If 0 ≤[x]

Question

If 0 ≤[x]<2,-1 ≤[y] < 1 and 1 ≤[z]<3 where [ . ] denotes greatest integral function then the maximum value of the determinant. D=|[x]+1 [y] [z] [x] [y]+1 [z] [x] [y] [z]+1| is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

| 1 -1 0 0 1 -1 [x] [y] [z]+1 | solving =[x]+[y]+[z]+1 taking maximum value we get 4, note that [x] is always an integer

Q. If $0 \leq [x] < 2, - 1 \leq [y] < 1$ and $1 \leq [z] < 3$ where $[.]$ denotes greatest integral function then the maximum value of the determinant. $D = ∣ ∣ [x] + 1 [x] [x] [y] [y] + 1 [y] [z] [z] [z] + 1 ∣ ∣$ is

$∣ ∣ 10 [x] - 1 1 [y] 0 - 1 [z] + 1 ∣ ∣$
solving $= [x] + [y] + [z] + 1$
taking maximum value we get $4$ , note that $[x]$ is always an integer

Q. If 0≤[x]<2,−1≤[y]<1 and 1≤[z]<3 where [.] denotes greatest integral function then the maximum value of the determinant. D=∣∣​[x]+1[x][x]​[y][y]+1[y]​[z][z][z]+1​∣∣​ is

∣∣​10[x]​−11[y]​0−1[z]+1​∣∣​ solving =[x]+[y]+[z]+1 taking maximum value we get 4, note that [x] is always an integer

Q. If $0 \leq [x] < 2, - 1 \leq [y] < 1$ and $1 \leq [z] < 3$ where $[.]$ denotes greatest integral function then the maximum value of the determinant. $D = ∣ ∣ [x] + 1 [x] [x] [y] [y] + 1 [y] [z] [z] [z] + 1 ∣ ∣$ is

$∣ ∣ 10 [x] - 1 1 [y] 0 - 1 [z] + 1 ∣ ∣$
solving $= [x] + [y] + [z] + 1$
taking maximum value we get $4$ , note that $[x]$ is always an integer