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JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    Let \[\alpha \] and \[\beta \] be two real roots of the equation \[(k+1)ta{{n}^{2}}x-\sqrt{2}\cdot \lambda \tan x=(1-k)\], where \[k(\ne -1)\] and \[\lambda \] are real numbers. If \[{{\tan }^{2}}(\alpha +\beta )=50,\] then a value of \[\lambda \] is: [JEE MAIN Held on 07-01-2020 Morning]

    A) 10        

    B) \[10\sqrt{2}\]

    C) 5                     

    D) \[5\sqrt{2}\]

    Correct Answer: A

    Solution :

    [a]
    \[\tan \alpha \] and \[\tan \beta \] are roots of \[(k+1){{x}^{2}}-\sqrt{2}\lambda x\]
    \[-(1-k)=0\]
    \[\therefore \tan \alpha +\tan \beta =\frac{\sqrt{2}\lambda }{k+1}\]
    \[\tan \alpha +\tan \beta =\frac{k-1}{k+1}\]
    Now \[\tan (\alpha +\beta )=\frac{\frac{\sqrt{2}\lambda }{k+1}}{1-\left( \frac{k-1}{k+1} \right)}=\frac{\sqrt{2}\lambda }{2}=\frac{\lambda }{\sqrt{2}}\]
    \[\frac{{{\lambda }^{2}}}{2}=50\]
    \[\therefore \lambda =10\]          


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