Question

If $f\left(\theta \right)=5\cos \theta +3\cos (\theta +\frac{\pi }{3})+3$, then range of $f\left(\theta \right)$ it

• A. $(-5,11)$
• B. $(-3,9)$
• C. $(-2,10)$
• D. $(-4,10)$

Answer 1

Answer: (D)

$(-4,10)$

$f\left(\theta \right)=5\cos \theta +3\cos (\theta +\frac{\pi }{3})+3$

$=5\cos \theta +3(\cos \theta \cos \frac{\pi }{3}-\sin \theta \sin \frac{\pi }{3})+3$

$=5\cos \theta +3(\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta )+3$$

$=\frac{13\cos \theta }{2}-\frac{3\sqrt{3}\sin \theta }{2}+3$

As $-\sqrt{a^2+b^2}\le a\cos x-b\sin x\le \sqrt{a^2+b^2}$

Therefore

$-7\le \frac{13\cos \theta }{2}-\frac{3\sqrt{3}\sin \theta }{2}\le 7$

$\Rightarrow -7+3\le \frac{13\cos \theta }{2}-\frac{3\sqrt{3}\sin \theta }{2}+3\le 7+3$

$\Rightarrow -4\le f\left(\theta \right)\le 10$