The number of integral values of k for which the equation ^4 x + 4 ^2 x = k has at least one real solution is:

Step-by-step solution

Given equation: ^4 x + 4 ^2 x = k We know that ^2 x = 1 - ^2 x . Substituting this in the equation, we get: ^4 x + 4(1 - ^2 x) = k ^4 x - 4 ^2 x + 4 = k ( ^2 x - 2)^2 = k Let t = ^2 x . Since 0 ^2 x 1 , we have t [0, 1] . The equation becomes k = (t - 2)^2 . We need to find the range of the function f(t) = (t - 2)^2 for t [0, 1] . At t = 0 , f(0) = (-2)^2 = 4 . At t = 1 , f(1) = (-1)^2 = 1 . Since f(t) is strictly decreasing on [0, 1] , the minimum value is 1 and the maximum value is 4 . Thus, k [1, 4] . The integr