Question

If $\left[\cos x\right]+[\sin x+1]=0$, then value of $x$ satisfying $f\left(x\right)$ where $x\in [0,2\pi ]$, and $\left[\right]$ denotes greatest integer function,

• A. $(\frac{\pi }{2},\pi ]\cup [\frac{3\pi }{2},2\pi ]$
• B. $(0,\frac{\pi }{2})$
• C. $[\frac{\pi }{2},\pi ]\cup [\frac{5\pi }{2},2\pi ]$
• D. $[\pi ,2\pi ]$

Answer 1

The correct option is A $(\frac{\pi }{2},\pi ]\cup [\frac{3\pi }{2},2\pi ]$

$\left[\cos x\right]+[\sin x+1]=0$
The range of $\cos x$ is $[-1,1]$ and thus the range of $\left[\cos x\right]$ becomes $-1,0,1$
Similarly, since the range of $1+\sin x$ is $[0,2]$; the range of $[1+\sin x]$ becomes $0,1,2$
Now, since the R.H.S. is 0, one of the terms on L.H.S will be 1 and the other -1 and the other possibility says both terms to be 0!
a) If $\left[\cos x\right]=0,\cos x\in [0,1)$; also, $[1+\sin x]=0,1+\sin x\in [0,1)$
$\Rightarrow \sin x\in [-1,0)$
This is possible in the fourth quadrant.
b) If $\left[\cos x\right]=-1,\cos x\in [-1,0)$ and also, $[1+\sin x]=1$ so $1+\sin x\in [1,2)$ i.e. $\sin x\in [0,1)$
This is possible in the second quadrant.
Hence option A is correct!