Let f: R R be a function defined by f(x) = _m ( 2mx x^2+1 + 2m ) for some real number m > 1 . If the range of f is [1, 2] , then the value of m is

Step-by-step solution

Let the inner function be g(x) = 2mx x^2+1 + 2m . We first find the range of the rational expression h(x) = 2x x^2+1 . Let y = 2x x^2+1 yx^2 - 2x + y = 0 . For real x , the discriminant must be non-negative: 4 - 4y^2 0 y^2 1 -1 y 1 . Thus, the range of h(x) is [-1, 1] . Multiplying by m (since m > 1 , m is positive), the range of 2mx x^2+1 is [-m, m] . Adding 2m , the range of g(x) = 2mx x^2+1 + 2m becomes [-m + 2m, m + 2m] = [m, 3m] . Since m > 1 , the logarithmic function with base m is strictly increasing. There