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Hunan province 20 1 1 senior three? 12 school entrance examination game 1

Reference answers to liberal arts mathematics

First, multiple choice questions

Title number12345677 8

Answer B C A B A D D B

Second, fill in the blanks

9.69 1 10. 1 1 1. 13 12. 16 1207 13.(2,0) 14.2π

15.(-∞,-6]∪∪[6,+∞)

Third, answer questions.

16. Solution: ① 1600 households. (4 points)

② P = 790 1000 = 0.79。 (8 points)

③ Farmers whose annual household income is less than 25,000 yuan account for 3 1.5% > 25% and do not need the support of national policies. (12)

17. Solution: (1) ∫1-2sinbsinc = cos2b+cos2c-cos2a.

∴ 1-2sinbsinc= 1-2sin2b+ 1-2sin2c- 1+2sin2a

Sine theorem shows that -2bc =-2b2-2c2+2a2.

Rank: B2+C2-A2 = BC (3 points)

∴cosA=b2+c2-a22bc= 12

∴ A = 60。 (6 points)

(2)sin b+ sinC = sin b+ sin( 120-B)= sin b+ 32 cosb+ 12 sinB

= 32 cosb+32 sinb = 3( 12 cosb+32 sinb)

= 3 inches (b+30) (8 points)

∫0 & lt; B& lt; 120

∴30 <b+30 & lt; 150 ,

12 & lt; sin(B+30 )≤ 1,

∴32<; 3sin(B+30 )≤3

∴ sinb+sinc has no minimum value, and the maximum value is 3. (12)

18. solution: (1) proof: from the three views, we can see the surface PAC⊥ surface ABC, BO⊥AC.

∴BO⊥ Aircraft armored personnel carriers. (3 points)

(2) PE⊥AC and AC are made in the surface PAC of point P. From the top view, it can be seen that CE = 1 and AE = 3.

Bo = 3,AC = 4,∴ s △ ABC = 12× 4× 3 = 6。

∴ VP-ABC = 13× 6× 2 = 4。 (7 points)

(3)PC = PE2+EC2 = 5,BE=BO2+OE2= 10

∴pb=be2+pe2= 14,bc=bo2+oc2= 13

∴cos∠PBC=PB2+BC2-PC22PB? BC = 14+ 13-52 14? 13=222 14× 13

= 1 1 14× 13

∴sin∠pbc= 1- 12 1 14× 13=6 1 14× 13

∴S△PBC= 12PB? BC? sin∠PBC= 12 14? 13? 6 1 14? 13

=6 12

Set the distance from point A to surface PBC as h.

∵VA-PBC=VP-ABC,∴ 13h? S△PBC=4

∴h = 12s△PBC = 126 12 = 246 16 1。 (12)

19. Solution: (1) The monthly sales volume of Q car {an} takes the first item A 1 = a,

Geometric series of q =1+1%=1.01(2 points).

Total sales in the first n months of ≤24 are sn = a (1.01n-1)1.01=100a (1.0/kloc)

(2)∵Sn-Tn = 100 a( 1.0 1n- 1)-228 a( 1.0 12n- 1)

= 100 a( 1.0 1n- 1)-228 a( 1.0 1n- 1)( 1.0 1n+ 1)

=-228 a( 1.0 1n- 1)? ( 1.0 1n+3257)

And1.01n-1>; 0. 1.0 1n+3257 & gt; 0,∴sn<; Tn。 (8 points)

(3) Remember that the sales of Q car and R car in the nth month are an and bn respectively, so an = a×1.01n-1.

When n≥2, BN = TN-TN-1= 228a (1.012n-1)-228a (1.012n-2).

= 228 a×( 1.0 12- 1)× 1.0 12n-2 = 4.5828 a 1.0 12n-2。 (10.

B 1 = 4.5828a, which is obviously 20% × B 1

When n≥2, if

1.0 12(n- 1)>54.5828× 1.0 1n- 1. 1.0 1n- 1 & gt; 54.5828≈ 1.09,n- 1 & gt; LG 1.09 LG 1.0 1≈8.66。

∴n≥ 10, that is, from the first month of 10, the monthly sales of Q car is less than 20% of that of R car. (13)

20. solution: (1) ∫ f ′ (x) = (a-1)+ax-2 = (a-1) x-a+2x-2 (1point).

①a & lt; 0,f′(x)=(a- 1)(x-a-2a- 1)x-2。

∫a-2a- 1-2 =-aa- 1 & lt; 0,∴0<; a-2a- 1 & lt; 2,∴x>; 2,f′(x)< 0

∴f(x) decreases at (2, +∞). (3 points)

② When a = 0, f (x) =-x, which decreases in (2, +∞). (4 points)

③0 & lt; a & ltAt 1,a-2a- 1 >; 2

∴x∈(2,a-2a- 1),f′(x)>; 0, f(x) is added to (2, a-2a- 1);

When x ∈ (a-2a- 1, +∞), f ′ (x)

∴ To sum up, when a≤0, f(x) decreases on (2, +∞).

When 0

(2) When a < 0, f(x) decreases on (2, +∞);

Let x 1, x2∈(2, +∞) and X 1

f(x 1)-f(x2)x 1-x2 & lt; -4 can be changed to f (x1)-f (x2) >-4 (x1-x2).

f(x 1)+4x 1 & gt; f(x2)+4x2

∴ Let g (x) = f (x)+4x, and ∴g(x) decreases on (2, +∞).

∴g′(x)<; 0 remains unchanged (2, +∞)

∴a- 1+ax-2+4<; 0 remains unchanged (2, +∞).

A & lt-3+3x- 1 remains unchanged (2, +∞).

And -3

2 1. Solution: (1) A = 2b, C = 3, A2 = B2+C2.

The solution is a = 2 and b = 1.

The elliptic equation is X24+Y2 = 1. (4 points)

(2) From (1), we can know that A (-2,0) and the coordinates of point B are (x 1, y 1).

The equation of the straight line l is y = k (x+2).

So the coordinates of point A and point B satisfy the equation.

Remove y from the equation and arrange it.

( 1+4k 2)x2+ 16k2x+ 16k 2-4 = 0

X 1 = 2-8k2 1+4k2 from-2x1=16k2-41+4k2, thus y 1 = 4k 1+4k2.

Let the midpoint of line segment AB be m, then the coordinate of m is (-8k21+4k2,2k1+4k2) (7 points).

There are two situations as follows:

(1) When k = 0, the coordinate of point B is (2,0), and the perpendicular of line segment AB is the Y axis.

So QA = (-2, -m), QB = (2, -m),

Shit, QA? QB≤4

Score: -22 ≤ m ≤ 22. (9 points)

(2) When k≠0, the equation of the midline of line segment AB is

y-2k 1+4k 2 =- 1k(x+8k 2 1+4k 2)

Let x = 0 and get m =-6k 1+4k2.

Shit, QA? QB=-2x 1-m(y 1-m)

=-2(2-8 k2) 1+4k 2+6k 1+4k 2(4k 1+4k 2+6k 1+4k 2)

= 4( 16k 4+ 15 k2- 1)( 1+4k 2)2≤4

The solution is-147 ≤ k ≤ 147, k≠0( 10 minute).

∴m=-6k 1+4k2=-6 1k+4k

∴ when-147 ≤ k

When 0

∴-32 ≤ m ≤ 32, and m ≥ 0 (12 points)

To sum up, -32 ≤ m ≤ 32, and m≠0. (13)