The equation `x^(2)-2[sinx]`, where `[.]` denotes the greatest integer function, has

Options

A. 3
B. 4
C. 2
D. 1

Answer

C

Explanation

We have, `[sin x]=-1,0,1`
So, we have the following cases.
CASE I When [sin x]=-1
In this case, we have
`x^(2)-2=-1 impliesx=+-1`
`:. X=-1 ` is the solution in this case.
CASE II When [sin x]=0
In this case, we have
`x^(2)-2=0 impliesx=+-sqrt(2)`
But, `[sinsqrt(2)]=0` and `[sin(-sqrt(2))]=-1`.
`:. x=sqrt(2)` is the solution in this case.
CASE III When [sinx]=1
In this case, we have
`x^(2)-2=1 impliesx=+-sqrt(3)`
But, `=[sin sqrt(3)]=0` and `[sin(-sqrt(2))=-1`. Therefore, there is no solution in this case.
REMARK It can be easily seen from the graphs of `y=x^(2)-2` and y=[sin x] that the two curves intersect at x=-1 and `x=sqrt(2)` . Hence, the given equation has two solutions only.