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(C) Given that $f$ is an odd function,we have $f(-x) = -f(x)$ for all $x$ in the domain.For $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$.We need to find $f

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$\frac{3}{2} - 2\sqrt{3}$

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(C) Given that $f$ is an odd function,we have $f(-x) = -f(x)$ for all $x$ in the domain.For $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$.We need to find $f\left(-\frac{11\pi}{6}\right)$.Since $f$ is an odd function,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right)$.First,calculate $f\left(\frac{11\pi}{6}\right)$:$f\left(\frac{11\pi}{6}\right) = 3 \sin\left(\frac{11\pi}{6}\right) + 4 \cos\left(\frac{11\pi}{6}\right)$$f\left(\frac{11\pi}{6}\right) = 3 \sin\left(2\pi - \frac{\pi}{6}\right) + 4 \cos\left(2\pi - \frac{\pi}{6}\right)$$f\left(\frac{11\pi}{6}\right) = 3(-\sin\frac{\pi}{6}) + 4(\cos\frac{\pi}{6})$$f\left(\frac{11\pi}{6}\right) = 3(-\frac{1}{2}) + 4(\frac{\sqrt{3}}{2}) = -\frac{3}{2} + 2\sqrt{3}$.Therefore,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right) = -(-\frac{3}{2} + 2\sqrt{3}) = \frac{3}{2} - 2\sqrt{3}$.

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:

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