Math masters, high school functions, high school functions Math masters come quickly

I'm not a math whiz, but I hope the following helps! ~

Let the function f(x) (x 1)ln(x 1), if for all x 0, f(x) ax holds, and find the range of the value of the real number a (2006 national .).Monday, 26 March 2007 at 13:41

Let the function f(x) (x 1)ln(x 1), if for all x 0, f(x) ax holds, and find the range of the value of the real number a (2006 national .).rationale).

Solution 1: Let g(x) (x 1)ln(x 1) ax, and find the derivative of the function g(x): g (x) ln(x 1) 1 a

Let g (x) 0, solve x ea 1 1, i) when a 1, for all x 0, g (x) 0, so g(x) is on [0, on.

The multiplication function, again g(0) 0, so for x 0, there is g(x) g(0), i.e. when a 1, for all x 0, there is f(x) ax

ii) When a 1, for 0 x ea 1 1, g(x) 0, so g(x) in.

0, ea 1 1) is a subtraction function, and g(0) 0, so for 0 x ea 1 1, there is g(x) g(0), that is, when a 1, not for all x 0, there is f(x) ax true

In summary, the range of values for a is (1].

Solution 2: Let g(x) (x 1)ln(x 1) ax, so that the inequality f(x) ax becomes immediately g(x) g(0) true

Find the derivative of the function g(x): g (x) ln(x 1) 1 a

Let g (x) 0, solve x ea 1 1, when x ea 1 1, g (x) 0, g(x) is the increasing function, when 1 x ea 1 1, g (x) 0, g(x) is the subtraction function, so for all x 0 there is g(x) g(0) g(0) full and necessary condition is ea 1 1 0

This gives a 1, that is, the range of values of a is (1].

f(x) (x 1)ln(x 1), if for all x 0, f(x) ax holds,

You don't have an answer to this question.

Shilling x=1, there is f(1)*f(f(1)+1)=1Let x=f(1)+1, substituting: f(f(1)+1)*f(f(1)+1(f(1)+1)))=1

Divide f(f(1)+1), there is f(f(1)+1 (f(1)+1))=f(1), due to monotony, so f(1)+1(f(1)+1)=1, the solution is f(1)=0, but f(x)>1 x, that is, f(1)>1, there is no solution!

If you look at it again, there should be something wrong with the condition f(x)>1 x.

(1) x,y r always have f[x]+f[y]=f[x+y],...

Let x=y=0 get: f(0)+f(0)=f(0) so f(0)=0.

where y=-x yields: f[x]+f[-x]=f(0).

The f(0)=0 f[x]+f[-x]=0 function is an odd function.

Take any two real numbers x1, x2, and x1>x2, where x=x1,y=-x2, we get: f(x1)+f(-x2)=f(x1-x2).

x1-x2>0 f(x1-x2)<0 i.e. f(x1)+f(-x2)<0

And because the function is an odd function, f(x1)-f(x2)<0

Thus we can see that f[x] is a subtractive function on r.

2) From (1), we know that f(x) is a subtraction function on [-3,3], the minimum value is f(3), f(3)=f(1)+f(2)=f(1)+f(1)+f(1)=-2, and the maximum value is f(-3)=-f(3)=2

Let's start with parity ...

Later use. Since: f(x+y)=f(x)+f(y).

then let x=y=0

Then there is: f(0+0)=f(0)+f(0).

f(0)=2f(0)

Then: f(0)=0

Re-order: y=-x

Then there is: f[x+(-x)]=f(x)+f(-x)f(0)=f(x)+f(-x).

Since: f(0)=0

Then: f(x)+f(-x)=0

f(-x)=-f(x)

Then: f(x) is an odd function.

Take x1, x2 belongs to r, and x1 > x2

Then: f(x1)-f(x2).

f(x1)+f(-x2)

f(x1-x2)

Since: x1>x2

Then: x1-x2>0

When x>0 again, f(x) <0

Then: f(x1-x2)<0

That is, for any x1, x2 belongs to r

When x1 > x2, there is always f(x1), so f(x) decreases monotonically on r, which is a decreasing function.

(0,0) is not a point on an ellipse in the first place.

Substituting in does not satisfy the ellipse.

Nor is it within the ellipse.

So it's not a flaw in the original integral function.

That is, the original integral functions are all meaningful within this ellipse.

Then we pay attention to the integral function.

Whether the path-independent condition is met.

That's right, so the points are 0

It's easy. The second question is the hard part.

If I'm not mistaken, the second question should be 2pi

Okay, I'll tell you why now.

Let's start with the curve.

Is it (0,0) wrapped in a curve.

That is, inside the curve.

So the original integral function has a flaw on the star line.

Right, there are two solutions.

The first one we start at 0,0

Nearby open a reverse circle.

The reason is that the integral of the ring between the star line and the small circle is 0, because the flaw is removed, and then we will look at it.

Look at the integral in that reverse circle.

We take a small circle with a radius of r

So is x square plus y squared equal to r squared?

We are interested in this.

It is necessary to use the method that will be the best way to find the integral of the curve.

You remember.

The answer is -2pi

Then the integration of the star line.

It is the integral of the middle ring minus the integral of the inverted circle.

So it's 2pi

The second method.

Mathematical-physical methods, the integrand satisfies the Cauchy-Riemann condition.

For those who have a flaw, directly from the conclusion (in this chapter 4 leave a number there), that is, 2pi

It feels like the answer is wrong :

y=-1 2(t+1) 2+(a-5) 2 The answer starts from this step, and the following should be:

Because: liquid sun t>=0, so: y<=-1 2+(a-5) with reed2=7 2==>a=13

Answer 1, from 0<=x<=4 , x=4sin 2 , 0<=sin 2 <=1, here you can also take -2<= <=0, but note that the result f(x) -4sin +2cos expression will be a little different.

Answer2, f(x)max=2 5 is undoubtedly meaningful, but when =0 f(x)min=2 to check y, there may be a range of values.

sin is a periodic function, and the general definition domain defaults to (0, ) and the value range is (1, 1).

1:2 (0, )

2:min =2?

1.Let f(x)=kx+b

f(1-x)=k-kx+b

f(x-1)=kx-k+b

4f(1-x)-2f(x-1)=4k-4kx+4b-2kx-2k-2b=-6kx+2(k+b)=3x+18

k=-1/2,b=17/2

f(x)=-1/2x+17/2;

Monotonically decreasing. The maximum value f(-1) = 9 on [-1,1].

f（2007）>f（2008）

4r+2x-(x/r)

Define the field as 0 < x < r*root number 2

2.With O as the origin, a simple Cartesian coordinate system is established. (A and D are on the left side of the Y axis) cross the O point to make the perpendicular line of AD, and the perpendicular foot is E; The perpendicular line of AO is made at point D, and the vertical foot is an F triangle, and the row OAD is an isosceles triangle.

cosa=(ae ao)= x (2r)And because cosa=(af ad), so, af=r-x (2r)so, cd=2of=2(oa-af)=2r-x r perimeter y=2r+2x+cd=4r+2x-(x r) defines the domain: x minimum coincides with point a, is 0 (undesirable) at the maximum, c and d coincide at one point on the y axis, x is r*root number 2 (undesirable, because it is not a trapezoidal but a triangle).

1.Knowing that f(x) is a one-time function and satisfies 4f(1-x)-2f(x-1)=3x+18, find the maximum value of the function f(x) on [-1,1], and compare the magnitudes of f(2007) and f(2008).

Let f(x)=ax+b

4f(1-x)-2f(x-1)=-6ax+6a+2b=3x+18 corresponding coefficients should be equal: -6a=3, 6a+2b=18, so there are: a=-1 2, b=21 2

f(x) = because f'(x)=<0, is a subtraction function, so f(2007)>f(2008).

On [-1,1], the maximum value is f(-1)=112.There is a semicircular steel plate with radius R, which is planned to be cut into the shape of an isosceles trapezoidal ABCD, its lower bottom AB is the diameter of the circle O, and the end point of the upper bottom CD is on the circumference of the circle, write the functional relationship between the circumference Y of this trapezoid and the waist length X, and find its definition domain.

The high line of the trapezoidal intersection of C is AB in E, because BCE is similar to ABC, so there are: be BC=BC AB, BE=BC AB=X 2R, and the circumference Y=2X+2R+2R-2BE=2X+2R+2R-2X 2Rr is Y=(-1 R)X +2X+4R

It defines domain 0

1. Solution: Let x=0, y=2k-3, so at the intersection point on the y-axis (0,2k-3), then make y=0,x=(3-2k) (k-1), and the intersection point with the x-axis ((3-2k) (k-1),0).

Area=1 2*|(3-2k)/(k-1)|*2k-3|=1/2*|(2k-3)^2/(1-k)|=25/6

2k-3)^2/(1-k)|=25/3

2k-3) 2 (1-k) = 25 3, I won't solve this equation.

2. Let y=a(x-3) 2+10 =ax 2 -6ax +(9a+10), easy to know a<0

Let it intersect the x-axis at two points x1,x2 and since the symmetry, you might as well set x10,a<0

x1+x2=6,x1x2=(9a+10)/a

x2-x1 = under the root number [(x1+x2) 2-4x1x2] = under the root number (36-4(9a+10) a) = 4

36-[4(9a+10)/a]=16

9a+10)/a=5,9a+10=5a,a=

y=3, axis of symmetry x=(m-1) 4<=2 because he increases monotonically at x>2.

m-1<=8,m<=9

4 Solution: Make two y's equal 2x+b=3x-4,x=b+4>0

y=3b+8>0, and the solution is b>-8 3

5. If the slope does not change due to translation, it is set to y=3x+p

Since y=5x+10 crosses x spell at (-2,0) brings y=3x+p

0=-6+p,p=6,y=3x+6

6. Known parabola y=-x 2-(m+1)x+1 4m 2+1

1) If you want to find the value range of m for any two positive numbers x1 x2 and the corresponding function value y1 y2.

2) Under the condition of (1), if there are any two negative numbers x1 x2 at the same time, the corresponding function value y1 y2, find the value of m or the range of values.

Solution: (1) This means that the function decreases monotonically in this interval.

The axis of symmetry of the function is: (m+1) 2 because greater than 0 is a subtractive function, so, (m+1) 2 = 0, so m = -1

2) Similar to the first question, (m+1) 2>=0, m>=-1, and m=-1 in general

1. When x 0, y 2k 3, when y 0, x (3 2k) (k-1), known by the meaning of the question, xy 2=25 6, can be brought into the solution equation.

When k 2 3 or k = 1 4, the area of the triangle enclosed by the image and the two axes is 25 6

2. From the meaning of the question, we can set y=-a(x-3) 2+10 (a>0), when y 0, the equation is connected to get x1, x2(x2>x1), all are expressions of a, and x2-x1=4, the solution can get a.

Therefore, y=3, from the meaning of the question, the coordinate axis of the quadratic function < 2, that is, (m 1) (2 2) <=2, and the solution gives m<=9

4. The intersection of two points means that x and y in y=2x+b and y=3x-4 are equal.

That is, 2x+b=3x-4 solves x=b+4 and then substitutes y=3x-4 to get y=3b+8

Intersection of straight lines (b+4, 3b+8).

The intersection is in the first quadrant, so the coordinates of the intersection are all greater than 0

So b+4>0

3b+8>0

So b>-8 3

5. The intersection point a of the other line and the x-axis is (2,0), and the straight line y=3x-7 moves in parallel, so the new straight line can be set to y=3x+a, and the point a can be brought into the new straight line, so 6+a 0 can be obtained, so a=6, so the analytic formula of the new straight line is y=3x+6

6. (1) Because the parabola opening is downward, and for any two positive numbers x1 x2, the corresponding function value y1 y2 is a subtraction function, so the axis of the parabola <=0, that is, (m+1) 2<=0, and the solution is obtained, m>=-1

2) Combined with (1), it can be seen that the y-axis is the axis of the parabola, that is, (m+1) 2=0, and the solution is obtained, m=-1