Question

Let f be an odd function defined on the set of real numbers such that for $x\ge 0,f\left(x\right)=3sinx+4cosx.$
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
$f\left(\frac{-11\pi }{6}\right)=-\left[3sin\left(\frac{+11\pi }{6}\right)+4cos\left(\frac{+11\pi }{6}\right)\right]$
$=-\left[3sin\left(\frac{11\pi }{6}\right)+4cos\left(\frac{11\pi }{6}\right)\right]$
$=-\left[3sin\left(2\pi -\frac{\pi }{6}\right)+4cos\left(2\pi -\frac{\pi }{6}\right)\right]$
$=+3sin\pi /6-4cos\frac{\pi }{6}$
$=3\times \frac{1}{2}-\frac{4\sqrt{3}}{2}=\frac{3}{2}-2\sqrt{3}$