Consider a function f( x )= ( - 1 - x ) (4 – 3x 2 ) where ‘α’ is a positive parameter…
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
Consider a function \(f(\mathrm{x})=\left(\alpha-\frac{1}{\alpha}-\mathrm{x}\right)\)(4 – 3x2) where ‘α’ is a positive parameter
Least possible value of the absolute difference between local maximum and local minimum values of f (x) is
- \(\frac{\mathbf{2}}{9}\)
- \(\frac{6}{9}\)
- \(\frac{8}{9}\)
- \(\frac{1}{9}\)
Answer
(A) 2 9
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