Question

The number of solutions of $[\sin x+\cos x]=3+[-\sin x]+[-\cos x]$ (where $[\cdot ]$ denotes the greatest integer function), $x\in [0,2\pi ]$, is

• A. 0
• B. 4
• C. Infinite
• D. 1

Answer 1

Answer: (C)

$[\sin x+\cos x]=3+[-\sin x]+[-\cos x]=1$
The maximum value of the L.H.S. is 1 and the minimum of the R.H.S. is also 1. Thus,
$[\sin x+\cos x]=3+[-\sin x]+[-\cos x]=1$
or $[\sin x+\cos x]=1,[-\sin x]=-1,[-\cos x]=-1$
$[\sin x+\cos x]=1$
$\therefore 1\le \sin x+\cos x\le \sqrt{2}$
$\therefore x\in \left[0,\frac{\pi }{2}\right]$...(i)
$\therefore [-\sin x]=-1$
$\therefore -1\le -\sin x<0$
$\therefore 0<\cos x\le 1$
$\therefore x\in \left(0,\frac{\pi }{2}\right]$ .... (ii)
$\therefore [-\cos x]=-1$
$\therefore -1\le -\cos x<0$
$\therefore 0<\cos x\le 1$
$\therefore x\in \left(0,\frac{\pi }{2}\right]$ ....(iii)
From Eqs. (i), (ii), (iii), $x\in \left(0,\frac{\pi }{2}\right)$