Question

If $\left[\sin x\right]+\left[\sqrt{2}\cos x\right]=-3$, (where $[.]$ represents the greatest integer function), then $x$ belongs to

• A. $[\pi ,\frac{5\pi }{4})$
• B. $(\pi ,\frac{5\pi }{4})$
• C. $[\pi ,\frac{5\pi }{4}]$
• D. no solution

Answer 1

The correct option is B $(\pi ,\frac{5\pi }{4})$

$\because -1\le \sin x\le 1\Rightarrow \left[\sin x\right]=\{-1,0,1\}$
and $-\sqrt{2}\le \sqrt{2}\cos x\le \sqrt{2}\Rightarrow \left[\sqrt{2}\cos x\right]=\{-2,-1,0,1,2\}$
So, $\left[\sin x\right]+\left[\sqrt{2}\cos x\right]=-3$ is possible iff
$\Rightarrow \left[\sin x\right]=-1$ and $\left[\sqrt{2}\cos x\right]=-2$
$\Rightarrow -1\le \sin x<0$ and $-\sqrt{2}\le \sqrt{2}\cos x<-1$
$\Rightarrow x\in (\pi ,2\pi )$ and $x\in (\frac{3\pi }{4},\frac{5\pi }{4})$
$\therefore x\in (\pi ,\frac{5\pi }{4})$