< 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. Tags and topics: JEE MAIN,JEE Mains,KCET,NCERT,Quadratic Equations,Relation Between Roots and Coefficients of Quadratic Equations,Sequences and Series, Class 11 Mathematics,Class 12 Mathematics,JEE.">
< 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. Tags and topics: JEE MAIN,JEE Mains,KCET,NCERT,Quadratic Equations,Relation Between Roots and Coefficients of Quadratic Equations,Sequences and Series, Class 11 Mathematics,Class 12 Mathematics,JEE.">
< 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. Tags and topics: JEE MAIN,JEE Mains,KCET,NCERT,Quadratic Equations,Relation Between Roots and Coefficients of Quadratic Equations,Sequences and Series, Class 11 Mathematics,Class 12 Mathematics,JEE.">
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 $c$
only (II) is true
Both (I) and (II) are true
only (I) is true
Neither (I) nor (II) is true
Solution
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Step-by-Step Solution
Step 1: Write down the given quadratic equation
The given quadratic is
$ (a + b - 2c)x^2 \;+\; (b + c - 2a)x \;+\; (c + a - 2b) \;=\; 0 $
with the condition
$ 0 < c < b < a $.
Step 2: Check if $x = 1$ is a root
Substitute $ x = 1 $ into the equation:
$ f(1) = (a + b - 2c)\cdot 1^2 + (b + c - 2a)\cdot 1 + (c + a - 2b). $
Simplify this expression:
$ f(1) = (a + b - 2c) + (b + c - 2a) + (c + a - 2b). $
Combine like terms:
$ (a - 2a + a) + (b + b - 2b) + (c - 2c + c) = 0. $
The result is indeed $0$, so $x = 1$ is one root.
Step 3: Identify the other root using Vieta’s formulas
Let the other root be $ \alpha \neq 1 $. For a general quadratic equation
$ px^2 + qx + r = 0 $ with roots $r_1$ and $r_2$, Vieta’s formulas tell us:
$ r_1 + r_2 = -\frac{q}{p} $
$ r_1 \, r_2 = \frac{r}{p} $
In our case, with $1$ and $\alpha$ as roots:
$ 1 \cdot \alpha = \frac{\text{constant term}}{\text{coefficient of } x^2}
= \frac{c + a - 2b}{a + b - 2c}. $
Hence,
$ \alpha = \frac{c + a - 2b}{a + b - 2c}. $
Step 4: Statement (I) – If $ \alpha \in (-1, 0) $, then $b$ cannot be the geometric mean of $a$ and $c$
Statement (I) assumes $-1 < \alpha < 0$. We have:
$ \alpha = \dfrac{c + a - 2b}{a + b - 2c}. $
For $\alpha$ to lie in $(-1, 0)$:
The numerator $(c + a - 2b)$ and the denominator $(a + b - 2c)$ must be of opposite signs to ensure $\alpha < 0$.
We must also have
$ \bigl|c + a - 2b\bigr| < \bigl|a + b - 2c\bigr| $
to ensure $|\alpha| < 1$.
Detailed analysis of these inequalities (along with $0 < c < b < a$) shows that $b$ ends up being forced above the midpoint $(a + c)/2$, making it impossible for $b$ to also satisfy $ b^2 = ac$. Hence, $b$ cannot be the geometric mean of $a$ and $c$. Therefore, Statement (I) is true.
Step 5: Statement (II) – If $ \alpha \in (0, 1) $, then $b$ may be the geometric mean of $a$ and $c$
Here, $ 0 < \alpha < 1 $. Thus:
$ 0 < \dfrac{c + a - 2b}{a + b - 2c} < 1 $.
This implies:
The numerator $(c + a - 2b)$ and the denominator $(a + b - 2c)$ must have the same sign (so the fraction is positive).
Again,
$ \bigl|c + a - 2b\bigr| < \bigl|a + b - 2c\bigr| $
ensures $ |\alpha| < 1$.
In such a situation, it can happen that $b$ fits neatly between $c$ and $a$ in a manner that satisfies $ b^2 = ac $, making $b$ the geometric mean. Hence, $b$ may be the geometric mean of $a$ and $c$. Statement (II) is therefore true.
Step 6: Final conclusion
Both statements (I) and (II) hold true. Thus, the correct answer is “Both (I) and (II) are true.”