Let A = [ x + 1 ] [ x + 2 ] [ x + 3 ] [ x ] [ x + 3 ] [ x + 3 ] [ x ] [ x + 2 ] [ x + 4 ] , where [ x ] denotes the greatest integer less than or equal to x . If det ( A ) = 192 , then the set of values of x is in the interval:

Step-by-step solution

Given, A = [ x + 1 ] [ x + 2 ] [ x + 3 ] [ x ] [ x + 3 ] [ x + 3 ] [ x ] [ x + 2 ] [ x + 4 ] A = [ x ] + 1 [ x ] + 2 [ x ] + 3 [ x ] [ x ] + 3 [ x ] + 3 [ x ] [ x ] + 2 [ x ] + 4 R 1 → R 1 − R 3 , R 2 → R 2 − R 3 A = 1 0 − 1 0 1 − 1 [ x ] [ x ] + 2 [ x ] + 4 det ⁡ ( A ) = 1 ( [ x ] + 4 + [ x ] + 2 ) − 1 ( − [ x ] ) = 3 [ x ] + 6 Given, de t ( A ) = 192 192 = 3 [ x ] + 6 3 [ x ] = 186 [ x ] = 62 x ∈ [ 62 , 63 )