JEE Advanced Model Paper 8 with solutions for online practice

Ans:- D
Hint:- We have [sin x] = -1 , 0 , 1
So, we have the following cases
Case – I :- when [sin x] = -1, In this case , we have x2 - 2 = -1 ⇒ x ± 1
∴ x = - 1 is the solution in this case
Case – II :- when [sin x] = 0 , In this case, we have x2 - 2 = 0 ⇒ x ± √2
But , [sin √2] = 0 and [sin(−√2)] = - 1
∴ x = √2 is the solution in this case
Case – III :- when [sin x] = 1, In this case, we have x2 - 2 = 1 ⇒ x = ± √3
But , [sin √3] = 0 and [sin (−√3)] = - 1 ,
Therefore, there is no solution in this case.
Hence, the given equation has two solutions only, namely, x = - 1 and x = √2