Question:medium

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

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

Updated On: Apr 7, 2026

3
4
5
None of the above

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The Correct Option is A

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Questions Asked in MET exam

If \(f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2x - 1, & 1<x \end{cases}\), then

MET - 2019
Continuity

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