The maximum integral part of positive value of a for which, the least value of 4x 2 – 4 a…
Mathematics · JEE Advanced · NTA Exams — Complex Numbers and Quadratic Equations
The maximum integral part of positive value of a for which, the least value of 4x2 – 4ax + a2 – 2a + 2 on [0, 2] is 3, is
- \(5\)
- \(6\)
- \(8\)
- \(9\)
Answer
(C) 8
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