Question

Let $a,b,c$ be the sides of a triangle, where $a\ne b\ne c$ and $\lambda \in R.$ If the roots of the equation $x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ are real. Then

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

Answer 1

The correct option is A $\lambda <\frac{4}{3}$

$a,b,c$ are sides of a triangle and $a\ne b\ne c$

$\therefore |a-b|<|c|\Rightarrow a^2+b^2-2ab<c^2$

Similarly, we have $b^2+c^2-2bc<a^2$ and $c^2+a^2-2ca<b^2$

On adding, we get $a^2+b^2+c^2<2(ab+bc+ca)$

$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2$ .........(1)

Since the roots of the given equation are real, therefore

$(a+b+c{)}^2-3\lambda (ab+bc+ca)\ge 0$

$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}>3\lambda -2$ .........(2)

From (1) and (2), we get

$3\lambda -2<2\Rightarrow \lambda <\frac{4}{3}$