It is known that the function f (x) = the third power of X + the square of AX + 3bx + C (B is not equal to 0), and G (x) = f (x) - 2 is an odd function

It is known that the function f (x) = the third power of X + the square of AX + 3bx + C (B is not equal to 0), and G (x) = f (x) - 2 is an odd function

From the meaning of the question, G (x) = f (x) - 2 = x ^ 3 + ax ^ 2 + 3bx + C-2 is an odd function. From the definition of the odd function, we know that f (- x) = - f (x), and substitute it into the above formula to get - x ^ 3 + ax ^ 2-3bx + C-2 = - x ^ 3-ax ^ 2-3bx-c + 2. We can get 2aX ^ 2 + 2c-4 = 0, that is, ax ^ 2 + C-2 = 0

Given that f (x) = 2x ten a, G (x) = 1 / 4 (X & # 178; + 3), if G (f (x)) = x & # 178; + X + 1, find the value of A,

g(f(x))
=1/4[(2x+a)²+3]
=1/4(4x²+4ax+a²+3)
=x²+ax+a²/4+3/4
=x²+x+1
therefore
a=1

F (x / X-6) = x & # 178; + 2x-1, try to find the value of F (1 / 3)

Let x = - 3
Then x / (X-6) = 1 / 3
So,
F (1 / 3) = - 3 square + 2 · (- 3) - 1
=9-6-1
=2

If f (x) = log3a (x-1) defined on interval (1,2)
If the function f (x) = log3a (x-1) defined on the interval (1,2) satisfies f (x) > 0, then the value range of a is? 3a, which is the base number, and (x-1) is the true number

Known 1

Let f (x) be defined on [0,1]. If f (x-a) + F (x + a) is defined, then the value range of a is ()
A. (−∞，−12)B. [−12，12]C. (12，+∞)D. (−∞，−12]∪[12，+∞)

From the condition: 0 ≤ x + a ≤ 10 ≤ x − a ≤ 1, that is − a ≤ x ≤ 1 − AA ≤ x ≤ 1 + a  function y = f (x + a) + F (x-a) is the intersection of set {x | - a ≤ x ≤ 1-A} and {x | a ≤ x ≤ 1 + a}. (1) when a > 1 / 2, 1-A < A, the intersection of set {x | - a ≤ x ≤ 1-A} and {x | a ≤ x ≤ 1 + a} is empty