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

Solution

Maths sir · 4 steps

Call [x] = n, and every entry becomes n plus a whole number. Then it is an ordinary determinant in n.

1. Replace by n. Let n = [x]. When you add a whole number k to x, the greatest integer goes up by exactly k, so [x + k] = n + k. The rows become (n+1, n+2, n+3), (n, n+3, n+3) and (n, n+2, n+4).
2. Row operations. Apply R₁ → R₁ - R₃ and R₂ → R₂ - R₃. This does not change the determinant, and every n cancels from the top two rows.

   det A = |1, 0, -1; 0, 1, -1; n, n+2, n+4| = [(n+4) + (n+2)] - (0 - n) = 3n + 6

3. Solve for n. Set 3n + 6 = 192, so 3n = 186 and n = 62. Then [x] = 62 means 62 ≤ x < 63.
4. So. The set of values of x is [62, 63), which is option (b).

The idea to keep. When every entry is [x + integer], write [x] = n first. Subtract rows to clear out n, and the condition on [x] gives an interval of length 1 for x.

Answer from the NTA final answer key. Solution by PrepWiser.