Let f ′ x = 192 x 3 2 + sin 4 ⁡ π x for all x ∈ R with f 1 2 = 0 . If m ≤ ∫ 1 2 1 f x d x ≤ M , then the possible values of m and M are

Step-by-step solution

Assuming f ( x ) to more increasing or less increasing 192 x 3 3 f ′ x 192 x 3 2 64 x 3 f ′ x 96 x 3 ∫ 64 x 3 d x ∫ f ′ x d x ∫ 96 x 3 d x 16 x 4 - 1 f x 24 x 4 - 3 2 ∫ 1 2 1 16 x 4 - 1 d x ∫ 1 2 1 f x d x ∫ 1 2 1 24 x 4 - 3 2 d x 26 10 ∫ 1 2 1 f x d x 78 20 Hence option m = 1 , M = 12 is correct.