Let \(a, b\) and \(c\) be the sides of a scalane triangle. If \(\lambda\) is a real number such that the…
Let \(a, b\) and \(c\) be the sides of a scalane triangle. If \(\lambda\) is a real number such that the roots of the equation \(x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0\) are real, then the interval in which \(\lambda\) lies is
\(\left(-\infty, \frac{4}{3}\right)\)
\(\left(\frac{5}{3}, \infty\right)\)
\(\left(\frac{1}{3}, \frac{5}{3}\right)\)
\(\left(\frac{4}{3}, \infty\right)\)
Solution
It is given that roots of given quadratic equation \(x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0\) are real, so
\(\begin{gathered}
D \geq 0 \\
\Rightarrow 4(a+b+c)^2-4 \times 3 \lambda(a b+b c+c a) \geq 0 \\
\Rightarrow(a+b+c)^2-3 \lambda(a b+b c+c a) \geq 0 \\
\Rightarrow \quad \lambda \leq \frac{(a+b+c)^2}{3(a b+b c+c a)}
\end{gathered}\)
Now, for scalane triangle.
\(\begin{aligned}
& \because \text { For } \triangle A B C|b-c| < a,|c-a| < b \text { and }|a-b| < c \\
& \Rightarrow(b-c)^2+(c-a)^2+(a-b)^2 < a^2+b^2+c^2 \\
& \Rightarrow a^2+b^2+c^2 < 2(a b+b c+c a) \\
& \Rightarrow a^2+b^2+c^2+2(a b+b c+c a) < 4(a b+b c+c a) \\
& \Rightarrow \frac{(a+b+c)^2}{3(a b+b c+c a)} < \frac{4}{3} \\
& \therefore \lambda < \frac{4}{3}
\end{aligned}\)
Hence, option (1) is correct.