Transcript

S GIO DC V O TO

S GIO DC V O TOK THI CHN HC SINH GII CP TNH THPT BNH PHC

NM HC 2013 - 2014 THI CHNH THC

Mn: Ton

( thi c 01 trang)

Thi gian lm bi: 180 pht (khng k thi gian giao )

Ngy thi: 03/10/2013Cu I:(THPT:4,0 im; GDTX: 4,0 im) Cho hm s:

(1)

1. Kho st s bin thin v v th ca hm s (1).

2. Vit phng trnh tip tuyn ca , bit tip tuyn ct ng tim cn ng v tim cn ngang ln lt ti sao cho , vi .

Cu II:(THPT:5,0 im; GDTX: 6,0 im)

1. Gii h phng trnh:

2. Gii phng trnh:

Cu III:(THPT:4,0 im; GDTX:4,0 im)

1. Trong mt phng vi h trc ta , cho hnh ch nht c , im thuc vo ng thng c phng trnh: . ng thng i qua v trung im ca on c phng trnh: . Tm ta ca v , bit im c honh dng.2. Cho tam gic nhn ni tip ng trn . Gi ln lt l cc im di ng trn cung nh , sao cho thng hng. Gi , ln lt l hnh chiu vung gc ca ln cc ng thng tng ng v ln lt l hnh chiu vung gc ca ln cc ng thng . Gi l giao im ca hai ng thng v . Tm gi tr ln nht ca din tch tam gic (theo ).Cu IV:(THPT:3,0 im; GDTX:3,0 im) Cho hnh chp c y l hnh ch nht, tam gic u cnh v nm trong mt phng vung gc vi y. Gc gia mt phng v mt phng y bng .1. Tnh th tch khi chp theo .

2. Tnh khong cch gia hai ng thng v theo .

Cu V:(THPT:2,0 im; GDTX:3,0 im) Cho l ba s dng. Tm gi tr ln nht ca biu thc:

Cu VI:(THPT:2,0 im) Cho dy s c xc nh: .

Xt dy s . Tm .

------------------HT------------------

Th sinh khng c s dng ti liu. Gim th khng gii thch g thm.

Lu : i vi th sinh hc ti cc trung tm GDTX th khng lm cu VI.S GIO DC V O TO

HNG DN CHM THI CHN HC SINH GII BNH PHC

CP TNH THPT NM HC 2013 2014

(Hng dn chm c 06 trang)

MN: TON

Ngy thi: 03/10/2013I VI TH SINH THPTCuLi giiim

I1

Cho hm s: . Kho st s bin thin v v th (C) ca hm s .

2,0

TX:

0,25

( phng trnh ng TCN: y = 2

( phng trnh ng TC: x = 20,5

( Hm s nghch bin trn tng khong xc nh.

Hm s khng c cc tr.0,5

Bng bin thin:

0,25

Giao im vi trc tung: A(0; 3/2)

Giao im vi trc honh: B(3/2;0)0,25

th:

0,25

2Vit phng trnh tip tuyn ca (C), bit tip tuyn ct ng tim cn ng v tim cn ngang ln lt ti A, B sao cho , vi I(2;2).2,0

Gi

PTTT ca (C) ti M:

0,5

Do v tam gic AIB vung ti I ( IA = IB nn h s gc ca tip tuyn k = 1 hoc k = -1. v nn ta c h s gc tip tuyn k = -1.0,5

0,5

( c hai phng trnh tip tuyn:

;

0,5

II1Gii h phng trnh:

2,5

k:

0,5

Pt(2)

1,0

Pt(1)

1,25

H cho tng ng:

Vy h phng trnh c 2 nghim:

0,75

2Gii phng trnh:

2,5

k: (*)0,5

Pt tng ng:

0,75

0,75

Nghim tha mn (*)Phng trnh c 2 h nghim:

0,5

III1Trong mt phng vi h trc ta , cho hnh ch nht c , im thuc vo ng thng c phng trnh: . ng thng i qua v trung im ca on c phng trnh: . Tm ta ca v , bit im c honh dng.2,0

Gi , M l trung im AB, I l giao im ca AC v d2: 3x 4y 23 = 0.Ta c ng dng

0,5

M nn ta c:

Vy C(1;5).0,5

Ta c:

0,5

Do

0,5

2Cho tam gic nhn ni tip ng trn . Gi ln lt l cc im di ng trn cung nh , sao cho thng hng. Gi , ln lt l hnh chiu vung gc ca ln cc ng thng tng ng v ln lt l hnh chiu vung gc ca ln cc ng thng . Gi l giao im ca hai ng thng v . Tm gi tr ln nht ca din tch tam gic (theo ).

2,0

Chng minh gc

K KH vung gc vi BC (H thuc BC), ta c:

(t gic PEBD ni tip)

Suy ra:

Tng t, ta chng minh c:

Vy (do PQ l ng knh)0,5

Chng minh :Tht vy, xt hnh thang vung vung ti D v D nn , du = xy ra khi

0,5

Xt tam gic . Ta c:

Vy din tch ln nht ca tam gic bng khi

1,0

IV1Cho hnh chp c y l hnh ch nht, tam gic u cnh v nm trong mt phng vung gc vi y. Gc gia mt phng v mt phng y bng .

1. Tnh th tch khi chp theo .1,5

H, M ln lt l trung im ca AB v CD

Ta c:

0,5

Gc gia (SCD) v mt y l

0,25

Ta c

0,25

0,5

22. Tnh khong cch gia hai ng thng v theo .1,5

K ng thng d i qua A v d//BD. Trong mt phng (ABCD) k ng thng ( i qua H , ( ( d v ( ct d ti J, ( ct BD ti I. trong (SHI) k HK vung gc vi SI ti K.Khi :

0,5

Ta c ng dng

EMBED Equation.DSMT4 0,5

Xt vung ti H, ta c:

Vy

0,5

VCho l ba s dung. Tm gi tr ln nht ca biu thc:

2,0

0,75

0,75

Vy

= vi

0,75

Vy gi tr ln nht ca khi

0,75

VICho dy s uc xc nh: .

Xt dy s . Tm .

2,0

Ta c .

Khi :

t . Khi ta c dy mi c xc nh bi:

0,25

Chng minh l dy tng:Xt hiu:

Do nn suy ra dy l dy tng.0,25

Chng minh (xn) khng b chn hay :Gi s (xn) b chn, do dy tng v b chn nn tn ti gii hn hu hn.

Gi s dy (xn) c gii hn hu hn, t .

T cng thc truy hi

Ly gii hn hai v, ta c: (khng tha mn)

Do dy cho khng c gii hn hu hn.0,5

Ta c:

EMBED Equation.DSMT4 M:

0,5

Do , ta c:

M nn

0,5

Ch : Nu th sinh lm cch khc m ng th vn chm im ti a.S GIO DC V O TO

HNG DN CHM THI CHN HC SINH GII

BNH PHC

CP TNH THPT NM HC 2013 2014

(Hng dn chm c 06 trang)

MN: TON

Ngy thi: 03/10/2013I VI TH SINH HC TI CC TRUNG TM GDTXCuLi giiim

I1

Cho hm s: . Kho st s bin thin v v th (C) ca hm s .

2,0

TX:

0,25

( phng trnh ng TCN: y = 2

( phng trnh ng TC: x = 20,5

( Hm s nghch bin trn tng khong xc nh.

Hm s khng c cc tr.0,5

Bng bin thin:

0,25

Giao im vi trc tung: A(0; 3/2)

Giao im vi trc honh: B(3/2;0)0,25

th:

0,25

2Vit phng trnh tip tuyn ca (C), bit tip tuyn ct ng tim cn ng v tim cn ngang ln lt ti A, B sao cho , vi I(2;2).2,0

Gi

PTTT ca (C) ti M:

0,5

Do v tam gic AIB vung ti I ( IA = IB nn h s gc ca tip tuyn k = 1 hoc k = -1. v nn ta c h s gc tip tuyn k = -1.0,5

0,5

( c hai phng trnh tip tuyn:

;

0,5

II1Gii h phng trnh:

3,5

k:

0,5

Pt(2)

1,0

Pt(1)

1,25

H cho tng ng:

Vy h phng trnh c 2 nghim:

0,75

2Gii phng trnh:

2,5

k: (*)0,5

Pt tng ng:

0,75

0,75

Nghim tha mn (*)Phng trnh c 2 h nghim:

0,5

III1Trong mt phng vi h trc ta , cho hnh ch nht c , im thuc vo ng thng c phng trnh: . ng thng i qua v trung im ca on c phng trnh: . Tm ta ca v , bit im c honh dng.2,0

Gi , M l trung im AB, I l giao im ca AC v d2: 3x 4y 23 = 0.

Ta c ng dng

0,5

M nn ta c:

Vy C(1;5).0,5

Ta c:

0,5

Do

0,5

2Cho tam gic nhn ni tip ng trn . Gi ln lt l cc im di ng trn cung nh , sao cho thng hng. Gi , ln lt l hnh chiu vung gc ca ln cc ng thng tng ng v ln lt l hnh chiu vung gc ca ln cc ng thng . Gi l giao im ca hai ng thng v . Tm gi tr ln nht ca din tch tam gic (theo ).

2,0

Chng minh gc

K KH vung gc vi BC (H thuc BC), ta c:

(t gic PEBD ni tip)

Suy ra:

Tng t, ta chng minh c:

Vy (do PQ l ng knh)0,5

Chng minh :

Tht vy, xt hnh thang vung vung ti D v D nn , du = xy ra khi

0,5

Xt tam gic . Ta c:

Vy din tch ln nht ca tam gic bng khi

1,0

IV1Cho hnh chp c y l hnh ch nht, tam gic u cnh v nm trong mt phng vung gc vi y. Gc gia mt phng v mt phng y bng .

3. Tnh th tch khi chp theo .1,5

H, M ln lt l trung im ca AB v CD

Ta c:

0,5

Gc gia (SCD) v mt y l

0,25

Ta c

0,25

0,5

24. Tnh khong cch gia hai ng thng v theo .1,5

K ng thng d i qua A v d//BD. Trong mt phng (ABCD) k ng thng ( i qua H , ( ( d v ( ct d ti J, ( ct BD ti I. trong (SHI) k HK vung gc vi SI ti K.

Khi :

0,5

Ta c ng dng

EMBED Equation.DSMT4 0,5

Xt vung ti H, ta c:

Vy

0,5

VCho l ba s dung. Tm gi tr ln nht ca biu thc:

3,0

0,75

0,75

Vy

= vi

0,75

Vy gi tr ln nht ca khi

0,75

Ch : Nu th sinh lm cch khc m ng th vn chm im ti a.

0

0

1/4

-

+

0

4

+(

1

t

f(t)

f(t)

0

0

1/4

-

+

0

4

+(

1

t

f(t)

f(t)

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