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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If the roots a, Rot the equation \[\frac{{{x}^{2}}-bx}{ax-c}=\frac{\lambda -1}{\lambda +1}\]are such that\[\alpha +\beta =0,\]then the value of\[\lambda \]is:

    A)  \[\frac{a-b}{a+b}\]        

    B)         \[c\]

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

    D)         \[\frac{a+b}{a-b}\]

    E)  \[\frac{1}{ab}\]

    Correct Answer: A

    Solution :

    The given equation can be rewritten as \[{{x}^{2}}(\lambda +1)-(b\lambda +b+a\lambda -a)x+x(\lambda -1)=0\] \[\because \]     \[\alpha +\beta =\frac{(b\lambda +b+a\lambda +a)}{\lambda +1}\] Also            \[\alpha +\beta =0\] \[\therefore \]  \[b\lambda +b+a\lambda -a=0\] \[\Rightarrow \]               \[\lambda =\frac{a+b}{a+b}\]


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