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3 and 4 .Determinants and Matrices
hard
Let $A=\left(\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right),$ where $[t]$ denotes the greatest integer less than or equal to $\mathrm{t}$. If $\operatorname{det}(\mathrm{A})=192$, then the set of values of $\mathrm{x}$ is the interval
A
$[68,69)$
B
$[62,63)$
C
$[65,66)$
D
$[60,61)$
(JEE MAIN-2021)
Solution
$\left|\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right|=192$
$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3} \;and\; \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & -1 \\ {[x]} & {[x]+2} & {[x]+4}\end{array}\right]=192$
$2[x]+6+[x]=192 \Rightarrow[x]=62$
Standard 12
Mathematics
