JEE Main 2024 Mathematics question
For x ≥ 0, the least value of K, for which 4 + 4, , 16 + 16 are three consecutive terms of an A.P. is equal to:
Answer: (A)
Solution
K = + + 16ˣ + 16⁻ˣ. With u = 4ˣ + 4⁻ˣ ≥ 2 this is u² + 4u − 2, minimum at u = 2: 4 + 8 − 2 = 10.