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
Number of points of extrema of f (x) for a given value of α is
- 0
- 1
- 2
- 3
Answer
(D) 3
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