(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 4-2.Quadratic Equations and Inequations 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 $c$.

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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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4-2.Quadratic Equations and Inequations

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Mix Examples-Quadratic Equations and Inequations

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If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots$ to $12$ terms $=$

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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

Given the equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$ has roots $\sin \theta$ and $\cos \theta$ $(0 < \theta < \frac{\pi}{2})$,then the maximum value of $(a + \sin \theta)$ is-

The probability of selecting integers $a \in [-5, 30]$ such that $x^{2}+2(a+4)x-5a+64 > 0$ for all $x \in \mathbb{R}$ is:

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to

Let $p, q, r \in R^+$. If $27pqr \geq (p + q + r)^3$ and $3p + 4q + 5r = 12$,then find the value of $p^3 + q^4 + r^5$.

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots$ to $12$ terms $=$

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