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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    If the plane \[2xy+2z+3=0\]has the distances \[\frac{1}{3}\]and\[\frac{2}{3}\]units from the planes \[4x2y+4z+\lambda =0\]and\[2xy+2z+\mu =0,\]respectively, then the maximum value of \[\lambda +\mu \] is equal to : [JEE Main 10-4-2019 Afternoon]

    A) 15                               

    B) 5

    C) 13       

    D) 9

    Correct Answer: C

    Solution :

    \[4x2y+4z+6=0\]           \[\frac{|\lambda -6|}{\sqrt{16+4+16}}=\left| \frac{\lambda -6}{6} \right|=\frac{1}{3}\]           \[|\lambda -6|=2\] \[\lambda =8,4\]           \[\frac{|\mu -3|}{\sqrt{4+4+1}}=\frac{2}{3}\]           \[|\mu -3|=2\] \[\mu =5,1\] \[\therefore \]Maximum value of \[(\mu +\lambda )=13.\]                  


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