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

For $0 < c < b < a$, let $(a + b - 2c)x^{2} + (b + c - 2a)x + (c + a - 2b) = 0$ 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

Solution

Given: f(x)=(a+b–2c) x2+(b+c–2a) x+(c+a–2b)

⇒f(1)=a+b–2c+b+c–2a+c+a–2b=0

⇒f(1)=0

Using product of roots,

⇒α·1=c+a-2ba+b-2c

⇒α=c+a-2ba+b-2c

If, -1<α<0

Statement-I:

⇒-1<c+a-2ba+b-2c<0

⇒b+c<2a and b>a+c2

So, $b$ cannot be G.M. between $a$ and $c$ as $A.M \geq G.M$. So, statement I is correct. Statement-II: $\Rightarrow 0 < \alpha < 1$ $\Rightarrow 0 < $\frac{c + a - 2b}{a + b - 2c}$ < 1$ $\Rightarrow $\frac{c + a - 2b}{a + b - 2c}$ > 0$ and $\frac{c + a - 2b}{a + b - 2c} < 1$ $\Rightarrow $\frac{c + a - 2b}{a + b - 2c}$ > 0$ and $\frac{c + a - 2b - a - b + 2c}{a + b - 2c} < 0$ $\Rightarrow $\frac{c + a - 2b}{a + b - 2c}$ > 0$ and $\frac{3c - 3b}{a + b - 2c} < 0$ $\Rightarrow b < \frac{a + c}{2}$ and $b > c$ Therefore, $b$ may be the G.M. between $a$ and $c$ as $A.M \geq G.M$. So, statement II is also correct.

Asked in: JEE Main 2024 (31 Jan Shift 1)

Practice more Quadratic Equation questions on Aicharya