JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 2)
Let $$A = \left( {\matrix{
{[x + 1]} & {[x + 2]} & {[x + 3]} \cr
{[x]} & {[x + 3]} & {[x + 3]} \cr
{[x]} & {[x + 2]} & {[x + 4]} \cr
} } \right)$$, 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 :
[68, 69)
[62, 63)
[65, 66)
[60, 61)
Explanation
$$\left| {\matrix{
{[x + 1]} & {[x + 2]} & {[x + 3]} \cr
{[x]} & {[x + 3]} & {[x + 3]} \cr
{[x]} & {[x + 2]} & {[x + 4]} \cr
} } \right| = 192$$
R1 $$\to$$ R1 $$-$$ R3 & R2 $$\to$$ R2 $$-$$ R3
$$\left[ {\matrix{ 1 & 0 & { - 1} \cr 0 & 1 & { - 1} \cr {[x]} & {[x] + 2} & {[x] + 4} \cr } } \right] = 192$$
$$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$
R1 $$\to$$ R1 $$-$$ R3 & R2 $$\to$$ R2 $$-$$ R3
$$\left[ {\matrix{ 1 & 0 & { - 1} \cr 0 & 1 & { - 1} \cr {[x]} & {[x] + 2} & {[x] + 4} \cr } } \right] = 192$$
$$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$
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