The sum of all the integral values of p such that the equation 3 ^2 x + 12 x - 3 = p , x R , has at least one solution, is:

Step-by-step solution

Given equation: 3 ^2 x + 12 x - 3 = p Substituting ^2 x = 1 - ^2 x : 3(1 - ^2 x) + 12 x - 3 = p -3 ^2 x + 12 x = p Let x = t . Since x R , t [-1, 1] . The equation becomes p = -3t^2 + 12t . Let f(t) = -3t^2 + 12t . Differentiating with respect to t : f'(t) = -6t + 12 = 6(2 - t) For t [-1, 1] , f'(t) > 0 , which means f(t) is strictly increasing in the interval [-1, 1] . The minimum value of f(t) occurs at t = -1 : f(-1) = -3(-1)^2 + 12(-1) = -15 The maximum value of f(t) occurs at t = 1 : f(1) = -3(1)^2 + 12(1) = 9