Question

Let $A=\left(\begin{array}{ccc}[x+1]& [x+2]& [x+3]\\ \left[x\right]& [x+3]& [x+3]\\ \left[x\right]& [x+2]& [x+4]\end{array}\right),$ where $\left[t\right]$ denotes the greatest integer less than or equal to $t.$ If $det\left(A\right)=192,$ then the set of values of $x$ is the interval

• A. $[60,61)$
• B. $[65,66)$
• C. $[62,63)$
• D. $[68,69)$

Answer 1

The correct option is C $[62,63)$

$|A|=\left|\begin{array}{ccc}[x+1]& [x+2]& [x+3]\\ \left[x\right]& [x+3]& [x+3]\\ \left[x\right]& [x+2]& [x+4]\end{array}\right|$
$=\left|\begin{array}{ccc}\left[x\right]+1& \left[x\right]+2& \left[x\right]+3\\ \left[x\right]& \left[x\right]+3& \left[x\right]+3\\ \left[x\right]& \left[x\right]+2& \left[x\right]+4\end{array}\right|$
$C_3\to C_3-C_2,C_2\to C_2-C_1$
$=\left|\begin{array}{ccc}\left[x\right]+1& 1& 1\\ \left[x\right]& 3& 0\\ \left[x\right]& 2& 2\end{array}\right|$
(Expanding by $C_3$)
$=1\left(2\right[x]-3[x\left]\right)+2\left(3\right[x]+3-[x\left]\right)=-\left[x\right]+2\left(2\right[x]+3)$
$=3\left[x\right]+6$

Given, $|A|=192$
$\Rightarrow 3\left[x\right]+6=192$
$\Rightarrow \left[x\right]=62$
$\Rightarrow x\in [62,63)$