Question

Number of solutions of the equations $y=\frac{1}{3}\left[\right.sinx+\left[\right.sinx+\left[\right.sinx\left]\right.\left]\right.\left]\right.$ and $\left[\right.y+\left[\right.y\left]\right.\left]\right.=2cosx,$ where $\left[.\right]$ denotes the greatest integer function is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

Answer: (A)

$\because y=\frac{1}{3}\left[\right.sinx+\left[\right.sinx+\left[\right.sinx\left]\right.\left]\right.\left]\right.$
$y=\frac{1}{3}\left(\right.\left[\right.sinx\left]\right.+\left[\right.sinx\left]\right.+\left[\right.sinx\left]\right.\left.\right)$
$y=\left[\right.sinx\left]\right.$ and $\left[\right.y+\left[\right.y\left]\right.\left]\right.=2cosx$
$\Rightarrow 2\left[\right.y\left]\right.=\left(\right.2cosx\left.\right)$
$\Rightarrow \left[\right.y\left]\right.=cosx$
Case I: when $-1\leq sinx < 0$
$\Rightarrow \therefore \left[\right.sinx\left]\right.=-1$
Then, $y=-1$
i.e. $\left[- 1\right]=cosx$
$\Rightarrow cosx=-1$
$\Rightarrow sinx=a$ (impossible)
Case II: when $0\leq sinx < 1$
$\therefore \left[\right.sinx\left]\right.=0$
Then, $y=0$
i.e. $0=cosx$
$\Rightarrow sinx=1$ (impossible)
Case III: $sinx=1$
$\therefore y=1$
i.e. $cosx=1$
$\Rightarrow sinx=0$ (impossible)
Hence, no solution i.e. number of solutions is zero.