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:

  1. 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} $



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