Question
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?
JEE Advanced
Mathematics
Trigonometric Ratios & Identities
2022
JEE Advanced 2022 (Paper 2)
JEE Advanced Mathematics Question (2022) — Solution
Options
- A. The minimum value of g x is 2 7 6
- B. The maximum value of g x is 1 + 2 1 3
- C. The function g x attains its maximum at more than one point
- D. The function g x attains its minimum at more than one point
Answer
C. The function g x attains its maximum at more than one point
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 ∈ 0 , 1 Hence, minimum value of g x = g 1 2 = 2 1 6 + 2 1 6 = 2 7 6 ⇒ Option A is correct Now maximum value of g x is either equal to g 0 or g 1 . g 0 = 1 + 2 1 3 g 1 = 2 1 3 + 1 Hence (B) and (C) are also correct.
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