Let f(x) = ⁻¹ x and g(x) = x^4 + ax^2 + 2 3x^4 + 2x^2 + 2 . If the domain of the function f g is R , then the number of integral values of a is

Step-by-step solution

The domain of f(g(x)) = ⁻¹(g(x)) requires -1 g(x) 1 . Given that the domain is R , this inequality must hold for all x R . The denominator of g(x) is 3x^4 + 2x^2 + 2 . Let t = x^2 0 . The denominator is 3t^2 + 2t + 2 , which is strictly positive for all t 0 (and its discriminant is 4 - 24 = -20 Thus, we can multiply the inequality by the denominator without changing the sign: -3x^4 - 2x^2 - 2 x^4 + ax^2 + 2 3x^4 + 2x^2 + 2 This gives two inequalities that must hold for all x R : 1) x^4 + ax^2 + 2 3x^4 + 2x^2 + 2 2)