A) \[\frac{3}{2}+2\sqrt{3}\]
B) \[-\frac{3}{2}+2\sqrt{3}\]
C) \[\frac{3}{2}-2\sqrt{3}\]
D) \[-\frac{3}{2}-2\sqrt{3}\]
Correct Answer: C
Solution :
Given f be an odd function \[f(x)=3sinx+4cosx\] Now,\[f\left( \frac{-11\pi }{6} \right)=3\sin \left( \frac{-11\pi }{6} \right)+4\cos \left( \frac{-11\pi }{6} \right)\] \[f\left( \frac{-11\pi }{6} \right)=3\sin \left( -2\pi +\frac{\pi }{6} \right)+4\cos \left( -2\pi +\frac{\pi }{6} \right)\]\[f\left( \frac{-11\pi }{6} \right)=3\sin \left\{ -\left( 2\pi -\frac{\pi }{6} \right) \right\}+4\cos \left\{ -\left( 2\pi -\frac{\pi }{6} \right) \right\}\]\[\sin (-\theta )=-sin\theta \] \[\left\{ \text{For}\,\text{odd}\,\text{functions}\begin{matrix} \sin (-\theta )=-sin\theta \\ and\sin (-\theta )=-\cos \theta \\ \end{matrix} \right\}\] \[\therefore \]\[f\left( \frac{-11\pi }{6} \right)=-3\sin \left( 2\pi -\frac{\pi }{6} \right)-4\cos \left( 2\pi -\frac{\pi }{6} \right)\] \[\Rightarrow \]\[f\left( \frac{-11\pi }{6} \right)=+3\sin \left( \frac{\pi }{6} \right)-4\cos \frac{\pi }{6}\] \[\Rightarrow \]\[f\left( \frac{-11\pi }{6} \right)=3\times \frac{1}{2}-4\times \frac{\sqrt{3}}{2}\] or\[f\left( \frac{-11\pi }{6} \right)=\frac{3}{2}-2\sqrt{3}\]
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