7/25/2019 5dethithu_5

1/5

Ngy thi 24/06/2016

GV ging dy: Trn Quang Thnh

THSC TRC KTHI THPT QUC GIA 2016

MN TON

Thi gian: 180 pht Khng kthi gian pht

THI THLN V

Cu 1. (2,0 im)Cho hm s 3 2y x 6x 9x 1 .

a. Kho st sbin thin v vth(C) ca hm s cho.

b. Tm cc gi trca m phng trnh 3 21 9x 3x x m 02 2

c mt nghim duy nht.

Cu 2. (1,0 im)

a. Gii phng trnh 2(sinx cosx) 1 cosx .

b. Tm cc sthc x, y tha mn ng thc 3x(3 5i) y(1 2i) 9 14i.

Cu 3. (1,0 im)

a. Gii bt phng trnh x 1 x 29 8.3 1 0.

b. Trong k thi THPT Quc gia nm 2016, c 4 mn thi trc nghim v 4 mn thi tlun. Mt gio

vin c bc thm ngu nhin phtrch coi thi 5 mn. Tnh xc sut gio vin c

coi thi t nht 3 mn thi trc nghim.

Cu 4. (1,0 im)Tnh tch phn2 3

2

1

x 2lnxI dx

x

.

Cu 5. (1,0 im)Cho hnh lng trng ABC.ABC c y ABC l tam gic cn ti C, cnh y AB bng

2a v gc ABC bng 30. Mt phng (CAB) to vi mt y (ABC) mt gc 60. Tnh thtch khi lng

trABC.ABC v khong cch gia hai ng thng AC v CB.

Cu 6. (1,0 im)Trong khng gian vi htrc Oxyz , cho hai im A(7;2;1),B( 5; 4; 3) v mt phng

(P) : 3x 2y 6z 3 0. Vit phng trnh ng thng AB v chng minh rng AB song song vi (P).

Cu 7. (1,0 im)Trong mt phng vi h ta Oxy , cho tam gic ABC ni tip trong ng trn

tm I. Tip tuyn ti B v C ca ng trn tm I ct nhau ti im P. on IP ct BC ti im H.

Gi E,F ln lt l trung im ca PC v BH. Bit EF c phng trnh 3x y 3 0 , im F nm trn

trc honh, im H c ta 2;2 , im A nm trn ng thng c phng trnh 5x y 6 0 v

im A c ta nguyn. Tm ta cc nh ca tam gic ABC v vit phng trnh ca ng trntm I.

Cu 8. (1,0 im)Gii hphng trnh 2

34 2 3

xy 1 1 x 4 y y 8,

3x y 2x y 26x 2 x 14.

Cu 9. (1,0 im)Cho a, b, c l cc sthc thuc on 1;2 v tha mn a b c 4. Tnh gi trln

nht ca biu thc4 4 2

a b 5c 6abc 1P abc.

ab bc ca

-----HT-----

Th sinh lm bi nghim tc - Cn bcoi thi khng gii thch lng nhng!

7/25/2019 5dethithu_5

2/5

P N S5

Cu Ni dung im

1aKho st v vthhm s 3 2y x 6x 9x 1 .

1.0

2

Vit phng trnh tip tuyn ca thhm sx 3

yx 1

bit rng tip tuyn i qua

im M(4;2). Pt: 3 2

1 9x 3x x m 0

2 2 3 2x 6x 9x 1 2m 1 (*)

0.25Pt (*) l pt honh giao im ca (C) v ng thng d y 2m 1 (d cng phng trc

Ox) . Snghim ca phng trnh l sgiao im ca (C) v d. 0.25

Da vo th(C), pt c mt nghim duy nht th :2m 1 1

2m 1 3

0.25

m 0

m 2

0.25

3a

Gii phng trnh: 2(sinx cosx) 1 cosx .

Ta c2

(sinx cosx) 1 cosx 1 2sinxcosx 1 cosx

cosx(2sinx-1) 0 0.25

cosx 0

1sinx=

2

x k2

x= k2 (k Z).6

5x k2

6

0.25

3b

Tm cc sthc x, y tha mn ng thc 3x(3 5i) y(1 2i) 9 14i.

Vtri ca ng thc c vit li: 3x(3 5i) y(1 2i) (3x 11y) (5x 2y)i. 0.25x, y l cc sthc tha mn bi khi v chkhi x, y l nghim ca h:

3x 11y 9

5x 2y 14

Gii hta c:172

x61

v3

y61

0.25

4a

Gii bt phng trnh x 1 x 29 8.3 1 0 .x 2 x

PT (3 ) 8.3 9 0 0.25x

3 9 x 2. 0.25

4b

Trong k thi THPT Quc gia nm 2016, c 4 mn thi trc nghim v 4 mn thi tlun.Mt gio vin c bc thm ngu nhin phtrch coi thi 5 mn. Tnh xc sut gio vin c coi thi t nht 3 mn thi trc nghim.Scch bc thm ngu nhin 5 trong 8 mn thi l

0.25Scch bc thm c t nht 3 mn thi trc nghim (3TN+1TL; 4TN+1TL) l

Vy xc sut cn tm l

0.25

5 Tnh tch phn2 3

2

1

x 2lnxI dx

x

.

7/25/2019 5dethithu_5

3/5

22 2 2 22

2 2 2

1 1 1 11

lnx x lnx 3 lnxI xdx 2 dx 2 dx 2 dx

x 2 x 2 x 0.25

Tnh

2

2

1

lnxJ dx

x

t2

1u lnx,dv dx

x

. Khi 1 1

du dx,v

x x

Do 2 2

2

1 1

1 1J lnx dx

x x

0.25

2

1

1 1 1 1J ln2 ln2

2 x 2 2 0.25

Vy1

I ln22

0.25

6

Cho hnh lng trng ABC.ABC c y ABC l tam gic cn ti C, cnh y AB bng2a v gc ABC bng 30. Mt phng (CAB) to vi mt y (ABC) mt gc 60. Tnh th

tch khi lng trABC.ABC v khong cch gia hai ng thng AC v CB.Gi M l trung im AB. Lc

( ) V do tnh cht lng trng, AA l ng cao calng tr

0.25

0.25

0.25

0.25

7/25/2019 5dethithu_5

4/5

7

Trong khng gian vi htrc Oxyz , cho hai im A(7;2;1),B( 5; 4; 3) v mt phng

(P) : 3x 2y 6z 3 0. Vit phng trnh ng thng AB v chng minh rng AB

song song vi (P).

ng thng AB i qua A, VTCP 2 x 4 1 y 3 1 z 4 0 0.25

c PTTS l

x 7 12t

y 2 6t

z 1 4t

0.25

Xt hphng trnh

x 7 12t

y 2 6t

z 1 4t

3x 2y 6z 3 0

0.25

v CM c hVN. 0.25

8

Trong mt phng vi h ta Oxy , cho tam gic ABC ni tip trong ng trn tm

I. Tip tuyn ti B v C ca ng trn tm I ct nhau ti im P.on IP ct BC tiim H. Gi E,F ln lt l trung im ca PC v BH. Bit EF c phng trnh

3x y 3 0 , im F nm trn trc honh, im H c ta 2;2 , im A nm trn

ng thng c phng trnh 5x y 6 0 v im A c ta nguyn. Tm ta

cc nh ca tam gic ABC v vit phng trnh ca ng trn tm I. Ta c F EF Ox Ta ca F l nghim ca hphng trnh:

3x y 3 0 x 1

y 0 y 0

F 1;0 .

F l trung im BH

B H F B

B H F B

x x 2x x 2 2y y 2y y 2 0

.

BB

x 0B 0; 2 .

y 2

0.25

H l trung im BC B C H C CB C H C C

x x 2x 0 x 4 x 4C 4;6

y y 2y 2 y 4 y 6

.

0.25

Ta c tnh cht 0IFE 90 .

Tht vy ta c HBI CPI g g . M IF,IE ln lt l cc ng trung tuyn tng

ng nn IFH IEC Tgic ICEF ni tip. M 0 0ICE 90 IFE 90 .

Ta c IF i qua F 1;0 v IF EF IF: x 3y 1 0.

Ta c IH i qua H 2;2 v nhn FH 1;2 l php vect nn IH c phng trnh:x 2y 6 0.

Ta c I IF IH I 4;1 R IB 16 9 5.

Khi 2 2

I : x 4 y 1 25 . 0.25

E

F H

P

I

A

B C

7/25/2019 5dethithu_5

5/5

Ta c A l giao ca ng trn tm I v ng thng 5x y 6 0 . Do ta ca A

l nghim ca h

2 2 2 2

2

x 1x 4 y 1 25 x 4 y 1 25

26x 42x 16 0 8x5x y 6 0 y 5x 6

13

V im A c ta nguyn nn A 1;1 .

Kt lun: 2 2

A 1;1 , B 0; 2 ,C 4;6 , I : x 4 y 1 25.

Ch : Ngoi cch sdng tnh cht 0IFE 90 ta c th tham sha im E ri kthp vi ta im C suy ra ta P. Sau cho P thuc PH ta tm c ta E, t

vit c phng trnh CI v tm c ta im I l giao ca PH v CI. 0.25

9

Gii hphng trnh

2

34 2 3

xy 1 1 x 4 y y 8 1

3x y 2x y 26x 2 x 14 2

K: y 0

Ta c 4 y y y y 0 do tphng trnh (1) suy ra x>0; y>0

21 xy 1 1 x 4 y y 4 y y 8 4 y y

2 2 2 2 4xy 1 1 x 2 4 y y x x 1 x 1yy y

2

2 2 2 2x x 1 x 1

y y y

(3)

0.25

Xt hm s 2f t t t 1 t trn 0; .

C 2

2

2tf ' t 1 1 t 0 t 0;

1 t

.

Suy ra hm sf(t) ng bin trn 0; .

M phng trnh (3) c dng 2

2 2 4f x f x y

xy y

0.25

Thay2

4y

x vo phng trnh (2) ta c

3 32 3 2 3

3

33 3

12x 26x 8 2 x 14 6x 13x 4 x 14

x 2 x 2 x 14 x 14 4

0.25

Xt hm s 3g u u u trn R

C 2g' u 3u 1 0 u R Suy ra hm sg(u) ng bin trn R m phng trnh (4) c dng:

3 33 3 2

x 1 2 ng x 2 g x 14 x 2 x 14 6x 12x 6 0

x 1 2 l

=> y 12 8 2 .

Vy hc nghim duy nht

1 2;12 8 2

0.25