1. Continuity at $x=\dfrac{\pi}{2}$

For continuity, the left-hand limit, right-hand limit, and function value must be equal: $\lim_{x\to (\pi/2)^-} f(x)=\mu=\lim_{x\to (\pi/2)^+} f(x).$

2. Left-hand limit

For $0<x<\dfrac{\pi}{2}$, $f(x)=(1+|\cos x|)^{\frac{\lambda}{|\cos x|}}.$

As $x\to \left(\dfrac{\pi}{2}\right)^-$, we have $\cos x>0$ and $|\cos x|=\cos x\to 0^+$. So, $f(x)=(1+\cos x)^{\frac{\lambda}{\cos x}}.$

Let $t=\cos x\to 0^+$. Then $\lim_{t\to 0^+}(1+t)^{\lambda/t} = e^{\lambda}.$

Hence, $\lim_{x\to (\pi/2)^-} f(x)=e^{\lambda}.$

So continuity gives $\mu=e^{\lambda}. \qquad (1)$

3. Right-hand limit

For $\dfrac{\pi}{2}<x<\pi$, $f(x)=e^{\frac{\cot 6x}{\cot 4x}}.$

We need $\lim_{x\to (\pi/2)^+} e^{\frac{\cot 6x}{\cot 4x}}=e^{\lim\limits_{x\to (\pi/2)^+}\frac{\cot 6x}{\cot 4x}}.$

So first evaluate $L=\lim_{x\to (\pi/2)^+}\frac{\cot 6x}{\cot 4x}.$

Put $x=\frac{\pi}{2}+h, \quad h\to 0^+.$ Then $6x=3\pi+6h, \qquad 4x=2\pi+4h.$ Using periodicity of cotangent, $\cot(3\pi+6h)=\cot 6h, \qquad \cot(2\pi+4h)=\cot 4h.$ Thus $L=\lim_{h\to 0^+}\frac{\cot 6h}{\cot 4h}.$

Now for small $h$, $\cot kh\sim \dfrac{1}{kh}$. Therefore, $L=\lim_{h\to 0^+}\frac{1/(6h)}{1/(4h)}=\frac{4}{6}=\frac{2}{3}.$

Hence, $\lim_{x\to (\pi/2)^+} f(x)=e^{2/3}.$

By continuity, $\mu=e^{2/3}. \qquad (2)$

From (1), $e^{\lambda}=e^{2/3} \implies \lambda=\frac{2}{3},$ and $\mu=e^{2/3}.$

4. Evaluate the required expression

We need $9\lambda+6\log_e \mu+\mu^6-e^{6\lambda}.$

Substitute $\lambda=\dfrac{2}{3}$ and $\mu=e^{2/3}$:

• $9\lambda=9\cdot \frac{2}{3}=6$
• $6\log_e\mu=6\log_e(e^{2/3})=6\cdot \frac{2}{3}=4$
• $\mu^6=(e^{2/3})^6=e^4$
• $e^{6\lambda}=e^{6\cdot 2/3}=e^4$

Therefore, $9\lambda+6\log_e \mu+\mu^6-e^{6\lambda}=6+4+e^4-e^4=10.$

5. Final answer

The value is $\boxed{10}.$ So the correct option is B.