If the quadratic function f (x) = x & # 178; + CX (C is a constant) (1) If the function f (x) is an even function, find the value of C (2) Under the condition of (1), any positive real number m, N, K satisfying m + n = 2K (M is not equal to n) has f (m) + F (n) > TF (k). The value range of real number T is obtained

If the quadratic function f (x) = x & # 178; + CX (C is a constant) (1) If the function f (x) is an even function, find the value of C (2) Under the condition of (1), any positive real number m, N, K satisfying m + n = 2K (M is not equal to n) has f (m) + F (n) > TF (k). The value range of real number T is obtained

1、f(-x)=f(x),x^2+cx=(-x)^2+c(-x),c=0
2. Because f (x) = x ^ 2, f (m) + F (n) > TF (k), that is m ^ 2 + n ^ 2 > TK ^ 2,
T0, n > 0, Mn, so (M + n) ^ 2 = m ^ 2 + n ^ 2 + 2Mn > m ^ 2 + n ^ 2
So t

Function f (x) = CX + 1 (0

You just write out the expression of F (x). By calculation, you can get C = 1 / 2

The function f (x) = - 1 / 3x ^ 2 + BX ^ 2 + CX + BC about X is known
2) Let y = f (x) - C (x + b) be K when x belongs to (0,1). If K ≤ 1, the value range of real number B is obtained

Y = f (x) - C (x + b) = - 1 / 3x & # 178; + BX & # 178; y '= - 2 / 3x + 2bx = k ∵ K ≤ 1 ∵ B ≤ (3 + 2x) / 6x, let g (x) = (3 + 2x) / 6x be the minimum value of G (x) ∵ g (x)' = - 1 / 2x & # 178