }\limits_{x \to c^-} f(x) = \mathop {\lim }\limits_{x \to c^+} f(x) = f(c)$.
• Limit of a Trigonometric Function: The standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$.
• Greatest Integer Function: $[x]$ denotes the greatest integer less than or equal to $x$. For $x$ approaching 2 from the right ($x \to 2^+$), $x$ is slightly greater than 2, so $[x] = 2$. For $x$ approaching 2 from the left ($x \to 2^-$), $x$ is slightly less than 2, so $[x] = 1$.
• Absolute Value Function: $|a| = a$ if $a \ge 0$ and $|a| = -a$ if $a < 0$.

Step-by-Step Solution

Step 1: Analyze the function definition and the condition for continuity. The function $f(x)$ is defined piecewise. For continuity at $x=2$, we must ensure that the left-hand limit, the right-hand limit, and the function value at $x=2$ are all equal. f(x) = \left\{ {\matrix{ {{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr {\mu ,} & {x = 2} \cr } } \right. The condition for continuity at $x=2$ is $\mathop {\lim }\limits_{x \to 2^-} f(x) = \mathop {\lim }\limits_{x \to 2^+} f(x) = f(2)$. From the definition, $f(2) = \mu$.

Step 2: Calculate the right-hand limit, $\mathop {\lim }\limits_{x \to 2^+} f(x)$. For $x > 2$, $f(x) = e^{\frac{\tan(x-2)}{x-[x]}}$. As $x \to 2^+$, $x$ is slightly greater than 2. Therefore, $[x] = 2$. So, the exponent becomes $\frac{\tan(x-2)}{x-2}$. We know the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. Let $\theta = x-2$. As $x \to 2^+$, $\theta \to 0^+$. Thus, $\mathop {\lim }\limits_{x \to 2^+} \frac{\tan(x-2)}{x-2} = 1$. Therefore, the right-hand limit is: $\mathop {\lim }\limits_{x \to 2^+} f(x) = \mathop {\lim }\limits_{x \to 2^+} {e^{{{\tan (x - 2)} \over {x - [x]}}}} = {e^{\mathop {\lim }\limits_{x \to 2^+} \frac{\tan (x - 2)}{x - 2}}} = {e^1} = e$

Step 3: Calculate the left-hand limit, $\mathop {\lim }\limits_{x \to 2^-} f(x)$. For $x < 2$, $f(x) = \frac{\lambda |x^2 - 5x + 6|}{\mu (5x - x^2 - 6)}$. First, let's factor the quadratic expressions: $x^2 - 5x + 6 = (x-2)(x-3)$. $5x - x^2 - 6 = -(x^2 - 5x + 6) = -(x-2)(x-3)$. So, for $x < 2$, the expression becomes: $f(x) = \frac{\lambda |(x-2)(x-3)|}{\mu (-(x-2)(x-3))}$ As $x \to 2^-$, $x$ is slightly less than 2. This means $x-2$ is negative. Also, for $x$ close to 2, $x-3$ is negative (e.g., if $x=1.9$, $x-3 = -1.01$). Therefore, $(x-2)(x-3)$ is positive (negative multiplied by negative). So, $|(x-2)(x-3)| = (x-2)(x-3)$ for $x < 2$ and $x$ close to 2. Substituting this back into the expression for $f(x)$: $f(x) = \frac{\lambda (x-2)(x-3)}{\mu (-(x-2)(x-3))}$ For $x \ne 2$ and $x \ne 3$, we can cancel the term $(x-2)(x-3)$: $f(x) = \frac{\lambda}{\mu (-1)} = -\frac{\lambda}{\mu}$ Now, we take the limit as $x \to 2^-$: $\mathop {\lim }\limits_{x \to 2^-} f(x) = \mathop {\lim }\limits_{x \to 2^-} \left(-\frac{\lambda}{\mu}\right) = -\frac{\lambda}{\mu}$

Step 4: Apply the continuity condition to find the relationship between $\lambda$ and $\mu$. For continuity at $x=2$, we have: $\mathop {\lim }\limits_{x \to 2^-} f(x) = \mathop {\lim }\limits_{x \to 2^+} f(x) = f(2)$ From our calculations: $-\frac{\lambda}{\mu} = e = \mu$ From this, we have two equations:

1. $\mu = e$
2. $-\frac{\lambda}{\mu} = e$

Step 5: Solve for $\lambda$ and $\mu$ and calculate $\lambda + \mu$. From equation (1), we have $\mu = e$. Substitute $\mu = e$ into equation (2): $-\frac{\lambda}{e} = e$ Multiply both sides by $e$: $-\lambda = e^2$ $\lambda = -e^2$ Now, we need to find $\lambda + \mu$: $\lambda + \mu = -e^2 + e$ We can factor out $e$: $\lambda + \mu = e(-e + 1)$

Common Mistakes & Tips

• Handling the Greatest Integer Function: Carefully consider the behavior of $[x]$ as $x$ approaches the point from the left and right. For $x \to 2^+$, $[x]=2$. For $x \to 2^-$, $[x]=1$.
• Absolute Value Simplification: Pay close attention to the sign of the expression inside the absolute value as $x$ approaches the limit point. For $x \to 2^-$, $(x-2)(x-3)$ is positive, so $|(x-2)(x-3)| = (x-2)(x-3)$.
• Trigonometric Limit: Recall the fundamental limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. This is crucial for evaluating the right-hand limit.

Summary To determine the continuity of the function at $x=2$, we equated the left-hand limit, the right-hand limit, and the function value at $x=2$. We calculated the right-hand limit by recognizing the standard trigonometric limit for the exponent. For the left-hand limit, we simplified the expression involving the absolute value by considering the signs of the factors as $x$ approached 2 from the left. Equating these values allowed us to find expressions for $\lambda$ and $\mu$. Substituting the value of $\mu$ into the equation for $\lambda$ gave us $\lambda = -e^2$. Finally, we computed $\lambda + \mu$ to be $e(1-e)$.

The final answer is \boxed{e(1-e)}.