For 0 < c < b < a , let ( a + b - 2 c ) x 2 + ( b + c - 2 a ) x + ( c + a - 2 b ) = 0 and be one of its root. Then, among the two statements [2024]

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Solution

Q.
For $0 < c < b < a,$ let $(a + b - 2 c) x^{2} + (b + c - 2 a) x + (c + a - 2 b) = 0$ and $α \neq 1$ be one of its root. Then, among the two statements [2024]

(I) If $α \in (- 1, 0),$ then $b$ cannot be the geometric mean of $a$ and $c$

(II) If $α \in (0, 1),$ then $b$ may be the geometric mean of $a$ and $c$

1 only (II) is true

2 Both (I) and (II) are true

3 only (I) is true

4 Neither (I) nor (II) is true

Ans.
(2)

We have, $(a + b - 2 c) x^{2} + (b + c - 2 a) x + (c + a - 2 b) = 0$

Put $x = 1$

$a + b - 2 c + b + c - 2 a + c + a - 2 b = 0$

$0 = 0$

$∴$    $x = 1$ is another root

$∴$    $α \cdot 1 = \frac{c + a - 2 b}{a + b - 2 c} \Rightarrow α = \frac{c + a - 2 b}{a + b - 2 c}$

If $- 1 < α < 0,$ then $- 1 < \frac{c + a - 2 b}{a + b - 2 c} < 0$

$\Rightarrow 0 < \frac{c + a - 2 b + a + b - 2 c}{a + b - 2 c} < 1 \Rightarrow 0 < \frac{2 a - b - c}{a + b - 2 c} < 1$

Since, $a > b > c > 0 \Rightarrow a + b > c + c$

$\Rightarrow a + b > 2 c \Rightarrow a + b - 2 c > 0$

$∴$    $2 a - b - c > 0 \Rightarrow 2 a > b + c \Rightarrow a > \frac{b + c}{2}$

$∴$ $b$ can not be the geometric mean of $a$ and $c$

If $0 < α < 1,$ then $0 < \frac{c + a - 2 b}{a + b - 2 c} < 1$

$\Rightarrow 0 < c + a - 2 b and c + a - 2 b < a + b - 2 c$

$\Rightarrow c < b$

$2 b < c + a, b < \frac{c + a}{2}$

$∴$ $b$ may be the geometric mean of $a$ and $c .$

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