Let α = ∑ k = 1 ∞ sin 2 k π 6 . Let g : 0 , 1 → ℝ be the function defined by g x = 2 a x + 2 a 1 - x . Then, which of the following statements is/are TRUE?

Step-by-step solution

Given, α = ∑ k = 1 ∞ sin 2 k π 6 and g x = 2 a x + 2 a 1 - x . Now solving, α = ∑ k = 1 ∞ 1 2 2 k = ∑ k = 1 ∞ 1 4 k = 1 4 1 - 1 4 = 1 3 Now putting the value of α in g x we get, g x = 2 x 3 + 2 1 - x 3 Now, g ' x = ln 2 3 2 2 x 3 - 2 1 3 2 x 3 Now finding the critical point by g ' x = 0 ⇒ x = 1 2 And, derivative changes sign from negative to positive at x = 1 2 , hence x = 1 2 is point of local minimum as well as absolute minimum of g x for x &#8