Let a be a real number, the function ex x power of e 2x 2a, a be a real number, and verify when a I

Derivative of f'=ex-2

When ex-2=0 i.e. x=ln2 is the derivative of f'=0, when ex-2<0, i.e., x0, i.e., x>ln2 is the derivative f'>0 The original function f is an increment function.

The minimum is f(ln2)=2-2ln2+2a

Let g(x)=e x-(x 2-2ax+1) the derivative of the function g'=ex-(2x-2a) is the function f, when a>ln2-1, the original function minima f(ln2)=2-2ln2+2a>2-2ln2+2(ln2-1)=0

i.e. the derivative function g'>0

The function g is an increasing function over r.

g（0）=1-（0-0+1）=0

For any x>0

g(x)>g(0)=0 is constant.

ex-(x2-2ax+1)>0 means ex>x2-2ax+1.

1) Solution: f(x)=ex-2x+2a,x r,f(x)=ex-2,x r

Let f (x) = 0 give x = ln2

So when x changes, f(x), f(x) changes as follows:

x （-ln2） ln2 （ln2，+∞

f′（x） -0 +

f(x) monotonically decreasing 2(1-ln2+a) monotonically increasing.

Therefore, the monotonically decreasing interval of f(x) is (- ln2), and the monotonically increasing interval is (ln2, + f(x) to obtain a minimum at x=ln2, and the minimum is f(ln2)=eln2-2ln2+2a=2(1-ln2+a).

2) Proof : Let g(x)=ex-x2+2ax-1,x r, then g (x)=ex-2x+2a,x r

From (1), when a ln2-1, the minimum value of g (x) is g (ln2) = 2(1-ln2+a) 0

So for any x r, there is g (x) 0, so g (x) is monotonically increasing within r

So when a ln2-1, there is g(x) g(0) g(0) for any x (+).

And g(0) = 0, thus for any x (0, + g(x) 0

i.e. ex-x2+2ax-1 0, so exx2-2ax+1

f'=ex-2

When ex-2=0 i.e. x=ln2 is the derivative of f'=0 when ex-20 is the original function f is the increment function.

The minimum is f(ln2)=2-2ln2+2a

Let g(x)=e x-(x 2-2ax+1) the derivative of the function g'=ex-(2x-2a) is the function f, when a>ln2-1, the original function minima f(ln2)=2-2ln2+2a>2-2ln2+2(ln2-1)=0

i.e. the derivative function g'>0

The function g is an increasing function over r.

g（0）=1-（0-0+1）=0

For any x>0

g(x)>g(0)=0 is constant.

ex-(x2-2ax+1)>0 i.e. ex>x2-2ax+1 is proven, 1,(1) f(x)=ex-2x+2a,x r,f (x)=ex-2,x r

Let f (x) = 0 give x = ln2

So when x changes, f (x) is losing front, and f(x) changes as follows:

x (-ln2) ln2 (ln2, and calendar +) f (x) -1, let a be a real number, the function ex(x power of e)—2x+2a, a is a real number, verify: when a>(in2)—1 and x>0, ex(x power) > x2 (xsquared)—2ax+1

Disturb the gods: Let g(x)=e x-x 2+2ax-1

then g(x).'=e^x-2x+2a=f(x)

and by the slow section of the loss of the f(x).'=e x-2, let f(x).'

(1) Derivative, get f'(x)=e^x/^2

Because to find the extreme point, then x=or.

0, the solution is x=or.

So the extreme point is x=or.

2)f'(x)=e^x(ax^2-2ax+1)/(1+ax^2)^2

Because it is a monotonic function, as long as ax 2-2ax+1 is greater than 0 or constant less than 0, when a=0, the condition is satisfied.

When a>0, the minimum value of 4ac-b 2 4a>0 gives 0

Proof: Let g(x)=e x-x 2+2ax-1 then g(x).'=e x-2x+2a=f(x) is again composed of f(x).'=e x-2, let f(x).'<0 can be solved, 02-2ln2+2(ln2-1)=0 is f(x)>0, so g(x).'>0

Therefore, the function g(x) is a monotonic increasing function, and g(0)=0, so when x>0, g(x)>0, that is, there is e x>x 2-2ax+1 typing slowly, and the writing is a bit simple.

Analysis: f'(x)=a^2/x-2x+a=(-2x^2+ax+a^2)/x=-(2x^2-ax-a^2)/x=-(2x+a)(x-a)/x>0

Get 0aThat is, when a>0, the monotonic increase interval is (0,a), and the subtraction interval is (a,+infinity).

When x [1 ,e],e-1<=f(x) <=e 2 is always established a>0, the function f(x) takes the maximum value at x=a, f(a)=a 2lna=e 2==>a=e

On the interval [1,e], f(1) is the minimum value: f(1)=a-1=e-1==>a=e;

A=E that satisfies the question

f'(x)=(a²/x)－2x＋a=(－2x²＋ax＋a²)/(x)=[－(2x＋a)(x－a)]/(x)

Since a>0, then f(x) increases over (0,a) and decreases on (a,).

1. If a<1, then only f(e) e 1 and f(1) e are needed, and 0e is obtained, then only f(1) e 1 and f(e) e;

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f'(x)=(a²/x)－2x＋a=(－2x²＋ax＋a²)/(x)=[－(2x＋a)(x－a)]/(x)

Since a>0, then f(x) increases over (0,a) and decreases on (a,).

1. If a<1, then only f(e) e 1 and f(1) e are needed, and 0e is obtained, then only f(1) e 1 and f(e) e;

Since a 0, the increasing interval of f(x) is (0,a) and the decreasing interval of f(x) is (a,+).

Proof: From the question, n(1)=a-1 e-1, i.e., a e, from ( ) knows that f(x) increases monotonically in [1,e].

To make e-1 f(x) e2 constant against x [1,e], only f(8)=a-8 e-8

f(e)=a2-02+ae≤e2

The solution is a=e

1),1/2[f(x1)+f(x2)]

ax1 +ax2 +bx1+bx2+2c] 2[a(x1 +x2 ) 2+b(x1+x2) 2+cf(x1+x2) search next to 2).

a(x1+x2)/2)²+b(x1+x2)/2)+ca(x1+x2)²/4+b(x1+x2)/2+c2(x1x2)≤(x1²+x2²)

2(x1x2)+x1²+x2²≤2(x1²+x2²)x1+x2)²≤2(x1²+x2²)

x1+x2) 4 (x1 +x2 ) 2 when a>0.

a(x1+x2)²/4≤a(x1²+x2²)/2a(x1+x2)²/4+b(x1+x2)/2+c≤a(x1²+x2²)/2+b(x1+x2)/2+c

Therefore f((x1+x2) 2) 2

2) When x belongs to [-1,1], f(x) <1, is there a,b,c so that the sum of the liquid is f2) >36 5 is true? If so, write a set of values a, b, and c that meet the conditions; If not, please explain the reason for the omission.

It is known that the quadratic function f(x) = ax +bx+c

When x belongs to [-1,1], |f(x)|≤1

Let x=1,-1,0, respectively, get.

f(-1)|=a-b+c|≤1

f(1)|=a+b+c|≤1

f(0)|=c|≤1

f2)|=4a+2b+c|

3(a+b+c)

a-b+c)

3c|3|a+b+c

a-b+c|

3|c|f2)|≤3|a+b+c

a-b+c|3|c|

f2)|≤7<36/5

does not exist.