Let a, b, c be the sides of a triangle. No two of them are equal and $\lambda \in \mathbb{R}$. If the roots of the equation $x^{2}+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ are real, then

(C) $\lambda \in\left(\frac{1}{3}, \frac{5}{3}\right)$

(D) $\lambda \in\left(\frac{4}{3}, \frac{5}{3}\right)$

JEE Advanced · 2006 · 09 Apr 2006 · Quadratic Equations and Inequalities

Let a, b, c be the sides of a triangle. No two of them are equal and

λ \in R

. If the roots of the equation

x^{2} + 2 (a + b + c) x + 3 λ (ab + b c + c a) = 0

are real, then

Answer: (A)

For real roots, the discriminant

D \geq 0

4 (a + b + c)^{2} - 12 λ (ab + b c + c a) \geq 0

λ \leq \frac{( a + b + c ) ^{2}}{3 ( ab + b c + c a )} = \frac{a ^{2} + b ^{2} + c ^{2} + 2 ( ab + b c + c a )}{3 ( ab + b c + c a )} = \frac{a ^{2} + b ^{2} + c ^{2}}{3 ( ab + b c + c a )} + \frac{2}{3}

For a triangle with unequal sides, the triangle inequalities give:

Adding these:

a^{2} + b^{2} + c^{2} < 2 (ab + b c + c a)

, so

\frac{a ^{2} + b ^{2} + c ^{2}}{ab + b c + c a} < 2

.

Therefore,

λ < \frac{2}{3} + \frac{2}{3} = \frac{4}{3}

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option