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

 $f\left(x\right)=3\sin x+4\cos x$
$f\left(x\right)$ is an odd function
 $f(-x)=-f\left(x\right)$
 $f(-\frac{11\pi }{6})=-[3\sin \frac{11\pi }{6}+4\cos \frac{11\pi }{6}]$
 $=-3\sin (2\pi -\frac{\pi }{6})-4\cos (2\pi -\frac{\pi }{6})$
 $=3\sin \frac{\pi }{6}-4\cos \frac{\pi }{6}$
 $=\frac{3}{2}-4\frac{\sqrt{3}}{2}=\frac{3}{2}-2\sqrt{3}$