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Mathematics C_0
1. Because there is a maximum, let f (x) =-(x+b) 2+a, where a = 8;;

Substitute f(2)=- 1.

-(2+b)^2+8=- 1

B= 1 or -4

F (x) =-(x+ 1) 2+8 or f (x) =-(x-4) 2+8.

Write the standard formula as f (x) =-x 2-2x+7 or f (x) =-x 2+8x-8.

2.( 1)=0->2a+b+ 1=0①

F(x)+ 1=0 has a real root->; b-a^2+ 1<; 0②

① Substituting ② to get a 2+2a > =0.

A & gt=0 or a

A > brought into ① respectively; =0,b & lt=- 1; a & lt=-2,b & gt=3

3. If a=0, then f(x)=(b-8)x, no matter what the value of b is, f(x) can only be a straight line or a point, which does not meet the meaning of the question. So a is not equal to 0.

Because a is not equal to 0, f(x) is a quadratic equation, X =-3, and X = 2 is the solution of the equation.

Substituting x =-3 and x = 2 into the original equation, we get:

9a-3(b-6)-a-ab=0,

4a+2(b-8)-a-ab=0。

Solve a and b from the above two formulas.

The analytical formula of y=f(x) can be obtained by substituting into the original equation.

That's easy to understand.

4.( 1)F( 1)- 1 >; = 0 f( 1)& lt; There are two situations when = is a increasing function or a subtraction function, namely, f(0)=3 or f(2)=3, a = 1 radical, 2 or a = 5 radical, 10. Verify (using the symmetry axis formula) that a= 1- radical sign, 2, a. So a= 1- 2 under radical sign or a = 5 under radical sign 10.

6.

Axis of symmetry x =-a/2

7.( 1, positive infinity)

8. The question means: the square of F(X+T)=(X+T)+ 1 = the square of x+2 (1+t) x+(t+1). There is a real number t that makes the square of x +2( 1+T)X+(T+ 1) < =X constant. The square of x+(t+ 1) < =0 is obtained. Because the range of x is [1.m], that is, when X= 1, the square of x+(2t+1) x+(t+1) = 0 is T=- 1 or T=-. =0, get the square of X-X.

9. let f (x) = 3x-5x+A be -2 < m < 0, 1 < n < 3 to get f(-2)>0, f (0) <; 0,f( 1)& lt; 0,f(3)>0

Just solve the equation.