The known function f x a 1 x ,

1. Just ask for the derivative.

Second question. Bring in inequalities to get.

a-1/|x|<2x

a<2x+1/|x|

Easy to know. Let g(x)=2x+1 |x|（1，+∞x+x+1/x

On the definition field, it is an increment function (hook function).

When x=1 has a minimum value of 3

g(x)∈【3，+∞

aa (-3).

The third question. For f(x) is asked by 1.

0,+ is the increment function.

and f(x) = f(-x).

So it's an even function.

In (- 0) is the subtraction function.

1.When m,n is negative.

f(m)=n is obtained

f(n)=m

2.When m, n is positive.

We get f(m)=m

f(n)=n

After finding A, and then it will be fine

The answer is a=m+1 n or n+1 m, and only if m=1 n has a solution.

Bring in inequalities to get.

a-1/|x|<2x

a<2x+1/|x|

Easy to know. Let g(x)=2x+1 |x|（1，+∞x+x+1/x

On the definition field, it is an increment function (hook function).

When x=1 has a minimum value of 3

g(x)∈【3，+∞

aa (-3).

The third question. For f(x) is asked by 1.

0,+ is the increment function.

and f(x) = f(-x).

So it's an even function.

In (- 0) is the subtraction function.

1.When m,n is negative.

f(m)=n is obtained

f(n)=m

2.When m, n is positive.

We get f(m)=m

f(n)=n

After finding A, and then it will be fine

The answer is a=m+1 n or n+1 m, and only if m=1 n has a solution.

Solution: Axis of symmetry x21b 2a2(a11) 2 When a 1 the axis of symmetry is to the right of the x-axis origin, when a 21 the axis of symmetry is the y-axis, and when a 1 the axis of symmetry is to the left of the x-axis origin.

1.The domain is defined as r, i.e., no matter what real number x takes, (a-1) x + (a-1) x+1 4 constant "0

When a=1, (a-1)x + (a-1)x+1 4=1 4>0, which satisfies the topic.

a≠1, for the parabola y=(a-1)x +(a-1)x+1 4, the quadratic coefficient a-1>0, and the discriminant equation of the unary quadratic equation (a-1)x +(a-1)x+1 4 < 0

a-1>0 a>1

a-1)²-4(a-1)(1/4)<0

a-1)²-a-1)<0

a-1)(a-2)<0

10, unary quadratic equation (a-1) x + (a-1) x+1 4 discriminant 0a-1>0 a>1

Discriminant 0 (a-1)(a-2) 0

A 2 or A 1

To sum up, get a 2.

Let g(x)=(a-1)x+(a-1)x+(1 4)1)a-1=0, i.e., a=1.

g(x)=1 4>0 is true.

A-1≠0 is A≠1.

To make g(x)>0 constant on r.

Even A-1>0

(a-1)²-a-1)<0

10△=(a-1)²-a-1)≥0

Solution: (1) Because f(x)=x 2-x a 1>=0, so (-1) 2-4 1(a 1)=<0, so a>=-3 4; (2) Because when x = -(-1) 2 1 = 1 2, the minimum value of the function f(x) = a 3 4, because f(x) has a minimum value g(x) in (negative infinity, a], and a>=3 4>1 2, so g(a) = a 2 1.

Solution: (1)|x^2-1|+a|x-1|=0, x 2=1 and x=1, x=1, as long as a does not =0, the only zero point is empty Li x=1

2) When -2<=x<-1, f(x)=x 2-1+a(1-x)=x 2-ax+a-1, the axis of symmetry is x=a 2>=-3 2, so the maximum value = f(-2)=(2) 2-a(-2)+a-1=3a+3

When -1<=x<1, f(x)=1-x 2+a(1-x)=-x 2-ax+a+1, and the axis of symmetry is x=-a 2<=3 2

Therefore, it is best to be careful with the large value = f(1) = -1 - a + a + 1 = 0

When 1 so the maximum value = f(2) = 4 + 2 a - a - 1 = a + 3

Solution: (1)f(x)=|x²-1|+a|x-1|=|x-1|（|x+1|+a)

f(1)=0 Therefore, f(x) must have a zero point x=1

If the function f(x) has only one zero point, then ||x+1|+A>0 is constant, then A>02) Classification discussion: Divide X into three sections [-2,-1], [1,1], [1,2] remove the absolute value source and then discuss A (the process is more cumbersome, you have to be patient to do it, and leave it to the leakage to do it yourself).

When x a, the function f(x)=x2+x-a+1=(x+ 1 2 )2-a+ 3 4 ,a - 1 2 So the function f(x) increases monotonically on [a,+, so that the minimum value of the function f(x) on [a,+ is f(a) =a2+1 In summary, when - 1 2 a 1 2, the minimum value of the function f(x) is a2+1

Solution: From Eq. (i) f(1) =x+a = 1+ a from Eq. (ii) f(1) =x 2-2x = 1, so 1 + a = 1 => a = 2

2) f(f(2)) f(2^2-2*2) =f(0) =0+ a = 2

3) From Eq. (i) f(x) =x -2 , x 1 knows f(x) 1, so we can only apply Eq. (ii) so that m 2 -2m = 3 can be solved to get m1=3, m2 = -1 (rounded).