In 2008, the national unified examination for enrollment of ordinary colleges and universities (Hubei volume)

Reference answers to mathematics (science, engineering, agriculture and medicine)

First, multiple-choice questions: This question examines basic knowledge and basic operations. 5 points for each small question, out of 50 points.

1.C 2。 B 3。 B 4。 D 5。 A six. D 7。 C 8。 A nine. C 10。 B

Fill-in-the-blank question: This question examines the basic knowledge and basic operation, with 5 points for each small question, out of 25 points.

1 1. 1 12. 13. 14.-6 15.,0

Third, answer: This big question is ***6 small questions, ***75 points.

16. This small question mainly examines the basic knowledge of the definition range, value range and properties of trigonometric functions, and examines the transformation of trigonometric identities, algebraic simplification and transformation, and operational ability. (Full score: 12)

Solution: (1)

=

(ii) By

In terms of decreasing function, in terms of increasing function,

And (when),

that is

Therefore, the value range of g(x) is

17. This small question mainly examines the concepts of probability, distribution list of random variables, expectation and variance, and basic computing power. (Full score 12)

Solution: The distribution list of (i) is:

0 1 2 3 4

P

∴

(2) From, a2× 2.75 = 1 1, that is, again.

When a=2, from 1 = 2× 1.5+B, b =-2;

When a=-2, from 1 =-2× 1.5+B, b=4.

∴ Or what you want.

18. This little question mainly examines the right-angle prism, the angle formed by a straight line and a plane, the dihedral angle and the relationship between a line and a plane, and also examines the spatial imagination and reasoning ability. (Full score: 12)

(1) Proof: As shown in the right figure, the intersection point A is on the plane A 1ab 1.

So, D-key AD⊥A 1B

From A 1BC⊥ plane A 1BC 1, and plane A 1BC plane A 1 = A 1b, we get

AD⊥ plane A 1BC, and BC plane A 1BC,

So AD⊥BC.

Because the prism ABC-a1b1c1is a straight prism,

Then AA 1⊥ bottom ABC,

So AA 1⊥BC.

And AA 1 AD=A, so BC⊥ here a 1ab 1,

And AB terminal a 1ab 1, so AB⊥BC.

(2) Solution 1: Connecting CD, then (1) Knowing the angle formed by straight line AC and plane A 1BC,

Is the plane angle of dihedral angle A 1-BC-A, i.e.

So in Rt△ADC and Rt△ADB,

From ab < AC, you got it again.

Solution 2: According to (i), take point B as the coordinate origin and divide it on the straight line where BC, BA and BB 1 are located.

For X-axis, Y-axis and Z-axis, establish the spatial rectangular coordinate system as shown in the figure, and let AA 1=a, AC=b,

If AB=c, then B (0 0,0,0) and A (0 0,c,0), then

Let the normal vector of plane A 1BC be n=(x, y, z), then

allow

Let n=(0, -a, c), so the angle with n is acute, and then it is complementary to each other.

therefore

So from c < b, you get

That's why.

19. This small topic mainly examines the basic knowledge of plane analytic geometry such as lines, circles and hyperbolas, and examines the ability to solve trajectory equations and inequalities and comprehensively solve problems. (Full score: 13)

(1) Solution 1: With O as the origin and straight lines of AB and OD as the X axis and Y axis respectively, a plane rectangular coordinate system is established, and then A (-2,0), B (2 2,0), D (0 0,2) and P () are obtained according to the meaning of the question.

|MA|-|MB|=|PA|-|PB|= 10, 0 < t> 10, so 0 < t < 4.

② v (t) = 4 (t-10) (3t-41)+50 < 50 when10.