Question

The value of \(a \ge b\) for which the sum of the cubes of the roots of \(x^2 - (a - 2)x + (a - 3) = 0\) assumes the least value, is

• A. 3
• B. 4
• C. 5
• D. None of the above

Hint:

Sum of cubes = \((\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let roots be \(\alpha, \beta\), then \(\alpha + \beta = a - 2\), \(\alpha\beta = a - 3\).
Step 2: Detailed Explanation:
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
= \((a - 2)^3 - 3(a - 3)(a - 2)\)
= \((a - 2)[(a - 2)^2 - 3(a - 3)]\)
= \((a - 2)[a^2 - 4a + 4 - 3a + 9]\)
= \((a - 2)(a^2 - 7a + 13)\)
Minimize: differentiate and set = 0 gives \(a = 3\)
Step 3: Final Answer:
\(a = 3\).