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When the inverse scale function y = K / X (k > 0) is known, the image passes through point a (2, m), where ab ⊥ X axis passes through point B, and the area of △ AOB is 3 [1] Find the direct [2] point C [x, y] of K and m on the image of inverse scale function y = K / x, find the value range of function value y when 1 ≤ x ≤ 3, take d [1,0] on the y-axis to connect BD, make a square with BD as the edge in the first quadrant, bdef asks whether e, f fall on the image of y = K / X
(1) S △ AOB = 0.5 × 2 × M = 3, then M = 3, and the coordinates of point a are substituted into the equation to obtain M = K / 2, k = 6
(2) When JX = 1, y = 6, x = 3, y = 2, then the value range of Y is 2 ≤ y ≤ 6
(3) D on the y-axis should be (0,1)
When the inverse scale function y = K / X (k > 0) is known, the image passes through point a (2, m), where ab ⊥ X axis passes through point B, and the area of △ AOB is 3
[1] Finding the direct relation between K and M
Point a (2, m), passing through point a, making ab ⊥ X axis at point B,
B(2,0)
S△AOB=OB*AB/2=2*m/2=3
M=3
The image with inverse scale function y = K / X (k > 0) passes through point a (2,3)
K = xy = 2 * 3 = 60), and the area of △ AOB is 3
[1] Finding the direct relation between K and M
Point a (2, m), passing through point a, making ab ⊥ X axis at point B,
B(2,0)
S△AOB=OB*AB/2=2*m/2=3
M=3
The image with inverse scale function y = K / X (k > 0) passes through point a (2,3)
K=xy=2*3=6
[2] Point C [x, y] is on the image of inverse scale function y = K / x, and the value range of function value y is obtained when 1 ≤ x ≤ 3
y=6/
=6/y
When 1 ≤ x ≤ 3
1≤6/y≤3
1/6≤1/y≤1/2
2≤y≤6
[3] Take d [1,0] on the y-axis to connect BD and make a square with BD as the edge in the first quadrant. Bdef asks if e and f fall on the image of y = K / X
d(0,1)
Side length of square bdef = √ (2 ^ 2 + 1 ^ 2) = √ 5
Straight line BD: y = - X / 2 + 1
Straight line de: y = 2x + 1
e(x,2x+1)
de=√[x^2+(2x+1-1)^2]=√5
x^2=1
X = ± 1 (negative value rounded off)
X=1
e(1,3)
1*3=3≠k
E is not on the image of y = K / X
Straight line F: y = 2x + B
Substitution point (B, 2)
b=y-2x=0-2*2=-4
y=2x-4
f(x,2x-4)
bf=√[(x-2)^2+(2x-4-0)^2]=√(5x^2-20x＋20)=√5
x^2-4x+3=0
X = 1 (point E)
X=3
f(3,2)
3*2=6=k
On the image of y = K / x, f asks why be =: y = 2x + 1
If log2 (- x) < x + 1 holds, the value range of X is______ .
It can be judged by drawing that f (x) = log2 (- x) and G (x) = x + 1 intersect at (- 1,0). The former is monotonically decreasing, and the latter is monotonically increasing. Therefore, log2 (- x) < x + 1 holds only when - 1 < x < 0, so the answer is: (- 1,0)
The curve in the right figure is a branch of the image with inverse scale function y = n + 7x
(1) Which quadrant is the other branch of the inverse scale function image in? What is the value range of constant n? (2) If the image of the first-order function y = - 23x + 43 intersects with the image of the inverse scale function at point a, and intersects with the X axis at point B, the area of △ AOB is 2
(1) From N + 7 & lt; 0, we get n & lt; - 7, that is, the range of constant n is n & lt; -7. (2) let y = 0 in y = - 23x + 43, then we get x = 2, that is, OB = 2. Let a be the vertical line of X axis through a, and the perpendicular foot is C, as shown in the figure. ∵ s △ AOB = 2, that is, 12ob · AC = 2, ∵ 12 × 2 × AC = 2, then we get AC = 2, that is, the ordinate of point a is 2. Substituting y = 2 into y = - 23x + 43, we get x = - 1, that is, a (- 1, 2). So 2 = n + 7-1, then we get n = - 9
The logarithm of log with base 2 (x + 2) is less than 1. What is the value range of X
log2(x+2)
log2 (x+2)
It is known that a (- 3,0) B (0,6) divides △ OAB into two parts with an area ratio of 1:3 through the straight line at the origin o, and finds the direction of the straight line
Draw your own image
The intersection of the straight line of positive proportion function and △ OAB is set as C
Make CD perpendicular to OA and C through C
CE through C is perpendicular to ob and E
Then the abscissa of the line is (- ce, CD)
Because the area of △ OAB is 3 * 6 / 2 = 9, the area ratio of 2 triangles is 1:3 (there are two cases)
So the areas of the two triangles divided by the straight line are (large: 6.75, small: 2.25)
1、 Let △ OCA: △ OCB = 1:3
Then CD = 2.25 * 2 △ 3 = 1.5
Similarly, CE = 2.25
So the straight line goes through (- 2.25,1.5)
So the linear analytical formula can be y = - 1.5x
2、 Let △ OCA: △ OCB = 3:1
Then CD = 6.75 * 2 △ 3 = 4.5
CE=2.25*2÷6=0.75
So straight over (- 4.5,0.75)
So the linear analytical formula can be y = - 1 / 6 X
Let the line AB be y = - 3x + 6, let the line passing through the origin be y = KX, K be negative, and the coordinates of point C intersected with the line AB be (a, b). Because the area ratio is 1:3 △ OAC area: △ OBC area is 1:3 or 3:1 (1 / 2 × 6a): (1 / 2 × 3b) = 1:3 or (1 / 2 × 6a): (1 / 2 × 3b) = 3:1, B = 3A, B = 2 / 3a are solved by substituting y = KX, k = - 3, k = - 2 / 3, and the straight line is y = - 3x or y = - 2 / 3x
Let the line AB be y = - 3x + 6, let the line passing through the origin be y = KX, K be negative, and the coordinates of point C intersected with the line AB be (a, b). Because the area ratio is 1:3 △ OAC area: △ OBC area 1:3 or 3:1 (1 / 2 × 6a): (1 / 2 × 3b) = 1:3 or (1 / 2 × 6a): (1 / 2 × 3b) = 3:1, B = 3A, B = 2 / 3a are solved by substituting y = KX, k = - 3, k = - 2 / 3, and the straight line is y = - 3x or y = - 2 / 3x
Let this line intersect AB at M
The analytic expression of this line is y = KX (K ≠ 0)
Do MH ⊥ OA over h at 1 ° M
∵S△OAB=½ ×OA ×OB=9
The area ratio of △ OAB is 1:3
∴1°S△AOM=9／4 S△BOM=27／4
∵S△AOM=½AO×MH
∴9／4=½×3×MH
MH=3／2
∵∠ AOB = ∠ AHM... Expand
Let this line intersect AB at M
The analytic expression of this line is y = KX (K ≠ 0)
Do MH ⊥ OA over h at 1 ° M
∵S△OAB=½ ×OA ×OB=9
The area ratio of △ OAB is 1:3
∴1°S△AOM=9／4 S△BOM=27／4
∵S△AOM=½AO×MH
∴9／4=½×3×MH
MH=3／2
∵∠AOB=∠AHM=90°
∠BAO=∠MAH
The two acute angles of a right triangle are complementary to each other, and I'll just write them here
The AOB and AHM are similar triangles
∵BO=6 MH=3／2
∴BO：MH=4:1
∵AO=3
∴AH=3／4
∴HO=9／4
And ∵ MH = 3 / 2
∴M(-9／4，3／2）
Substituting m (- 9 / 4, 3 / 2) into y = kx
Y = - 2 / 3x
2°S△AOM=27／4 S△BOM=9／4
Similarly, you can get y = - 6x
I weakly ask a method that you can use Pythagorean theorem to do
Known 0
It is equivalent to ㏒ taking a as the base (X-2) > 0
∴0＜x-2＜1
∴2＜x＜3
Given a (- 3,0), B (0,6), the area of △ OAB is divided into two parts of 1:3 by the straight line of origin o, and the function analytic formula of this straight line is obtained
Draw your own image
The intersection of the straight line of positive proportion function and △ OAB is set as C
Make CD perpendicular to OA and C through C
CE through C is perpendicular to ob and E
Then the abscissa of the line is (- ce, CD)
Because the area of △ OAB is 3 * 6 / 2 = 9, the area ratio of 2 triangles is 1:3 (there are two cases)
So the areas of the two triangles divided by the straight line are (large: 6.75, small: 2.25)
1、 Let △ OCA: △ OCB = 1:3
Then CD = 2.25 * 2 △ 3 = 1.5
Similarly, CE = 2.25
So the straight line goes through (- 2.25,1.5)
So the linear analytical formula can be y = - 1.5x
2、 Let △ OCA: △ OCB = 3:1
Then CD = 6.75 * 2 △ 3 = 4.5
CE=2.25*2÷6=0.75
So straight over (- 4.5,0.75)
So the linear analytical formula can be y = - 1 / 6 X
. there may be mistakes. It is estimated that the input will be sloppy, but the smart LZ should be able to find out (hope it is correct)... This problem must be solved in 2/
LZ will understand the idea
Solving inequality: log with X as the base, 1 / 2 as logarithm greater than 1
Analysis:
If x > 0 and X ≠ 1, then:
When x > 1, the original inequality can be reduced to log with X as the base 1 / 2 logarithm > log with X as the base x logarithm
Because the base x > 1, the solution is: X
Log with X as the base, 1 / 2 = 1, x = 1 / 2
Then the solution set greater than 1 is (0,1 / 2)
Given a (- 3,0), B (0,6), we divide △ OAB into two parts with an area of 1:3 through the straight line of origin O. the analytic function of the straight line of ball is given
In the first case, the line intersects AB (- 3 / 4,9 / 2), and the line is y = - 6x
In the second case, the line intersects AB (- 9 / 4,3 / 2), and the line is y = - 2 / 3x
Log with a as the base (1-1 / x) and a ≠ 1)
According to the meaning of the title:
Zero
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