Verified Solution • 2025–26 Session moderate

QJEEM24Ma6919: JEE Main 2024 Mathematics - Class 12 JEEM24Ma6919 Solution: Consider the system of linear equations where . Which one of…

Consider the system of linear equations x+y+z=4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15 where λ,μ∈R\lambda, \mu \in \mathbf{R} . Which one of the following statements is NOT correct ?

Multiple Choice Options

  • A The system has unique solution if λ≠12\lambda \neq \frac{1}{2} and μ≠1,15\mu \neq 1,15
  • B The system has infinite number of solutions if λ=12\lambda=\frac{1}{2} and μ=15\mu=15
  • C The system is consistent if λ≠12\lambda \neq \frac{1}{2}
  • D The system is inconsistent if λ=12\lambda=\frac{1}{2} and μ≠1\mu \neq 1

Step-by-Step Verified Solution

x+y+z=4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda{ }^2 z=\mu^2+15 ,

Δ=∣1amp;1amp;11amp;2amp;2λ1amp;3amp;4λ2∣=(2λ−1)2\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\1 & 2 & 2 \lambda \\1 & 3 & 4 \lambda^2\end{array}\right|=(2 \lambda-1)^2

For unique solution Δ≠0,2λ−1≠0,(λ≠12)\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)

Let Δ=0,λ=12\Delta=0, \lambda=\frac{1}{2}

amp;Δy=0,Δx=Δz=∣4μamp;1amp;110μamp;2amp;1μ2+15amp;3amp;1∣amp;=(μ−15)(μ−1)\begin{aligned}& \Delta_y=0, \Delta_x=\Delta_z=\left|\begin{array}{ccc}4 \mu & 1 & 1 \\10 \mu & 2 & 1 \\\mu^2+15 & 3 & 1\end{array}\right| \\& =(\mu-15)(\mu-1)\end{aligned}

For infinite solution λ=12,μ=1\lambda=\frac{1}{2}, \mu=1 or 15

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