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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If the roots of the equation\[\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r},\] \[(x\ne -p,x\ne -q,r\ne 0)\]are equal in magnitude but opposite in sign, then\[p+q\]is equal to

    A)  \[r\]                                     

    B)  \[2r\]

    C)  \[{{r}^{2}}\]                  

    D)         \[\frac{1}{r}\]

    E)  \[\frac{2}{r}\]

    Correct Answer: B

    Solution :

    Given, \[\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\] \[\Rightarrow \]               \[r(2x+p+q)=[{{x}^{2}}+(p+q)x+pq]\] \[\Rightarrow \]               \[{{x}^{2}}+(p+q-2r)x+pq-r(p+q)=0\] As we know, if roots are equal in magnitude but opposite in sign, then coefficient of\[x\]will be zero. \[\therefore \]  \[p+q-2r=0\Rightarrow p+q=2r\]


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