Let a, b, c, p, q be real numbers. Suppose α, β are the roots of the equation x² + 2px + q = 0 and α, 1/β are the roots of the equation ax² + 2bx + c = 0, where β² ∉ {-1, 0, 1} Statement-I : (p² - q)(b² - ac) ≥ 0 and Statement-II : b ≠ pa or c ≠ qa

Let a, b, c, p, q be real numbers. Suppose α, β are the roots of the equation x² + 2px + q = 0 and α, 1/β are the roots of the equation ax² + 2bx + c = 0, where β² ∉ {-1, 0, 1} Statement-I : (p² - q)(b² - ac) ≥ 0 and Statement-II : b ≠ pa or c ≠ qa α, β roots of x² + 2px + q = 0 α, 1/β roots of ax² + 2bx + c = 0 β² ∉ -1, 0, 1 Statement-I: (p² - q)(b² - ac) ≥ 0 Statement-II: b ≠ pa or c ≠ qa Choose the correct option. Plan: Vieta ⇒ p²-q, b²-ac ⇒ Statement-I then test b=pa and c=qa ⇒ Statement-II α + β = -2p, αβ = q α + 1/β = -2b/a, α/β = c/a If α ∉ R: β = ᾱ and 1/β = ᾱ ⇒ β = 1/β ⇒ β² = 1, contradiction ⇒ α, β ∈ R p = -(α+β)/2, q = αβ p² - q = ((α+β)²)/4 - αβ = ((α-β)²)/4 ≥ 0 b = -a/2(α + 1/β), c = a α/β b² - ac = a²/4(α - 1/β)² ≥ 0 (p²-q)(b²-ac) = a²/16(α-β)²(α-1/β)² ≥ 0 ⇒ Statement-I is TRUE b = pa ⇔ α + 1/β = α + β ⇔ β² = 1 Given β² ≠ 1 ⇒ b ≠ pa c = qa ⇔ α/β = αβ ⇔ α(1/β - β) = 0 ⇔ α = 0 or β² = 1 β² = 1 ⇔ the two equations coincide Statement-II rules this out ⇒ α,β real ⇒ Statement-I Statement-I TRUE, Statement-II TRUE, and II explains I ⇒ Option (1)

Let $a$ , $b$ , $c$ , $p$ , $q$ be real numbers. Suppose $\alpha$ , $\beta$ are the roots of the equation $x^{2} + 2px + q = 0$ and $\alpha$ , $\frac{1}{\beta}$ are the roots of the equation $ax^{2} + 2bx + c = 0$ , where $\beta^{2} \notin \{-1, 0, 1\}$ Statement-I : $(p^{2} - q)(b^{2} - ac) \geq 0$ and Statement-II : $b \neq pa$ or $c \neq qa$

Answer: B. statement-I is True, statement-II is True; statement-II is NOT a correct explanation for statement-I.

Complex Numbers and Quadratic Equations Let

a

b

c

p

q

be real numbers. Suppose

\alpha

\beta

are the roots of the equation

x^{2} + 2px + q = 0

and

\alpha

\frac{1}{\beta}

are the roots of the equation

ax^{2} + 2bx + c = 0

, where

\beta^{2} \notin \{-1, 0, 1\}

Statement-I :

(p^{2} - q)(b^{2} - ac) \geq 0

and Statement-II :

b \neq pa

c \neq qa

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Written solution

$\alpha, \beta \text{ roots of } x^{2} + 2px + q = 0$ $\alpha, \tfrac{1}{\beta} \text{ roots of } ax^{2} + 2bx + c = 0$ $\beta^{2} \notin \{-1,\, 0,\, 1\}$ $\text{Statement-I: } (p^{2} - q)(b^{2} - ac) \geq 0$ $\text{Statement-II: } b \neq pa \ \text{ or } \ c \neq qa$ $\text{Choose the correct option.}$ $\text{Plan: Vieta} \Rightarrow p^{2}-q,\ b^{2}-ac \Rightarrow \text{Statement-I}$ $\text{then test } b=pa \text{ and } c=qa \Rightarrow \text{Statement-II}$ $\alpha + \beta = -2p, \qquad \alpha\beta = q$ $\alpha + \dfrac{1}{\beta} = -\dfrac{2b}{a}, \qquad \dfrac{\alpha}{\beta} = \dfrac{c}{a}$ $\text{If } \alpha \notin \mathbb{R}: \ \beta = \bar{\alpha} \text{ and } \tfrac{1}{\beta} = \bar{\alpha}$ $\Rightarrow \beta = \tfrac{1}{\beta} \Rightarrow \beta^{2} = 1, \text{ contradiction}$ $\Rightarrow \alpha, \beta \in \mathbb{R}$ $p = -\dfrac{\alpha+\beta}{2}, \qquad q = \alpha\beta$ $p^{2} - q = \dfrac{(\alpha+\beta)^{2}}{4} - \alpha\beta = \dfrac{(\alpha-\beta)^{2}}{4} \geq 0$ $b = -\dfrac{a}{2}\!\left(\alpha + \dfrac{1}{\beta}\right), \qquad c = a\,\dfrac{\alpha}{\beta}$ $b^{2} - ac = \dfrac{a^{2}}{4}\!\left(\alpha - \dfrac{1}{\beta}\right)^{2} \geq 0$ $(p^{2}-q)(b^{2}-ac) = \dfrac{a^{2}}{16}(\alpha-\beta)^{2}\!\left(\alpha-\dfrac{1}{\beta}\right)^{2} \geq 0$ $\Rightarrow \text{Statement-I is TRUE}$ $b = pa \ \Leftrightarrow\ \alpha + \dfrac{1}{\beta} = \alpha + \beta \ \Leftrightarrow\ \beta^{2} = 1$ $\text{Given } \beta^{2} \neq 1 \ \Rightarrow\ b \neq pa$ $c = qa \ \Leftrightarrow\ \dfrac{\alpha}{\beta} = \alpha\beta \ \Leftrightarrow\ \alpha\!\left(\dfrac{1}{\beta} - \beta\right) = 0$ $\Leftrightarrow \alpha = 0 \ \text{ or } \ \beta^{2} = 1$ $\beta^{2} = 1 \ \Leftrightarrow\ \text{the two equations coincide}$ $\text{Statement-II rules this out} \Rightarrow \alpha,\beta \text{ real} \Rightarrow \text{Statement-I}$ $\text{Statement-I TRUE, Statement-II TRUE, and II explains I}$ $\Rightarrow \text{Option (1)}$

Written solution

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