1. When a=- 1, find the range of function y=f(x).
2. If the function y=f(x) is
Find the range of a.
3. Find the maximum and minimum values of the function y=f(x) when x belongs to (0, 1), and find out.
When the value of x.
When a=- 1, f(x)=2x+ 1/x, f(x) >; =2 roots (2x* 1/x)=2 roots 2, which is obtained when x= 2/2 roots.
When x tends to 0, f(x) tends to
Then the range of f(x) is (2 times the root number 2,
)
2 f' = 2+a/x 2, as can be seen from the title, on (0, 1), f'
If a & gt=0, f'>0, this is obviously irrelevant.
If a<0, f' decreases monotonically, then 2+A/ 1
3 f'=2+a/x^2
If a>0, f'>0 and f(x) increase monotonically, f(x) has no minimum value.
If a=0, then f(x)=2x, and the minimum value cannot be obtained.
If a<0, there is a root sign (-a/2) in order to make the maximum and minimum values exist.
When x= 1, the maximum value is 2-a.
When x= root number (-a/2), the minimum value is 2 root numbers (-a/2).
Let x function f(x)=cos2x-4acosx+2a, where 0≤x≤π/2.
① express the minimum value m of f(x) as a function m=g(a).
② Whether there is a real number A, so that f (x)>0 remains at [0,2/π].
③ Is there a real number A that makes the function f(x) monotonically increase on x ∈ [0,2/π]? If it exists, write down the set of all A's; If it does not exist, explain why.
1.f (x) = cos2x-4acosx+2a = 2cos squared x-4acosx+2a- 1=2(cosx-a) squared -2a-1.
Because 0≤x≤π/2, 0≤cos≤ 1.
So when 0≤a≤ 1, m=g(a)=-2a squared +2a- 1.
When a<0, m=g(a)=2a- 1.
When a> is at 1, m=g(a)=-2a+ 1.
2.0 ≤ A ≤ 1 from 1, m = g (a) =-2A square +2a- 1=-2(a- 1/2) square-1/2.
a & lt0，m = g (a) = 2a- 1
A> at 1, m = g (a) =-2a+ 1
So there is no real number a, so f (x) >; 0 remains at [0, π/2]
3. let x 1, x2∈[0, π/2] make x1>; x2
F (x1)-f (x2) = squared x 1-4acosx 1-2cos squared x2+4acosx2.
= 2(cosx 1-cos x2)(cosx 1+cos x2)-4a(cosx 1-cos x2)
=[2(cosx 1+cos x2)-4a](cosx 1-cos x2)
Cosx monotonically decreases at [0, π/2], so cosx 1
If f(x) monotonically increases on [0, π/2], f (x1) >; f(x2)
Then 2 (cosx 1+cosx2)-4a
Because cosx 1+cosx2
Therefore, the existence of a≥ 1 makes the function f(x) monotonically increase on x∈[0, π/2].
The first problem is to build a rectangle with a volume of 8000 cubic meters and a depth of 6 meters.
The cost of the pool wall is one yuan per square meter, and the cost of the pool bottom is two yuan per square meter. The total cost y element is expressed as a function of the length of one side of the bottom x meters, the function expression is obtained, and its definition domain is pointed out.
The second question is, if a commodity is priced in 60 yuan, there is no extra charge.
In that year, the annual sales volume was about 800 thousand pieces, if the government levied.
If every sales 100 yuan is taxed at X yuan (that is, the tax rate is x%), the annual sales volume will be reduced by (20/3)x million pieces.
(1) express the total tax revenue y (ten thousand yuan) collected by the government for this product every year as a function of x, and point out the definition and maximum value of this function.
(2) When X belongs to [4,8], find the maximum sales volume of the manufacturer.
First, given the volume V=8000m3, the depth H=6m and the bottom area 8000/6, the bottom cost is (8000*2a)/6.
One side of the bottom is X, then the other side is 8000/(6X). Then the total lateral area is {X+[8000/(6X)]}*6. The total cost of the side
{X+[8000/(6X)]}*6a .
So y = [(8000 * 2a)/6]+{x+[8000/(6x)]} * 6a.
The domain of x is 0.
We can see that the expression of y consists of two parts: the bottom surface and the side surface. The area of the bottom surface is constant, so finding the domain of y is actually finding the maximum and minimum area of the side surface.
Find the maximum and minimum values of {X+[8000/(6X)]}*6a. Obviously there is no maximum.
The minimum value under the root sign is 10, which is equal to 40 times. Due to the limitation of words, let me explain how to calculate it.
Anonymous telephone number 2009-065438
1. Find the tangent of the X axis, the center of which is on the line 3x-y=0, and is cut by the line y = X..
Equation of a circle equal to twice the root number of 7
Let the circle be (x-a) 2+(y-b) 2 = c 2.
The center of the circle is on the straight line 3x-y = 0, so b = 3a.
Tangent to the x axis means that only one root is associated with Y = 0.
Get (x-a) 2+(3a) 2-c 2 = 0.
X 2-2ax+( 10a 2-c 2) = 0。
△=4a^2-4( 10a^2-c^2)=0
c^2=9a^2
Cyclic equation (x-a) 2+(y-3a) 2 = 9a 2
Then combine the above equation with the straight line y = X.
Simplification can get 2x 2-8ax+a 2 = 0.
because
The radical number 7 equal to 2
So the above equation must have two roots set to x 1 x2.
You can get (x1-x2) 2+(y1-y2) 2 = (2 roots 7) 2.
Here y 1 = x 1y2 = x2, which needs no explanation and continues to be simplified.
(x 1+x2)^2-4x 1x2=0
pass by
A 2 = 1 can be found by bringing it in, so A = 1.
So the equation of a circle is (X- 1) 2+(Y-3) 2 = 9.
Or (x+ 1) 2+(y+3) 2 = 9.
17. (Full score for this small question)
As shown in the figure, in the cube, the side length is
(1) Verification: Straight line plane
(2) verification: plane plane;
Solution: (1) connection, so the quadrilateral is a parallelogram.
(2)
18. (Full score for this small question)
As shown in the figure,
The graphic area of OABC on the right side of the straight line is.
(1) Try to find the analytical formula of the function; (2) Draw the image of the function.
Solution: (1) Let the intersection of the straight line and the trapezoid be D and E. At that time,
At that time,
therefore
(2) Image (omitted)
19. (Full score for this small question 10)
Given the coordinates of the endpoint B of the line segment AB, the endpoint A moves on the circle.
(1) Find the locus of the point m in the line segment AB;
(2) The straight line L passing through point B and the circle have two intersections A and B. When OAOB, find the slope of L.
Solution: (1) Hypothesis, derived from the midpoint formula.
Because a is on the circle c, so
The locus of point M is a circle with the center and radius of 1.
(2) Let the slope of L be, then the equation of L is
Because CACD, △CAD is
The distance from the center of the circle C (- 1, 0) to l is
Distance formula from point to straight line
17. (The full mark of this small question is 12) If, find the value of the number.
Solution:
or
or
When the time comes, the conditions are right;
When the time comes, the conditions are right.
Therefore, or
18. (Full score for this small item 12) Set complete sets of,,,,,
Solution:
19. (The full score of this small question is 12) Set the complete works, set the set, and seek,
Solutions:, and
20. (Full score for this small question 12) The range of values of the set known and realistic numbers.
Solution:
At that time,
At that time,
or
Therefore, the range of real numbers is
2 1. (the full mark of this small question is 12) The range of known sets, and real numbers.
Solution:
At that time,,;
When it is a single element set,
At this time;
When it is a collection of two elements,
So the range of real numbers is
22. (The full score of this small question is 14) The range of values of known sets, if and real numbers.
Solution: Method 1
, containing at least one negative number, that is, the equation has at least one negative root.
When the equation has two negative roots,
When the equation has a negative root and a positive root,
When the equation has negative roots and zero roots,
Or
So the range of real numbers is
Method 2
A that contains at least one negative number.
Take the complete work,
All the elements in it are
When,
So the range of real number A at that time was
Therefore, the value range of real number A at that time was