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WBJEE · Maths · Basic of Mathematics

If \(a, b\) and \(c\) are positive numbers in a GP, then the roots of the quadratic equation \(\left(\log _{e} a\right) x^{2}-\left(2 \log _{e} b\right) x+\left(\log _{e} c\right)=0\) are

  1. A -1 and \(\frac{\log _{e} c}{\log _{e} a}\)
  2. B 1 and \(-\frac{\log _{e} c}{\log _{e} a}\)
  3. C 1 and \(\log _{a} c\)
  4. D -1 and \(\log _{c} a\)
Verified Solution

Answer & Solution

Correct Answer

(C) 1 and \(\log _{a} c\)

Step-by-step Solution

Detailed explanation

Since, \(a, b\) and \(c\) are in \(\mathrm{GP}\). \(\therefore b^{2}=a c\) Given equation is \((\log _{e} \text { a) } x^{2}-\left(2 \log _{e} \text { b) } x+\log _{e} c\right)=0\) Put \(x=1,\) we get…