If the function f given by f x = x 3 - 3 a - 2 x 2 + 3 a x + 7 , for some a ∈ R is increasing in 0 , 1 and decreasing in 1 , 5 , then a root of the equation, f x - 14 x - 1 2 = 0 , x ≠ 1 is :

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At x = 1 , f ′ x = 0 ∴ f ′ x = 3 x 2 - 6 a - 2 x + 3 a Now, f ' 1 = 0 ⇒ 3 - 6 a - 2 + 3 a = 0 ⇒ a = 5 ∴ f x - 14 x - 1 2 = 0 ⇒ x 3 - 9 x 2 + 15 x - 7 x - 1 2 = 0 ⇒ x - 7 x 2 - 2 x + 1 x - 1 2 = 0 ⇒ x - 7 = 0 ⇒ x = 7

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