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
- A -1 and \(\frac{\log _{e} c}{\log _{e} a}\)
- B 1 and \(-\frac{\log _{e} c}{\log _{e} a}\)
- C 1 and \(\log _{a} c\)
- D -1 and \(\log _{c} a\)
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…
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