First slide

Theory of equations

Question

lf a, b and c are the sides of $△ ABC$ such that $a \neq b \neq c and x^{2} - 2 (a + b + c) x + 3 λ$ $(ab + bc + ca) = 0$ has real roots, then

Difficult

A

$λ < \frac{4}{3}$
B

$λ > \frac{5}{3}$
C

$λ \in (\frac{4}{3}, \frac{5}{3})$
D

$λ \in (\frac{1}{3}, \frac{5}{3})$

Solution

since' $a, b and c are the sides of a △ ABC .$
$\begin{array}{lr} ∴ | a - b | < c \\ \Rightarrow a^{2} + b^{2} - 2 ab < c^{2} \\ Similarly, b^{2} + c^{2} - 2 bc < a^{2} \\ and c^{2} + a^{2} - 2 ca < b^{2} \end{array} On adding, we get \begin{aligned} (c^{2} + b^{2} + c^{2}) < 2 (ab + bc + ca) \\ \Rightarrow \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} < 2 . . . . . . . . . (i) \\ Also, D \geq 0 \Rightarrow (a + b + c)^{2} - 3 λ (ab + bc + ca) \geq 0 \\ \Rightarrow \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} \geq 3 λ - 2 . . . . . . . . . . . (ii) \end{aligned}$
From Eqs. (i) and (ii),
$3 λ - 2 < 2 \Rightarrow λ < \frac{4}{3}$

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