Question

Let a,b,c be the sides of a triangle. No two of them are equal and $\lambda \in R.$ If the roots of the equation

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

Answer 1

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

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

Given, roots of equation $x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ are real,

So,$D\ge 0$

$\Rightarrow {\left[2(a+b+c)\right]}^2-4\times 1\times 3\lambda (ab+bc+ca)\ge 0$

$\Rightarrow 4{(a+b+c)}^2-12\lambda (ab+bc+ca)\ge 0$

$\Rightarrow 4{(a+b+c)}^2\ge 12\lambda (ab+bc+ca)$

$\Rightarrow \lambda \le \frac{4{(a+b+c)}^2}{12\lambda (ab+bc+ca)}$

$\Rightarrow \lambda \le \frac{4(a^2+b^2+c^2)}{12\lambda (ab+bc+ca)}+\frac{8\lambda (ab+bc+ca)}{12\lambda (ab+bc+ca)}$

$\Rightarrow \lambda \le \frac{4(a^2+b^2+c^2)}{12\lambda (ab+bc+ca)}+\frac{8}{12}$

$\Rightarrow \lambda \le \frac{4(a^2+b^2+c^2)}{12\lambda (ab+bc+ca)}+\frac{2}{3}$ .......$\left(1\right)$

Now, we know that,

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

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

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

On adding,

$a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac<a^2+b^2+c^2$

$a^2+b^2+c^2<2ab+2bc+2ca$

$\Rightarrow dfrac{a}^2+b^2+c^{2}ab+bc+ca<2$

So, eqn$\left(1\right)$ becomes,

$\lambda <\frac{2}{3}+\frac{2}{3}$

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