Let f : ℝ → ℝ be a function defined by f x = 4 x 4 x + 2 and M = ∫ f a f 1 − a x sin 4 x 1 − x d x , N = ∫ f a f 1 − a sin 4 x 1 − x d x ; a ≠ 1 2 . If α M = β N , α , β ∈ ℕ , then the least value of α 2 + β 2 is equal to ______

Step-by-step solution

Given, f x = 4 x 4 x + 2 ⇒ f 1 - x = 4 1 - x 4 1 - x + 2 ⇒ f 1 - x = 4 4 + 2 · 4 x = 2 2 + 4 x ⇒ f x + f 1 − x = 1 Now, solving M = ∫ f a f 1 − a x · sin 4 x 1 − x d x Now, using the formula ∫ a b f x d x = ∫ a b f a + b - x d x we get, ⇒ M = ∫ f a f 1 − a f a + f 1 - a - x · sin 4 f a + f 1 - a - x 1 − f a + f 1 - a - x d x ⇒ M = ∫ f a f 1 − a 1 - x · sin 4 1 - x x d x ⇒ M = ∫ f a f 1 − a sin 4 1 - x x d x - ∫ f a f 1 − a x · sin 4 1 - x x d x ⇒ M = N − M ⇒ 2 M = N So, on comparing we get, α = 2 , β = 1 Hence, α 2