Q48 / 90 · JEE Main 2024 (30 Jan Shift 1) · Mathematics

Consider the system of linear equation x+y+z= 4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15, where λ,μ∈R. Which one of the following statements is NOT correct?

  • A
    The system has unique solution if λ≠12 and μ≠1,15
  • Correct answer
    B
    The system is inconsistent if λ=12 and μ≠1
  • C
    The system has infinite number of solutions if λ=12 and μ=15
  • D
    The system is consistent if λ≠12

Solution

Given: 

System of linear equations,

x+y+z=4μ,

x+2y+2λz=10μ,

x+3y+4λz2=μ2+15.

⇒Δ=111122λ134λ2

⇒Δ=8λ2-6λ-4λ2-2λ+3-2

⇒Δ=8λ2-6λ-4λ2+2λ+1

⇒Δ=4λ2-4λ+1

⇒Δ=(2λ-1)2

For unique solution, Δ≠0.

⇒2λ-1≠0

⇒λ≠12

Let, Δ=0, λ=12

We know that, for infinite solutions, Δx=Δy=Δz=0

⇒Δx=4μ1110μ21μ2+1531

⇒Δx=4μ2-3-10μ-μ2-15+30μ-2μ2-30

⇒Δx=-4μ-10μ+μ2+15+30μ-2μ2-30

⇒Δx=-μ2+16μ-15

⇒(μ-15)(μ-1)=0

So, for infinite solution λ=12, μ=1 or 15.

Hence, the correct answer is option: 'B'.

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