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)