The slope of the tangent to a curve y = f(x) at any point x is given by dy dx = 12 x^2 + 2 - a . If the curve is non-decreasing for all x [1, 2] and A is the maximum possible value of a , then the value of 10 m_N , where m_N is the slope of the normal to the curve at x = 6 when a = A , is

Step-by-step solution

For the curve to be non-decreasing on [1, 2] , we must have dy dx 0 for all x [1, 2] . 12 x^2 + 2 - a 0 a 12 x^2 + 2 This inequality must hold for all x [1, 2] , which means a must be less than or equal to the minimum value of g(x) = 12 x^2 + 2 on this interval. Since x^2 + 2 is increasing for x > 0 , g(x) is a decreasing function on [1, 2] . Thus, the minimum value of g(x) occurs at x = 2 : g(2) = 12 2^2 + 2 = 12 6 = 2 Therefore, a 2 , and the maximum possible value is A = 2 . Substitute a = 2 into the derivative