Let f x = − x 3 + b 3 − b 2 + b − 1 b 2 + 3 b + 2 ; 0 ≤ x < 1 2 x - 3 ; 1 ≤ x ≤ 3 . Find all possible real values of b such that f x has the smallest value at x = 1 .

We have, f x = − x 3 + b 3 − b 2 + b − 1 b 2 + 3 b + 2 ; 0 ≤ x < 1 2 x - 3 ; 1 ≤ x ≤ 3 For f x to have smallest value at x = 1 , f x must be decreasing in 0 , 1 and increasing in 1 , 3 . Here, f 1 = - 1 is the smallest value at x = 1 . So, lim x → 1 f x = lim x → 1 − x 3 + b 3 − b 2 + b − 1 b 2 + 3 b + 2 ⇒ lim x → 1 f x = - 1 + b 3 − b 2 + b − 1 b 2 + 3 b + 2 In order this value is not less than - 1 , we must have b 3 − b 2 + b − 1 b 2 + 3 b + 2 ≥ 0 ⇒ b 2 b − 1 + b − 1 b + 1 b + 2 ≥ 0 ⇒ b 2 + 1 b − 1 b 2 + 3 b + 2 ≥ 0 ⇒ b − 1 b + 1 b + 2 ≥ 0 ∵ b 2 + 1 > 0 ∀ b ∈ R ⇒ b ∈ − 2 , − 1 ∪ [ 1 , ∞ )

A cubic f ( x ) vanishes at x = - 2 & has relative minimum/maximum at x = - 1 and x = 1 3 If ∫ - 1 1 f x d x = 14 3 find the cubic f ( x ) .

Since, f x is a cubic polynomial, so f ' x must be s quadratic polynomial. Since, f x has maxima-minima at x = - 1 , 1 3 , therefore f ' - 1 = 0 and f ' 1 3 = 0 Let f ' ( x ) = k x + 1 3 x - 1 ⇒ f ' ( x ) = k 3 x 2 + 2 x - 1 ⇒ ∫ f ' ( x ) d x = k ∫ 3 x 2 + 2 x - 1 d x ⇒ f x = k x 3 + x 2 - x + C Now, f - 2 = k - 8 + 4 + 2 + C = 0 ⇒ C = 2 ∴ f x = k x 3 + x 2 - x + C Now, ∫ - 1 1 f x d x = 14 3 ⇒ k ∫ - 1 1 x 3 + x 2 - x + 2 d x = 14 3 ⇒ k ∫ - 1 1 x 2 + 2 d x = 14 3 ∵ ∫ - a a f x d x = 2 ∫ 0 a f x d x if f - x = f x 0 if f - x = - f x ⇒ 2 k ∫ 0 1 x 2 + 2 d x = 14 3 ⇒ 2 k x 3 3 + 2 x 0 1 = 14 3 ⇒ k 1 3 + 2 = 7 3 ⇒ 7 k 3 = 7 3 ⇒ k = 1 ∴ f x = x 3 + x 2 - x + 2