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The first chapter is the concept of set and function.

First, multiple choice questions

1. Let the complete set u = {(x, y)| x∈R, y∈R} and the set m =,

P = {(x, y) | y ≠ x+ 1}, then CU(M∪P) is equal to ().

A.B.{(2，3)}

C.(2，3)d . {(x，y)| y=x+ 1}

2. If a = {a, b}, B A, then the number of elements in set b is ().

A.0b. 1 c.2d.0 or 1 or 2.

3. The common point of the image of the function y = f (x) and the straight line x = 1 is ().

A. 1b.0c.0 or 1d. 1 or 2

4. Let the function f (x) = 2x+3 and g(x+2) = f (x), then the expression of g(x) is ().

a . 2x+ 1b . 2x- 1c . 2x-3d . 2x+7

5. Given the image of function f (x) = ax3+bx2+CX+d, then ().

a . b∑(-∞，0)b . b∑(0， 1)

c . b∑( 1，2)d . b∑(2，+∞)

6. Let the function f (x) =, if f (-4) = f (0) and f (-2) =-2, then the number of solutions of the equation f (x) = x is f (x) =.

a . 1b . 2c . 3d . 4

7. Let the set a = {x | 0 ≤ x ≤ 6} and b = {y | 0 ≤ y ≤ 2}, and the following corresponding rules f from a to b do not map ().

a . f:x→y = XB . f:x→y = xc . f:x→y = xd . f:x→y = x

8. There are the following four propositions:

① The image of even function must intersect with the Y axis;

② The image of odd function must pass through the origin;

③ The image of even function is symmetrical about Y axis;

④ The function that is both odd function and even function must be f (x) = 0 (x ∈ r).

The number of correct propositions is ().

a . 1b . 2c . 3d . 4

9. The function y = x2-6x+ 10 is () in the interval (2,4).

A. Decreasing function B. Increasing function

C. decrease first and then increase D. increase first and then decrease

10. If the image symmetry axis of the quadratic function y = x2+bx+c is x = 2, there is ().

a . f( 1)< f(2)< f(4)b . f(2)< f( 1)< f(4)

c . f(2)< f(4)< f( 1)d . f(4)< f(2)< f( 1)

Second, fill in the blanks

1 1. In the set {3, x, x2-2x}, the condition that x should meet is.

12. if there is only one element a in the set A = {x | x2+(A- 1) x+b = 0}, then A = _ _ _, and B = _ _ _.

13. Build a cuboid uncovered pool with a volume of 8 m3 and a depth of 2 m. If the cost per square meter of the bottom and wall of the pool is 120 yuan and 80 yuan respectively, the minimum total cost of the pool is RMB.

14. If f (x+ 1) = x2-2x, then f (x) =; f(x-2)=。

15.Y = (2A- 1) X+5 is a subtraction function, so find the range of a 。

16. let f(x) be the odd function on R. when x∈ [0, +∞], f (x) = x (1+x3), then when x∈.

(-∞，0)，f (x) =。

Third, answer questions.

17. the known set A = {x ∈ r | AX2-3x+2 = 0}, where a is a constant and A ∈ r 。

① If A is an empty set, find the range of A;

② If there is only one element in A, find the value of A;

③ If there is only one element in A at most, find the range of A. 。

18. Given m = {2, a, b}, n = {2a, 2, b2}, m = n, find the values of a and b. 。

19. It is proved that f (x) = x3 is a increasing function on R. 。

20. Determine the parity of the following function:

( 1)f(x)= 3 x4+； (2)f(x)=(x- 1)；

(3)f(x)=+； (4)f(x)= +

The first chapter is the concept of set and function.

Reference answer

First, multiple choice questions

1.B

Analysis: Set M is a set composed of points on the straight line Y = X+ 1 except point (2,3). The set p is a set composed of points on the coordinate plane that are not on the straight line Y = X+ 1, so M P is a set composed of all points on the coordinate plane that do not contain the point (2,3). Therefore, copper (melting point)

Cu (MP) = {(2，3)}。 So, B.

2.D

Analysis: ∫ A subset includes, {a}, {b}, {a, b} ∴ Set B may be one of {a}, {b}, {a, b}, ∴ Select D.

3.C

Analysis: According to the definition of function, it is possible that the image of function y = f (x) does not intersect with the straight line x = 1. If there is intersection, then x = 1 has only one function value.

4.B

Analysis: ∫g(x+2)= 2x+3 = 2(x+2)- 1, ∴ g (x) = 2x- 1.

5.A

Analysis: Be good at analyzing the characteristics of a function from its image.

Solution 1: Let f (x) = ax (x-1) (x-2) = ax3-3ax2+2ax, and the comparison coefficients are b =-3a, c = 2a and d = 0. From the image of f (x), we can know that f (3) > 0.

F (3) = 3a (3- 1) (3-2) = 6a > 0, that is, a > 0, so B < 0. So the correct answer is a.

Solution 2: substitute X = 0, X = 1 and X = 2 into F (X) = AX3+BX2+CX+D to get D = 0 and A =

- b，c =-b . ∴f(x)=b(- x3+x2-x)=-[(x-)2-]。

According to the function image, when x ∈ (-∞, 0), f (x) < 0, and [(x-) 2-] > 0, ∴ b < 0.

When x ∈ (0, 1), f (x) > 0, and [(x-) 2-] > 0, ∴ b < 0.

When x ∈ (1, 2), f (x) < 0, and [(x-) 2-] < 0, ∴ b < 0.

When x∈(2, +∞), f (x) > 0, and [(x-) 2-] > 0, ∴ b < 0.

So b ∈ (-∞, 0).

6.C

Solution: f (-4) = f (0), f (-2) =-2,

Ok, Ⅷ

∴f(x)=

X =- 1 or x =-2; X = 2。

To sum up, the number of solutions of equation f (x) = x is three.

7.A

Solution: Take element 6 in set A, and under the action of F: X→ Y = X, you should get image 3, but 3 is not in set B =

{y | 0 ≤ y ≤ 2}, so the answer is a.

8.A

Tip: ① No; ② No, because the domain of even function or odd function may not contain 0; 3 correct; ④ No, functions that are both odd function and even functions can also be f (x) = 0, x ∈ (-a, a). So the answer is a.

9.C

Analysis: This problem can give a vivid explanation of the function y = x2-6x+ 10. According to the image, the function first decreases and then increases in (2,4). The answer is C.

10.B

Analysis: ∫ axis of symmetry x = 2, ∴ f (1) = f (3). ∫y monotonically increases in [2, +∞],

∴ f (4) > f (3) > f (2), so f (2) < f (1) < f (4) The answer is B.

Second, fill in the blanks

1 1.x ≠ 3 and x≠0 and x ≠- 1.

Analysis: according to the different elements that make up the set, X is satisfied.

The solution is x≠3 and x≠0 and x ≠- 1.

12.a=，b=。

Analysis: According to the meaning of the question, if the two roots of the equation X2+(A- 1) X+B = 0 are equal and X = A, then △ = (A- 1) 2-4b = 0 ①, and substitute X = A into the original equation to get A2+(a-1).

13. 1 760 yuan.

Analysis: Let the length of the bottom of the pool be x meters and the total cost of the pool be y yuan. It is known that the bottom area is 4 m2 and the bottom width is m 。

The bottom cost y 1 = 120× 4 = 480.

Shell cost Y2 = (2× 2x+2× 2x )× 80 = (4x+)× 80.

The total cost of the pool is y = y 1+y2 = 480+(4x+) × 80.

That is, y = 480+320 (x+)

=480+320 .

When =, that is, x = 2, the minimum value of y is 480+320× 4 = 1 760 yuan.

14.f(x)=x2-4x+3，f(x-2)=x2-8x+ 15。

Analysis: let x+ 1 = t, then x = t- 1, so f (t) = (t-1) 2-2 (t-1) = t2-4t+3, that is, f (x).

15.(-∞, ).

Analysis: y = (2a- 1) x+5 is a decreasing function, 2a- 1 < 0, a.

16.x( 1-x3)。

Analysis: Let x ∈ (-∞, 0) and use -x ∈ [0, +∞).

∴f(-x)=-x[ 1+(-x)3]=-x( 1-x3)，

∵f(x) is odd function, ∴ f (-x) =-f (x). ∴ f (x) =-f (-x) = x ( 1-x3)，

That is, when x ∈ (-∞, 0], the expression of f(x) is x∈(-∞ 1-x3).

Third, answer questions.

17. Solution: ① A is an empty set,

∴ Equation AX2-3x+2 = 0 has no real root.

Get a >.

② There is only one element in ∵ A,

The equation AX2-3x+2 = 0 has only one real root.

When a = 0, the equation becomes -3x+2 = 0, and the real number root x =；;

When a≠0, let δ = 9-8a = 0 and get a =. At this time, the unary quadratic equation AX2-3x+2 = 0 has two equal real roots, that is, there is only one element in A. 。

As can be seen from the above, when a = 0 or a =, there is only one element in a.

(3) If there is only one element in A at most, there are two situations: there is only one element in A; A is an empty set. From the results of ① and ②, we can get A = 0 or A≥0.

18. Solution: According to the mutual dissimilarity of elements in the collection, there are

Solve or or or

According to the mutual dissimilarity of elements in the set, or is obtained.

19. proof: let x 1, x2∈R and x 1 < x2, then

f(x 1)-f(x2)=-=(x 1-x2)(+x 1x 2+)。

+x 1x2+= (x 1+x2) 2+。

X 1-x2 is less than 0 from X 1 < x2, and X 1+x2 and x2 cannot be 0 at the same time.

Otherwise, x 1 = x2 = 0 contradicts x 1 < x2.

So +x 1x2+> 0.

Therefore, f (x 1)-f (x2) < 0, that is, f (x 1) < f (x2),

F (x) = x3 is the increasing function of R 。

20. Solution: (1)∵ The function domain is {x | x∈R and x≠0},

F (-x) = 3 (-x) 4+= 3x4+= f (x), ∴f(x)= 3 x4+ is an even function.

(2)- 1 ≤ x < 1 from ≥0。

The definition domain of ∴ function is x ∈ [- 1), which is asymmetric about the origin, and ∴ f (x) = (x-1) is a nonsingular non-even function.

(3) The field of f (x) =+is x = 1,

∴ The function is f (x) = 0 (x = 1), and the domain is asymmetric about the origin.

∴f(x)=+ is an odd or even function.

(4) What is the domain of f (x) =+? x∈{ 1}，

The function ∴ is transformed into f (x) = 0 (x = 1), and ∴f(x)=+ is both a odd function and an even function.