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hina Mathematicalompetition

2 8he Popularization Committee of Chinese Methematica Society and

Chongqing Mathematical Society were responsible for theassignment of the competition problems in the first round test andthe supplementary test

Part I Multiple-choice Questions Questions 1--6. six IIUll kseach)Theminimumvalueoff x) = 5 4 ~ forx E -=t2) is ) .

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2 Mathematical Olympiad in China

A) 0 B ) C) 2 D) 3Solution Let x < 2=> 2 - x > o Then

f x ) = 1 + 4 - 4 x + x 2 ) = _ 1 _ +C2 -x2 -x 2 -x~ 2 XJ ~ X 2 -X = 2

The equality holds if and only i f2

~ x = 2 - x , and it is sowhen x = 1 E =, 2). This means that f reaches theminimum value 2 at x = 1

Answer: C

LetA = [ -2 , 4), B = { x I x - a x - 4 0 } I f B c:Athen the range of real a is ) .A) [ -1 , 2)C) [0, 3]

B) [ - 1 , 2]D) [0, 3)

Solution x -ax- 4 = 0 has two roots:

We have B c ~ x ~ 2 and xz < 4 This means that

From the above we get 0 ~ a < 3Answer: D

and B are playing ping-pong, with the agreement thatthe winner of a game will get 1 point and the loser 0 point;the match ends as soon as one of the players is ahead by 2points or the number of games reaches six. Suppose that

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China Mathernatir: l Competition 3

the probabilities of and B winning a game are and, respectively, and each game is independent. Then the

expectation E ~ for the match ending with ~ games 1s) .

A) 24181 B) 26681 C) 27481 D) 670243olution I I t is easy to see that ~ c a n only be 2, 4 or 6. Wedivide the six games into three rounds, each consisting of two

consecutive games. I f one of the players wins two games in thefirst round, the match ends and the probability is

(1_ 2 l)2 = _i3 + 3 9.Otherwise the players tie with each other, earning one pointeach, and the match enters the second round; this probability is

5 41 9 = 9We have similar discussions for the second and third

rounds. So we getP ~ = 2) = ~

4 5 20p ~ = 4) = 9 X 9 = 81 'p ~ = 6) = ) = ~ ~ .

ThenE ~ = 2 X _i + 4 X 20 + 6 X 16 = 2669 81 81 81.

Answer: Bolution Let ~ denote the event that wins the k th

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game, while .A- means that B wins the game. Since Ai and A-are incompatible, and are independent of the other events,we have

5P C ~ 2) = PCA1Az) +PCA1Az) = gP C ~ = 4) = PCA1AzA3A4) + PCA1AzAaA4) +

PCA1AzAaA4 + PCA1AzA3A4)= 2[ f ( ) ( f ( )

P ~ = 6) = PCA1AzA3A4) +PCA1AzAaA4) +PCA1AzA3A4) + PCA1AzA3A4)

Then

168r

~ = 2 X___ +4 X 20 +6 X 16 = 2669 81 81 81.Answer: B

Given three cubes with integer edge lengths, i the sum oftheir surface areas is 564 cm2 , then the sum of theirvolumes is ) .A) 764 cm3 or 586 cm3C) 586 cm3 or 564 cm3

B) 764 cm 3D) 586 cm3

Solution Denote the edge lengths of the three cubes as a , band c , respectively. Then we have

i.e. a +b 2 = 94. We may assume that1 ~ ~ b ~ c < 10.

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China Mathernatir:4l Competition 5

Then

t follows that c > 31 So 6 c < 10, and this means thatc can only be 9, 8, 7 or 6

f c = 9, then

t is easy to see that a = 2, b = 3 So we get the solutiona , b c) = 2, 3, 9).

f c = 8, thena 2 b = 94 - 82 = 30

1his means that b 4 and b 30; it follows thatb = 4 or 5,so a = 5 or 14; in both cases a has no integer solution.f c = 7, then

t is easy to see that a = 3, b = 6 is the only solution.f c = 6, then

So 2b 2 58, or b 29. This means that b 6, but bc = 6, sob = 6 Then a = 22 and a cannot be an integer.

In summary, there are two solutions: a, b c) = 2, 3, 9)and a , b, c) = 3, 6, 7). Then the possible volumes are

v = 23 33 93 = 764 cm3 = 33 63 73 = 586 cm3

Answer: A

0 The number of rational solutions to the system of equations

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{x y +z = 0,xyz + z = 0, isxy +yz +xz + y = 0 .A) 1 B) 2 C) 3 D) 4

{X+ y = 0,olution I f z = 0, then I t follows thatxy y = 0.

{x 0, or{x 1,y=O y= l .I f z I 0, from xyz + z = 0 we get

xy = 1 . Dromx + y +z = 0 we have

z = - x y.Substituting into xy +yz +xz + y = 0, we obtain

xz +Yz +xy - Y = 0.From D we have x _ _. We substitute it into , andy

then make a simplification:y -1 y 3 - y - 1 = 0

I t is easy to see that y - y 1 has no rational solution, andso y = 1 Then from D and we get 1 and z = 0,contradicting z I 0.

In summary, the system has exactly two solutions:

{X0, { 1,

y = o, y = 1,z = 0, z = 0.

Answer: B

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Suppose that the sides a , b, c of MBC correspondingto the anglesA B , C respectively, constitute a geometricseqreore. hen the range of cotC +cos is C )smBcotC cosB A) 0, + oo) B) o, J5 t1)

D) ( J5 2-1 + =Solution Suppose that the common ratio of a , b, c is q.Thenb = aq, c = aq 2 We have

sinAcot C +cossin Bcot C +cos B sinAcos C + cosAsin Csin Bcos C +cos Bsin CsinCA +C)sin B +CsinE b

sinC1e-Bsin 7C-A)

= sin = - ; = q So we only need to determine the range of q. As a , b, c

are the sides of a triangle, they satisfy a +b > c and b +c >a .That is to say,

I t follows that{

a aq >aq 2 ,aq aq2 >a.

{q 2 - q - 1 < o,q2 q - 1 > 0.

Their solutions are

{JS-1 < 2 orq < - 2

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t is only possible that. /S -1 < < . /5 12 q 2

Answer: C

Part II Short-Answer Questions Questions 7-12 nine markseach

Letf (x) = a x +b, with a, b real nwnbers; 1Cx)f ( x ) , fn+l(x) = f ( f , . (x ) ) , n = 1, 2, f f r (x )12Bx +381, then a + b =

Solution From the definitions we getfn (x) = a x (a"-1 a..--2 + a 1)b

a - 1= a x Xb.a - 1a 7 - 1As f 7 (x) = 128x +381, we have a 7 = 128 and a _ 1 Xb =

381. Then a = 2, b = 3 The answer is a b = 5

0 Suppose that the minimwn of f ( x ) = cos 2x - 2a(lcosx) i s ~ . Then a=Solution We have

f ( x ) = 2cos x 2 - 1 - 2a - 2acos x

For > 2, f x ) takes the minimwn value of 1 - 4a whencos x = 1 ; for

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China Mathernatir: l ompetition 9

when cos x = ~ . I t is easy to see that x ) will never be - ~for a > 2 or a < 2. So it is only possible that 2 2. Thenf rom- ~ az 2a 1 = ~ we get a = 2 +../3or a = 2 -../3discarded). Therefore, the correct answer is a 2 +../3.

Twenty-four volunteers will be allocated to three schools.The rule is that each school will accept at least onevolunteer and all the schools will accept different numbersof volunteers. Then there are different ways ofallocating volunteers.

olution We may use each space between every two consecutivebars I to represent a school and each asterisk * to represent avolunteer, as seen in the following example: the first. second andthird schools receive 4 18 and 2 volunteers, respectively.

1 1 1 1Then the allocation problem may be regarded as a permutationand-combination problem of 4 bars and 24 asterisks.

Since the two ends of the line must be occupied by a barrespectively, there are 22

3) = 253 ways to insert the other 2bars into the 23 spaces between the 24 asterisks such that thereis at least 1 asterisk between every two consecutive bars inwhich there are 31 ways that at least two schools have the samenumber of volunteers. So the number of allocating wayssatisfying the conditions is 253 - 31 = 222.

4] Let S . denote the sum of the first n terms in a numbersequence {a,.}, satisfying

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n 1Sn +a,. = n (n1

) , n = 1, 2, .Then a,.=

olution As

n n I= (n + I)(n +2) n(n 1) +a,.

we haven 2 - 2 I I2a..-t-l = (n +I ) n + 2 ) - n + I + n(n +I) +a,.

-2 I(n + l ) (n +2) +a, .+ n(n + 1)Therefore,

a 1 1 + n + I)I n + 2) = a,. n n I+ 1) ) Define b . = a,. n(n 1 I) I t is easy to see that b .

I 12..-1h1, b1 = a 1 + 2 . Ontheotherhand, fromS1+a1 =2a1 = 0I 1we geta1 = 0. So b1 = 2 , b . = 2,.. Therefore,

1 I Ia,. = b , . - n(n + I ) = 2 - n(n + I ) '

4] Suppose that x ) is defined on R, satisfying j O)2008, and for any x E R

f x 2) - f x ) 3 X 2- ,f x + 6 ) - f x ) 63 X 2z.

Then /(2008) =

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China Mathernatir:4l Competition

olution I We havef x +2 - f x )

f x +4 - f x + 2 ) ) - f x +6)- f x +4) ) + f x +6 - f x ) )

~ 3 X 2x 2 - 3 X 2x t 63 X 2x = 3 X 2xThis means that f x 2 - f x ) = 3 X 2x So we have

/ 2008) = / 2008) - / 2006) + / 2006) - / 2004) + .../ 2 ) - / 0 ) 0 )

= 3 X (22006 + 22004 + .. + 22 + 1) + / 0 )41003-t-1 - 1= 3 X +20084 - 1

= 22008 + 2007.olution We define g x ) = x ) - 2x Then we have

g x 2 - g x ) = f x 2 f x ) - 2x 2 2:::;;; 3 X 2x -3 X 2x = 0,

g x + 6) - g x ) = f x + 6) - f x ) - 2x+-6 + 2~ 63 X 2 ' - 63 X 2.r = 0.

11

This means thatg x) -:::;;;g x +6) -:::;;;g x +4) -:::;;;g x +2 -:::;;;g x) , and it implies that g x ) is a periodic function with 2 as aperiod. So

/ 2008) = g 2008) + 2 2008 = g O) + 2 2008= 2007 + 22008.

Cf Suppose that a ball with radius 1 moves freely inside aregular tetrahedron with edge length 4./6. Then the areaof the inner surface of the container, which the ball cannever touch, is

olution As shown in Fig. 1, consider the situation where the

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ball is in a corner of the container. Drawthe planeA1B1C1 ABC tangent to theball at point D. Then the ball center 0 isalso the center of the tetrahedronP-A1B1C1, with PO ..lA1B1C1 and thefoot point D being the center of M1B1C1.

Since

p

BFig. I

= 4 X Vo-A1B1c 11= 4 X 3 X S L -AIBICI X OD

we have PD = 40D = 4r where r is the radius of the ball. tfollows that P = PD - OD = 3r

Suppose that the ball is tangent to the plane PAB at pointP1 Then we have

As shown in Fig. 2 it is easy to seethat the locus of the ball on the planePAB is also a regular triangle, denoted byP1EF Through P1 draw P1M ..l PA with

p

point M on PA. Then L.MPP1 = , and A t : t t ~ ~ ~ ~ ~ Q 8

PM =PP1XcosL viPP1 = 2t/Zr X =J6r.Fig. 2

t follows that P1E = PA 2PM a 2J6r where aPA. Now, the space on PAB which the ball will never touch isthe shaded part of Fig. 2 and its size is equal to

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China Mathematical Competition

SM B S L >P1EF = 1 2 - a - 2/6r) 2 )= 3 /2a r - 6 /3r2= 24../3 - 6 /3= 18../3

13

since r = 1 and a = 4/6 under given conditions. Then the totaluntouched area is

4 X 18../3 = 72../3.

Part III ord Problems Questions 13-15 20 marks each)G I t is known that the curve f x) = I sin x I intercepts theline y = kx k > 0) at exactly three points, the maximumx coordinate of these points being a. Prove that

cos asina+ sin aolution The image of the three intercepting points of j ( x ) and

y = kx is shown in the figure. I t iseasy to see that the curve and the lineare tangent to each other at pointA a , - sin a) , and a E ( 7 , 327 ) .

y

As j x ) = - cos x for x E ( rr 327 ) , we have - cos a- sma 1. e. a =tan a. Thena

cos asin a+ sin 3a cos a2sin 2acos acos2a sin2a4sinacos a1 +a 24a

14sinacos a1 tan2a4tana

.r

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m Solve the inequalitylog2 x12 3x 1 +5x 8 3x 6 +1) < 1 +log2 x4 +1).

olution I As

and o ~ y is monotonically increasing over (0, = , the giveninequality is equivalent to

orx 12 3x 10 5x8 3x6 1 < 2x4 2

I t can be rewritten as

2x 10 2x8 - 2x 6

That is to say,

+4x 8 +4x 6 -4 x 4+x6 +x4 - x z

x 8 + 2x 6 + 4x4 + x 2 + 1) x4 + x 2 -1 < 0.Then we have x 4 x 2 - 1 < 0 I t follows that x 2 x 6 + 4 + 3x +1+ Zx + Z= xz + 1) 3 + Z xz + 1).

Defineg t) = t +Zt. Then we have

Obviously, g t is a monotonically increasing function;then we have

That is to say,

We obtain x < - 1 +15z So the solution set is

D As is shown in the figure, is amoving point on the parabola y =Zx , points B , C are on the y axis,and the circle x - 1) 2 + y = 1 is c

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internally tangent to h.PBC. Find the minimum value ofthe area of h.PBC.

olution Denote P, B, by P x 0 , y 0 , B O, b), C O, c),and assume that b > c . The equation for the line PB is

Yo -by-b=--x.ot can be rewritten as

yo - b x - x o y +xob = 0.Since the distance between the circle center 1 , 0) and the

line PB s 1, we haveI Yo - b xob I = 1

V yo - b 2 x ~That s to say,

t is easy to see that xo > 2 . Then the last equation can besimplified as

x 0 - 2 b 2 2yob - o = 0.In a similar way,

xo - 2)c 2 2yoc - o = 0.Therefore,

-2yo -x 0b + c = -- be= -- .Xo - 2 Xo - 2

Then we getx ~ y ~ - 8x0

xo - 2 2

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China Mathernatir: l Competition

As P xo, yo) is on the parabola, ~ = 2xo. So we haveb z 4x5- c = (xo - 2) 2

orb - c = ~ T h e n we haveXo - 21S B C = 2 b -c XxoXo X x Xo - 2

4= x - 2 2 +4Xo-~ +4 = 8

17

The equality holds when xo -2 = 2; this means that xo = 4and Yo = 2./2. So the minimum of S oc is 8.

2009 n:mt ut JifhtiThe Popularization Committee of MS and the MathematicalSociety of Heilongjiang Province were responsible for theassignment of the competition problems in the first round test andthe extra test

Part I Short-Amwa- uestiom{ uestiom 1-8 seven IIIIUiai each)0 Suppose that f x ) = X

f f f . . . f x )]] . Then 99 0 =yn

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Solution We havepn x ) = f x ) = x./I +x

J 9 9 ) x ) = x/1 99x2

Therefore '99) 1) = 110

Given the lineL: x 9 = 0 and the circleM: 2x 2y - B x - By - 1 = 0, point A is on L and points B, Care on M; LBAC = 45 o and the line AB is through thecenter of M. Then the range of the x coordinate of point

isSolution Suppose that A a , 9 - a ) . Then the distance fromthe center of M to the line AC is

d = IAM IX sinLBAC= ../ a - 2) 2 9 - a - 2) 2 X sin 45= 2a -IBa +53 X

On the other hand, since the line AC intercepts M, itfollows that d s ;;; the radius of M = J- i e.

2 /2 IT./2a -IBa +53 x 2 s ; ; ; ~ z The solution is 3 s ;;; a s ;;; 6

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China Mathematical Competition 19

On a coordinate plane there are two regions M and N:{ y ?0M is confined by y S ;; x , and N is determined by they :S:;;2 x;

inequalities t S ;; x S ;; t 1 0 S ;; t S ;; 1 Then the size of thecommon area of M and N is given by f t )

Solution As shown in the figure we haveJ t ) = Sshadedarea y

= S L: AO - S L: OCD - S L: B F1 1= 1 t 2 - - 1-t 22 2

The inequality1 1 1 1n 1 n 2 2n 1 < a - 2007 3

holds for every positive integer n Then the least positiveinteger of a is

Solution Obviously1 1 1f n ) = n 1 n 2 2n 1

is monotonically decreasing. Therefore j l) reaches themaximum of f n) . From

1 1 1f l ) = 2 3 2008. Therefore the least positive integer of ais 2009.

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2 zGiven points P, Q on an ellipse:2 = 1 a >b >O),satisfyingOP l_ OQ, the minimum of I OP IX OQ I is

Solution DefineP I P I cos ), IOP I sin )),

Q I OQ I cos o ; ) , I OQ I sin ; ) ) .We have

1 _ cos2 sin2 )IOPI 2 - a 2 b2 1 = sin28 cos28

I OQ 12 a 2 b2 Then

1 1 1 1I P 1 I OQ 1 = a 2 b2 . . 2a 2b 2Therefore, I P I X I 0 Q I reaches the mmunum a 2 bz

when I P I =I OQ I =Suppose that the equation lg kx = 2lg x 1) has exactlyone real root. Then the range of k is

Solution We havekx >0,

X +1 >O,k x = C x + l 2

The expression can be standardized asx 2 2 - k x 1 = 0

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The two roots of areX1 X 2 = [k - 2 ./k 2 - 4k],

wheret = k 2 - 4k 0 : k 0 or k 4.

( i) When k < 0, it is easy to see from that x i 1 > 0,x2 1 < 0, and kx1 > 0 Then the equation has one real root,X I = [k 2 /k2 -4k] .

(ii) Whenk = 4, the equation has one real root, x = -1 = 1.

(iii) When k > 4, the two roots x i x2 are both positive, aswell as xi x2 Discarded.

Therefore, the range of k is k < 0, k = 4Consider a pattern of numbers in the shape of a triangle:the first row consists of numbers from 1 to 100 arrangedin order; each number in the second row is the sum of thetwo numbers directly below it in the first row; eachnumber in the third row is the sum of the two numbersdirectly below it in the second row;

The number in the last row isolution It is easy to see that

( i) There are 100 rows in the pattern;(ii) The numbers in the ith row constitute an arithmetic

sequence with the common differencedi = 2i-l i = 1 2 . . . 99

(iii) We have

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22 Mathematical Olympiad n China

a,. = a.. 1 Can 1 2.. 2 )= 2an-1 2 -2= 2[2an-2 + 2 3J + 2,_2= 22 [2an-3 2..-4] 2 X 2 -2

= 2n-1a1 n -I X 2 -2= n 1)2n-2.

Therefore, the number in the last row is a 100 = 101 X 298

Every day at a railway station, there is just one trainarriving between 8: 00 am and 9: 00 am and between 9: 00am and 10: 00 am respectively. The arrival times andtheir probabilities for the two trains are shown in thefollowing table:

TrainA 8110 8130 8150Arrival time Train B 9=10 9=30 9=50Probability 1 1 16 2 3

Suppose that these random events are independent ofeach other. Now, a traveler comes into the station at8: 20. Then the mathematical expectation of his waitingtime is (round to minute).

olution The distribution table for the waiting times of thetraveler is shown below.

Waiting 10 30 50 70 90time/minProbability 1 1 x x x2 3 6 6 2 6 3 6

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Therefore, the mathematical expectation of his waiting time is1 1 1 1 1 .10 X z + 30 X J 50 X 36 + 70 X 12 + 90 X 18 27 rnm),

Part II Word Problems 14 marks for Question 9, 15 markseach for Questions 10 and 11, 44 marks in total}

Suppose that the line l : y =lex m k , m are integers)intercepts an ellipse 2 = 1at two different pointsA

xz yzB and intercepts the hyperbola -4 - - = 1 at two12 -ifferent points C, D. Can the line be such that AC-D = O f yes, how many different possibilities are therefor the line l? f no, explain the reason.

{y = kx m ,

olution For xz yz _ by eliminating y and simplifying16 12 - 1.it, we get

= 8km) - 4 3 +4k ) 4m -48) > 0 Dby eliminating y and simplifying it,

we get

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2 = - 2 k m ) 2 +4 3 2 ) m 2 +12) >O. _____ ____FromAC +BD = 0 we get x 4 - x 2 ) + x 3 - x 1 ) =0, which

implies that X t xz = xs X4.Then

Bkm3 +4k 2

2km3 2 4Therefore, km = 0 or - 3 4k2 13 _ k 2 discarded).

Then a possible solution is either k = 0 or m = 0.When k = 0, from D and we have -2/3 < m < 2/3. As

m is an integer, it can be -3 -2 - 1 0, 1 , 2, 3.Whenm = O from

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Let b., = a,H - pa,. Then b..-1-1 = ab,. n = 1 2, ). Thismeans that {b,} is a geometric sequence with the common ratioa. The first term of {b,} is

Therefore, b, =a 2 Xa -1 =a +1. Thena,H -pa, =a H.By rewriting

a..-1-1 = a..-t-1 +fh , n = 1, 2, ).When D = p 2 - 4 q = 0, we have a = fJ 0, a1 = p = 2a.

Th CD ecomes .. 1 1 + . a..-1-1 a, 1expression a..-1-1 = a aa, , 1. e. - ---;;- = .a aThen {::} is an arithmetic sequence with the commondifference 1 , whose first term is a1 = 2a = 2. Therefore,a a

a ; = 2 1 X n -1 = n 1.aAs a result, the general expression of {a,.} is

a, = n +Da .When D.> 0, a fJ, we havea..-1-1 = a..-t-1 pa,

= pa, +fJ__f _a..-t-1 - f J ~ a . . - l - n = 1, 2, ).- a - aBy rewriting,

.. 1 2 .. 1 )a..-t-1 +;-a =fJ a,+ ; a n = 1, 2, ).{

n+l }Then a, +;_a becomes a geometric sequence with the

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common ratio {3 whose first term is2 2 aa l -a-= a +a - a - = - ~ - - .{3-a {3-a {3-a

Therefore,

Then the general expression of {a,.} is_ pn+l _an+l _a, - { n - 1, 2, . . . ).- a

(2) Given p = I , q = , e have 6. = p - 4q = 0. Thena = { = . By the expression @, the general expression of{a,.} is

( 1 ) n +1a, .= n +1 2 = n = 1, 2, . . . ).Therefore, the sum of the first n terms of {a,.} is

Then1 2 3 n I2s.. = 22 + 2a + + 2n+l

(4)-@,1 3 +32 5 = 2 - zn+l

We finally getS = 3 _ n + 32 .

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olution Cl) By Vieta s theorem, we have a Xf3 = q 0,a +{3 = p Then

@The character equation of {a,. } is ). 2 - p .. q = 0, which

has roots a , f3 Then we can write down the general expressionof {a,.} according to the following different situations:

When a = f 0, a,. = CA1 A 2n)a . From@, we haveCA1 +Az)a = 2a,

CA1 +2Az)a 2 = 3a 2 Then we getA1 = Az = 1. Therefore,

a,. = (n Da .When a 3, a,. = A 1a A 2 f3 . From@, we have

A1a +Azf3 = a +f3,A1a 2 +Azf32 =a 2 +af3+f32

- - a - ___ ___Then we get A1 - rJ-- Az - rJ Therefore,t . J a t . J a

C2 The solution is the same as Solution I .Find the maximum and minimum of the function

y = /x 27 + /13 -x ...fx.olution The domain of y is x E [0, 13]. We have

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y = X 27 ../13 X rx= Jx +27 Jl3 +2../x l3 - x

./27 + ./IT = 3 /3 + ./IT.The equality holds whenx = 0 Therefore, the minimum of

y is 3 /3 +./IT.On the other hand, by the Cauchy inequality we have

y = rx Jx +27 /13 - x 2+1 + )C2x + x +27) +303 - x ]

= 121.The equality holds when 4x = 9 13 - x) = x 27 t is so

for x = 9 Therefore, the maximum of y is 11.

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8389_chap01