Theory of Equations 1 Question 23
24. If $a, b, c$ 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
$(2006,3 M)$
(a) $\lambda<\frac{4}{3}$
(b) $\lambda>\frac{5}{3}$
(c) $\lambda \in \frac{4}{3}, \frac{5}{3}$
(d) $\lambda \in \frac{1}{3}, \frac{5}{3}$
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Answer:
Correct Answer: 24. (a)
Solution:
- Since, roots are real, therefore $D \geq 0$
$ \begin{array}{lc} \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 & a^{2}+b^{2}+c^{2} \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 { Also, } & \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}<1 \end{array} $
$ \begin{array}{ll} \Rightarrow & b^{2}+c^{2}-a^{2}<2 b c \\ \text { Similarly, } & c^{2}+a^{2}-b^{2}<2 c a \\ \text { and } & a^{2}+b^{2}-c^{2}<2 a b \\ \Rightarrow & a^{2}+b^{2}+c^{2}<2(a b+b c+c a) \\ \Rightarrow & \frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}<2 \end{array} $
From Eqs. (i) and (ii), we get
$ 3 \lambda-2<2 \Rightarrow \lambda<\frac{4}{3} $
