Review the examination papers at the end of the first semester of the 2007-2008 school year.
Senior one math test questions
(Examination time: 120 minutes, total score 160 points)
Precautions:
1, this paper is divided into two parts, the first volume is multiple-choice questions, and the second volume is fill-in-the-blank questions and solutions.
2. The answers to all questions are filled in the answer sheet (in schools that use the answer sheet for multiple-choice questions, the answers to multiple-choice questions are directly filled in the answer sheet), and the answers written on the test paper are invalid.
Formula: cone volume v = surface area of SH sphere s = 4πR2；; The lateral area of the cone S=πrl.
First, fill in the blanks:
1. Given the coordinates of the three vertices of the parallelogram ABCD are A (- 1, 2,3), B (2 2,2,3), C (1, 5, 1), and the coordinates of the fourth vertex D are.
2. Use "
.
3. Evaluation: (LG5) 2+LG2× LG50 = _ _ _ _ _ _ _ _ _ _ _ _ _ _
4. It is known that a = {(x, y) | x+y-2 = 0}, b = {(x, y) | x-2y+4 = 0}, c = {(x, y) | y = 3x+b}, and if (a ∩ b).
5. Assuming that the function is an even function and is a subtraction function at (0, +∞), the integer value is.
6. As shown in the figure, assume that ⊥, ⊥ and vertical feet are b and d respectively. If you add a condition, you can export BD⊥EF. There are three situations:
① ⊥ ;
(2) On the same straight line with the projection of the inner side;
③ ‖ .
One of the conditions that can be added is. (Fill in the serial numbers of all conditions that you think are correct)
7. The maximum value of (1) is
(2) The minimum value of the function is
8. If, are two vectors that are not * * * and are known and have three * * * lines, then the real number is =
9. If () is known and || = || (), then.
10. For the function, the following four propositions are given: ① existence of (0,) make; (2) The existence of (0,) makes it constant; (3) There is r, which makes the image of the function symmetrical about the axis; ④ The image of the function is symmetrical about (,0). The serial number of the correct proposition is
The minimum positive period of the 1 1. function is.
12. It is known that if it is a parallelogram OACB, the included angle with is _ _ _ _ _ _ _.
Second, solve the problem: (the answer should write the necessary text description, prove the process or calculus steps. )
13.( 14 points) Known function f (x) = (a >; 0, a≠ 1, a is a constant, x∈R).
(1) If f(m)=6, find the value of f (-m);
(2) If f( 1)=3, find the sum of f(2).
14.( 18 points) Known function.
(1) Judge the monotonicity of f(x) in the world and prove your conclusion;
(2) If A={y | y=f(x),}, B=[0, 1], try to judge the relationship between A and B;
(3) If there are real numbers A and B (A
15. It is known that the function period defined on R is
(1) Write the expression of f(x);
(2) Write the monotonic increasing interval of the function f(x);
(3) Explain how the image of f(x) is transformed from the image of function y=2sinx.
16. Known vectors.
(1) If A, B and C cannot form a triangle, be realistic from the conditions that M should meet;
② If △ABC is a right triangle, find the value of m 。
17. Known functions
(1) Find the minimum positive period and maximum value of the function;
(2) The image of the function can be translated from the image of the function according to a certain vector A, and the vector A satisfying the conditions can be found.
18.( 1) If the sum of two right angles of a right triangle is 12, find the minimum value of its perimeter p;
(2) If the inner angle of the triangle is 0 and the perimeter is constant p, find the maximum value of the area s;
(3) In order to study that the side lengths A, B and C satisfy 9? Answer? 8? b？ 4? c？ Does the triangle of 3 have the largest area? The existing solutions are as follows: 16S2? (a？ b？ c)(a？ b？ c)(a？ b？ c)(？ Answer? b？ c)
[(a？ b)2？ c2][c2？ (a？ b)2]c4？ 2(a2？ b2)c2？ (a2？ b2)2
【c2？ (a2？ b2)]？ 4a2b2
And then what? 【c2？ (a2？ b2)]？ 0，a2？ 8 1，b2？ 64, what about S? However, the condition of the equal sign is c2? a2？ b2，a？ 9，b？ 8, what about c2? 145, and 3? c？ 4 contradiction, so this triangle has no maximum area.
Is the above answer correct? If not, please give the correct answer.
(Note: 16S2? (a？ b？ c)(a？ b？ c)(a？ b？ c)(？ Answer? b？ C) Helen's formula is called triangular area, which has been proved to be correct)
Reference answer:
1.(-2,9, 1) 2.log0.53 & lt& ltlog23 & lt0.5- 1 3. 1
4.2 5. 1 or 3 6. ①②
7.( 1) (2) 8.-8 9. 10.①,③,④
1 1.3 12.
13. 1)∵f(-x)= =f(x)
∴f(x) is an even function
∴f(-m)=f(m)=6(2)∫f( 1)= 3 ∴a+ = 6
∴ =36 ∴ =34
∴f(2)=34/2= 17∶=8,∴
∴ ,
14. 1) f (x) at the top is increasing function.
∫x≥ 1，f (x) = 1-
For any x 1, x2, when 1 ≤ X 1
f(x 1)-f(x2)=( 1-)-( 1-)=
∵x 1x 2 & gt； 0，x 1-x2 & lt； 0
∴
∴f(x 1)<； f(x2)
F (x) is playing an increasingly important role in the world.
(2) It is proved that f(x) monotonically decreases and monotonically increases in [1, 2].
It is found that a = [0, 1] indicates that a = b (3) ∫ a.
∵f(x)≥0，∴ma≥0，a≠0，∴a>； 0
1 0 & lt； A & ltB≤ 1, from the image, f(x) decreases when x [a, b],
∴ and a < B contradiction 20; 0
This is also inconsistent with the topic; 3 1≤a & lt； B, f(x) When x [a, b] increases.
It is known that mx2-x+ 1=0 has two unequal real roots.
By, by
All in all,
15. Solution: (1)
(2) in each closed interval
(3) Move the image of function y=2sinx to the left by one unit, then keep the vertical coordinates of all points on the obtained function image unchanged and shorten the horizontal coordinates to the original position.
16. Solution ① Known vector
If a, b and c cannot form a triangle, the lines of these three points,
old friend
∴ Real number, conditions to be met
② If △ABC is a right triangle and (1)∠A is a right angle, then,
solve
17. Solution: (1)
that is
(2) Assuming that the function image can be obtained by vector translation of the image,
Then there is
All the required vectors can be written as,
18. Solution: (1) Let two right-angled sides of a right triangle be X and Y, then x+y= 12. So the hypotenuse length z is satisfied.
Therefore, when x=6 and zmin=, then the minimum value of the perimeter of the right triangle is
(2) Let the length of the middle side of the triangle be X, and the included angle between the two sides of Y be
So the perimeter of this triangle
If and only if x=y, the equal sign holds, then,
Therefore, the maximum area of a triangle is
③ Incorrect
And, then, that is, the condition for the equal sign to be established is
, b=8, c=4, then, satisfied, so when the triangle is a right triangle with a side length of 4,8, its area reaches the maximum 16.