Let f(x) be a quadratic polynomial such that f(– 1) + f(2) = 0 . If one of the roots of f(x) = 0 is 3 , then its other root lies in

Correct answer: A. (- 1, 0) Let f(x) = a x ^ 2 + bx + c be a quadratic equation with roots α, β f(2) + f(- 1) = 0 ⇒ 5a + b + 2c = 0 Also f(3) = 0 ⇒ 9a + 3b + c = 0 ⇒ (a)/(- 5) = (b)/(13) = (c)/(6) Product of roots αβ = (c)/(a) = - (6)/(5) We know α = 3 ⇒ β = - (2)/(5) ∈ (- 1, 0)

JEE Main 20202 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2020 · 2 September · Evening Shift · Mathematics · Quadratic Equation

Let f(x) be a quadratic polynomial such that f(– 1) + f(2) = 0 . If...

Let

f\left(x\right) be a quadratic polynomial such that

f\left(– 1\right) + f\left(2\right) = 0

. If one of the roots of

f\left(x\right) = 0

is

3$, then its other root lies in

Options

A $\left(- 1, 0\right)$
B $\left(1, 3\right)$
C $\left(– 3, – 1\right)$
D $\left(0, 1\right)$

Correct answer

A. $\left(- 1, 0\right)$

Step-by-step solution

Let

f\left(x\right) = a{x}^{2} + bx + c

be a quadratic equation with roots

\alpha, \beta

$f\left(2\right) + f\left(- 1\right) = 0 ⇒ 5a + b + 2c = 0$

Also $f\left(3\right) = 0 ⇒ 9a + 3b + c = 0$

$⇒ \dfrac{a}{- 5} = \dfrac{b}{13} = \dfrac{c}{6}$

Product of roots $\alpha\beta = \dfrac{c}{a} = - \dfrac{6}{5}$

We know $\alpha = 3 ⇒ \beta = - \dfrac{2}{5} \in \left(- 1, 0\right)$

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