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JEE Main202117 Mar 2021Evening ShiftMathematicsApplication of DerivativesActual

Let f : [ - 1 , 1 ] → R be defined as f ( x ) = a x 2 + b x + c for all x ∈ [ - 1 , 1 ] , where a , b , c ∈ R such that f ( - 1 ) = 2 , f ' ( - 1 ) = 1 and for x ∈ ( - 1 , 1 ) the maximum value of f " ( x ) is 1 2 . If f ( x ) ≤ α , x ∈ - 1 , 1 , then the least value of α is equal to

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Step-by-step solution

Given f : [ - 1 ,   1 ] → R , f ( x ) = a x 2 + b x + c ,   f ' ( x ) = 2 a x + b and f ' ' ( x ) = 2 a ⇒ f ( - 1 ) = a - b + c = 2       . . . 1 , ⇒ f ' ( - 1 ) = - 2 a + b = 1       . . . 2 and f ' ' ( x ) = 2 a Given the maximum value of f ' ' ( x ) = 2 a = 1 2 ⇒ a = 1 4 , From the equations 1 and 2 we get b = 3 2 and c = 13 4 . ∴   f ( x ) = x 2 4 + 3 x 2 + 13 4 We know that, the vertex of the quadratic A

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