1. Let 0 ＜ x ＜ π / 2, then x (SiNx) ^ 2 ＜ 1 is what condition of xsinx ＜ 1. 2. It is known that the function f (x) satisfies f (1) = 1 / 4, 4f (x) f (y) = f (x + y) + F (X-Y) (x, y belongs to R), then f (2010) =?

1. Let 0 ＜ x ＜ π / 2, then x (SiNx) ^ 2 ＜ 1 is what condition of xsinx ＜ 1. 2. It is known that the function f (x) satisfies f (1) = 1 / 4, 4f (x) f (y) = f (x + y) + F (X-Y) (x, y belongs to R), then f (2010) =?

1.∵0＜x＜π/2,∴sinX＜1
∴（sinx）^2＜sinX
Based on this, we can roughly draw the image of F (x) = (SiNx) ^ 2 (0 < x < π / 2)
The image of F (x) = (SiNx) ^ 2 is always below f (x) = SiNx
From X (SiNx) ^ 2 ＜ 1 to (SiNx) ^ 2 ＜ 1 / X
From xsinx < 1 to SiNx < 1 / X
It is easy to prove that the image of y = 1 / X passes the image of F (x) = (SiNx) ^ 2 and f (x) = SiNx
Let X1 and X2 be two parts of (SiNx) ^ 2 = 1 / X and SiNx < 1 / X
From the above results, it can be concluded that X1 > x2
Therefore, it is necessary and insufficient
2. Let y = 1, we get 4f (1) f (x) = f (x) = f (x + 1) + F (x-1) ①
Let x = 1, f (y) = f (1 + y) + F (1-y)
By replacing y with X, we get f (x) = f (1 + x) + F (1-x) 2
Combining with ①, f (1-x) = f (x-1)
That is, f (x) is an even function
From ②, f (2010) = f (2011) + F (1-2010) = f (2010) + F (2009)
∴f(2011)=f(2010)-f(2009)
Similarly, Yizheng f (2010) = f (2009) - f (2008)
∴f(2010)=f(2009)-f(2008)
=f(2008)-f(2007)-f(2008)
=-f(2007)
Similarly, f (2007) = - f (2004)
∴f(2010)=f(2004)
That is, take 6 as the cycle
∴f(2010)=f(0)
From ①, f (0) = f (1) + F (- 1) = 2F (1) = 0.5
So f (2010) = 0.5
Typing is sour, give more points!
Given that the function y = f (x) is odd, the domain of definition is & nbsp; (- ∞, 0) ∪ (0, + ∞), and y = f (x) is an increasing function on (0,, + ∞), and f (- 1) = 0, then the value range of X satisfying & nbsp; f (x) > 0 & nbsp; is ()
A. （1，+∞）B. （0，1）C. （-∞，-1）∪（-1，+∞）D. （-1，0）∪（1，+∞）
From the fact that the function y = f (x) is an odd function, we can get f (- x) = - f (1) = - f (- 1) = 0. Because y = f (x) is an increasing function on (0, + ∞) and the image of the odd function is symmetric about the origin. The general image of the function is shown in the figure. When - 1 < x < 0 or 0 < x < 1, f (x) > 0
(- 1,0) and up 1 to positive infinity... Process, it's hard to say, according to the property graph of odd function, we want to be symmetrical about the origin, so f (1) = 0, and zero to positive infinity is an increasing function, so x belongs to 1 to positive infinity, and (- 1,0)
F (x) is an odd function
So f (- x) = - f (x)
F (x) is an increasing function in the interval (0, + ∞)
So if 0
If the function f (x) is quadratic differentiable, find the second derivative of y = f (SiNx)
y'=f'(sinx)cosx
y''=f''(sinx)cosxcosx+f'(sinx)(-sinx)
It is known that the domain of definition of odd function f (x) is [- 2,2] and decreases in the interval [- 2,0]. The value range of real number m satisfying f (1-m) + F (1-m2) < 0 is obtained
The definition domain of ∵ f (x) is [- 2, 2], ■ − 2 ≤ 1 − m ≤ 2 − 2 ≤ 1 − M2 ≤ 2, and the solution is - 1 ≤ m ≤ 3. ① ------ (4 points) and ∵ f (x) are odd functions, and they decrease on [- 2, 0], ∵ f (x) decreases on [- 2, 2], --- - (6 points) then f (1-m) ＜ - f (1-m2) = f (M2-1) is transformed into: 1-m ＞ M2-1, and the solution is - 2 ＜ m ＜ 1 In conclusion, 1 ≤ m ＜ 1
The n-th derivative of F (x) = xsinx
N = odd number = 2N-1 derivative = (- 1) of N + 1 power * n * SiNx + (- 1) of N + 1 power * (cosx)
N = even number = 2n derivative = (- 1) n + 1 power * n * cosx + (- 1) n power * (SiNx)
First order SiNx + xcosx
Second order 2cosx - xsinx
It seems that the order n should be asinx + bcosx + X (csinx + dcosx)
It's very complicated to get the N + 1 order equation by calculating its first derivative, and then calculate the iterative formula of ABCD
It is known that the function f (x) is an odd function in the domain of definition (&; - 2,2)
It is known that the function f (x) is an odd function in the domain of definition (&; - 2,2) and increases monotonically in the interval [0,2]. The solution to the inequality f (x-1) + F (x ^ 2-1) < 0
First, define the domain requirements: - 2
Find f (x) = xsinx
F '(x) = x'sinx + xcos'x = SiNx + xcosx how does this come from and how to find its origin
Derivation formula
If the functions u = u (x) and V = V (x) are differentiable at point x, then their quotient is also differentiable at point X and (UV) '= u'v + v'u
If f (1-A) + F (1-a2) > 0, the range of a is obtained
∵ f (x) is an odd function, ∵ f (1-A) + F (1-a2) > 0 can be transformed into f (1-A) > - f (1-a2) = f (A2-1), and f (x) is increasing in the domain (- 1, 1), ∵ 1 − a > A2 − 1 − a < 1 − a < 1 − A2 < 1, that is − 2 < a < 10 < a < 2 − 2 < a < 0 or 0 < a < 2, the solution is 0 < a < 1
The first derivative of y = e ^ xsinx f (x) = e ^ x (x + 1)
1、y=e^xsinx
y ' =e^xsinx+e^xcosx
=e^x(sinx+cosx)
2、f(x)=e^x(x+1)
f ' (x) =e^x(x+1) +e^x
=e^x(x+2)
If the odd function f (x) is a decreasing function over the domain (- 1,1) and f (1-m) + F (1 + m square) is less than 0, then the value range of the real number m is determined?
1 + m square? You mean 1 + m * m?
If so, m does not exist
Because 1 + m * m < 1, such m does not exist
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