JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 5)
For $$0 < c < b < a$$, let $$(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$$ and $$\alpha \neq 1$$ be one of its root. Then, among the two statements
(I) If $$\alpha \in(-1,0)$$, then $$b$$ cannot be the geometric mean of $a$ and $$c$$
(II) If $$\alpha \in(0,1)$$, then $$b$$ may be the geometric mean of $$a$$ and $$c$$
Explanation
$$\begin{aligned} & f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\ & f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\ & f(1)=0 \\ & \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\ & \alpha=\frac{c+a-2 b}{a+b-2 c} \\ & \text { If, }-1<\alpha<0 \\ & -1<\frac{c+a-2 b}{a+b-2 c}<0 \\ & b+c<2 a \text { and } b>\frac{a+c}{2} \end{aligned}$$
therefore, b cannot be G.M. between a and c.
$$\begin{aligned} & \text { If, } 0<\alpha<1 \\ & 0<\frac{c+a-2 b}{a+b-2 c}<1 \\ & b>c \text { and } b<\frac{a+c}{2} \end{aligned}$$
Therefore, $$\mathrm{b}$$ may be the G.M. between $$\mathrm{a}$$ and $$\mathrm{c}$$.
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