Only function FX = AX2 + 1 / BX + C (a, B, C ∈ z) is odd, and F1 = 2, F2 < 3, find a, B, C

Only function FX = AX2 + 1 / BX + C (a, B, C ∈ z) is odd, and F1 = 2, F2 < 3, find a, B, C

Why is b greater than 0?

Given the function FX = AX2 + BX + C, if F1 > 0, F2 < 0, then the number of zeros of FX on (1,2) is ()
A. At most one B. one or two
C. There is and only one D. none
Seeking answers and analysis~~

B

Function FX = AX2 + BX + C, if F1 > 0, F2 < 0, then the number of zeros of FX on (1,2) is several
Is there only one,

There are one or two
If (1,2) is monotonically unique, there is a

Let f (x) be a function whose domain is symmetric with respect to the origin, then f 1 (x) = f (x) + F (- x) is an even function and F 2 (x) = f (x) - (- x) is an odd function

F1(-x)=f(-x)+f（x）=F1(x)
F2(-x)=f(-x)-f（x）=-[f(x)-f（-x）]=-F2(x)
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Let f (x) = 1 + x1 − x, and let F 1 (x) = f (x), f K + 1 (x) = f (f K (x)), k = 1, 2 Then f2011 (x) = ()
A. -1xB. xC. 1+x1−xD. x−1x+1

F1 (x) = 1 + x1 − x, F2 (x) = f (F1 (x)) = - 1 x, F3 (x) = f (F2 (x)) = 1 − 1x1 + 1 x − 1 x + 1, F4 (x) = f (F3 (x)) = 1 + X − 1 x + 11 − x − 1 x + 1 = x, F5 (x) = f (F4 (x)) = 1 + x1 − x ℅ the analytic expression of the function takes 4 as a period and appears as a period ∵ f2011 (x)

Let f (x) = 1 + x1 − x, and let F 1 (x) = f (x), f K + 1 (x) = f (f K (x)), k = 1, 2 , then f2009 (x) = ()
A. -1xB. xC. x−1x+1D. 1+x1−x

Because f (x) = 1 + X (x) = 1 + x1-x (f (x) = f (x), and F1 (x) = f (x), FK + 1 (FK (x) = f (FK (x) (FK (x)), we have: F2 (x) = f (f (x) = f (f (f (x) = f (FK (FK (x) (FK (FK) (FK (FK (x)), because f (x) = f (x) f (x) = f (x) (FK (FK (FK (FK (FK (x) = f (F1 (F1 (F1 (x) (f (F1 (x) (f (F1 (F1 (F1 (x)) = f (F1 (F1 (F1 (F1 (F1 (x) (f (F1 (F1 (x)) (f (f (F1 (F1 (F1 (F1 (F1 (F1 (x) = f (F1 (F1 (F1 (F1 (F1 (x))) (f (F1 (F1 (F1 (F1 (F1 (x))) = f (F1 (F1 the period is 4, and 2009 = 4 × 1002 + 1 F2009 (x) = F1 (x) = 1 + x1 − x, so D

Among the following functions, the function which is not only even but also monotonically decreasing on the interval (0, + ∞) is ()
A. y=2|x|B. y=x3C. y=-x2+1D. y=cosx

In a, y = 2 | x | is even function, but monotonically increasing on (0, + ∞), excluding a; in B, y = X3 is odd function, excluding B; in C, y = - x2 + 1 is even function, and monotonically decreasing on (0, + ∞); in D, y = cosx is even function, but not monotone on (0, + ∞), excluding D; so select C

Among the following functions, which are even functions and the decreasing functions in the interval (0, + ∞), is a.y = log2x2b. Y = cosx

A
y=log2(x²)
Is it based on two? Is it true?
If yes, it is even function, but not minus function in (0, + ∞)
B
y=cosx
It is an even function, but not a decreasing function in (0, + ∞)
C. What is option D?

If f (x) is an even function defined on R and monotonically increasing on interval (a, b), then what is the relationship between a and B?

Even function, the domain is symmetric about the origin
Because it's even, it's impossible to cross the origin
ab>0

If the even function f (x) increases monotonically in the interval [0, + ∞), then f (x) must decrease monotonically in the interval (- ∞, 0). Why?

Yes
Even function f (x) = f (- x)
Take 0 ＜ x1 ＜ x2
Because it is an increasing function in the interval [0, + ∞), f (x1) < f (x2)
And 0 ＞ - x1 ＞ - x2
f(-x1)=f(x1)＜f(x2)=f(-x2)
In the interval (- ∞, 0), f (- x1) < f (- x2), so it must be monotonically decreasing
In addition, we can see that the function image is symmetrical about y axis