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 values of $x$ lie in the interval

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

Answer 1

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

$\left|\begin{array}{ccc}[x+1]& [x+2]& [x+3]\\ \left[x\right]& [x+3]& [x+3]\\ \left[x\right]& [x+2]& [x+4]\end{array}\right|=192$

$\Rightarrow \left|\begin{array}{ccc}\left[x\right]+1& \left[x\right]+2& \left[x\right]+3\\ \left[x\right]& \left[x\right]+3& \left[x\right]+3\\ \left[x\right]& \left[x\right]+2& \left[x\right]+4\end{array}\right|=192$

$R_1\to R_1-R_3\&R_2\to R_2-R_3$
$\left|\begin{array}{ccc}1& 0& -1\\ 0& 1& -1\\ \left[x\right]& \left[x\right]+2& \left[x\right]+4\end{array}\right|=192$

$\Rightarrow 2\left[x\right]+6+\left[x\right]=192$
$\Rightarrow \left[x\right]=62$
$\therefore x\in [62,63)$