Let [a, b] be the set of all values of for which the equation 4 ^6 x + 2 ^4 x - 3 ^2 x = has at least one real solution. The value of b - 54a is

Step-by-step solution

Let the given equation be expressed in terms of a single trigonometric ratio. Let t = ^2 x . Since x is real, t [0, 1] . We know ^2 x = 1 - t and ^4 x = (1 - t)^2 . Substituting these into the given equation: = 4t^3 + 2(1 - t)^2 - 3(1 - t) = 4t^3 + 2(1 - 2t + t^2) - 3 + 3t = 4t^3 + 2t^2 - t - 1 Let f(t) = 4t^3 + 2t^2 - t - 1 for t [0, 1] . We need to find the range of f(t) . Differentiating with respect to t : f'(t) = 12t^2 + 4t - 1 Setting f'(t) = 0 for critical points: 12t^2 + 6t - 2t - 1 = 0 6t(2t + 1) - 1(2t +