Let A = [ cases [ x + 1 ] & [ x + 2 ] & [ x + 3 ] [ x ] & [ x + 3 ] & [ x + 3 ] [ x ] & [ x + 2 ] & [ x + 4 ] cases ] , where [ x ] denotes the greatest integer less than or equal to x . If det ( A ) = 192 , then the

Correct answer: A. [ 62, 63 ) Given, A = [ cases [ x + 1 ] & [ x + 2 ] & [ x + 3 ] [ x ] & [ x + 3 ] & [ x + 3 ] [ x ] & [ x + 2 ] & [ x + 4 ] cases ] A = [ cases [ x ] + 1 & [ x ] + 2 & [ x ] + 3 [ x ] & [ x ] + 3 & [ x ] + 3 [ x ] & [ x ] + 2 & [ x ] + 4 cases ] R _ 1 → R _ 1 - R _ 3 , R _ 2 → R _ 2 - R _ 3 A = [ cases 1 & 0 & - 1 0 & 1 & - 1 [ x ] & [ x ] + 2 & [ x ] + 4 cases ] det ( A ) = 1 ( [ x ] + 4 + [ x ] + 2 ) - 1 ( - [ x ] ) = 3 [ x ] + 6 Given, det ( A ) = 192 192 = 3 [ x ] + 6 3 [ x ] = 186 [ x ] = 62 x ∈ [ 62, 63 )

JEE Main 202127 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2021 · 27 August · Evening Shift · Mathematics · Determinants

Let A = [ cases [ x + 1 ] & [ x + 2 ] & [ x + 3 ] [ x ] & [ x + 3 ]...

Let

A = \left[

\begin{cases}[ x + 1 ] & [ x + 2 ] & [ x + 3 ] \\ [ x ] & [ x + 3 ] & [ x + 3 ] \\ [ x ] & [ x + 2 ] & [ x + 4 ]\end{cases}

\right]

, where

[ x ]

denotes the greatest integer less than or equal to

x

. If

det ( A )

= 192

, then the set of values of

x

is in the interval:

Options

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

Correct answer

A. $[ 62, 63 )$

Step-by-step solution

Given, $A = \left[$ \begin{cases}[ x + 1 ] & [ x + 2 ] & [ x + 3 ] \\ [ x ] & [ x + 3 ] & [ x + 3 ] \\ [ x ] & [ x + 2 ] & [ x + 4 ]\end{cases} $\right]$

$A = \left[$ \begin{cases}[ x ] + 1 & [ x ] + 2 & [ x ] + 3 \\ [ x ] & [ x ] + 3 & [ x ] + 3 \\ [ x ] & [ x ] + 2 & [ x ] + 4\end{cases} $\right]$

${R}_{1} \to {R}_{1} - {R}_{3}, {R}_{2} \to {R}_{2} - {R}_{3}$

$A = \left[$ \begin{cases}1 & 0 & - 1 \\ 0 & 1 & - 1 \\ [ x ] & [ x ] + 2 & [ x ] + 4\end{cases} $\right]$

$det ( A ) = 1 ( [ x ] + 4 + [ x ] + 2 ) - 1 ( - [ x ] )$

$= 3 [ x ] + 6$

Given, $det ( A ) = 192$

$192 = 3 [ x ] + 6$

$3 [ x ] = 186$

$[ x ] = 62$

$x \in [ 62, 63 )$

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