If λ ∈ R is such that the sum of the cubes of the roots of the equation, x^2+(2-λ) x+(10-λ)=0 is minimum, then the magnitude of the difference of the roots of this equation is

Correct answer: B. 2 5 Let, the roots of the equation, x^2+(2-λ) x+(10-λ)=0 are α and β . Also roots of the given equation are λ-2 ± 4-4 λ+λ^2-40+4 λ 2 = λ-2 ± λ^2-36 2 The magnitude of the difference of the roots is | λ^2-36 | So, α^3+β^3=((λ-2)^3)/(4)+(3(λ-2)(λ^2-36))/(4) aligned =((λ-2)(4 λ^2-4 λ-104))/(4) =&(λ-2)(λ^2-λ-26)=f(λ) aligned As f(λ) attains its minimum value at λ=4 . Therefore, the magintude of the difference of the

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Quadratic Equation

If λ ∈ R is such that the sum of the cubes of the roots of the...

\lambda \in \mathrm{R}

is such that the sum of the cubes of the roots of the equation,

x^2+(2-\lambda) x+(10-\lambda)=0

is minimum, then the magnitude of the difference of the roots of this equation is

Options

A20
B $2 \sqrt{5}$
C $2 \sqrt{7}$
D $4 \sqrt{2}$

Correct answer

B. $2 \sqrt{5}$

Step-by-step solution

Let, the roots of the equation,

x^2+(2-\lambda) x+(10-\lambda)=0

are

\alpha and

\beta

. Also roots of the given equation are

\frac{\lambda-2 \pm \sqrt{4-4 \lambda+\lambda^2-40+4 \lambda}}{2}=\frac{\lambda-2 \pm \sqrt{\lambda^2-36}}{2}

The magnitude of the difference of the roots is \left|\sqrt{\lambda^2-36}\right|

So,

\alpha^3+\beta^3=\frac{(\lambda-2)^3}{4}+\frac{3(\lambda-2)\left(\lambda^2-36\right)}{4}

\begin{aligned} =\frac{(\lambda-2)\left(4 \lambda^2-4 \lambda-104\right)}{4}\\ =&(\lambda-2)\left(\lambda^2-\lambda-26\right)=f(\lambda) \end{aligned}

As

f(\lambda)

attains its minimum value at

\lambda=4$. Therefore, the magintude of the difference of the

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