    Question

# If a, b, c are positive numbers such that a>b>c and the equation (a+b−2c)x2+(b+c−2a)x+(c+a−2b)=0 has a root in the interval (-1, 0), then

A

b cannot be the G.M. of a, c

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B

b may be the G.M. of a, c

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C

b is the G.M. of a, c

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D

none of these

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Solution

## The correct option is A b cannot be the G.M. of a, c Let f(x)=(a+b−2c)x2+(b+c−2a)x+(c+a−2b) According to the given condition, we have f (0) f (-1) < 0 i.e. (c +a -2b) (2a - b - c) < 0 i.e. (c +a -2b) (a - b + a - c) < 0 i.e. c + a - 2b < 0 [a > b > c, given ⇒ a - b > 0, a - c >0] i.e b > a+c2 ⇒ b cannot be the G.M. of a, c, since G.M < A.M. always.  Suggest Corrections  0      Similar questions  Related Videos   Relation of Roots and Coefficients
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