If the functions f (x) and G (x) are odd and even functions on R respectively and satisfy f (x) + G (x) = ex, where e is the base of natural logarithm, then () A. f（e）＜f（3）＜g（-3）B. g（-3）＜f（3）＜f（e）C. f（3）＜f（e）＜g（-3）D. g（-3）＜f（e）＜f（3）

If the functions f (x) and G (x) are odd and even functions on R respectively and satisfy f (x) + G (x) = ex, where e is the base of natural logarithm, then () A. f（e）＜f（3）＜g（-3）B. g（-3）＜f（3）＜f（e）C. f（3）＜f（e）＜g（-3）D. g（-3）＜f（e）＜f（3）

In F (x) + G (x) = ex (1), let x = - x, then f (- x) + G (- x) = E-X, and functions f (x), G (x) are odd and even functions on R respectively, so there is - f (x) + G (x) = E-X (2). From the solution of (1) and (2), f (x) = 12 (ex-e-x), G (x) = 12 (ex + E-X)

Even function f (x) defined on R, when x > = 0, f (x) = e ^ x + A, e is the base of natural logarithm
1) When x > = 0, f (x) > = Xe is constant, the value range of a is obtained
2) For the minimum value of a in 1), f (X-2) is constant for X

1) Let g (x) = xe-e ^ x, G '(x) = E-E ^ x, when G' (x) = 0, x = 1, and x > 1, G '(x)

Given that f (x) = x ^ 2-2ax + 3 (x belongs to [0,1]), if f (x) has an inverse function, then the value of real number a

There is an inverse function, that is, the function f (x) is monotonic (increasing or decreasing) in the interval [0,1]. 1) the derivation of F (x) is: 2x-2a. In order to make the function increase in the interval, that is, the derivative of F (x) is always greater than 0, the value range of a is a = 1