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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    Let a, b, c be positive real numbers. If \[\frac{{{x}^{2}}-bx}{ax-c}=\frac{m-1}{m+1}\]has two roots which are numerically equal but opposite in sign, then the value of m is

    A)  \[c\]                     

    B)         \[\frac{1}{c}\]

    C)  \[\frac{a+b}{a-b}\]        

    D)         \[1\]

    E)  \[\frac{a-b}{a+b}\]

    Correct Answer: E

    Solution :

    Given, \[\frac{{{x}^{2}}-bx}{ax-c}=\frac{m-1}{m+1}\] \[\Rightarrow \]\[(m+1){{x}^{2}}-b(m+1)x\]                                 \[=a(m-1)x-c(m-1)\] \[\Rightarrow \]\[(m+1){{x}^{2}}-(bm+b+am-a)x\]                                                 \[+c(m-1)=0\] \[\Rightarrow \]\[(m+1){{x}^{2}}-\{(a+b)m+(b-a)\}x\]                                                 \[+c(m-1)=0\] Let the roots of the equation is\[(a,-a)\]. Then sum of the roots                 \[=\frac{(a+b)m+(b-a)}{(m+1)}\]                 \[\alpha -\alpha =\frac{(a+b)m+(b-a)}{(m+1)}=0\] \[\Rightarrow \]               \[(a+b)m-(a-b)=0\] \[\Rightarrow \]               \[m=\frac{a-b}{a+b}\]


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