Question

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

• A. $[-2,-1]$
• B. $(-2,-1)$
• C. $(-1,\frac{-1}{2})$
• D. None of these

Answer 1

The correct option is B

$(-2,-1)$

$\left[2sinx\right]+\left[cosx\right]=-3$, only if $\left[2sinx\right]=-2\&\left[cosx\right]=-1$
$-2\le 2sinx<-1$ and $-1\le cosx<0$
$-1\le sinx<-\frac{1}{2}$ and $-1\le cosx<0$
$\frac{7\pi }{6}<x<\frac{11\pi }{6}$ and $\frac{\pi }{2}<x<\frac{3\pi }{2}$
common values of x is $\frac{7\pi }{6}<x<\frac{3\pi }{2}$
$\left\{\text{function is periodic so consider the interval}\right\}0\le x\le 2\pi$
For $f\left(x\right)=sinx+\sqrt{3}cosx=2sin(\frac{\pi }{3}+x)$
Now $\frac{7\pi }{6}<x<\frac{3\pi }{2}$
$\frac{3\pi }{2}<\frac{\pi }{3}+x<\frac{11\pi }{6}-1<sin(\frac{\pi }{3}+x)<\frac{-1}{2}-2<2sin(\frac{\pi }{3}+x)<-1$
Range is (-2,-1)