Answers to the mathematics examination questions of Tianjin senior high school entrance examination in 2004.
1 b . 2 a . 3 c . 4 c s . d . 6d . 7 c . 8 a . 9 b . 1 o . b
1 1 . x≤6. 1 2.2 . 1 3.7 . 1 4.3 . 1 5. 1 6。 For example, 1+
19. Among these 20 data, 80 appears 7 times, which is the most frequent, that is, the mode of this set of data is 80; The 20 data in the table can be regarded as arranged in order from small to large.
The middle two data are both 70, that is, the median of this set of data is 70; The average value of this data set is 72 (minutes).
Answer: The scores of 20 students are 80, 70 and 72 respectively.
20. After testing, x=2, x=- 1 and X = 1 are the roots of the original equation. The roots of the original equation are x 1=2, x2=-l, x3= 1+, X4 = 65438.
2 1.( 1) The parabola y=x2+k+c has only one intersection with the X axis, and the equation x2+bx+c=O has two equal real roots, that is, B2-4c = o① (1) The coordinates of the intersection a are (2,0), ∴ 4+. (2) According to (1), the analytical formula of parabola is y = x2-4x+4. When x=O, y=4 and the coordinate of point b is (o, 4). At Rt△OAB, from OA=2, OB=4, AB = = 2.
22.( 1) ∵ point p (x0,3) is on the image of linear function y=x+m ∴3=x0+m, that is, m = 3-x0. And the point p (x0 x0,3) is on the inverse proportional function y = (m+65438.
(2) From (1), it is concluded that m=3-x0=2, the analytical formula of ∴ linear function is y=x+2, and the analytical formula of inverse proportional function is y = 3/x. 。
23.( 1) connect OC∴C as the tangent point, OC ∴ PC, △POC as the right triangle ∴OC=OA= 1, PO=PA+AO=2, ∴ SIN ∠ P =
(2)∵BD⊥PD, ∴ in Rt△PBD, from ∠ P = 30, PB=PA+AO+OB=3, BD = 3/2. The diameter of the connection AE∴AB is \ o.
24. in Rt△ABC, BC=d 1, ∠ACB=∠θ, A B=BC? tan∠ACB，AB=d 1？ Tan θ 1 = 4 Tan 40。 In Rt△ABD, BD==d2, ∠ ADB = ∠ θ. AB = D2？ Tan θ 2 = D2TAN 36。 Therefore, 4TAN 40 = D2TAN 36, D2 ≈ 4× 1. 155 = 4.620.
∴ D2-D1≈ 4.620-4 = 0.620 ≈ O.62. Answer: The floor occupied by stairs has increased by 0.62m. 。
25.( 1) As shown in the figure, in ⊙O, extend the intersection of AO and ⊙O at point D and connect DM.
∵AD is the diameter ⊙ O, ∴∠ AMD = 90. ∴∴ AB is the radius of⊙⊙⊙A, MN is the tangent of⊙⊙⊙A, and B is the tangent point.
∴A B⊥MN, where ∠ abm = 90.
In Rt△ABM and Rt△AMD, ∠BAM=∠DAM, ∴Rt△ABM∽Rt△AMD, AM2=AB? A D.
From the vertical diameter theorem, am = an, ab = r.ad = 2r ∴ am? AN = 2Rr
(2) As shown in the figure, a P? AQ=2Rr hold.
Extend the intersection of A0 and ⊙O at point D, connect DQ and AC∴PQ as the tangent of ⊙A, and C as the tangent point.
∴∠∠ Feijiatai = 90. If the AD is ⊙∣∣∣∣∣∣,
∴Rt△ADQ∽Rt△APC,AP？ AQ = a.d.? AC∴AD=2R,AC=r。 ∴AP？ AQ=2Rr。
26.( 1) Fill in the form as follows:
x-(-3)\(-2)\(- 1)\(0)\( 1)\(2)\(3)
y 1 = 2x-(-6)\(-4)\(-2)\(0)\(2)\(4)\(6)
y2 = x2+ 1-( 1O)\(5)\(2)\( 1)\(2)\(5)\( 10)
(2) It is proved that yl-y2=-(x- 1)2≤O, and y 1≤y2 holds when the independent variable x takes any real number;
(3) If the solution is known, the image of quadratic function y3=ax2+bx+c passes through point (-5,2), 25a-5b+C = 2. (1) was obtained.
When x=l, y 1=y2=2, y3 = a+b+c; If yl≤y3≤y2 holds for the independent variable X, then 2≤a+b+c≤2, ∴ a+b+c = 2.
From ① ②, b=4a, c=2-5a, ∴ y3 = ax2+4ax+(2-5a) are obtained.
When y 1≤y3, there is 2x≤ax2+4ax+(2-5a), that is, ax2+(4a-2)x+(2-5a)≥0. If the quadratic function y=ax2+(4n-2)x+(2-5a) for all real numbers x, 0 {4a-2) 2-4a (2-5a) ≤ 0. That is, A >；; 0，(3a- 1) 2 ≤ 0。 A = 1/3。 When y3≤y2, there is ax2+4ax+(2-5a)≤x2+ 1, that is, (1-a) x2-4ax. 0( 1-4a)2-4( 1-a)(5a- 1)≤0。 That is, a < 1 and a= 1/3. To sum up, a = 1/3, b=4a=4/3, c=2-5a= 1/3.
∴ There is a quadratic function y3=x2/3+4x/3+ 1/3. In the range of real numbers, y 1≤y3≤y2 holds for the same value of x.