Question

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

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

Answer 1

Answer: (A)

$\because y=\frac{1}{3}[sinx+[sinx+[sinx]]]$
$y=\frac{1}{3}([sinx]+[sinx]+[sinx])$
$y=[sinx]$ and $[y+[y]]=2cosx$
$\Rightarrow 2[y]=(2cosx)$
$\Rightarrow [y]=cosx$
Case I: when $-1\le sinx<0$
$\Rightarrow \therefore [sinx]=-1$
Then, $y=-1$
i.e. $\left[-1\right]=cosx$
$\Rightarrow cosx=-1$
$\Rightarrow sinx=a$ (impossible)
Case II: when $0\le sinx<1$
$\therefore [sinx]=0$
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.