The domain of the function f ( x ) = sin − 1 x + 5 x 2 + 1 is − ∞ , − a ∪ a , ∞ , then a is equal to

Step-by-step solution

f x = sin − 1 x + 5 x 2 + 1 ⇒ - 1 ≤ x + 5 x 2 + 1 ≤ 1 ,   ∵ x ∈ - 1 , 1   for   sin - 1 x to exist. Case 1 - x 2 - 1 ≤ x + 5 ⇒ x 2 + 1 + x + 5 ≥ 0 , true for all x ∈ R . Case 2 x + 5 ≤ x 2 + 1 ⇒ x 2 - x - 4 ≥ 0 Let x = t ⇒ t 2 - t - 4 ≥ 0 ⇒ x + 17 - 1 2 x - 17 + 1 2 ≥ 0 ⇒ x ∈ - ∞ , - 17 + 1 2 ∪ 17 + 1 2 , ∞ ∴ a = 1 + 17 2