Quadratic Equation Question 7
Question 7 - 2024 (31 Jan Shift 1)
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 $c$
(1) Both (I) and (II) are true
(2) Neither (I) nor (II) is true
(3) Only (II) is true
(4) Only (I) is true
Show Answer
Answer (1)
Solution
$f(x)=(a+b-2 c) x^{2}+(b+c-2 a) x+(c+a-2 b)$
$f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0$
$f(1)=0$
$\therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c}$
$\alpha=\frac{c+a-2 b}{a+b-2 c}$
If $-1<\alpha<0$
$-1<\frac{c+a-2 b}{a+b-2 c}<0$
$b+c<2 a$ and $b>\frac{a+c}{2}$
therefore, $b$ cannot be G.M. between $a$ and $c$.
If, $0<\alpha<1$
$0<\frac{c+a-2 b}{a+b-2 c}<1$
$b>c$ and $b<\frac{a+c}{2}$
Therefore, $b$ may be the G.M. between $a$ and $c$.
