1. Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  • $f(c)$ is defined.
  • $\mathop {\lim }\limits_{x \rightarrow c} f(x)$ exists.
  • $\mathop {\lim }\limits_{x \rightarrow c} f(x) = f(c)$. For a piecewise function, this implies that the left-hand limit, the right-hand limit, and the function value at the point must all be equal.
  • Left-hand limit: $\mathop {\lim }\limits_{x \rightarrow c^-} f(x)$
  • Right-hand limit: $\mathop {\lim }\limits_{x \rightarrow c^+} f(x)$
  • Function value: $f(c)$
• Limit of the form $(1+u)^{1/u}$: A standard limit form is $\mathop {\lim }\limits_{u \rightarrow 0} (1+u)^{1/u} = e$. This can be generalized to $\mathop {\lim }\limits_{u \rightarrow 0} (1+u)^{k/u} = e^k$.
• Trigonometric Identities and Limits:
  • $\cot x = \frac{\cos x}{\sin x}$
  • $\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}$ and $\lim_{x \to 0} \frac{\tan(ax)}{bx} = \frac{a}{b}$.
  • When dealing with limits of trigonometric functions as $x \to \pi/2$, it's often helpful to consider the substitution $x = \pi/2 + h$ or $x = \pi/2 - h$ and use small angle approximations or standard limits.

2. Step-by-Step Solution

Step 1: Evaluate the left-hand limit at $x = \frac{\pi}{2}$. For $0 < x < \frac{\pi}{2}$, the function is given by $f(x) = (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}}$. As $x \to \frac{\pi}{2}^-$, $x$ approaches $\frac{\pi}{2}$ from values less than $\frac{\pi}{2}$. In this interval, $\cos x > 0$, so $|\cos x| = \cos x$. Let $u = \cos x$. As $x \to \frac{\pi}{2}^-$, $u = \cos x \to \cos(\frac{\pi}{2}) = 0$. The left-hand limit becomes: $\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{-}}{2}} (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}} = \mathop {\lim }\limits_{u \rightarrow 0^+} (1+u)^{\frac{\lambda}{u}}$ Using the standard limit form $\mathop {\lim }\limits_{u \rightarrow 0} (1+u)^{k/u} = e^k$, we get: $\mathop {\lim }\limits_{u \rightarrow 0^+} (1+u)^{\frac{\lambda}{u}} = e^\lambda$ Thus, the left-hand limit is $e^\lambda$.

Step 2: Evaluate the right-hand limit at $x = \frac{\pi}{2}$. For $\frac{\pi}{2} < x < \pi$, the function is given by $f(x) = e^{\frac{\cot 6x}{\cot 4x}}$. As $x \to \frac{\pi}{2}^+$, $x$ approaches $\frac{\pi}{2}$ from values greater than $\frac{\pi}{2}$. We need to evaluate: $\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6x}{\cot 4x}}$ Let's first evaluate the exponent: $\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} \frac{\cot 6x}{\cot 4x}$. We can write $\cot x = \frac{\cos x}{\sin x}$. So the limit is: $\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} \frac{\cos 6x}{\sin 6x} \cdot \frac{\sin 4x}{\cos 4x}$ As $x \to \frac{\pi}{2}^+$, let $x = \frac{\pi}{2} + h$, where $h \to 0^+$. Then: $6x = 6(\frac{\pi}{2} + h) = 3\pi + 6h$ $4x = 4(\frac{\pi}{2} + h) = 2\pi + 4h$

$\cos(6x) = \cos(3\pi + 6h) = \cos(\pi + 6h) = -\cos(6h)$ $\sin(6x) = \sin(3\pi + 6h) = \sin(\pi + 6h) = -\sin(6h)$ $\cos(4x) = \cos(2\pi + 4h) = \cos(4h)$ $\sin(4x) = \sin(2\pi + 4h) = \sin(4h)$

The exponent limit becomes: $\mathop {\lim }\limits_{h \rightarrow 0^+} \frac{-\cos 6h}{-\sin 6h} \cdot \frac{\sin 4h}{\cos 4h} = \mathop {\lim }\limits_{h \rightarrow 0^+} \frac{\cos 6h}{\sin 6h} \cdot \frac{\sin 4h}{\cos 4h}$ We can rewrite this as: $\mathop {\lim }\limits_{h \rightarrow 0^+} \left( \frac{\cos 6h}{\cos 4h} \right) \cdot \left( \frac{\sin 4h}{1} \right) \cdot \left( \frac{1}{\sin 6h} \right)$ $= \mathop {\lim }\limits_{h \rightarrow 0^+} \left( \frac{\cos 6h}{\cos 4h} \right) \cdot \left( \frac{\sin 4h}{4h} \cdot 4h \right) \cdot \left( \frac{6h}{\sin 6h} \cdot \frac{1}{6h} \right)$ $= \mathop {\lim }\limits_{h \rightarrow 0^+} \left( \frac{\cos 6h}{\cos 4h} \right) \cdot \left( \frac{\sin 4h}{4h} \right) \cdot \left( \frac{6h}{\sin 6h} \right) \cdot \frac{4h}{6h}$ As $h \to 0$: $\cos 6h \to \cos 0 = 1$ $\cos 4h \to \cos 0 = 1$ $\frac{\sin 4h}{4h} \to 1$ $\frac{6h}{\sin 6h} \to 1$ $\frac{4h}{6h} = \frac{2}{3}$

So, the limit of the exponent is: $(1/1) \cdot 1 \cdot 1 \cdot \frac{2}{3} = \frac{2}{3}$ Therefore, the right-hand limit is $e^{2/3}$.

Step 3: Evaluate the function value at $x = \frac{\pi}{2}$. At $x = \frac{\pi}{2}$, the function is defined as $f(\frac{\pi}{2}) = \mu$.

Step 4: Apply the continuity condition. For the function to be continuous at $x = \frac{\pi}{2}$, the left-hand limit, the right-hand limit, and the function value must be equal: $\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{-}}{2}} f(x) = \mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} f(x) = f(\frac{\pi}{2})$ $e^\lambda = e^{2/3} = \mu$ From this equality, we can determine the values of $\lambda$ and $\mu$: $\lambda = \frac{2}{3}$ $\mu = e^{2/3}$

Step 5: Calculate the required expression. We need to find the value of $9\lambda + 6\log_e\mu + \mu^6 - e^{6\lambda}$. Substitute the values of $\lambda$ and $\mu$: $9\lambda = 9 \times \frac{2}{3} = 6$ $6\log_e\mu = 6\log_e(e^{2/3}) = 6 \times \frac{2}{3} = 4$ $\mu^6 = (e^{2/3})^6 = e^{(2/3) \times 6} = e^4$ $e^{6\lambda} = e^{6 \times (2/3)} = e^{12/3} = e^4$

Now, substitute these values into the expression: $9\lambda + 6\log_e\mu + \mu^6 - e^{6\lambda} = 6 + 4 + e^4 - e^4$ $= 10$

3. Common Mistakes & Tips

• Sign errors with trigonometric functions: Be careful with the signs of trigonometric functions in different quadrants. For $x \to \frac{\pi}{2}^-$, $\cos x > 0$, and for $x \to \frac{\pi}{2}^+$, $\cos x < 0$. However, in the first part of the limit, $|\cos x|$ simplifies to $\cos x$ as $x \to \frac{\pi}{2}^-$.
• Handling indeterminate forms: Limits of the form $e^{\text{limit of exponent}}$ are common. Ensure the limit of the exponent is correctly evaluated, especially for trigonometric functions. Using substitutions like $x = \frac{\pi}{2} + h$ can simplify the evaluation of limits as $x \to \frac{\pi}{2}$.
• Algebraic manipulation of limits: When evaluating limits involving trigonometric functions like $\frac{\cot 6x}{\cot 4x}$, rewriting them in terms of $\sin$ and $\cos$ and then using standard limits like $\lim_{y \to 0} \frac{\sin(ay)}{by} = \frac{a}{b}$ is a robust strategy.

4. Summary

The problem requires us to find the value of an expression involving $\lambda$ and $\mu$ by ensuring the continuity of a piecewise function at $x = \frac{\pi}{2}$. We first evaluated the left-hand limit as $x$ approaches $\frac{\pi}{2}$ from the left, which resulted in $e^\lambda$. Next, we computed the right-hand limit as $x$ approaches $\frac{\pi}{2}$ from the right, which involved evaluating a limit of the form $e^{\text{exponent}}$ and found the exponent's limit to be $\frac{2}{3}$, leading to a right-hand limit of $e^{2/3}$. The function value at $x = \frac{\pi}{2}$ is given as $\mu$. By equating these three values for continuity, we determined $\lambda = \frac{2}{3}$ and $\mu = e^{2/3}$. Finally, we substituted these values into the given expression $9\lambda + 6\log_e\mu + \mu^6 - e^{6\lambda}$, simplifying it to $6 + 4 + e^4 - e^4 = 10$.

5. Final Answer

The final answer is $\boxed{10}$.