QJEEM21Ma8aad: JEE Main 2021 Mathematics - Class 12 JEEM21Ma8aad Solution: Let A = ( [ x + 1 ] [ x + 2 ] [ x + 3 ] [ x ] [ x + 3 ] [ x + 3 …

Question

Verified Solution • 2025–26 Session moderate

Let

A = (\begin{matrix} [x + 1] & [x + 2] & [x + 3] \\ [x] & [x + 3] & [x + 3] \\ [x] & [x + 2] & [x + 4] \end{matrix})

, 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 :

Accepted Answer

| \begin{matrix} [x + 1] & [x + 2] & [x + 3] \\ [x] & [x + 3] & [x + 3] \\ [x] & [x + 2] & [x + 4] \end{matrix} | = 192

R₁

\to

R₁

-

R₃ & R₂

\to

R₂

-

R₃

[\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ [x] & [x] + 2 & [x] + 4 \end{matrix}] = 192

2[x] + 6 + [x] = 192 \Rightarrow [x] = 62

Step-by-Step Verified Solution