Question

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

• A. $(\pi ,\frac{5\pi }{4})$
• B. $[\pi ,\frac{5\pi }{4}]$
• C. $(\frac{5\pi }{4},2\pi )$
• D. $[\frac{5\pi }{4},2\pi ]$

Answer 1

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

$\begin{array}{c}Wehave\\ \sin x\in \left[-1,1\right]\Rightarrow \left[\sin x\right]\in -1,0,1\\ \cos x\in \left[-1,1\right]\Rightarrow \sqrt{2}\cos x\in \left[-\sqrt{2},\sqrt{2}\right]\Rightarrow \left[\sqrt{2}\cos x\right]\in -2,-1,0,1\\ Clearlytheonlywaytoget-3asthesumis-1+-2\\ For\left[\sin x\right]=-1,x\in \left(\pi ,2\pi \right)\\ For\left[\sqrt{2}\cos x\right]=-1,x\in \left(\frac{3\pi }{4},\frac{5\pi }{4}\right)\\ Hence,bothofthemtogetherx\in \left(\pi ,\frac{5\pi }{4}\right)\\ Hence,theoptionAiscorrectanswer.\end{array}$