Quadratic Equation Question 7

Question 7 - 2024 (31 Jan Shift 1)

For $0<\mathrm{c}<\mathrm{b}<\mathrm{a}$, let $(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^{2}+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root.

Then, among the two statements

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

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

(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, $\mathrm{b}$ cannot be G.M. between $\mathrm{a}$ and $\mathrm{c}$.

If, $0<\alpha<1$

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

$\mathrm{b}>\mathrm{c}$ and $\mathrm{b}<\frac{\mathrm{a}+\mathrm{c}}{2}$

Therefore, $\mathrm{b}$ may be the G.M. between $\mathrm{a}$ and $\mathrm{c}$.