Question

If \(\lambda\in\mathbf{R}\) is such that the sum of the cubes of the roots of the equation, \(x^2+(2-\lambda)x+(10-\lambda)=0\) is minimum, then the magnitude of the difference of the roots of this equation is :

• A. \(4\sqrt{2}\)
• B. \(2\sqrt{5}\)
• C. \(2\sqrt{7}\)
• D. 20

Hint:

Use \(\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)\), minimise in \(\lambda\), then compute \(|\alpha-\beta|=\sqrt{|(\alpha+\beta)^2-4\alpha\beta|}\).

Answer 1

The Correct Option is B

\(\alpha+\beta=\lambda-2\), \(\alpha\beta=10-\lambda\).
\(S=\alpha^3+\beta^3=(\lambda-2)^3-3(10-\lambda)(\lambda-2)=(\lambda-2)(\lambda^2-\lambda-26)=\lambda^3-3\lambda^2-24\lambda+52\).
\(S'=3\lambda^2-6\lambda-24=3(\lambda-4)(\lambda+2)=0\Rightarrow\lambda=4\) or \(-2\). \(S''=6\lambda-6\) is positive at \(\lambda=4\), so the minimum is at \(\lambda=4\).
The equation becomes \(x^2-2x+6=0\), with roots \(1\pm i\sqrt5\).
Magnitude of the difference \(=|2i\sqrt5|=2\sqrt5\) (equivalently \(\sqrt{|D|}=\sqrt{|4-24|}\)).
Answer: (2) \(2\sqrt{5}\)