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Take any two numbers P and Q (P ≠ q) from the four numbers 2, 3, 4 and 5 to form the functions y = px-2 and y = x + Q, and make the image of these two functions If the intersection point is on the right side of the straight line x = 3, how many pairs of ordinal numbers are there in total?
Let the two equations be equal, px-2 = x + Q, then the X coordinate of the intersection of the two functions is (Q + 2) / (p-1), as long as the equation is greater than 3, then there are (P, q) = 2,4; 2,5; 3,5; three pairs
Take any two of the four numbers 2345, P and Q (and P is not equal to q), form the functions y = px-2 and y = x + Q, and make the intersection of the two function images on the right side of the straight line x = 2, then such ordinal number pairs (P, Q) are_____ ?
If y = px-2 and y = x + Q cancel y, x = (Q + 2) / (p-1) > 2, because P > 1, we can get P
Take any two numbers P and Q (P ≠ q) from the four numbers 2, 3, 4 and 5 to form the functions Y1 = px-2 and y2 = x + Q, so that the intersection of the two function images is on the left side of the straight line x = 2, then such ordered array (P, q) has ()
A. 4 groups B. 5 groups C. 6 groups D. uncertain
Let px-2 = x + Q, the solution is x = q + 2p − 1, because the intersection point is on the left side of the line x = 2, that is, q + 2p − 1 ＜ 2, then q ＜ 2p-4 is sorted out. By substituting P = 2, 3, 4, 5 respectively, the corresponding value of Q can be obtained. The ordinal number pairs are (4, 2), (4, 3), (5, 2), (5, 3), (5, 4), (5, 5), and because P ≠ Q, so (5, 5) is rounded off, and there are 5 pairs that satisfy the condition
From the four numbers 2.3.4.5, take any two numbers P and Q (P is not equal to q) to form the function, Y1 = px-2 and y2 = x + Q, so that the image intersection is on the left side of the line x = 2, and ask PQ group
Let's make it clear that they are all increasing functions, upward. Intersection point, Y1 = Y2. Then, the intersection point is on the left side, indicating that when x = 2, Y1 > Y2 (the increase of Y1 is greater than Y2), substitute x = 2 into the equation, and let Y1 > Y2, that is, 2p-2 > 2 + Q, after simplification, P > 2 + Q / 2. The answer is very simple, q = 2, P = 4 or 5; q = 3, P = 4 or 5; q = 4, P = 5; q = 5
RELATED INFORMATIONS
- 1. Triangle ABC is an equilateral triangle, PA is perpendicular to plane ABC, and D is the midpoint of BC
- 2. In equilateral triangle ABC, is de / / BC. Angle ade an equilateral triangle?
- 3. It is known that △ ABC and △ ade are equilateral triangles, and points a and E are on the same side of BC (1) As shown in Figure 1, if point D is on BC, write the quantitative relationship among line segments AC, CD and CE, and prove it; (2) as shown in Figure 2, if point D is on the extension line of BC, and other conditions remain unchanged, directly write the quantitative relationship among AC, CD and CE
- 4. As shown in the figure, D in △ ABC is any point on edge BC, de and DF are bisectors of △ ADB and △ ADC respectively, connecting EF. Try to judge the shape of △ def and explain the reason
- 5. In the triangle ABC, the angle BAC = 90 ° AB = AC, take a point D in the triangle to make the angle abd = 30 ° and BD = Ba, AE perpendicular to BD and e to find the bisector angle EAC of AD
- 6. As shown in the figure, in the triangle ABC, the angle ABC is 90 degrees and the angle c is 30 degrees. Rotate the triangle ABC counterclockwise around point a to the position of triangle ab'c ', and point B falls on AC The value of tancdb 'is?
- 7. In RT △ ABC, ∠ C = 90 degree, ∠ a = 30 degree, then BC: AC: ab = what
- 8. Δ ABC ≌ Δ CDA, AB and CD, BC and DA are corresponding edges. Write other corresponding edges and corresponding angles
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- 10. As shown in the figure, a, B and C are three points that are not on the same straight line, AA ′‖ BB ′‖ CC ′, and AA ′ = BB ′ = CC ′. Prove: plane ABC ‖ plane a ′ B ′ C ′
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- 12. If vector AP = 1 / 3 vector Pb and vector AB = Λ vector BP, then the value of Λ is?
- 13. Proof: as shown in Figure 9, in the triangle ABC, ab = AC,
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- 16. It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o
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- 19. In ABC, AB is equal to AC, BF is the height on the side of AC, and the angle FBC is equal to 1 / 2 angle BAC
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