Q48 / 125 · BITSAT 2009 · Mathematics

For the equation \(\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}\), if the product of roots is zero, then the sum of roots is

  • A
    0
  • B
    \(\frac{2 \mathrm{ab}}{\mathrm{b}+\mathrm{c}}\)
  • C
    \(\frac{2 b c}{b+c}\)
  • Correct answer
    D
    \(\frac{-2 b c}{b+c}\)

Solution

\(\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}\) or \(x^2+(a+b) x+a b=(b-a) x+(b-a) c\) or \(\mathrm{x}^2+2 \mathrm{ax}+\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0\) Since product of the roots \(=0\) \(a b+c a-b c=0 ; a=\frac{b c}{b+c}\) Thus sum of roots \(=-2 a=\frac{-2 b c}{b+c}\)

Hence, the correct answer is option: 'D'.

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