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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    If the system of equations \[2x+3yz=0,\]\[x+ky2z=0\]and \[2xy+z=0\]has a non-trival solution \[\left( x,y,z \right),\]then \[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k\]is equal to:-             [JEE Main 9-4-2019 Afternoon]

    A) \[\frac{3}{4}\]                        

    B) \[-4\]

    C) \[\frac{1}{2}\]

    D) \[-\frac{1}{4}\]

    Correct Answer: C

    Solution :

    \[\left| \begin{matrix}    2 & 3 & -1  \\    1 & K & -2  \\    2 & -1 & 1  \\ \end{matrix} \right|=0\]           By solving\[K=\frac{9}{2}\] \[2x+3y2y=0\]                                    ?(1) \[x+\frac{9}{2}y-2z=0\]                                            ?(2) \[2x-y+z=0\]                                       ?(3) \[(1)-(3)\Rightarrow 4y-2z=0\] \[2y=z\]                                                           ?(4) \[\]                                                        ?(5) Put z form eqn. (4) into (1) \[2x+3y-2y=0\] \[2x+y=0\] \[\]                                            ?(6) \[\frac{(6)}{(5)}\] \[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+K=\frac{1}{2}\]             


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