13 points The function has a maximum value in the interval , and the real number is found in

Solution: The axis of symmetry <>

When <> is a decreasing interval of <>, <>

When <> is an increasing interval of <>, <>

When <><> contradicts <>; So <>

Or <> omitted.

Since the coefficient of the quadratic term is -1, the opening direction of the image of the function f(x)=-x2+2ax+1-a is downward, and the axis of symmetry is x=a, so it needs to be classified and discussed according to the relationship between the axis of symmetry and the interval

Solution: The axis of symmetry of the function f(x)=-x2+2ax+1-a is x=a, and the image opening is downward.

When a 0, the function f(x)=-x2+2ax+1-a is a subtraction function in the interval [0,1].

fmax(x)=f(0)=1-a, from 1-a=2, a=-1, when 0 a 1, the function f(x)=-x2+2ax+1-a is an increasing function in the interval [0,a], and on [a,1] is a subtracting function.

fmax(x)＝f(a)＝−a2+2a2+1−a=a²-a+1

From a2-a+1=2, a=(1+ 5) 2 or =(1- 5) 2 is obtained

0 a 1, neither value is satisfied;

When a 1, the function f(x)=-x2+2ax+1-a is an increasing function in the interval [0,1].

fmax（x）=f（1）=-1+2a+1-a=a

a=2 In summary, a=-1 or a=2

So the answer is: a=-1 or a=2

a="-2,"In line with the topic. Because the function has a minimum value of -2 in the interval, then the function is formulated, and the symmetry axis is discussed in the interval, so that we can know where the minimum value of the spine beat type is obtained, and the corresponding value of a is solved, 1 point (1) When , that is, the function is an increasing function in the interval, and the minimum value of the function is , which is not in line with the topic, and is rounded off....... 4 points (2) Immediately, the function function is a subtraction function in the interval , and the minimum value of the function is a= , which contradicts . If it doesn't match the topic, give up the .......

7 points (3), i.e., , the minimum value of is =-2A="-2,"In line with the topic. 10 points.

Analysis] Since the coefficient of the quadratic term is a parameter, it is solved in three cases. When a=0, the function y=-x-3 is a primary function and is a subtraction function, and the maximum value is discarded because it does not match the meaning of the topic, when a>0 and a<0 the function is a quadratic function, the symmetry axis is then classified quadratically according to the position relationship between the symmetry axis and the fixed interval, and then the maximum value is obtained from the opening direction of the quadratic function and the position relationship between the symmetry axis and the fixed interval, and the maximum value is verified by the range of the shed If a=0, the maximum value of the function y=-x-3 in the interval [ , 2] is , which does not conform to the topic, so it is discarded; If a>0, then the axis of symmetry of the function image is , nWhen -1+, i.e., a, the maximum value of the function is f(2)=a 2 2 +(2a-1) 2-3=3, and the solution is a=1;When <-1+ <2, ie

a="-2,"

In line with the topic. The examination of the maximum-value problem of quadratic functions in the closed-interval problem on the dynamic axis determines the maximum-value problem, which embodies the ideological method of classification discussion and motion change, and is a mid-range problem because of the function.

In the intervals. If there is a minimum value of -2, then the function is formulated, and whether the axis of symmetry is in the interval is discussed, so that we can know where the function predecessor obtains the minimum value, and solve the corresponding value of a

Solution: ......1 point.

1) When. Namely.

Function. In the intervals.

is the increment function, and in this case, the minimum value is .

If it doesn't fit the topic, regret it, and give up the .......4 points.

2) When. Namely.

Functions. In the interval Wuzi.

is the subtraction function, and the minimum value is .

A=This with.

Contradiction; If it doesn't match the topic, give up the .......7 points.

Namely. , the minimum value is .

2.A="-2,"

In line with the topic. 10 points.

<> (2) increases in the range of <>

The minus interval is <>

In the first question, because of <>

is a function <>

an extreme point. Therefore, there is.

Get to a conclusion. In the second question, on the basis of the first question, the progressive relationship is further solved for the derivative of the function, and it is reduced to .

Determine monotonic intervals.

Solution: (1) Because.

is a function <>

an extreme point.

Proven <>

In line with the topic. 2) Since the first question has always determined that the analytic formula of the function is <><

Let the derivative be greater than zero to give an increase interval of <>

Let the derivative be less than zero to get the subtraction interval of <>

<><3) omitted. Solution: ......

When <> "Monotonous Decline, When <>

<> monotonically increasing ......1 point.

t no solution; ......2 points.

That is, when <>, <>

…3 points. That is, when <>, <>

Monotonically increasing on the <>, <>

So <>

…5 points. > is <> set <> is <> "monotonous increase, <>."

<> monotonically decreasing, so <>

Because of the <> of everything

Constant was established, so <>

…9 points. The question is equivalent to proving <>

It is known from <>

The minimum value is <>

If and only if <>

When you get it, set <>

then <> easy to get <>

If and only if <>

when to take, thus <> on everything

Both have <>

Establishment of ......13 points.

a="-2,"In line with the topic.

The examination of the maximum-value problem of quadratic functions in the closed-interval problem on the dynamic axis determines the maximum-value problem, which embodies the ideological method of classification discussion and motion change, and is a mid-range problem because of the function.

In the intervals.

On the minimum value -2, then the function is formulated, and whether the symmetry axis is in the interval is discussed, so that we can know where the function obtains the minimum value and solve the corresponding value of a

Solution: ......

…1 point. 1) When.

That is<> function <>

In the intervals.

On is the increment function, at this time, <

The minimum value of is <>

If it doesn't match the topic, give up the .......4 points.

2) When.

That is<> function function <>

In the intervals.

on is the subtractive function, <

The minimum value of is <>

A=

This is not the same as <>

Contradiction; If it doesn't match the topic, give up the .......7 points.

That is, when <>, <>

The minimum value of is <>

2.A="-2,"In line with the topic. 10 points.

And not our hot dogs are too hot.

Solution: (1) Function <>

The definition domain is (1,+<

When a=1, <>

So <>

In <> is a subtraction function. In <>

is an increment function, so the function <>

The minimum value of is <>

If it is <>, it will be <>

0 in (1,<>

Constant was established, so <>

(1,<>

If <> should be <>

When <>, <>

So a>0 <>

The reduction interval is (<>

The increase interval is <>

, known by ( ) <>

In (1,+<

The minimum value of is <>

Order <> at 1,+<

On the monotonic decreasing, so <>

then <> therefore there is a real <>

Make the minimum value of the <> greater than <>

Therefore, there is a real number <>

Make y=<>

The image is <> with y=

There are no public spots. Outline.

Solution: ( Because <>, x >0, then <>, when <>

time, <>

When <>

time, <>

So <>

monotonically increasing on (0,1); In <>

on monotonically decreasing, so the function <>

Maximums are achieved at <>

Because the function <>

In the intervals.

<> of them

There are extremums on it, so <>

The solution is <>

Inequality <>

That is, <>

Remember<> so <>

If the order is <>, it will be <>

Monotonously increasing on the <>, >, thus <>, so <>

It's also monotonically increasing on <>, so it's <>, so it's <> omitted.