Question

Left $f$ 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 :

• A. $-\frac{3}{2}-2\sqrt{3}$
• B. $\frac{3}{2}-2\sqrt{3}$
• C. $\frac{3}{2}+2\sqrt{3}$
• D. $-\frac{3}{2}+2\sqrt{3}$

Answer 1

Answer: (B)

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

$f\left(x\right)=3\sin x+4\cos x$ $x\ge 0$
$f\left(x\right)=5\sin (x+\alpha )$
$\alpha ={\sin }^{-1}\left(\frac{4}{5}\right)$
$f\left(\frac{-11\pi }{6}\right)=-f\left(\frac{11\pi }{6}\right)=-[3\sin \left(\frac{11\pi }{6}\right)+4\cos \left(\frac{11\pi }{6}\right)]$
$=-\left[3\sin \right(2\pi -\frac{\pi }{6})+4\cos (2\pi -\frac{\pi }{6}\left)\right]$
$=-[-\frac{3}{2}+2\sqrt{3}]=\frac{3}{2}-2\sqrt{3}$