Let λ ≠ 0 be in R . If α and β are the roots of the equation, x ^ 2 - x + 2λ = 0 and α and γ are the roots of the equation, 3 x ^ 2 - 10x + 27λ = 0, then (βγ)/(λ) is equal to:

Correct answer: B. 18 Satisfying α we have, 3 α ^ 2 - 10α + 27λ = 0 (i) and 3 α ^ 2 - 3α + 6λ = 0 (ii) Subtract - 7α + 21λ = 0 3λ = α by (ii) , 9 λ ^ 2 - 3λ + 2λ = 0 ⇒ λ = 0, (1)/(9)butλ ≠ 0 ∴ Given equation are x ^ 2 - x + (2)/(9) = 0 and 3 x ^ 2 - 10x + 3 = 0 ∴ α = (1)/(3), β = (2)/(3), α = (1)/(3), γ = 3 ∴ (βγ)/(λ) = ((2)/(3) · 3)/((1)/(9)) = 18

JEE Main 20204 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2020 · 4 September · Evening Shift · Mathematics · Quadratic Equation

Let λ ≠ 0 be in R . If α and β are the roots of the equation, x ^ 2...

Let

\lambda \ne 0

be in

R

. If

\alpha

and

\beta

are the roots of the equation,

{x}^{2} - x + 2\lambda = 0

and

\alpha

and

\gamma

are the roots of the equation,

3{x}^{2} - 10x + 27\lambda = 0,

then

\dfrac{\beta\gamma}{\lambda}

is equal to:

Options

A $27$
B $18$
C $9$
D $36$

Correct answer

B. $18$

Step-by-step solution

Satisfying $\alpha$ we have,

$3{\alpha}^{2} - 10\alpha + 27\lambda = 0 \ldots \left(i\right) and $ 3{\alpha}^{2} - 3\alpha + 6\lambda = 0 \ldots \left(ii\right) $ Subtract$ - 7\alpha + 21\lambda = 0 $ $ 3\lambda = \alpha $ by \left(ii\right)$ , $9{\lambda}^{2} - 3\lambda + 2\lambda = 0$
$⇒ \lambda = 0, \dfrac{1}{9}\text{but}\lambda \ne 0$
$∴$ Given equation are ${x}^{2} - x + \dfrac{2}{9} = 0$ and 3{x}^{2} - 10x + 3 = 0
$∴ \alpha = \dfrac{1}{3}, \beta = \dfrac{2}{3}, \alpha = \dfrac{1}{3}, \gamma = 3$
$∴ \dfrac{\beta\gamma}{\lambda} = \dfrac{\dfrac{2}{3} \cdot 3}{\dfrac{1}{9}} = 18$

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