If the function f (x) = (e ^ x) - ax (E is the bottom of the natural logarithm). For any real number x, f (x) is greater than or equal to 1, find the value of the real number a

If the function f (x) = (e ^ x) - ax (E is the bottom of the natural logarithm). For any real number x, f (x) is greater than or equal to 1, find the value of the real number a

e^x-ax>=1
e^x>=1+ax
Take logarithm
x>=ln(1+ax)
Let g (x) = x
h(x)=ln(1+ax)
g'(x)=1
h'(x)=a/(1+ax)
To make g '(x) > = H' (x)
1>=a/(1+ax)
When a (1-x) is 0, it is X1 / (1-x)
When x = 1
a

The even function f (x) defined on R, when x > = 0, f (x) = e ^ x + A, in fact, e is the base of the natural logarithm 1) When x > = 0, f (x) > = Xe is constant, find the value range of A 2) For the minimum value of a in 1), for X, it belongs to [1, M] constant f (X-2)

1) That is, when Xe = 0 is constant, that is, a > = xe-e ^ x is constant, so that G (x) = xe-e ^ x, G '(x) = E-E ^ x, when G' (x) = 0, x = 1, and x > 1, G '(x)

It is known that the minimum value of the even function f (x) defined on R is 1, and when x > = 0, f (x) = e ^ x + A, where e is the base of the natural logarithm (1) Find the analytical formula of function f (x) (2) If the function f (x) = f (x) - BX ^ 2 has exactly two different zeros, find the value of B

(1) We can judge that x > = 0, f (x) is a monotonically increasing function, because f (x) is an even function,
x=0,f(x)=e^x x

For the even function f (X-2) defined on R, when x > 2, f (x) = ex + 1-2 (E is the base of natural logarithm). If there is k ∈ Z, so that the real root x0 ∈ (k-1, K) of equation f (x) = 0, the value set of K is () A. {0} B. {-3} C. {-4，0} D. {-3，0}

∵ the graph of even function f (X-2) is symmetric about the Y axis
The image of function y = f (x) is symmetric about x = - 2
∵ when x ＞ - 2, f (x) = ex + 1-2
∵ f (x) = ex + 1-2 monotonically increases at (- 2, + ∞), and f (- 1) < 0, f (0) = E-2 > 0
According to the existence theorem of zeros, the function f (x) = ex + 1-2 has zeros on (- 1, 0)
According to the symmetry of the function image, when x < - 2, there is a unique zero x ∈ (- 4, - 3)
From the real root x0 ∈ (k-1, K) of equation f (x) = 0, then k-1 = - 4 or k-1 = - 1
K = - 3 or K = 0
So choose D

If the maximum value of F (x) = e − (x − U) 2 is m and f (x) is an even function, then M + U = __

∵ f (x) is an even function,
∴f（-1）=f（1），
∴u=0
∴f（x）=e−x2，
When x = 0, the function f (x) gets the maximum value, and the maximum value is 1,
∴m+ μ= 1．
So the answer is: 1

Known function f (x) = (2x + a) e ^ x (e ^ x is the base of natural logarithm) If for all real numbers x in the interval [- 1,1], there is - 2 ≤ f (x) ≤ E ² True, find the value range of real number a

f'(x)=(2x+2+a)*e^x
Let f '(x) = 0 x = - (2 + a) / 2
(1) - (2 + a) / 2 > = 1, i.e. a