Let f be an odd function defined on the set of real numbers such that for $$x \geq 0$$, $$f(x) = 3\sin x + 4\cos x$$. Then $$f(x)$$ at $$x = -\frac{11\pi}{6}$$ is equal to:

We are told that the function $$f$$ is odd. By definition of an odd function, we have the fundamental property

$$f(-x) = -\,f(x)\,.$$

For all non-negative arguments, the rule is

$$f(x)=3\sin x + 4\cos x \qquad\text{for } x \ge 0.$$

We have to find $$f\!\left(-\dfrac{11\pi}{6}\right)$$. The argument $$-\dfrac{11\pi}{6}$$ is negative, so we first use the odd-function property. Putting $$x=\dfrac{11\pi}{6}$$ in the identity $$f(-x)=-f(x)$$ gives

$$f\!\left(-\dfrac{11\pi}{6}\right)= -\,f\!\left(\dfrac{11\pi}{6}\right).$$

The quantity $$\dfrac{11\pi}{6}$$ is positive, so we may apply the explicit formula $$f(x)=3\sin x+4\cos x$$ to it. Hence

$$f\!\left(\dfrac{11\pi}{6}\right) = 3\sin\!\left(\dfrac{11\pi}{6}\right) + 4\cos\!\left(\dfrac{11\pi}{6}\right).$$

Now we evaluate the standard trigonometric ratios at $$\dfrac{11\pi}{6}$$ (which corresponds to $$330^\circ$$, a fourth-quadrant angle):

$$\sin\!\left(\dfrac{11\pi}{6}\right) = -\dfrac{1}{2}, \qquad \cos\!\left(\dfrac{11\pi}{6}\right) = \dfrac{\sqrt3}{2}.$$

Substituting these values we obtain

$$\begin{aligned} f\!\left(\dfrac{11\pi}{6}\right) &= 3\left(-\dfrac{1}{2}\right) + 4\left(\dfrac{\sqrt3}{2}\right) \\ &= -\dfrac{3}{2} + 2\sqrt3. \end{aligned}$$

Finally we return to our target value using the odd-function relation:

$$\begin{aligned} f\!\left(-\dfrac{11\pi}{6}\right) &= -\,f\!\left(\dfrac{11\pi}{6}\right) \\ &= -\left(-\dfrac{3}{2} + 2\sqrt3\right) \\ &= \dfrac{3}{2} - 2\sqrt3. \end{aligned}$$

Hence, the correct answer is Option C.