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
Absolute difference between local maximum and local minimum values of f (x) in terms of α is
- \(\frac{4}{9}\left(\alpha+\frac{1}{\alpha}\right)^{3}\)
- \(\frac{2}{9}\left(\alpha+\frac{1}{\alpha}\right)^{3}\)
- \(\left(\alpha+\frac{1}{\alpha}\right)^{3}\)
- independent of α
Answer
(A) 4 9 ( + 1 )^ 3
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