Question

If$a,b$and $c$are the sides of $∆ABC$ such that $a\ne b\ne c$and $x^2-2(a+b+c)x+3\lambda (ab+bc+ca)=0$ has real roots, then

• A. $\lambda <(4/3)$
• B. $\lambda >(5/3)$
• C. $\lambda \in (4/3,5/3)$
• D. $\lambda \in (1/3,5/3)$

Answer 1

The correct option is A

$\lambda <(4/3)$

Step1. The given quadratic equation :

$x^2-2(a+b+c)x+3\lambda (ab+bc+ca)=0$

The roots are real, then the discriminant $D\ge 0$

$$\begin{gathered} 4{(a+b+c)}^2–12\lambda (ab+bc+ca)\ge 0 \\ \Rightarrow {(a+b+c)}^2\ge 3\lambda (ab+bc+ca) \\ \Rightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge 3\lambda (ab+bc+ca) \\ \Rightarrow a^2+b^2+c^2\ge (ab+bc+ca)(3\lambda –2) \\ \Rightarrow 3\lambda –2\le \frac{(a^2+b^2+c^2)}{(ab+bc+ca)}.............\left(i\right) \end{gathered}$$

Step2. Find the condition of $\mathit{\lambda }$:

We know that

$$\begin{gathered} \Rightarrow cosA<1 \\ \Rightarrow \frac{({\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2)}{2\mathrm{bc}}<1 \\ \Rightarrow {\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2<2\mathrm{bc} \end{gathered}$$$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{}\mathbf{=}\frac{\mathbf{(}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}}{\mathbf{2}\mathbf{b}\mathbf{c}}$

Similarly, $c^2+a^2–b^2<2ca;a^2+b^2–c^2<2ab$

$$\begin{gathered} \Rightarrow a^2+b^2+c^2<2(ab+bc+ca) \\ \Rightarrow \frac{(a^2+b^2+c^2)}{(ab+bc+ca)}<2..........\left(ii\right) \end{gathered}$$

From equations (i) and (ii), $3\lambda –2<2$

$\Rightarrow \lambda <4/3$

Hence, the correct option is (A).