Explain the relationship between the value of f()+f()+f () and 0.
The solution is +> 0, and the solution is >-.
∵f(x) is a monotonically decreasing function on r, ∴ f () < f (-).
And ∵f(x) is odd function, ∴ f ().
Similarly, f ()+f () < 0, f ()+f () < 0,
∴f()+f()+f() 0, Y > 0, and += 1, find the minimum value of x+y;
Given x
(3) If x, y∈(0, +∞) and 2x+8y-xy=0, find the minimum value of x+y. 。
Solution (1) ∵ x > 0, y > 0, += 1,
∴x+y=(x+y)
=++ 10≥6+ 10= 16.
The above equation holds if and only if =.
When+=1,∴ x = 4 and y =12, (x+y)min= 16.
(2)∵x0，
∴y=4x-2+=-+3≤-2+3= 1，
The above equation holds if and only if 5-4x=, that is, x= 1.
Therefore, when x= 1, ymax= 1.
(3) From 2x+8y-xy=0, 2x+8y=xy,∴+= 1,
∴x+y=(x+y)= 10++
= 10+2≥ 10+2×2×= 18,
Take the equal sign if and only if =, that is, x=2y,
2x+8y-xy=0,∴x= 12,y=6，
When x = 12 and y = 6, the minimum value of x+y is 18.
A paper mill plans to build a rectangular plane.
And the area is 162 square meters, and the depth of the tertiary sewage treatment tank is certain.
(The scheme is shown in the figure). If the unit price of the wall around the swimming pool is 400 yuan /m,
The unit price of the middle two partition walls is 248 yuan /m, and the unit price of the bottom of the pool.
80 yuan /m2, ignoring all the wall thickness of the swimming pool.
(1) Try to design the length and width of the sewage treatment tank to minimize the total cost and find out the one with the lowest total cost;
(2) If the length and width of the tank cannot exceed 16m due to the terrain limitation, try to design the length and width of the sewage tank to minimize the total cost and calculate the lowest total cost.
Solution (1) If the width of the sewage treatment tank is x meters and the length is m. 1 min.
The total cost f(x)=400×+248×2x+80× 162.
= 1 296x++ 12 960
= 1 296+ 12 960 3 points
≥1296× 2+12 960 = 38 880 (yuan),
If and only if x = (x > 0),
That is, when x= 10, take the equal sign. Five points.
∴ When the length is16.2m and the width is10m, the total cost is the lowest, and the lowest total cost is 38 880 yuan.
6.( 1) Suppose 0 < x
(2) The point (x, y) moves on the straight line x+2y=3 to find the minimum value of 2x+4y.
The solution (1) is called 0 < x.
∴x(4-3x)=(3x)(4-3x)≤=
"=" holds if and only if 3x=4-3x, that is, x=.
When x=, the maximum value of x(4-3x) is.
(2) It is known that the point (x, y) moves on the straight line x+2y=3, so x+2y=3.
∴2x+4y≥2=2=2=4.
"=" holds if and only if, that is, x=, y= and y =.
When x=, y= and y =, the minimum value of 2x+4y is 4.
8. Solve the inequality ≥(x2-9)-3x.
Solving the original inequality can be transformed into -x2+≥ x2-3x,?
That is 2x2-3x-7≤0.
Solve equation 2x2-3x-7=0 and get x=. ?
So what is the solution set of the original inequality?
9. known inequality (a∈R).
(1) Solve this inequality about x; ?
(2) If the inequality holds when x=-a, find the value range of a..
Solving the original inequality (1) is equivalent to (ax- 1) (x+ 1) > 0.
(1) when a=0, from -(x+ 1) > 0, x.
② when a > 0, the inequality becomes (x+ 1) > 0,?
Get x; ?
③ When a < 0, the inequality becomes (x+1) < 0; ?
if
If =- 1, that is, a=- 1, the inequality solution set is empty; ?
If >-1, i.e.
To sum up,?
When a man
When a=- 1, the original inequality has no solution; ?
When-1 < a < 0, the solution set is; ?
When a=0, the solution set is {x | x.
When a > 0, the solution set is. ?
(2) When ∵ x =-a,?
That is, -a+ 1 < 0,?
∴ a > 1, that is, the range of a is a > 1.
Straight line: It is known that real numbers X and Y satisfy y=x2-2x+2 (- 1≤x≤ 1).
Try to find the maximum and minimum values of.
The geometric meaning of the solution shows that it represents the slope k of a straight line passing through a fixed point P(-2, -3) and any point (x, y) on the curve segment AB, as shown in the figure: kPA≤k≤kPB,
It is known that a (1, 1), b (- 1, 5),
∴≤k≤8，
Therefore, the maximum value of is 8 and the minimum value is.
13 (1) crosses point P (3 3,2), and the intercepts on the two coordinate axes are equal;
(2) After passing through point A (-1, -3), the inclination angle is equal to twice of the straight line y=3x.
Solution (1) Method 1 Let the intercept of a straight line L on the X and Y axes be a,
If a=0, that is, l passes through points (0,0) and (3,2),
The equation of ∴l is y=x, which means 2x-3y=0.
If a≠0, let the equation of L be,
*∴l border crossing points (3, 2),
∴a=5, ∴l's equation is x+y-5=0,
To sum up, the equation of the straight line L is 2x-3y=0 or x+y-5=0.
The second method knows from the meaning of the question that the slope k of the straight line exists and k≠0,
Let the linear equation be y-2=k(x-3),
Let y=0, x=3-, x=0, y=2-3k,
From the known 3-=2-3k, we can get k=- 1 or k=,
∴ The equation of the straight line L is:
Y-2=-(x-3) or y-2=(x-3),
That is x+y-5=0 or 2x-3y=0.
(2) After passing through point A (-1, -3), the inclination angle is equal to twice of the straight line y=3x.
(2) It is known that the inclination angle of a straight line y=3x,
Then the inclination of the straight line is 2.
∵tan=3,∴tan2==-.
Go straight to point a (-1, -3),
Therefore, the linear equation is y+3=-(x+ 1),
That is 3x+4y+ 15=0.
15: It is known that the coordinates of both ends of the straight line PQ are (-1, 1) and (2,2) respectively. If the straight line L: x+my+m = 0 intersects with the straight line PQ, find the range of m. 。
Solution: A straight line x+my+m=0 passes through point A (0,-1).
kAP==-2，kAQ==，
Then-≥ or -≤-2,
∴-≤m≤ and m≠0.
When ∵m=0, the straight line x+my+m=0 intersects with the straight line PQ.
The value range of ∴m is -≤m≤
Method 2 The linear equation through point P and point Q is as follows
Y- 1=(x+ 1), which means y=x+,
Substitute x+my+m=0,
Finishing, x=-.
It is known that-1≤-≤2,
The solution is -≤m≤
Two points A (- 1 2) and B (m, 3) are known in 16.
(1) Find the equation of straight line AB;
(2) Given the real number m∈, find the inclination range of the straight line AB.
Solution (1) When m=- 1, the equation of straight line AB is x=- 1.
When m≦- 1, the equation of straight line AB is y-2=(x+ 1).
(2)① When m=- 1, =;
② When m≦- 1, m+ 1∈,
∴k=∈(-∞,-〕∪，
∴∈.
Comprehensive ① ② Know the inclination of straight line AB ∈.
17 Find the equation of line l 1:y=2x+3 symmetry about line L: y=x+ 1.
A solution
It is known that the coordinates where the straight line l 1 intersects with l are (-2,-1).
∴ Let the equation of l2 line be y+ 1=k(x+2),
That is kx-y+2k- 1=0.
Take any point on the straight line L (1, 2),
The distance from the topic knowledge point (1, 2) to the straight line L 1 is equal to L2.
Distance formula from point to straight line
=,
The solution is k=(k=2),
The equation of l2 line is x-2y=0.
Method 2 sets a point P(x, y) on a straight line,
Then there must be a point P 1(x0, y0) on the straight line l 1 that is symmetrical with the point p 。
The topic is: the straight line PP 1 is perpendicular to the straight line L, and the midpoint of the line segment PP 1.
P2 is on the straight line L 。
∴, deformity,
Substituting the straight line l 1:y=2x+3, we get x+ 1 = 2x (y- 1)+3.
X-2y=0。
So the linear equation is x-2y=0.
18 two kinds of steel plates with different sizes can be cut into three specifications: a, b and c according to the table below.
Specification b specification c specification
The first steel plate 2 1 1
Second steel plate 1 2 3
A site needs 15, 18 and 27 pieces of finished products respectively. How to cut these two kinds of steel plates can get three kinds of finished products, and the number of steel plates used is the least.
The solution needs x first steel plates and y second steel plates, and the total number of steel plates is z, z=x+y,?
Constraints include:
Make a feasible domain as shown in the figure:?
Let z=0 as a reference line l:y=-x, and find a parallel moving line l in the feasible region. The intersection point A of the straight line x+3y=27 and the straight line 2x+y= 15 can minimize z, because they are not integers, and in the optimal solution (x, y), both x and y must be integers, and the point A in the feasible region is not. ?
Draw a grid in the feasible region, and find that the straight line passing through the whole point in the feasible region and closest to point A is x+y= 12, and the whole points passing through are B (3,9) and C (4 4,8), which are all optimal solutions.
There are two ways to cut three kinds of steel plates and minimize the number of two kinds of steel plates:
The first cutting method is to cut three first steel plates and nine second steel plates; ?
The second cutting method is to cut four first steel plates and eight second steel plates; ?
Both methods require at least two kinds of steel plates *** 12.
The curve 19 is shown in the figure. It is known that p (4 4,0) is a point x2+y2=36 on the circle, and A and B are two moving points on the circle.
And satisfy ∠ APB = 90, and find the trajectory equation of the vertex Q of the rectangular APBQ.
Let the midpoint of AB be R, the coordinates be (x 1, y 1), and the coordinates of Q point be (x, y).
Then in rt delta ABP,
|AR|=|PR|，？
And because r is the midpoint of the chord AB, according to the vertical diameter theorem, there is?
In Rt△OAR, | AR | 2 = | AO | 2-| or |2=36- ().
|AR|=|PR|=，？
So there is (x 1-4)2+=36- ().
That is -4x 1- 10=0.
Because r is the midpoint of PQ.
So x 1=, y 1=. ?
Substitute in the equation -4x 1- 10=0, what?
- 10=0.？
After sorting, x2+y2=56.
This is the trajectory equation of Q point.
20. Given circle C 1: (x+3) 2+Y2 = 1, circle C2: (x-3) 2+Y2 = 9, the moving circle m is circumscribed with circle C 1 and circle C2 at the same time, and the trajectory equation of the moving center m is found.
The solution is shown in the figure. Let the moving circle m and circle C 1 and circle C2 be circumscribed at point A and point B, respectively. According to the necessary and sufficient conditions of circumscribed two circles, we can get?
|MC 1|-|AC 1|=|MA|，？
|MC2|-|BC2|=|MB|。 ?
Because |MA|=|MB|,?
So | mc2 |-| MC1| = | bc2 |-AC1| = 3-1= 2.
This shows that the difference between the distances from the moving point m to the two fixed points C2 and C 1 is a constant of 2.
According to the definition of hyperbola, the trajectory of moving point M is the left branch of hyperbola (the distance from point M to C2 is large, but the distance from point M to C 1 is small), where a = 1, c = 3, b2=8, the coordinate of point M is (x, y), and its trajectory equation is x2-=/kloc-0.
Lines and circles
17.( 13 point) A straight line passes through a point, and the triangle enclosed by two coordinate axes has an area of 5. Find the equation of this straight line.
18.( 13 minutes) The circle crosses a point and is tangent to a straight line with the center of the circle on it. Find its equation.
19.( 13 minutes) Find the equation of the circle with the smallest intersection area between a straight line and a circle.
20.( 13 point) Two points are known, which makes,, a arithmetic progression with a tolerance less than zero.
What is the trajectory of (1) point?
(2) If the coordinate of the point is, record it as the included angle with, and find.
2 1.( 12 minutes) As shown in the figure, the radius of the circle and the radius of the circle are both equal to 1, and the passing points are the tangent points of the circle and the circle respectively, so as to try to establish a plane rectangular coordinate system and find out the trajectory equation of the departure point.
22.( 12 points) Given a straight line and a circle, is there a real number that makes the self-luminescence tangent to the circle after being reflected by the straight line, and if so, the value obtained; If it does not exist, please explain why.
answer
17. Solution: Let the linear equation, and then
or
A linear equation is or
18. Solution: Let the center of the circle be here, and then
Or 9
∴ or
∴ The equation of a circle is or
19. Solution: The equation that can determine a circle is
that is
The center is
Obviously, when the center of the circle is on the chord of the intersection, that is, on the straight line, the radius of the circle is the smallest, so the area is the smallest.
∴
The equation for finding a circle is
20. Solution: If (1) is set, then,
that is
Tolerance is less than 0
∴
So the trajectory of a point is the right half of a circle with the origin as the center and the radius (excluding the endpoint).
(2)
∴
∴
∴
2 1. Solution: Take the straight line as the axis and the middle vertical line as the axis, and establish the coordinate system.
rule
That is all right
that is
22. solution: suppose there is such a real number, and the symmetry point about is
∴ The equation of the straight line where the reflection line is located is
that is
The reflection line is tangent to circle VIII.
Finishing: Ⅷ
∴ There is a real number that meets the conditions.
If the curve and have two common points, find the range of numbers.
Analysis: "the curve has two common points" is transformed into "the equation has two different solutions", so as to study the number of solutions of a quadratic equation. If we give the approximate shapes of the two curves, we may get some enlightenment.
Solution 1: from:
∵,∴,
Namely.
So that the above equation has two different non-negative real roots.
There are:
It's also VIII
∴ The solution is:
The range of real numbers is.
The curve of Solution 2 is a polyline with the axis symmetry and the vertex at the origin, representing a straight line with a slope of 1 passing through this point. As can be seen from the following figure, the right branch of the polyline did not intersect with the straight line at that time, so there was only one intersection point between the two curves. At that time, both branches of the straight line and the broken line crossed, so the two straight lines had two different intersections.
Note: the better solution to this kind of problem is the second scheme, that is, to explore by combining numbers and shapes. If the condition of the question is changed to "",please explore it yourself.
Ellipse:
Make two perpendicular straight lines at the intersection point. If the intersection axis exists and the intersection axis exists, find the trajectory equation of the midpoint of the line segment.
Solution: Connect, set, and then.
∵
∴ is a right triangle.
Proceed from that nature of right triangle
that is
P is a point on the square of the ellipse X /a +y /b = 1, and F 1 is its focus. It is proved that the circle with the diameter PF 1 is tangent to the circle with the diameter of the major axis.
The simplified trajectory equation is
Ellipse x 2/a 2+y 2/b 2 =1(a > b > 0) two focal points are F 1, F2, and point p is on ellipse C. Find │PF 1│=4/3, │PF2│= 14/3, PF1⊥ F1.
│PF 1│+│PF2│=2a, so 2a=6 a=3.
pf 1⊥f 1f2 so(2c)2+(4/3)2 =( 14/3)2。
C^2=45/9
a^2=b^2+c^2
b^2=4
So the equation is x 2/9+y 2/4 =1.