Given a ∈ R, the function f (x) = SiNx - | a | and X ∈ R is an odd function, then a = __

Given a ∈ R, the function f (x) = SiNx - | a | and X ∈ R is an odd function, then a = __

∵ f (x) = SiNx - | a |, X ∈ R is an odd function,
∴f（0）=-|a|=0，
∴a=0．
So the answer is: 0

Let function f (x) = (x + 1)   2+sinx x   If the maximum value of 2 + 1 is m and the minimum value is m, then M + M = _

The function can be reduced to f (x) = (x + 1)   2+sinxx   2 + 1 = 1 + 2x + sinxx2 + 1, let g (x) = 2x + sinxx2 + 1, then G (x) = 2x + sinxx2 + 1 is an odd function, and the sum of the maximum and minimum values of G (x) = 2x + sinxx2 + 1 is 0. The function f (x) = (x + 1)   2+sinxx   Maximum value of 2 + 1

The monotonicity of function f (x) = (a + 1) INX + ax ^ 2 + 1 is discussed

First find the derivative, f '(x) = (a + 1) / x + 2aX
The domain is a positive number
So the above formula = (1 / x) (a + 1 + 2aX ^ 2)
1, a > = 0, the derivative is always greater than 0, and f (x) increases monotonically
2,-1

Given the function f (x) = INX ax + (1-A) / X-1 (a = R), discuss the monotonicity of the function

F (x) = INX ax + (1-A) / X-1 (a = R) domain (0, + ∞) f '(x) = 1 / x-a + (A-1) / X ² = [-ax ²+ x+(a-1)]/x ²=- (x-1)[ax+(a-1)]/x ² When a = 0, f '(x) = (x-1) / X ² ≥ 0f (x) is a decreasing function on (0,1) and an increasing function on (1, + ∞). When a ≠ 0

Let a > 0, function f (x) = x ^ 2 + a|inx-1|. When a = 2, find the monotonic increasing interval of the function If x ≥ 1, the inequality f (x) ≥ A is constant, and the value range of real number a

If inx-1 > = 0, i.e. x > = e, yes
F (x) = x ^ 2 + a (inx-1), and the derivative is: F '(x) = 2x + A / X
When a = 2, it is 2x + 2 / X. because x > = e, f '(x) > 0 is constant, and f (x) monotonically increases on x > = E
If inx-1

The function f (x) = inx-a / x, G (x) = f (x) + ax-6inx, where a ∈ R (1) discusses the monotonicity of F (x) (2) if G (x) is an increasing function in its definition domain, find the value range of positive real number a; (3) Let the function f (x) = x Λ 2-mx + 4, when a = 2, if there is x1 ∈ (0,1) to give x2 ∈ [1,2], there is always g (x1) ≥ H (x2), and find the value range of the real number M. if it is a positive solution,

0