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Q.

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:

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a

[68, 69]

b

[60, 61]

c

[62, 63]

d

[65, 66]

answer is C.

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Detailed Solution

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$\begin{array}{rcl} A & = & [\begin{matrix} [x] + 1 & [x] + 2 & [x] + 3 \\ [x] & [x] + 3 & [x] + 3 \\ [x] & [x] + 2 & [x] + 4 \end{matrix}] \\ R_{1} & \to & R_{1} - R_{3}, R_{2} \to R_{2} - R_{3} \\ A & = & [\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ [x] & [x] + 2 & [x] + 4 \end{matrix}] \\ | A | & = & [x] + 4 + [x] + 2 + [x] \\ 192 & = & 3 [x] + 6 \\ 3 [x] & = & 186 \\ [x] & = & 62 \\ ∴ n & \in & (62, 63) \end{array}$

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