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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 [ λ ] be the greatest integer less than or equal to λ . The set...

Let [λ][ \lambda ] be the greatest integer less than or equal to λ\lambda. The set of all values of λ\lambda for which the system of linear equations x+y+z=4,3x+2y+5z=3,9x+4y+(28+[λ])z=[λ]x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + ( 28 + [ \lambda ] ) z = [ \lambda ] has a solution is:

Options

  1. ARR
  2. B(−∞,−9)∪[−8,∞)( - \infty , - 9 ) \cup [ - 8, \infty )
  3. C(−∞,−9)∪(−9,∞)( - \infty , - 9 ) \cup ( - 9, \infty )
  4. D[−9,−8)[ - 9, - 8 )

Correct answer

A. RR

Step-by-step solution

Given, the system of linear equations,

x+y+z=4........................(i)x + y + z = 4 . . . . . . . . . . . . . . . . . . . . . . . . \left(i\right)

3x+2y+5z=3................(ii)3x + 2y + 5z = 3 . . . . . . . . . . . . . . . . \left(ii\right)

9x+4y+(28+[λ]]z=[λ].....................(iii)9x + 4y + ( 28 + [ \lambda ] ] z = [ \lambda ] . . . . . . . . . . . . . . . . . . . . . \left(iii\right)

D = \left|\begin{cases}1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28 + [ \lambda ]\end{cases}\right|

=1{56+2[λ]−20}−1{84+3[λ]−45}+1(12−18)= 1\left\{56 + 2\left[\lambda\right] - 20\right\} - 1\left\{84 + 3\left[\lambda\right] - 45\right\} + 1\left(12 - 18\right)

D=−([λ]+9)D = - ( [ \lambda ] + 9 )

If D=−([λ]+9)≠0D = - \left([ \lambda ] + 9\right) \ne 0 then unique solution
If D=−([λ]+9)=0D = - \left([ \lambda ] + 9\right) = 0 then D1=D2=D3=0{D}_{1} = {D}_{2} = {D}_{3} = 0 infinite solutions

Hence, λ∈R\lambda \in R

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