Question

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

• A. $(\frac{\pi }{4},\frac{7\pi }{4})$
• B. $(\frac{5\pi }{4},\frac{7\pi }{4})$
• C. $(\frac{3\pi }{4},\frac{5\pi }{4})$
• D. $(\pi ,\frac{5\pi }{4})$

Answer 1

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

$\left[\sin x\right]+\left[\sqrt{2}\cos x\right]=-3$
We know that,
$\sin x\in [-1,1]\Rightarrow \left[\sin x\right]\in \{-1,0,1\}\sqrt{2}\cos x\in [-\sqrt{2},\sqrt{2}]\Rightarrow \left[\sqrt{2}\cos x\right]\in \{-2,-1,0,1\}$
So, solution is possible only when
$\left[\sin x\right]=-1,\left[\sqrt{2}\cos x\right]=-2\Rightarrow -1\le \sin x<0\Rightarrow x\in (\pi ,2\pi )\cdots \left(1\right)\left[\sqrt{2}\cos x\right]=-2\Rightarrow -\sqrt{2}\le \sqrt{2}\cos x<-1\Rightarrow -1\le \cos x<-\frac{1}{\sqrt{2}}\Rightarrow x\in (\frac{3\pi }{4},\frac{5\pi }{4})\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$x\in (\pi ,\frac{5\pi }{4})$