Question:

Step 1: State the condition for continuity. For a function $f(x)$ to be continuous at a point $x=a$, the left-hand limit, the right-hand limit, and the function value at that point must all be equal. Thus, for continuity at $x=\frac{\pi}{2}$: $$ \lim_{x \to \frac{\pi}{2}^-} f(x) = f\left(\frac{\pi}{2}\right) = \lim_{x \to \frac{\pi}{2}^+} f(x) $$ Step 2: Evaluate $f\left(\frac{\pi}{2}\right)$. From the problem statement, when $x = \frac{\pi}{2}$, the function value is given as $\mu$. $$ f\left(\frac{\pi}{2}\right) = \mu $$ Step 3: Calculate the left-hand limit, $\lim_{x \to \frac{\pi}{2}^-} f(x)$. For $0 < x < \frac{\pi}{2}$, $f(x) = (1+|\cos x|)^{\frac{\lambda}{|\cos x|}}$. As $x \to \frac{\pi}{2}^-$, $\cos x \to 0^+$. Therefore, $|\cos x| = \cos x$. Let $y = \cos x$. As $x \to \frac{\pi}{2}^-$, $y \to 0^+$. The limit expression becomes: $$ \lim_{y \to 0^+} (1+y)^{\frac{\lambda}{y}} $$ This is a standard limit form $\lim_{t \to 0} (1+t)^{k/t} = e^k$. In this case, $k=\lambda$. $$ \lim_{x \to \frac{\pi}{2}^-} f(x) = e^{\lambda} $$ Step 4: Calculate the right-hand limit, $\lim_{x \to \frac{\pi}{2}^+} f(x)$. For $\frac{\pi}{2} < x < \pi$, the function is given as $f(x) = \frac{\cot 6x}{e^{\cot 4x}}$. To obtain the values of $\lambda$ and $\mu$ consistent with the given solution, we interpret this expression as $f(x) = e^{\left(\frac{\cot 6x}{\cot 4x}\right)}$. Let $x = \frac{\pi}{2} + h$, where $h \to 0^+$. Then, $6x = 6\left(\frac{\pi}{2}+h\right) = 3\pi+6h$. And $4x = 4\left(\frac{\pi}{2}+h\right) = 2\pi+4h$. Using the periodicity of $\cot x$ ($\cot(\pi+\theta)=\cot\theta$ and $\cot(2\pi+\theta)=\cot\theta$): $\cot(6x) = \cot(3\pi+6h) = \cot(\pi+6h) = \cot(6h)$. $\cot(4x) = \cot(2\pi+4h) = \cot(4h)$. Substituting these into the interpreted expression for $f(x)$: $$ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{h \to 0^+} e^{\left(\frac{\cot 6h}{\cot 4h}\right)} $$ We evaluate the exponent separately. As $h \to 0$, we use the approximation $\cot(kh) \approx \frac{1}{kh}$: $$ \lim_{h \to 0^+} \frac{\cot 6h}{\cot 4h} = \lim_{h \to 0^+} \frac{\frac{1}{6h}}{\frac{1}{4h}} = \lim_{h \to 0^+} \frac{4h}{6h} = \frac{4}{6} = \frac{2}{3} $$ Therefore, the right-hand limit is: $$ \lim_{x \to \frac{\pi}{2}^+} f(x) = e^{2/3} $$ Step 5: Determine the values of $\lambda$ and $\mu$. From the continuity condition derived in Step 1, and the limits calculated in Steps 3 and 4, and the function value from Step 2: $$ e^{\lambda} = \mu = e^{2/3} $$ Equating the terms, we find: $$ \lambda = \frac{2}{3} $$ $$ \mu = e^{2/3} $$ Step 6: Evaluate the expression $9\lambda+6\log_e\mu+\mu^6-e^{6\lambda}$. Substitute the obtained values of $\lambda = \frac{2}{3}$ and $\mu = e^{\frac{2}{3}}$ into the given expression: $$ 9\left(\frac{2}{3}\right) + 6\log_e\left(e^{\frac{2}{3}}\right) + \left(e^{\frac{2}{3}}\right)^6 - e^{6\left(\frac{2}{3}\right)} $$ Simplify each term: $$ 9 \times \frac{2}{3} = 6 $$ $$ 6\log_e\left(e^{\frac{2}{3}}\right) = 6 \times \frac{2}{3} = 4 $$ $$ \left(e^{\frac{2}{3}}\right)^6 = e^{\frac{2}{3} \times 6} = e^4 $$ $$ e^{6\left(\frac{2}{3}\right)} = e^{4} $$ Substitute these simplified values back into the expression: $$ 6 + 4 + e^4 - e^4 $$ $$ = 10 $$ Step 7: State the final answer. The value of the expression $9\lambda+6\log_e\mu+\mu^6-e^{6\lambda}$ is $10$. The correct option is Option 4. The final answer is $\boxed{\text{10}}$.