Let the function f(x) = 3 _e(x-1) - x^2 2 + 3x be defined for x > 1 . If the equation f(x) = k has exactly two distinct real roots in the interval (1, 5) , then the number of integral values of k is (Given _e 2 0.693 , _e 3 1.098 )

Step-by-step solution

Given f(x) = 3 _e(x-1) - x^2 2 + 3x Differentiating with respect to x , we get: f'(x) = 3 x-1 - x + 3 = 3 - x(x-1) + 3(x-1) x-1 = 4x - x^2 x-1 = x(4-x) x-1 For x (1, 5) , f'(x) = 0 at x = 4 . For 1 0 , so f(x) is strictly increasing. For 4 Thus, f(x) has a local maximum at x = 4 . Maximum value f(4) = 3 _e(3) - 16 2 + 12 = 4 + 3 _e(3) 4 + 3(1.098) = 7.294 Now, let us evaluate the function at the boundaries of the interval (1, 5) . As x 1^+ , f(x) - . At x = 5 , f(5) = 3 _e(4) - 25 2 + 15 = 2.5 + 6 _e(2) 2.5 + 6(0.6