If a > 0 ,  b > 0 and a 2 + b = 2 , then the maximum value of the term independent of x in the expansion of a x 1 6 + b x - 1 3 9 is

Step-by-step solution

Let k + 1 t h term be independent of x T k + 1 = 9 C k a x 1 6 9 - k b x - 1 3 k = 9 C k   a 9 - k   b k   x 9 - k 6 - k 3 For this to be independent of x , 9 - k 6 - k 3 = 0 ⇒ 9 6 = k 6 + k 3 = k 2 ⇒ k = 3 . ⇒ T k + 1 = 9 C 3 a 6 b 3 = 84 a 2 b 6 Using A . M . ≥ G . M . we get a 2 + b 2 ≥ a 2 b ⇒ a 2 b ≤ 1 ⇒ T k + 1 ≤ 84