If a, b and c are the sides of a A B C such that x^2-2(a+b+c) x+3 λ(a b+b c+c a)=0 has real roots, then

Correct answer: A. λ < (4)/(3) Since, roots are real, D ≥ 0 array ll & 4(a+b+c)^2-12 λ(a b+b c+c a) ≥ 0 & (a+b+c)^2 ≥ 3 λ(a b+b c+c a) & a²+b²+c² ≥(a b+b c+c a)(3 λ-2) & 3 λ-2 ≤ (a^2+b^2+c^2)/(a b+b c+c a) where, a^2+b² ≥ 2 a b, b²+c² ≥ 2 b c and c^2+a² ≥ 2 c a & a²+b²+c² ≥(a b+b c+c a) & (a^2+b^2+c^2)/(a b+b c+c a) ≥1 array Also, A=(b^2+c^2-a^2)/(2 b c) 1 b²+c²-a² < 2 b c Similarly, c^2+a^2-b^2 2 c a and a^2+b^2-c^2 2 a b a²+b²+c² < 2(a b+b c+c a) or (a^2+b^2+c^2)/(a b+b c+c a) < 2 From Eqs. (i), (ii) and (iii) array lrl & 3 λ-2 & < 2 & λ < (4)/(3) array

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Properties of Triangles

JEE Advanced 2006 · Mathematics · Properties of Triangles

If a, b and c are the sides of a A B C such that x^2-2(a+b+c) x+3...

a, b

and

c

are the sides of a

\triangle A B C

such that

x^2-2(a+b+c) x+3 \lambda(a b+b c+c a)=0

has real roots, then

Options

A $\lambda < \frac{4}{3}$
B $\lambda>\frac{5}{3}$
C $\lambda \in\left(\frac{4}{3}, \frac{5}{3}\right)$
D $\lambda \in\left(\frac{1}{3}, \frac{5}{3}\right)$

Correct answer

A. $\lambda < \frac{4}{3}$

Step-by-step solution

Since, roots are real,

D \geq 0

\begin{array}{ll} \Rightarrow& 4(a+b+c)^2-12 \lambda(a b+b c+c a) \geq 0 \\ \Rightarrow & (a+b+c)^2 \geq 3 \lambda(a b+b c+c a) \\ \Rightarrow & a²+b²+c² \geq(a b+b c+c a)(3 \lambda-2) \\ \Rightarrow & 3 \lambda-2 \leq \frac{a^2+b^2+c^2}{a b+b c+c a}\\ \text{ where, }a^2+b² \geq 2 a b, b²+c² \geq 2 b c \text{ and }c^2+a² \geq 2 c a \\ \Rightarrow & a²+b²+c² \geq(a b+b c+c a) \\ \therefore & \frac{a^2+b^2+c^2}{a b+b c+c a} \geq1 \end{array}

Also,

\quad \cos A=\frac{b^2+c^2-a^2}{2 b c} \lt 1

\Rightarrow \quad b²+c²-a² < 2 b c Similarly,

c^2+a^2-b^2 \lt 2 c a

and

a^2+b^2-c^2 \lt 2 a b

\Rightarrow a²+b²+c² < 2(a b+b c+c a)

\text{ or } \frac{a^2+b^2+c^2}{a b+b c+c a}

< 2 From Eqs. (i), (ii) and (iii)

\begin{array}{lrl} \therefore& 3 \lambda-2 & < 2 \\ \Rightarrow & \lambda < \frac{4}{3} \end{array}

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