Question

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

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

Answer 1

The correct option is A

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

Since, roots are real, therefore $D>0$
$\Rightarrow 4(a+b+c{)}^2-12\lambda (ab+bc+ca)>0\Rightarrow (a+b+c{)}^2>3\lambda (ab+bc+ca)\Rightarrow a^2+b^2+c^2>(ab+bc+ca)(3\lambda -2)\Rightarrow 3\lambda -2<\frac{a^2+b^2+c^2}{ab+bc+ca}....\left(i\right)$
Also, $cosA=\frac{b^2+c^2-a^2}{2bc}\le 1.$
$\Rightarrow b^2+c^2-a^2\le 2bc$
Similarly, $c^2+a^2-b^2\le 2ca$
and $a^2+b^2-c^2\le 2ab$
$\Rightarrow a^2+b^2+c^2\le 2(ab+bc+ca)\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}\le \mathrm{2.....}\left(ii\right)$
From Eqs. (i)and (ii), we get
$3\lambda -2<2\Rightarrow \lambda <\frac{4}{3}$