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…...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

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Page 1: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

1

������������������������ ������������ � �� �� �� �

��������������������

������������ ���������������� -��������������������������������

(1) (2) 217 (2) (3) { }1R − −

(3) (2) 2

(4) (1 ) sec 1x <

(5) (1) 0

(6) (1) 45

(7) (3) 1

(8) (4) 4

(9) (2) ( )1, 2−

(10) (1) 3

(11) (4) 25

(12) (2) 60˚

(13) (2) ˆˆ ˆ2i j k− −

(14) (4) ���� ��� (15) (3) 0

(16) (3) 0

(17) (1) 6

(18) (2) log1

x

x +

(19) (4) 3 52

3log 2

x

c+

+

(20) (4) 2

3

(21) y = cos x

(22) log log logx y z

ky z z x x y

= = =− − −

��.

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Page 2: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

2

log x = ( ) ( )k y zk y z x e

−− ⇒ = ... (1)

log y = ( ) ( )k z xk z x y e

−− ⇒ = ... (2)

log z = ( ) ( )k x yk x y z e

−− ⇒ = ... (3)

( ) ( ) ( )1 2 3× ×

xyz = ( ) ( ) ( )k y z k z x k x ye e e

− − −× ×

= ( )k y z z x x ye

− + − + −

= 0 1e = xyz = 1

(23) ( )tan 45 A° − = tan 45 tan

1 tan 45 tan

A

A

°−+ °

= 1 tan

1 tan

A

A

−+

(24) ������� ������������� ����� ����� = 6!

2!

= 6 5 4 3 2!

2!

× × × ×

= 360

(25) S = 4 7 10

15 25 125

+ + + +L

a = 1

1, 3,5

d r= =

s∞ = ( )21 1

a dr

r r+

− −

= 2

13

1 5 3 25 3551 4 5 16 1611 15 5

×+ = + × = − −

(26) OAuuur

= ˆˆ ˆ2 3 5i j k+ −

OBuuur

= ˆˆ ˆ3 2i j k+ −

OCuuur

= ˆˆ ˆ6 5 7i j k− +

ABuuur

= ˆˆ ˆ2 3OB OA i j k− = − +uuur uuur

BCuuur

= ˆˆ ˆ3 6 9OC OB i j k− = − +uuur uuur

= ( )ˆˆ ˆ3 2 3i j k− +

BCuuur

= 3ABuuur

���, BCuuur

�!"# ABuuur

��$��� %��&'�(. )��� B ��� %������ �(�

)�����

A,B,C *�+ ���,-� .���� )/#.

(27) ( )2 16

4

xf x

x

−=

+

( )f x ��� 4x = − -�� ��+�"����&1��.

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Page 3: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

3

���, ( )f x ��� { }4x R∈ − − , �2 *3%��4 �(�5# %��&'67��/#.

(28) y = 10 10log log logx x e

e=

dy

dx = 10

1loge

x

dy

dx = 10loge

x

(29) sin

1 cos

xdxx+∫

= logdu

u cu

−= − +∫

= log 1 cos x c− + +

(30) 4 2

1A

x

= −

( )( )2 3A I A I− − = 4 2 2 0 4 2 3 0

1 0 2 1 0 3x x

− × − − −

= 2 2 1 2

1 2 1 3x x

− − − −

( )( )2 3A I A I− − = 0 �� %�������,�(8�

2 2 1 2

1 2 1 3x x

− − − −

= 0 0

0 0

( )

( ) ( )( )2 2 4 2 3

1 2 2 2 3

x

x x x

− + − − − − − + − −

= 0 0

0 0

2 2x − = 0

x = 1

(31) .���� a S∈ -�/# ( , )a a R∈ ����� R )�� �!9,&�/#.

( , )b a R∉ ������� .��5��" ( , )a b R∈ � :� ;"��# ���. ��"

1����� <=��� *3%��4 ;"�� ( , )a b R∈ -�/# ( , )b a R∈ �� <2 >-5#. �����

R )�� ��6?+�/#.

( , )a c R∉ @��" ( , )a b �!"# ( , )b c )A� �4;"����8 R –� ��$ ����.

����� R)�� “�&�� .�” ��� ;�����". ��� “ R )�� �&�� )/#”

�2 <!" ;�����/#. ��� R )�� �&�� )/#.

����� R )�� �!9,�. ��6?'. �&��� %��&'� )/#. ��� %�������,&

%��&'� ����� %��&'��/#.

(32) �.�.

= cot(180 ).sin(90 ) cos( )

sin(270 ). tan( ).cosec(360 )

θ θ θθ θ θ+ − −

+ − +

= cot cos cos

( cos ) ( tan ) cosec

θ θ θθ θ θ× ×

− × − ×

1 cos x u+ = ��. sin x dx du− =

sin xdx du= −

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Page 4: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

4

= 2cos cos sincot cos

sinsin

cos

θ θ θθ θ

θθθ

×× =

= �.�.

��� CDE����,&�

(33) 9 ��(1��� 5 ��(1�F�/ 1�&������& ����#. .��

2 ��(1�F�/ ��-���� 1�&���� ����#

��� 1�&�����!���

����� ����� = 37C

= 7 6 5

3 2 1

× ×× ×

= 35 .

(34) 2

3

1, , 10a x b n

x= = =

1rT + = n r r

rnC a b− ×

= 2 10

3

110 ( )

r

r

rC xx

− ×

= 20 2 310 r r

rC x x− −×

1rT + = 20 510 r

rC x −

20 5r− = 15

5r− = 5−

1r =

1 1T +∴ = 20 5

110C x −

= 15

110C x

��� 15x -� /$�# 10 .

(35) ��+�&���4:�,

1m = tan 60 3° =

2m = tan120 tan(180 60 )° = − °

= 3−

∴ ������� �������(

y = 1m x c+ ,

3 7y x= + , 3 7y x= − +

(36) �.�. 2 2 2

1 1 1

x y z

x y z

=

60°

30°

120°

O

y

x

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Page 5: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

5

2 2 1c c c→ −

3 3 1c c c→ −

=2 2 2 2 2

1 0 0

x y x z x

x y x z x

− −

− −

= 2

1 0 0

( )( ) 1 1y x z x x

x y x z x

− −

+ +

( )y x− �!"# ( )z x− G 2c �!"#

3c -�4:� >�2�� ���

= ( )( )[ ]y x z x z x y x− − + − −

= ( )( )( )x y y z z x− − −

=�.�.

(37) | | 3 | | 4 | | 5a b c= = =rr r

| |a b c+ +rr r

= 0 �� %�������,�(8���

2| |a b c+ +rr r

= 2 2 2| | | | | | 2( )a b c a b b c c a+ + + ⋅ + ⋅ + ⋅r r rr r r r r r

0 = 9 16 25 2( )a b b c c a+ + + ⋅ + ⋅ + ⋅r rr r r r

50− = 2( )a b b c c a⋅ + ⋅ + ⋅r rr r r r

25 .a b b c c a− = ⋅ + ⋅ + ⋅r rr r r r

(38) logx x dx∫

logu x= dv∫ = x dx∫

du =1dx

x v =

2

2

x

udv∫ = uv vdu− ∫

logx x dx∫ = 2 21log | |

2 2

x xx dx

x− ×∫

= 2 2

log | |2 4

x xx c− +

(39) ( )P A = 3

8

1( )

8P B =

( )P A = 1 ( )P A−

= 3 5

18 8− =

( )P A B∪ = ( ) ( )P A P B+ ( AQ �!"# B *��2%���" 1�/# C�H67�( ( ) 0P A B∩ = )

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Page 6: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

6

= 3 1 4 1

8 8 8 2+ = =

( )P A B∩ = ( ) ( )P B P A B− ∩

= 1 1

08 8− = .

(40) 0

2 2 2 2lim

2 2x

x x

x x→

+ − + +×

+ +

=( )0

2 2 1 1lim

2 2 2 22 2x

x

x x→

+ −= =

++ + .

(41) (a) | |x =

;

;

x

x

0x ≤

0x > I�

I�

( )f x =

x x

x x

− ++

0x ≤

0x > I�

I�

( )f x =

0

2x

0x ≤

0x > I�

I�

( )g x =

x x

x x

− −−

0x ≤

0x > I�

I�

( )g x =

2

0

x−

0x ≤

0x > I�

I�

0x ≤ @#%��J�

( )( )f g xo = [ ( )] ( 2 )f g x f x= −

0x > @#%��J� ( )f g xo = [ ( ) (0) 0f g x f= =

0x > @#%��J�

( )( )g f xo = [ ( )] [2 ]g f x g x= =0

0x ≤ @#%��J�

( )( )g f xo = [ ( )] (0) 0g f x g= =

(b) 3

3

x y

x y

+ ≥

+ =

2 5

2 5

x y

x y

− ≤

− =

2 3

2 3

x y

x y

− + ≤

− + =

x 0 3 x 0 2.5 x 0 -3

y 3 0 y -5 0 y 1.5 0

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Page 7: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

7

(0,3), (3,0)A B ��� 3x y+ = �2 ���,-K(8 �4 �(��(. (0,0) , 3x y+ ≥

�2 .�����,-� L'M� �/�N� ���.

(0, 5), (2.5,0)C D− ��� 2 5x y− = �2 ���,-K(8 �4 �(��(. (0,0) , 2 5x y− ≤

�2 .�����,-� L'M� �/�N� ;(8�.

(0,1.5), ( 3,0)E F − ��� 2 3x y− + = �2 ���,-K(8 �4 �(��(. (0,0) ,

2 3x y− + ≤ �2 .�����,-� L'M� �/�N� ;(8�.

���∆ GHI ��� L'M �$��/#.

(42) A B C π+ + = ��%�������,�(8�.

(a) �.� cos cos cosA B C= + +

= 2cos cos cos2 2

A B A BC

+ − ⋅ +

2 2 2

A B Cπ+ = − Q

= 2cos cos cos2 2 2 2

C A BC

π − ⋅ − +

= 22sin cos 1 2sin2 2 2 2

C A B C ⋅ − + −

= 1 2sin cos sin2 2 2 2

C A B C + − −

= 1 2sin cos cos2 2 2 2 2

C A B Cπ + − − −

= 1 2sin cos cos2 2 2 2 2

C A B A B + − − +

= 1 2sin 2sin sin2 2 2

C B A + ⋅ ⋅

= 1 4sin sin sin2 2 2

A B C + ⋅ ⋅

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Page 8: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

8

=�.�.

(b) �.�. cos cos cosa A b B c C= + +

= 2 sin cos 2 sin cos 2 sin cosR A A R B B R C C+ +

sin

a

AQ = 2

sin sin

b cR

B C= = .

= [sin 2 sin 2 sin 2 ]R A B C+ +

= 2 2 2 2

2sin cos sin 22 2

A B A BR C + − ⋅ +

= [ ]2sin( ) cos( ) sin 2R A B A B C+ ⋅ − +

= [ ]2sin( ) cos( ) sin 2R c A B Cπ − − +

[2sin .cos( ) 2sin cos ]R C A B C C= − +

= 2 sin [cos( ) cos ]R C A B C− +

= 2 sin [cos( ) cos[ ( )]R C A B A Bπ− + − +

= 2 sin [cos( ) cos( )]R C A B A B× − − +

= 2 sin 2sin sinR C A B×

= 2 sin 2sin sinR A B C

= 2 sin sina B C

�.�. = �.�.

��� CDE����,&�.

(43) (a) 2( ) 2 4 6 2P n n n n= + + + + = +L n N∀ ∈ ��.

�-�-�-�- (1)

n = 1 ��.

(1)P = 22 1 1 1× = +

2 = 1

(1)P∴ ��� ;���.

�-�-�-�- (2)

n = k ��.

( )P k = 22 4 2k k k+ + + = +L ��� ;��� ��

�-�-�-�- (3)

n = 1k + ��.

( 1)P k + = 22 4 2( 1) ( 1) ( 1) ( 1)( 2)k k k k k+ + + + = + + + = + +L

�.�. = 2 4 2 2( 1)k k+ + + + +L

= ( 1) 2( 1)k k k+ + +

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Page 9: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

9

= ( 1)( 2)k k+ +

( 1)P k∴ + ��� ;���.

��� ���� %��/��=��� O# ( )P n ��� n N∈ -� �� ���E!/#

;���.

(b) ���P���1� )+#� ����� 30a = ��.

*3%��4 �� �Q+># ���P���1� ����� �+�� �&R������.

���P���1� �����N� %��&'>�2

1 230,30 2 ,30 2× × L

�� *4 %�4�/� %��&'>�2��/#. ��

2 �� �Q+ >-1� = 30 2 1202× =

4 �� �Q+ >-1� = 430 2 480× =

��� n�� �� �Q+ >-1� = 30 2n× .

(44) (a) �.�.

log log log

log 2 log 2 log 2

log3 log3 log3

x y z

x y z

x y z

2 2 1 3 3 1,R R R R R R→ → → −

log log log

log 2 log 2 log 2

log3 log3 log3

x y z

=

log log log

log 2 log3 1 1 1

1 1 1

x y z

× 2 3R R∴ ≅

= 0

�.�. = �.�.

��� CDE����,&�.

(b) ar = ˆˆ ˆ2 3i j k− +

br = ˆˆ ˆ2 3 4i j k− + −

cr = ˆˆ 2j k− +

ar = lb mc+

r r

ˆˆ ˆ2 3i j k− + = ˆ ˆˆ ˆ ˆ( 2 3 4 ) ( 2 )l i j k m j k− + − + − +

1 = 2l− 1

2l⇒ = −

2− = 3l m− ... (2)

3 = 4 2l m− + ... (3)

(1) �!"# (2)G L'��

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Page 10: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

10

1

32

m − −

= 2= −

3

22

− + = m

1

2m =

l �!"# m -� ����� ������ (3)-� E+�N&

1 1

4 22 2

− ×− + × = 3

2 1+ = 3

��� *4 %��&' )�� �!2 �4 %��&'��� *4 �-���� �������/#.

��� �3%��&'�( *4 �8 %��&'�8�/#.

(45) (a) ( )h x �2 ��'� ( , )R = −∞ ∞ -�� ;(8 ��� �(���K#

��+�"����,�(8�.

0

lim ( )x x

h x→

=0

1lim sinx x

xx→

= 0 0

0

1( )

sinx h x

x=

0x = 0 ���!/1

( ) sinh x xx

=

x− ≤ 1

sinx xx≤

( )g x = ,x− 1 1

( ) sin , ( )f x h x xx x

= =

0

lim ( )x

g x→

= 0 , 0

lim ( ) 0x

h x→

= �!"# 0

1lim sin 0x

xx→=

��&U,�� ��!2����-

0

limx→

1

sinxx

= 0 (0)h=

( )h x∴ ��� �:� >J ��&%��NK# %��&'67����

(b) sin y = sin( )x a y+

x = sin

sin( )

y

a y+

���N&

1 = 2

sin( ) cos sin cos( )

sin ( )

dy dya y y y a y

dx dx

a y

+ ⋅ ⋅ − × + ⋅

+

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Page 11: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

11

2sin ( )a y+ = [sin( ) cos cos( ) sin ]dy

a y y a y ydx

+ ⋅ − + ⋅

= [sin( )]dy

a y ydx

+ −

2sin ( )

sin

a y dy

a dx

+=

(46) (a) I = 6

2 1

xdx

x⋅

+∫ .

= 2 1

62

tt dt

t

−×∫

= 3 2( 1)t dt−∫

= 3

33

tt

= 3 3

33

t tc

−+

= 2( 3)t t c− +

= 2 1(2 1 3)x x c+ + − +

= 2 1(2 2)x x c+ − +

= 2( 1) 2 1x x c− + +

(b) 1A �!"#

2A ��� %��=��8'�( I �!"# II %�V5# ������ /=�/#

C�H67�( ��.

B ��� ���N� E�W C�H��!��� C�H67 ��.

1( )P A = 60

100 1

3( / ) 0.03

100P B A = =

2( )P A = 40

100 2

4( / ) 0.04

100P B A = =

��UX-� ��!2����-,

1( / )P A B = 1 1

1 1 2 2

( ) ( / )

( ) ( / ) ( ) ( / )

P A P B A

P A P B A P A P B A

⋅⋅ + ⋅

=

60 3 18

100 100 100060 3 40 4 18 16

100 100 100 100 1000 1000

×=

× + × +

= 18 1000 9

1000 34 17× =

2 1x + = t ��

2 1x + = 2t

2dx = 2tdt

��%��J� x = 2 1

2

t −

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Page 12: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

12

2( / )P A B =

40 4

100 10034

1000

×

= 16 1000 8

1000 34 17× =

1( / )P A B > 2( / )P A B

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���N� Y!�,& ���2 .�' ��W��4��� <�#.

(47) (a) ,a v �!"# s ��� >���#. ������# �!"# Z+# ��.

v = ds

dt

a = 28 /dv

m sdt= −

dv∫ = 8 dt−∫

1

8v t c= − +

E�+���� �������#���� 0t = , 24 /v m s=

24 = 18 0 c− × +

1c = 24

v∴ = 8 24t− +

ds

dt = 8 24t− +

ds∴ = ( 8 24)t dt− +∫

s = 2

2

824

2

tt c− + +

s = 2

24 24t t c− + +

t = 0 @#%��J�, 0s =

0 = 2c

s∴ = 24 24t t− +

���,&�' ���A�8 C"��#%��J� , v = 0

8 24t− + = 0

3t =

t = 3 @#%��J� 24 3 24 3s = − × + ×

= 36 72− +

s = 36 [,&'

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Page 13: …...Feb 11, 2019  · 4 = 2 cos cos sin cot cos sin sin cos θ θ θ θ θ θ θ θ × × = = . . CDE˜ ,&ˇ (33) 9 (1 ˝ 5 (1 F˜/ 1& ˝˜ & ˛ #. . 2 (1 F˜/ ˛- 1& ˝˜ ˛ #

13

���,&�' ���A( �����/ 4 [,&4�/ >��� C"�����#.

(b) 2 22 3 6 19 20 0x xy y x y− − − + − = .

2 22 3x xy y− − = (2 3 )( )x y x y− +

2 22 3 6 19 20x xy y x y− − − + − = (2 3 )( )x y l x y m− + + +

;x y %�J���85# �!"# ��=���85# ��� %�V�

6− = 2l m+ ... (1)

19 = 3l m− ... (2)

20− = lm ... (3)

(1), (2)G L'��,

2l m+ = 6−

3l m− = 19

5m = 25−

5m = −

( 10)l∴ + − = 6−

l = 4

��� �I�� �������(

2 3 4x y− + = 0

5x y+ − = 0

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