Question

If $\left[2\cos x\right]+\left[\sin x\right]=-3$, then the range of the function, $f\left(x\right)=\sin x+\sqrt{3}\cos x$ in $[0,2\pi ]$ Where $\left[\right]$ denotes the greatest integer function.

• A. $[-\sqrt{3},\sqrt{3})$
• B. $[-2,\sqrt{3}]$
• C. $[-3,-1]$
• D. $[-2,-\sqrt{3}]$

Answer 1

The correct option is B $[-2,\sqrt{3}]$

Since, $\sin x,\cos xϵ[1,-1]$

So, for equality to hold true in

$\left[2\cos x\right]+\left[\sin x\right]=-3$,

$\cos xϵ[-1,\frac{-1}{2})$ and $\sin xϵ[-1,0)$ in $xϵ[0,2\pi ]$

For $\sin xϵ[-1,0)$

$xϵ[\frac{2\pi }{3},\frac{4\pi }{3}]$

$\left[f\right(x)=\sqrt{3}\cos x+\sin x$$[=2[\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x]$

$f\left(x\right)=2\left[\sin x\cos \right(\pi /3)+\sin (\pi /3\left)\cos x\right]=2\sin (x+\frac{\pi }{3})$

for

$\pi <x<40/3$

$\pi +\frac{\pi }{3}<x+\frac{\pi }{3}<4\pi +\frac{\pi }{3}$

$\frac{4\pi }{3}<x+\frac{\pi }{3}<\frac{5\pi }{3}$

$\sin (x+\frac{\pi }{3})ϵ[-1,-\frac{\sqrt{3}}{2})$

$2\sin (x+\frac{\pi }{3})ϵ[-2,-\sqrt{3})$