Question

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 :

• 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: (C)

$f\left(-x\right) = - f\left(x\right)$ as $f\left(x\right)$ is odd function
$f\left(\frac{-11\pi}{6}\right) = -\left[3\,sin\left(\frac{+11\pi }{6}\right)+4\,cos\left(\frac{+11\pi }{6}\right)\right]$
$= -\left[3\,sin\left(\frac{11\pi }{6}\right)+4\,cos\left(\frac{11\pi }{6}\right)\right]$
$= -\left[3\,sin\left(2\pi-\frac{\pi }{6}\right)+4\,cos\left(2\pi -\frac{\pi }{6}\right)\right]$
$= +\, 3 \,sin\, \pi/6 - 4 \,cos \frac{\pi}{6}$
$= 3\times \frac{1}{2}-\frac{4\sqrt{3}}{2} = \frac{3}{2}-2\sqrt{3}$