Question

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

• A. $(2,-1)$
• B. $(-1,-\frac{1}{2})$
• C. $(-2,-1)$
• D. $(-2,1)$

Answer 1

The correct option is C $(-2,-1)$

We have $\left[2\sin x\right]+\left[\cos x\right]=-3$
only if $\left[2\sin x\right]=-2$ and $\left[\cos x\right]=-1$

$\Rightarrow -2\le 2\sin x<-1$ and $-1\le \cos x<0$
$\Rightarrow -1\le \sin x<-\frac{1}{2}$ and $-1\le \cos x<0$
$\Rightarrow \frac{7\pi }{6}<x<\frac{11\pi }{6}$​ and $\frac{\pi }{2}<x<\frac{3\pi }{2}$
$\Rightarrow \frac{7\pi }{6}<x<\frac{3\pi }{2}$​

For the above values of $x,$
$\sin x+\sqrt{3}\cos x=2\sin (\frac{\pi }{3}+x)$ lies between $-2$ and $-1.$
$\therefore$ Range of $f\left(x\right)=(-2,-1)$