(A) Let $f(x) = (a+b-2c)x^2 + (b+c-2a)x + (c+a-2b)$.Sum of coefficients: $f(1) = (a+b-2c) + (b+c-2a) + (c+a-2b) = 0$.Since $f(1) = 0$,$x = 1$ is one r

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(A) Let $f(x) = (a+b-2c)x^2 + (b+c-2a)x + (c+a-2b)$.Sum of coefficients: $f(1) = (a+b-2c) + (b+c-2a) + (c+a-2b) = 0$.Since $f(1) = 0$,$x = 1$ is one root of the equation.Let the roots be $1$ and $\alpha$. From the product of roots formula,$1 \cdot \alpha = \frac{c+a-2b}{a+b-2c}$.Thus,$\alpha = \frac{c+a-2b}{a+b-2c}$.Case $(I)$: If $-1 \frac{a+c}{2}$. Since the geometric mean of $a$ and $c$ is $\sqrt{ac}$ and $\sqrt{ac} Case $(II)$: If $0 Therefore,both statements are true.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Mix Examples-Quadratic Equations and Inequations

Difficult

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 roots. 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 $c$.

JEE MAIN 2024 All JEE MAIN

A
Both $(I)$ and $(II)$ are true
B
Neither $(I)$ nor $(II)$ is true
C
Only $(II)$ is true
D
Only $(I)$ is true

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Mix Examples-Quadratic Equations and Inequations

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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 roots. 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 $c$.

Similar Questions

Let $\alpha, \beta$ be the roots of the equation $x^2 - |a|x - |b| = 0$ such that $|\alpha| < |\beta|$. If $|a| < \beta - 1$,then the positive root of $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) - 1 = 0$ is

The curves $y=x^2+9x+20$ and $y=x^2+bx+c$ intersect the $X$-axis at the points $(\alpha_i, 0)$ for $i=1, 2, 3, 4$. If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4$ are such that $|\alpha_1-\alpha_3|=|\alpha_2-\alpha_4|=8$,then the sum of all possible values of $b$ and $c$ is:

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