30
. กก 4 122 2 x 3 lim 3 - 9 x 3x - 3 = x 3 lim d dx (3 - 9x -1 29 d dx ( 3x -329 = x 3 lim 9x -2 1 2 3x (329 = x 3 lim 6 3x x 2 = 18 9 = 2 2 x 3 lim 3 - 9 x 3x - 3 = x 3 lim 3x - 9 x( 3x -329 = x 3 lim ( 3x -329 ( 3x + 329 x( 3x -329 = x 3 lim 3x + 3 x = 6 3 = 2 6 122 2 x 0 lim c + x 0 lim x x + 4 - 2 = 6 c + x 0 lim 1 1 2x + 4 (129 = 6 c = 2 c + 1 1 4 = 6 8 124 1 Con x = 0 0 () : 3 = x = 0 0 (ก) : 3 Con x = 2 2 (ก) : 3 2 () : x = 2 3 0 0 diff 2x - 1 1 x = 2 9 125 1 f(x) x 0 - lim f(x29 = x 0 + lim f(x29 x 0 lim f(x29 = x 0 lim 4 + x - 2 x = x 0 lim 1 24 + x (129 1 = 1 4 = 1

Math quota-cmu-g-455

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Page 1: Math quota-cmu-g-455

������������ �������� ��.����������������ก� �!��"#$�%ก�!&����!��"

��'��" (�)�*���� 4 ���� 122 ��� 2

������x→ 3lim

3− 9x3x − 3

=x→3lim

ddx

(3− 9x−1)

ddx

( 3x −3)=x→3lim

9x−21

2 3x(3)

=x→ 3lim

6 3x

x2= 18

9= 2

������ 2x→ 3lim

3− 9x3x − 3

=x→3lim

3x− 9x( 3x −3)

=x→ 3lim

( 3x −3)( 3x +3)

x( 3x −3)

=x→ 3lim

3x +3x = 6

3= 2

��� 6 ���� 122 ��� 2

x→ 0lim c +

x→ 0lim

x

x+ 4 − 2= 6 → c +

x→ 0lim

11

2 x+4(1)

= 6

∴ c = 2c + 114

= 6

��� 8 ���� 124 ��� 1Con � x = 0 0 (��) : 3

= ∴ ���������� x = 00 (ก���) : 3

Con � x = 2 2 (ก���) : 32 (����) : ��� � x = 2 ��� 30

0→ diff

2x− 11

∴ ���������� x = 2��� 9 ���� 125 ��� 1

��! �"#�$%&� f(x) �()��x→ 0 −lim f(x) =

x→ 0 +lim f(x)

x→ 0lim f(x) =

x→ 0lim

4+ x − 2x =

x→ 0lim

1

2 4+x(1)

1= 1

4

=

1

Page 2: Math quota-cmu-g-455

��� 10 ���� 125 ��� 2

x→ 1−lim f(x) =

x→ 1 −lim

1

3x+ 1 = 1

4

x→ 1+lim f(x) =

x→ 1+lim

2− 5− xx− 1 =

x→ 1+lim

0− 1

2 5− x

(−1)

1= 1

4

∴x→ 1−lim f(x) =

x→ 1+lim f(x)

�� ∴ f �*����������� x = 1x→ 1lim f(x) ≠ f(1)

��� 11 ���� 126 ��� 41 (��) : � x = 1 ��� 4(x− 1)2

( x −1)2= [( x −1)( x +1)]2

( x − 1)2= ( x −1)2( x + 1)2

( x − 1)2

1 (����) : k ∴ k = 4��� 14 ���� 127 ��� 3

4

g : Con � x = 0g(0) = a

x→ 0−lim

f(x)x =

x→ 0−lim

f (x)1

= f (0) = 1

4

x→ 0+lim

b

x+ 4= b

2

/$��� ∴a = 1

4= b

2→ a = 1

4, b = 1

2a + b = 3

4

��� 17 ���� 129 ��� −1

x→ 1−lim f(x) =

x→ 1−lim

2x2− x− 1x− 1 =

x→1−lim

4x− 11

= 3

x→ 1+lim f(x) =

x→ 1+lim [a(x − 2) + 2] = − a + 2

/$���)�� ∴ −a + 2 = 3 a = −1

��� 18 ���� 129 ��� 8

x→ 2+lim f(x) =

x→ 2+lim x2 − 5x − 6 = 12

�(��(1� ∴ k = 8x→ 2−lim f(x) =

x→ 2−lim (10x − k) = 20 − k 20 − k = 12

=

2

Page 3: Math quota-cmu-g-455

��� 20 ���� 130 ��� 3f (x) = (x+ 2)g (x) − g(x)(1)

(x+ 2)2

f (−3) = (−1)g (−3) −g(−3)(1)(−1)2

= (−1)(−6) − (8)(−1)2

= − 2

��� 21 ���� 131 ��� 7.5g(x) = x2

f(x) → g (x) = f(x)⋅(2x) − x2(f (x))(f(x))2

g (3) = f(3)(6) − 9f (3)(f(3))2

= (2)(6) − (9)(−2)22

= 30

4= 7.5

��� 22 ���� 131 ��� 456$ :x = −1 f(x) = x(−x) − x + 1 = − x2 − x + 1

∴ f (x) = − 2x − 1 f (−1) = 1

56$ :x = 1, 3 f(x) = x(x) − x + 1 = x2 − x + 1

f (x) = 2x − 1 → f (x) = 2 → f (x) = 0

∴ f (1) = 2, f (3) = 0

��� 23 ���� 132 ��� 4= g (x) f (x + f(x + f(x))) ⋅ d

dx(x + f(x + f(x)))

= f (x + f(x + f(x))) ⋅ [1 + f (x + f(x)) ⋅ ddx

(x + f(x))]

= f (x + f(x + f(x))) ⋅ [1 + f (x + f(x))(1 + f (x))]

=g (1) f (1 + f(1 + f(1))) ⋅ [1 + f (1 + f(1))(1 + f (1))]

= f (1 + f(2)) [1 + f (2)(1 + 1)]

= f (3) [1 + (2)(2)] = (3)(5) = 15

3

Page 4: Math quota-cmu-g-455

��� 29 ���� 135 ��� 4h : ������������ x = 2

h(2) = x→ 2−lim h(x) f(x) = ax+ 1

x2 + 1

f(2) = x→ 2−lim g(x) = g(2) f (x) = (x2 + 1)(a) − (ax+ 1)(2x)

(x2 + 1)2

=2a+ 15

5 f (2)

=2a+ 15

55a− (2a+ 1)(4)

25

=2a+ 15

5a− 8a− 45

∴ a = −1

�(��(1� f(x) = 1− xx2 + 1

, f (x) = (x2 + 1)(−1) − (1− x)(2x)(x2 + 1)2

8/9:;�* 2h(−2) − h(2) = 2g(−2) − f(2) = 2(5)(f (−2)) − f(2)

= = 310f (−2) − f(2) = 10

(5)(−1)−(3)(−4)25

−15

��� 30 ���� 136 ��� 0.3f(x) = 7x+ 1

2x+ 1 → f (x) = (2x+ 1)(7)−(7x+ 1)(2)(2x+ 1)2

(gof) (x) = d

dx(g(f(x)) = g (f(x)) ⋅ f (x)

(gof) (2) = g (f(2)) ⋅ f (2) = g 15

5

35− 3025

= (1.5)

1

5 = 0.3

��� 31 ���� 136 ��� 26. /<�=(> m=(> = m8@��

dy

dx

(2, 1) m=(> = 1

2 5− x2(−2x)

m=(> = −2

∴ >*ก�#�>��>(*=(> @�� =y − 1 −2(x − 2)

y = −2x + 5

�C�/<�%� choice (1� 4 *� ��F/$"�)�� *5�� 2 �"9�5����9)��GH�/#I�

y = 5 - x2

(2, 1)m=(>

4

Page 5: Math quota-cmu-g-455

��� 35 ���� 138 ��� y = 46 /<�=(> m=(> = m8@��

dy

dx

(1, 4) m=(> = f (x) = f (1)

/�ก g(x) = f(x)x → g (x) = x f (x) − f(x)(1)

x2

� x = 1 ��� ∴ g (1) = f (1) − f(1) → −4 = f (1) − 4 f (1) = 0

∴ �>��>(*=(>=��� (1, 4) �$ @)�*&(� = 0 *>*ก�#@�� y = 4��� 36 ���� 139 ��� 15

F(x) = x2f(x) → F (x) = x2f (x) + f(x)(2x)

F (3) = 9f (3) + f(3)(6) (1)

6. /<�=(> m=(> = m8@��dy

dx

(3, f (3)) 1 = f (x)

1 = f (3) (2)

�$/�ก �9F��� /$��� (3, f(3)) y = x − 2 f(3) = 3 − 2 → f(3) = 1 (3)

� (2) �$ (3) %� (1), F (3) = 9(1) + (1)(6) = 15

��� 37 ���� 139 ��� 4y = x2 + 4

x = x + 4x = x + 4x−1

6 /<�=(> m=(> = m8@�� dy

dx

(1,5) m=(> = 1 − 4x−2

m=(> = 1 − 4 = − 3

�(��(1� >*ก�# l @�� y − 5 = − 3(x − 1)

y = − 3x + 8 (1)

�(�ก(��>��8@�� y = x2 − 10 (2)

∴ (2) − (1), 0 = x2 + 3x − 18→ 0 = (x + 6)(x − 3) x = − 6, 3

��8/9:��ก/<��(��9F�@)�� #��:� 4 ∴ ��� 5�� 4x = 3→

y = f(x)

(1, 4)

f (1) = 4

y = f(x)

y = x - 2(3, f(3))

(1, 5) m=(>l

5

Page 6: Math quota-cmu-g-455

��� 38 ���� 140 ��� 36. /<�=(> m=(> = m8@��

dy

dx

m=(> =(1, 12) (x+ 1)(1)−(x)(1)

(x+ 1)2= 1

4

∴ =m⊥ −41

"I/�#6�/�ก choice /$"�)��*5�� 3 5����9)�@)�*&(� = −4

��� 40 ���� 141 ��� 2

/�ก f (x) = ∫ f (x)dx = ∫(2x + 1)dx = x2 + x + c

∴ f (2) = 2 → 22 + 2 + c = 2 c = −4

∴ f (x) = x2 + x − 4 → f (1) = −2

�(��(1� m=(> = −2 , m⊥ = 1

2

>*ก�#�>���#�@�� y − 3 = 1

2(x − 1)

∴ y = 1

2x + 5

2

��� 42 ���� 142 ��� 1y = 2x3 − x2 → dy

dx= 6x2 − 2x

6. /<�=(> m=(> = m8@��(a, b) 4 = 6x2 − 2x

4 = 6a2 − 2a

= 03a2 − a − 2

= 0(3a + 2)(a − 1)

∴ a = −23, 1

���������� m=(>L

m

y = xx + 1

(1, )12

�C�/<� (1, 3) �G check choice "�)��* 5�� 2 �GH�/#I��"9�5����9)

f (x)

(1, 3)m

m = f (1)=(>

m = 4(a,b)

x + 4y = 10

6

Page 7: Math quota-cmu-g-455

��� 43 ���� 142 ��� 2.46. /<�=(> m=(> = m8@��

dy

dx

(1, k) =k− 122

2k(x − 2)

=k− 122

−2k

m=(> = = k− 121− (−1) = k− 12

2k − 12 −4k

∴ k = 12

5= 2.4

��� 48 ���� 145 ��� 9m = @���C��><�5�� f(x) = x − 6 + 5

/�ก f(x) "�)�� ��! ∴ m = 5x − 6 ≥ 0

/�ก g(x) = −x4

4+ x3 − x2 + 4 → g (x) = −x3 + 3x2 − 2x = 0 → − x(x2 − 3x + 2) = 0

∴ −x(x − 2)(x − 1) = 0 x = 0, 1, 2

Max Min Max

∴ @��>F�><�>(*"(O: P#�� �$ ∴ M = 4=M = g(0) g(2) → g(0) = 4 g(2) = 4

��� 49 ���� 145 ��� 4f(x) = x3

3− 3x2 + 8x − 2 → f (x) = x2 − 6x + 8 = 0

(x − 4)(x − 2) = 0 → x = 2 , 4

Max Min

��� 1 @)�*&(�� x = 1 @�� f (1) = 1 − 6 + 8 = 3

��� 2 @�� Min = f(4) = 10

3

��� 3 @�� Max = f(2) = 14

3

��� 4 /�ก f (x) = (x − 4)(x − 2)

��� 51 ���� 146 ��� 4 y = f(x) = 1 − 5x = (1 − 5x)

12

dydx

= 1

2(1 − 5x)−1

2 ⋅ ddx

(1 − 5x) = − 5

2(1 − 5x)−1

2

d2y

dx2= 5

4(1 − 5x)−3

2 ⋅ ddx

(1 − 5x) = − 25

4(1 − 5x)−3

2

(-1,12) (1,k)m=(>

f(x) = k(x-2)2

�"I�* �"I�*��+ - +

2 4

7

Page 8: Math quota-cmu-g-455

@)�*&(��>��8@��� (0, 1) @�� f (0) = − 5

2(1 − 0)−1

2 = − 5

2

>*ก�#�>��>(*=(>� (0, 1) @�� y − 1 = − 5

2(x − 0)

2y − 1 = − 5x

5x + 2y − 1 = 0

��� 52 ���� 147 ��� 2 �� �"I�* ��/�ก dy

dx= −(x + 1)(x − 2) = 0

x = −1, 2

Min Max

��� 54 ���� 148 ��� 5y = f(x) = 4x

32 − 9x

23 + 6

f (x) = 6x12 − 6x−1

3

∴ � x = 1 ��//$�กI� Max, Min �G��9��)��f (1) = 6 − 6 = 0

f (x) = 3x−12 + 2x−4

3

"�)�� f (1) = 3 + 2 = 5 f (1) > 0

∴ � x = 1 �กI�/<��C��><�>(*"(O: ��!��� 57 ���� 151 ��� 7

!"ก�$%�$�&�'�(x − h)2 = 4c(y − k)

(x − 3)2 = 4(5)(y − 1)

(x − 3)2 = 20(y − 1)

∴y = (x− 3)2

20+ 1

� (a, 6) ��%�>*ก�#"�#�8��� /$���6 = (a− 3)2

20+ 1 → (a − 3)2 = 100 → a − 3 = 10,−10

�� (a, 6) �9F� ∴ a = 13a = 13,−7 Q1

+- --1 2

ก#�S

x = -1 x = 2

x

y

c = 5 (a,6)

(b,0)

F(3,6)

V(3,1)

m=(>

8

Page 9: Math quota-cmu-g-455

/�ก � x = 13 /$��� y = (x− 3)220

+ 1 → y = (x− 3)10

y = 1

�(��(1� m8@�� � (13, 6) @�� 1 ∴ m=(> � (13, 6) = 1P�@)�*&(�#$P)��� (13, 6) ก(� ∴ b = 7(b, 0) , 6− 0

13− b = 1

��� 60 ���� 152 ��� 30/�ก = 180 %P�a + b + c (1) f(b) = abc = (60 − b)b(120)

/�ก =a+ bc

1

2f(b) = 7200b − 120b2

=a + b c

2(2) f (b) = 7200 − 240b = 0

� (2) %� (1) , ∴c

2+ c = 180 b = 7200

240= 30

∴ c = 120 → a + b = 60

��� 61 ���� 153 ��� 7 f(x) = x2

x3 + 200

f (x) = (x3 + 200)(2x) − (x2)(3x2)(x3 + 200)2

= 0

2x4 + 400x − 3x4 = 0→ 400x − x4 = 0

∴ ก)��!x(400 − x3) = 0 x = 0, 3 400 → x = 0, 7

��8/9:��ก)�� x �GH�/C��)���V*�)ก��� � x = 7 : f(7) = 72

73 + 200= 49

543= 0.090

��� � x = 8 : f(8) = 82

83+ 200= 64

712= 0.089

��� 62 ���� 153 ��� 3�(�#�ก�#�G��9� G��@)�*&(� = f (x) = 2x − 6

/�ก8/9: "@)�*&(�5���>��8@��� ���ก(� 8" >��)�� (0,−6) f(0) = − 6, f (0) = 8f (x) = ∫ f (x)dx = ∫(2x − 6)dx = x2 − 6x + c

/�ก ∴ f (0) = 8 → f (0) = 02 − 6(0) + c = 8 c = 8

f(x) = ∫ f (x)dx = ∫(x2 − 6x + 8)dx = x3

3− 3x2 + 8x + c

/�ก f(0) = −6 → f(0) = c = −6

∴ f(2) = 8

3− 12 + 16 − 6 = 2

3

9

Page 10: Math quota-cmu-g-455

��� 64 ���� 154 ��� 3.5/�ก ∴ f(x) = A(x − 2)3 + B → f (x) = 3A(x − 2)2 → f (1) = 3A = 2 A = 2

3

∫ f(x)dx = ∫(23(x − 2)3 + B)dx = ∫ 23(x − 2)3dx + ∫Bdx (1)

"I/�#6� %P� ∫ 23(x − 2)3dx u = x − 2 → du

dx= 1 → du = dx

�(��(1� ∫ 23(x − 2)3dx = ∫ 23u3du = 2

3

u4

4+ c = 1

6(x − 2)2 + c

�%� (1), ∴ 0

1

∫ f(x)dx = 1

6(x − 2)4 + Bx 1

0= (1

6+ B) − (16

6) = 1 B = 3.5

��� 66 ���� 155 ��� 4F(x) =

x

0∫ (t2 + t − 2)dt = t3

3+ t2

2− 2t x

0= x3

3+ x2

2− 2x

F (x) = x2 + x − 2 = 0 → (x + 2)(x − 1) = 0 → x = − 2, 1

�@�� x = − 2 , F(−2) = 10

3∗ ∗

x = 1 , F(1) = − 7

6

∴ @��>F�><�>(*�F#6:@�� 103

x = − 3 , F(−3) = 3

2

x = 2 , F(2) = 2

3

��� 70 ���� 157 ��� 18/�ก8/9: "@)�*&(�5���>��>(*=(>�>��8@�� � (1, 2) ���ก(� 4"y = f(x)

/$��� �$ f (1) = 4 f(1) = 2

>**<�I f(x) = Ax2 + Bx +C→ f(1) = A + B +C = 2 (1)

f (x) = 2Ax + B → f (1) = 2A + B = 4 (2)

−1

2

∫ f(x)dx = 12→2

−1∫ (Ax2 + Bx +C)dx = 12→ Ax3

3+ Bx2

2+Cx 2

−1 = 12

(83A + 2B + 2C) − (−A

3+ B

2−C) = 12

3A + 3

2B + 3C = 12→ 2A + B + 2C = 8 (3)

10

Page 11: Math quota-cmu-g-455

ก�>*ก�# (1), (2), �$ (3) ��� A = 4,B = − 4, C = 2

�(��(1� f(x) = 4x2 − 4x + 2 → f(−1) = 10

f (x) = 8x − 4→ f (x) = 8→ f (−1) = 8

∴ f (−1) + f(−1) = 18

��� 71 ���� 158 ��� 1005

%P� F(a) =a+ 1

a∫ (2011x − x2)dx = 2011x2

2− x3

3

a + 1a

F(a) = 2011

2(a + 1)2 − (a+ 1)3

3−

2011a2

2− a3

3

F (a) = 2011(a + 1) − (a + 1)2 − 2011a + a2 = 0

2011 − (a2 + 2a + 1) + a2 = 0

∴ a = 10052a = 2010

��� 73 ���� 159 ��� 3F(x) = ∫ f(x)dx = ∫(px2 + qx + r)dx

F(x) = p

3x3 + q

2x2 + rx + c

/�ก ∴F(0) = 0 → c = 0 F(x) = p

3x3 + q

2x2 + rx

F(1) = p

3+ q

2+ r , F(−1) = − p

3+ q

2− r

∴ F(1) + F(−1) = q

��� 75 ���� 160 ��� 2512

=A1

1

−1∫ (x2 − x3)dx = x3

3− x4

4

1

−1

= 1

3− 1

4 −

−13

− 1

4 = 2

3

=A2 −2

1∫ (x2 − x3)dx = x3

3− x4

4

1

2

= 1

3− 1

4 −

8

3− 4

= 17

12

∴ ".. #��� = A1 +A2 = 2

3+ 17

12= 25

12

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

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

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

y

0-1 1

A1

A2

2 x

11

Page 12: Math quota-cmu-g-455

��� 77 ���� 161 ��� 4��!"ก�$%�$�&�'�

(x − h)2 = 4c(y − k)

(x − 3)2 = 4c(y − 9)

(�� (1, 5) , (1 − 3)2 = 4c(5 − 9)

4c = − 1

∴ >*ก�# @�� (x − 3)2 = − (y − 9)

��,-.�/.(ก� x � y = 0 , (x − 3)2 = 9

∴ x = 6, 0(x − 3) = 3,−3

∴ "�1��"�#�8��� ��#��P��)9= 2

3(6)(9) = 36

��� 80 ���� 162 ��� 2/�ก �$ f */<� f (x) = 2x + 2 = 0 → x = −1 Min (−1,−3) → f(−1) = −3

f(x) = ∫ f (x)dx = ∫(2x + 2)dx = x2 + 2x + c

∴ f(−1) = (−1)2 + 2(−1) + c = −3 c = −2

P�/<��(� ก� x ( � y = 0) f(x) = x2 + 2x − 2 → → x2 + 2x − 2 = 0

x = −2± 4+ 82

= −2± 2 3

2= −1 + 3 ,−1 − 3

A = −−1

0

∫ (x2 + 2x − 2)dx = 8

3

��� 82 ���� 163 ��� 2

−2

1

∫ 4x3dx =−2

0

∫ 4x3dx+0

1

∫ 4x3dx = −A2 +A1

x

y

V(3,9)

9(1,5)

60

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

-1-1- 3 -1+ 3

x

y

12

Page 13: Math quota-cmu-g-455

������ ��ก�� ���ก�����������������ก�� ���ก������

��� 2 ���� 166 ��� 32

��� (22)x+ x2 − 2 − 5(2−1)(2x+ x2 − 2 ) = 6 , A = 2x+ x2 − 2

A2 − 5

2A − 6 = 0 → 2A2 − 5A − 12 = 0 → (2A + 3)(A − 4) = 0

∴ A = − 3

2, 4 → 2x+ x2 − 2 = − 3

2, 4

x + x2 − 2 = 2 → x2 − 2 = 2 − x → ( x2 − 2 )2 = (2 − x)2

∴x2 − 2 = 4 − 4x + x2 → 4x = 6 x = 3

2

��� 3 ���� 167 ��� 3��� ���� A = 2x 1

A − A2 − A

− 1

A + A2 − A

= 7

2

(A + A2 − A ) − (A − A2 − A )

A2 − (A2 − A)= 7

2→ 2 A2 − A

A= 7

2

4(A2 − A)A2

= 7

2→ 8A2 − 8A = 7A2 → A2 − 8A = 0 → A(A − 8) = 0

∴ x = 3A = 0, 8 → 2x = 0, 8

��� 5 ���� 168 ��� 4� ก 27x+ y = 36 → (33)x+ y = 36 → 33x + 3y = 36

���� 3x + 3y = 6 (1)

� ก ���� 23x+ y = 1 3x + y = 0 (2)

����� (2) ���� (1) − (2) , 2y = 6 → y = 3 → x = − 1

∴ 3x+ 1 + 3y − 1 = 3−1 + 1 + 33− 1 = 30 + 32 = 10

��� 6 ���� 168 ��� 2 ⋅ 5

3x ⋅ 4x − 2(3x) − 9(4x) + 18 = 0 → 3x(4x − 2) − 9(4x − 2) = 0

���� ∴ (4x − 2)(3x − 9) = 0 → 4x = 2 3x = 9 → x = 1

2, 2

��������

13

Page 14: Math quota-cmu-g-455

��� 9 ���� 170 ��� 2� ก����� 2x+ 2 − 9 2x + 2 = 0 → 4 ⋅ 2x − 9 2x + 2 = 0 → (4 2x − 1)( 2x − 2) = 0

∴ 2x = 1

4, 2 → 2x = 1

16, 4 x = −4, 2

��� 1 ��� x = 2 � �!�����"#��� 2 logx+ 6(2x2 + 14x + 28) = 2 → 2x2 + 14x + 28 = (x + 6)2 → x2 + 2x − 8 = 0

∴ check () *�+� �!������,-#(.�(x − 4)(x − 2) = 0 → x = −4, 2

��� 3 ∴ x = 23

2

−x3

2

2(x− 1)= 1 →

3

2

x − 2= 1

��� 4 ∴ log5logx+62 = 0 → logx+ 62 = 50 → 2 = x + 6 x = −4

��� 10 ���� 170 ��� 4log3(x(log

535)log553) = 0 → log3(x (1

3)3) = 0

∴ x = 27x

1

27 = 30

��� 11 ���� 171 ��� 3log (0.006) = log(6 × 10−3) = log 6 + log 10−3 = log 6 − 3 log 10

= log (2 × 3) − 3 = log 2 + log 3 − 3 = A + B − 3

��� 12 ���� 171 ��� 3log 45 = log (9 × 5) = log(32 × 5) = log 32 + log 5 = 2 log 3 + log 5

= 2 log 3 + log 10

2 = 2 log 3 + log 10 − log 2 = 2b + 1 − a

��� 17 ���� 173 ��� 11log2(x − 9) + 4 log2(x − 9)2 = 3 → log2(x − 9) + 4

log22(x−9) = 3

��� ���� log2(x + 9) + 4

1 + log2(x− 9) = 3 A = log2(x − 9) A + 4

1+A= 3

A(1 + A) + 4 = 3(1 + A) → A + A2 + 4 − 3 − 3A = 0 → A2 − 2A + 1 = 0

→ (A − 1)2 = 0

∴ x = 11A = 1 → log2(x − 9) = 1 → x − 9 = 2

��� 18 ���� 174 ��� 1logb

xx − c

= a → x

x− c = ba → x = bax − bac

∴bac = bax − x → bac = x(ba − 1) x = bac

ba − 1

14

Page 15: Math quota-cmu-g-455

��� 19 ���� 174 ��� 100 ��� 9logx − 31+ logx − 54 = 0 → 32 log x − 3 ⋅ 3logx − 54 = 0 A = 3logx

���� A2 − 3A − 54 = 0 → (A − 9)(A + 6) = 0 → A = 9, −6 → 3logx = 9, −6

∴ log x = 2 x = 100

��� 23 ���� 176 ��� 1log10

log 5+

log34−1

log35+ (log 2)2 ⋅

log e

log 5

log e= 1

log 5+

− log 4

log 3

log 5

log 3

+ (log 2)2 ⋅ 1

log 5

= 1

log5− log4

log5+ (log2)2

log5= 1 − 2 log2 + (log2)2

log 5= (1 − log 2)2

1− log2

= 1 − log 2 = 1 − 0.3010 = 0.699

��� 24 ���� 177 ��� 4a = 7 + 4 3 = 7 + 2 12 = 4 + 3 = 2 + 3

1+!� b < c < ab = 2 2 2...

∴b2 = 2 2 2...1

b> 1

c > 1a

b2 = 2b

b2 − 2b = 0

b(b − 2) = 0

∴ b = 0, 2

�*� ���5 ∴b ≠ 0 b = 2

��� 25 ���� 177 ��� 2ก. � ก x2 + y2 = 11xy → x2 − 2xy + y2 = 9xy → (x − y)2 = 9xy

∴ x − y = 3 xy

,#�,-� log 1

3(x − y)

= log 1

3(3 xy

= log xy = log (xy)1

2

= 1

2log (xy) = 1

2(log x + log y)

�. 2(log 125 + log 27 − log 1000 ) = 2(log 53

2 + log 33 − log 103

2 )

= 23

2log 5 + 3 log 3 − 3

2 = 3 log 5 + 6 log 3 − 3 = 6 log 3 − 3(1 − log 5)

= 6 log 3 − 3 log 2 = − 3(log 2 − 2 log 3)

4+3 4 3x

15

Page 16: Math quota-cmu-g-455

��� 27 ���� 178 ��� 6logyx + 4

logyx= 4 → (logyx)2 − 4 logyx + 4 = 0 → (logyx − 2)2 = 0

∴ logyx = 2 logyx3 = 3 logyx = 6

��� 29 ���� 180 ��� 2(log

11

12

113 + log773

2 )x2 − [(log223)(log33

4) + 4 log 202 − log 28] x = 15

2

312

+ 3

2

x2 − [(3)(4) + log 208 − log 28] x = 15

2

15

2x2 − [12 + log

20

2

8] x = 15

2→ 15

2x2 − (12 + log 108) x = 15

2

15

2x2 − 20x = 15

2→ 15x2 − 40x − 15 = 0 → 3x2 − 8x − 3 = 0

∴(3x + 1)(x − 3) = 0 x = − 1

3, 3 → a = − 1

3, b = 3

∴ 3a + b = 3−1

3 + 3 = 2

��� 31 ���� 181 ��� 3log3(x + 1) + 2 log

32x = log3323 + log3(x + 6) → log3[(x + 1)(x)] = log3[2(x + 6)]

∴ x2 + x = 2x + 12 → x2 − x − 12 = 0 → (x − 4)(x + 3) = 0 x = 4, −3

*�!�() *�+� �! �������� ∴ x = 4x = −3

��� 32 ���� 181 ��� 16.52x+ 2y = 1 → x + 2y = 0 → x = −2y (1) , y = −x

2(2)

� ก ��� �� 5x+ y = 10 x = − 2y 5− 2y+ y = 10 → 5−y = 10

∴ 5y = 1

10→ 5y+ 3 = ( 1

10)(53) → 5y+ 3 = 12.5

� ก ��� �� 5x+ y = 10 y = − x

25

x− x

2 = 10 → 5x

2 = 10

∴ 5x = 102 → 5x − 2 = 102 × 5−2 → 5x− 2 = 4

∴ 5x− 2 + 5y + 3 = 4 + 12.5 = 16.5

16

Page 17: Math quota-cmu-g-455

��� 33 ���� 182 ��� 6log2(64x − 56) − 2 log2(x + 1) = 3 → log2

64x− 56

(x+ 1)2 = 3

64x − 56

x2 + 2x+ 1= 8 → 8x − 7 = x2 + 2x + 1 → x2 − 6x + 8 = 0

*�!�() *�+� �!������,-#(.�(x − 4)(x − 2) = 0 → x = 4, 2

∴ : +!ก() *�+ = 6��� 34 ���� 182 ��� −1

(1 − log 2)log5x + log(x + 1) = log 12 → log 5 ⋅ logx

log5+ log(x + 1) = log 12

log(x(x + 1)) = log 12 → x2 + x = 12 → (x + 4)(x − 3) = 0

*�!�() *�+� �! �������� ∴ x = 3x = − 4, 3 x = −4

� ก log3log2(2y + 10) = 0 → log2(2y + 10) = 1 → 2y + 10 = 2

*�!�() *�+� �!����� ∴ ∴ y = −4 y = −4 x + y = −1

��� 37 ���� 184 ��� 2log 5 + log((22)x − 2 + 1) − log 2x− 2 = 1

��� ����log

(2x− 2)2 + 1

2x− 2

= 1 − log 5 , 2x− 2 = A

logA2 + 1

A

= log 2 → A2 + 1

A= 2 → A2 + 1 = 2A → A2 − 2A + 1 = 0

∴ (A − 1)2 = 0 → A = 1 → 2x− 2 = 1 → x − 2 = 0 x = 2

��� 38 ���� 184 ��� 1� ก log

x2 x = logx2x

12 = 1

2× 1

2logxx = 1

4

,#�,-� 16(log

x2 x ) = 161

4 = (24)1

4 = 2

� ก��������� x4 − 5x2 + 6 = 2 → x4 − 5x2 + 4 = 0 → (x2 − 4)(x2 − 1) = 0

(x − 2)(x + 2)(x − 1)(x + 1) = 0 → x = 2, −2, 1, −1

�*�������; ,#�,-� x > 0 � � ∴ ��=� �� () *�+=;�!logx2 x x ≠ 1 x = 2

17

Page 18: Math quota-cmu-g-455

��� 39 ���� 185 ��� 3log4(2 log3(1 + log2a)) = 1

2→ 2 log3(1 + log2a) = 4

1

2

∴ log3(1 + log2a) = 1 → 1 + log2a = 3 → log2a = 2 → a = 4

� ก 22x− a = a2 + 3a + 4 → 22x − 4 = 42 + 3(4) + 4 → 22x − 4 = 32

∴2x − 4 = 5 x = 9

2= 4.5

��� 40 ���� 185 ��� 22(log4x)3 − (log4x)2 + log4(x−2) + 1 = 0

� ���� 2(log4x)3 − (log4x)2 − 2 log4x + 1 = 0 A = log4x

2A3 − A2 − 2A + 1 = 0 → A2(2A − 1) − (2A − 1) = 0 → (2A − 1)(A2 − 1) = 0

∴(2A − 1)(A − 1)(A + 1) = 0 A = 1

2, 1, −1

∴ log4x = 1

2, 1, −1 → x = 4

1

2 , 41 , 4−1 → x = 2, 4,1

4

∴ : +!ก() *�+ = 2 + 4 + 1

4= 6.25

��� 42 ���� 186 ��� b��� a � � b =>?��) �!�=*@�+!ก � � b = ka

log ba

b

2 + log a

b

9a

2 = 1 → log

ba

b

2 × a

b

9a

2

= 1

��� b = ka ����ba

b

2 × ba

−9a

2 = 101 → ba

b

2− 9a

2 = 10

� �=��A�#� ก kaa

ka2

− 9a2 = 10 → (k)

a2

(k− 9) = 101 a, b, k ∈ I+

,#�,-� k = 10 � � =�� �,-�a

2(k − 9) = 1

���� ∴k = 10, a = 2, b = ka = 20 b2 − a2 = 202 − 22 = 396

18

Page 19: Math quota-cmu-g-455

��� 44 ���� 187 ��� 3A : B : log( x + 1 + 5) = log x log2(3x) + 1

2log2(9x) + 1

3log2(27x) − 1

3log2x = 3

x + 1 + 5 = x log2

(3x)(9x)12 (27x)

13

x

13

= 3

x + 1 = x − 5 (3x)(3x12 )(3) = 23

x + 1 = x2 − 10x + 25 x3

2 = 2

3

3

x2 − 11x + 24 = 0x

32

23

= 2

3

3

2

3 → x = 4

9

(x − 8)(x − 3) = 0

∴ x = 8, 3 ∴ : (.G = A ∪ B = 8,

4

9→ 32

9

��� 45 ���� 188 ��� 3� ก x + 2y = 8 → x = 8 − 2y (1)

� ก ��� ����32x− 2y − 6 ⋅ 3−2x − 3−y = 0 (2) x = 8 − 2y

32(8− 2y) − 2y − 6 ⋅ 3−2(8 − 2y) − 3−y = 0 → 316− 6y − 6 ⋅ 3−16+ 4y − 3−y = 0

(.G* ��!� �Hก*,!���� 3y 316− 5y − 6 ⋅ 3−16+ 5y − 1 = 0

��� ���� A = 316− 5y A − 6

A− 1 = 0 → A2 − 6 − A = 0 → A2 − A − 6 = 0

1+!� ��������(A − 3)(A + 2) = 0 → A = 3, −2 → 316− 5y = 3, −2 −2

��� y = 3 ��J�ก � (1) ���� x = 2316− 5y = 3 → 16 − 5y = 1 → y = 3 →

∴ log6y9x + logy(x3 + 1) = log1818 + log3(9) = 1 + 2 = 3

��� 46 ���� 188 ��� 1024log

22a − log25b3 = 19 → 1

2log2a − 3

5log2b = 19 (1)

log22b − log25a3 = 8 → 1

2log2b − 3

5log2a = 8 (2)

(1) − (2) ,11

10(log2a) − 11

10(log2b) = 11 → 1

10(log2a − log2b) = 1

∴log2a − log2b = 10 → log2a

b = 10

a

b= 210 = 1, 024

19

Page 20: Math quota-cmu-g-455

��� 49 ���� 190 ��� 3log(3x + 4) > log(x − 1) + log 10 → log(3x + 4) > log[10(x − 1)]

x < 2 ____ (1)3x + 4 > 10x − 10 → 14 > 7x →

� � ____(2)3x + 4 > 0 → x > −4

3

� � x > 1 ____(3)x − 1 > 0 →

�) ,(1) ∩ (2) ∩ (3)

∴ =L*() *�+(�� (1, 2)��� 50 ���� 190 ��� 4

� ก 4a − 9 ⋅ 2a − 1 + 2 = 0 → 22a − 9 ⋅ 2a

2+ 2 = 0

2(22a) − 9(2a) + 4 = 0 → (2 ⋅ 2a − 1)(2a − 4) = 0 → 2a = 1

2, 4

∴ � ������+�ก a > 0 ∴ a = 2a = − 1, 2

�J�ก � 2 log2(x + 2) − log2(x − 1) < 4 → log2

(x+ 2)2

(x − 1) < 4

x2 + 4x+ 4

x− 1< 24 → x2 + 4x + 4 < 16(x − 1) → x2 − 12x + 20 < 0

____ (1)(x − 10)(x − 2) < 0 → x ∈ (2, 10)

� � x + 2 > 0 ____ (2)→ x > − 2

� � x > 1 ____ (3)x − 1 > 0 →

�) ,(1) ∩ (2) ∩ (3)

∴ =L*() *�+(�� (2, 10)

1 2-43

1-2 102

20

Page 21: Math quota-cmu-g-455

��� 51 ���� 191 ��� 23

5

5x2 − 23x + 3>

3

5

−(x+ 5)

52 − 23x + 3 < −x − 5

5x2 − 22x + 8 < 0

(5x − 2)(x − 4) < 0

1"� �G *,!= ��ก��� 2(4x − 1)(x − 4) < 0

+ +_42

5

+ +_41

4

21

Page 22: Math quota-cmu-g-455

������ ������� ������������ก���������, ���������������� 4 ���� 199 ��� 1

������ก��� A ��� ���������������������������� � �!�"#��"����$�� %��&���ก��� B ก�( C *+�ก�$,-���$�

��� 7 ���� 200 ��� 2� ��ก�/�� 2 ��, ��% 3 �� ����$� !�2�

3

2

5

3 = 30

��� 12 ���� 202 ��� 2!����& 1 ���%��$� 2 !�2� ��� A → Z , Z → A

!����& 2 ���%��$� 4! = 24 !�2�!����& 3 ��$���$���$ก�� __ __ __ __ __ ���%��$� 3!3! = 36 !�2�

∴ !�2����%�������$ = !�2�2 × 24 × 36 = 1, 728

��� 13 ���� 202 ��� 360- � ��ก��ก���%��� (ก�+$�� ����$� 10 !�2�- � ��ก��ก���%�������!���+��$ ����$� !�2�

9

2 = 36

∴ !�2�ก��� ��ก������$ = !�2�10 × 36 = 360

��� 14 ���� 203 ��� 4!�2���&� ��ก�$���ก9-ก:���:%9���� ��������� 1 ��

= !�2�� ��ก������$ - !�2���&�������ก9-ก:���:%9���� � %= !�2�

10

3 −

4

3 = 120 − 4 = 116

��� 15 ���� 203 ��� 4- "�+2�� ��&��$� 1 !�2�- ��� � +� ,� ��&��$� 2! = 2 !�2�- ����&�� �� ��&��$� 6! !�2�∴ !�2�*�$����&� = 2(6!) !�2�

� "

22

Page 23: Math quota-cmu-g-455

��� 18 ���� 205 ��� 3=�� F ���?�(� , B ��� (���ก�(�

n(F∪B) = n(F) + n(B) − n(F∩B)

∴ ��30 = 17 + 18 − n(F∩B) n(F∩B) = 5

- � ��ก"�+2�� ����$� !�2�5

1 = 5

- � ��ก���"�+2�� *�ก����&�� ����&���=��"�+2�� = !�2�29

1 = 29

∴ *���!�!�2�� ��ก = !�2�5 × 29 = 145

��� 19 ���� 205 ��� 3���%� ,��&���$�!% 3 ���! �� 3, 6, 9, 12, 15���%� ,��&���$�!% 5 ���! �� 5, 10, 15ก�����& 1 Bก��ก�$� 3, 6, 9, 12 Bก��& 2 �$� 5, 10, 15

����$� !�2�4 × 3 = 12

ก�����& 2 Bก��ก�$� 15 Bก��& 2 �$� 5, 10����$� !�2�1 × 2 = 2

∴ �!� 2 ก��� �$� 12 + 2 = 14 !�2���� 20 ���� 206 ��� 3

���ก���� 1 ���ก���� 2 ���ก���� 3!�2�

12

7 ×

5

3 ×

2

2 = 7, 920

��� 29 ���� 210 ��� 1D*�% ����ก��=�� 3 �� E�ก(���� ���$�%!ก�� �(���"#� 2 ก���ก�� ! " 1 : 3 ��� $%�ก�������� 3 �� �����(�� 6 ����&�� ���"#� 2, 4

*�$�$� !�2�6!

2!4!= 15

ก�� ! " 2 : 3 ��� $%�ก�������� 4 �� �����(�� 6 ����&�� ���"#� 2, 3, 1*�$�$� !�2�6!

2!3!1!= 60

∴ �!�� �!*�$�$�������$ !�2�15 + 60 = 75

23

Page 24: Math quota-cmu-g-455

��� 32 ���� 211 ��� 528

!�2���& ก � + , ����$�%����$ก��=���!�$�%!ก��= !�2�������$ - !�2���& ก � + , %����$ก��=���!�$�%!ก��= 6! − 4 × 2! × 4! = 720 − 192 = 528

��� 34 ���� 211 ��� 1�� ��� 2 Bก ,�! 2 Bก

P(������ก��) = (��������ก��) 1 − P = 1 −22

+

32

62

= 1 − 4

15= 11

15

��� 37 ���� 214 ��� 4n(s) =

7

3 = 35

E : ��& 2 , ()� 1 ()� 2 , ��& 1n(E) =

3

2

4

1 +

4

2

3

1 = 12 + 18 = 30

P(E) = 30

35= 6

7

��� 38 ���� 214 ��� 3 ����ก�� 2 Bก��������ก�� �(���"#� 3 ก���n(s) = 7 × 7 = 49,E :

�,�%!������ก�� �$�������ก�� ,�!������ก��n(E) = 3 × 3 + 2 × 3 + 2 × 1 = 17

∴ P(E) = 17

49

ก, , ��$ก���$� 4 !�2� (�����&!�$���(�)

� �( ก,,� �( 4 ����&�� ��

24

Page 25: Math quota-cmu-g-455

��� 39 ���� 215 ��� 3!�2�n(S) =

10

3 = 120

E : � ,�B��� 0, 2, 4, 6, 8 �%�( 3 =( J �!���กก!�� 10 ��� (0, 4, 8) (0, 6, 8) (2, 4, 6) (2, 4, 8) (2, 6, 8) (4, 6, 8) ∴ n(E) = 6

∴ P(E) = 6

120= 1

20

��� 40 ���� 215 ��� 2� ,��&��&�ก!�� 5 ��� 1, 2, 3, 4 ∴ n(E) =

4

3

6

5 = 24

∴ n(S) = 10

8 = 45 P(E) = 24

45= 8

15

��� 41 ���� 216 ��� 4�� Bกก!�$���$� 24 ��L$ , ��,�! x ��L$ , ���,�%! y ��L$*�กD*�% P(,�!�����,�%!) = x+ y

24+ x+ y = 5

6→ 6x + 6y = 120 + 5x + 5y

x + y = 120 ____ (1)P(�,�%!�����$�) = 24+ y

24+ x+ y = 3

4→ 96 + 4y = 72 + 3x + 3y

____ (2)3x − y = 24

, ____ (3)(1) × 3 3x + 3y = 360

, ∴ y = 84(3) − (1) 4y = 336 →

��� 42 ���� 216 ��� 3n(S) =

n

2 = n(n− 1)

2

∴ E : (1, 3) (2, 3) n(E) = 2

∴ P(E) = 2n(n−1)2

= 4

n(n− 1)

25

Page 26: Math quota-cmu-g-455

��� 43 ���� 217 ��� 2������� Bก(� ������$ n Bก , �%�( �� 2 Bก P(w 2 Bก) = → 2

15

P(w 2 Bก) = ∴ n = 1042

n2

=4×32×1n(n−1)2×1

= 12

n(n− 1) = 2

15→ n(n − 1) = 90

$������ Bก(� =�ก ����"#� ��,�! (w) 4 Bก, ���$� (R) 3 Bก, ���,�%! (G) 3 Bก����%�( 4 Bก !�2�n(s) =

10

4 = 10× 9× 8×7

4× 3× 2× 1 = 210

E : G 1 Bก , Bก Bก = R ≥ 1 → All −w 3 7

3 −

4

3 = 31

3

1 = 3

∴ ∴ n(E) = 3 × 31 = 93 P(E) = 93

210= 31

70

��� 46 ���� 218 ��� 0.9n(S) =

5

3 = 10

n(E) = n(�%�(�$� ก ���� ,) = n(ก ,) = n(UUUU) - n(ก ,)∪ ∪

= ก ,n(S) − n( ∩ ) = 10 − 1 = 9

�%�(����$� ก � +����$� , *+�� 1 !�2� ��� �%�( ,, �, �

∴ P(E) = 9

10= 0.9

��� 47 ���� 219 ��� 3�%�(�� + Bก 4 ����� (�%�(� �!=�����) → n(s) = 10 × 10 × 10 × 10 = 104

E : �%�(�$� 4 ��T + 1 Bก n(E) = 4

1

3

1

2

1

1

1 × 4!

� �( ��$�(,����� +���$�

∴ P(E) = 4×3×2×1×4 !104

= 36

625

26

Page 27: Math quota-cmu-g-455

��� 48 ���� 219 ��� 34

s q r s ∧ q (s ∧ q) ∨ r [(s ∧ q) ∨ r] → s

T T T T T TT T F T T TT F T F T TT F F F F TF T T F T FF T F F F TF F T F T FF F F F F T

∴ P(E) = 6

8= 3

4

��� 50 ���� 220 ��� 2D%� Bก��Y� 2 Bก → n(s) = 6 × 6 = 36

�����!� 4, 7 : (1, 3) (3, 1) (2, 2) (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) ∴E1 : n(E1) = 9

�����!� 6, 11 : (1, 5) (5, 1) (2, 4) (4, 2) (3, 3) (5, 6) (6, 5) ∴E2 : n(E2) = 7

∴ J ���� ∴ P(����) = 36 − 9 − 7 = 20 = 20

36

���D%� 72 ����� ��$!��*+���� �����20

36(72) = 40

��� 52 ���� 221 ��� 3����%/ I �� ��!, ก��% ����%/ II �� ��!, ��! ����%/ III ��ก��%, ก��%*�กD*�% D%�� �!�$���! ��$�!�� ∴ S = {I, II}, E = {II} P(E) = 1

2

��� 53 ���� 222 ��� 3

A B C !�2�←4

→4←5

→5 n(S) = 4 × 5 × 5 × 4 = 400

E : A B C !�2�←3

→4←4

→5 n(E) = 4 × 5 × 4 × 3 = 240

∴ P(E) = 240

400= 0.6

+, ก���

27

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��� 54 ���� 222 ��� 4n(S) =

14

3 = 364

∴ n(E) = 5

3 +

6

3 +

3

3 = 31 P(E) = 31

364

��� 55 ���� 223 ��� 2�� 4 �� ����%�(� ��$ 4 ,!$ !�2�n(s) = 4! = 24

E : �%������% 2 ���$�� ��$,������� �(���"#� 2 ก�������� 2 �� �$�� ��$,������� � ��ก�$� !�2�

4

2 = 6

�� 4 �� �$�� ��$,������� � ��ก�$� !�2�4

4 = 1

∴ !�2� ∴ n(E) = 7 P(E) = 7

24

��� 58 ���� 224 ��� 2 =�� A ����$���9����, B ����$�E�\\�� ∴ P(A) = 1

10, P(B) = 1

15

�E��+*+�$� 2 �%���E����ก������$� P(A∪B) = P(A) + P(B) − P(A∩B), P(A∩B) = 0

∴ P(�$����!� ) , P(����$����!� ) P(A∪B) = 1

10+ 1

15= 1

6= 1

6= 5

6

P(����$����!� � %���� 3 !��) = 5

6

5

6

5

6 = 125

216

∴ P(�$��%������% 1 ���!� ) = 1 − 125

216= 91

216= 0.42

��� 59 ���� 225 ��� 3%�� 24 ���� $� 23, ���% 1 → n(S) =

24

4

$� 3 ���% 1

∴n(E) = 23

3

1

1 P(E) =

233

244

= 1

6

28

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��� 60 ���� 225 ��� 2��� Bก(�9ก ,��$ ���������� 6 ���� *�ก������$��ก�"#� 64 Bก(�9ก %��%T 4 × 4 × 4

� �!���� ��ก�� 1 Bก � �!D%� *����!�����*+�"#���&*+��L�������&�����E�%� 1 ������������

=� 64 Bก *+�(���"#�A : �������� % �� 8 BกB : ���� 1 ���� �� 24 BกC : ���� 2 ���� �� 24 BกD : ���� 3 ���� �� 8 Bก

ก����&D%�� �!*+��L�������&�����E�%� 1 ���� *+�ก�$�$� 2 ก���ก�����& 1 : ����$��(( B � +D%�� �!�������������&���� �

P(E1) = 24

64× 5

6= 5

16

ก�����& 2 : ����$��(( C � +D%�� �!�������������&���� �P(E2) = 24

64× 2

6= 2

16

∴ �!� 2 ก��� P(E) = 5

16+ 2

16= 7

16

��� 66 ���� 228 ��� 0.2��� P(A ∩C ) = P(A∪C) = b

*�ก = 1 a + b + c (1)

= 0.7a + b (2)

= 0.5b + c (3)

= 0.7 + 0.5(2) + (3) , (a + b + c) + b

= 1.2 ∴ b = 0.21 + b

a b

cC

A B

0

29

Page 30: Math quota-cmu-g-455

��� 67 ���� 228 ��� 4P(E) = 93

100= 0.93

��(,����&=ก ����%���&�$��� 0.95

��� 70 ���� 230 ��� 84*�ก Tr+ 1 =

nr a

n− rbr → T2+ 1 = n

2 (x2)n− 2 ⋅

1x

2

�.".�. ∴ n = 9T3 = n

2 = 36 → n(n− 1)

2= 36 → n(n − 1) = 72

*�ก Tr+ 1 = 9r (x2)9− r ⋅

1x

r

= 9rx18−2r

xr=

9r

x

18− 3r

���E*� �������� x ��$�!�� 18 − 3r = 0 → r = 6

∴ �.".�. T6+ 1 = 9

6 = 84

��� 71 ���� 230 ��� 23x + 5 = x + 2 → ( 3x + 5 )2 = (x + 2)2 → 3x + 5 = x2 + 4x + 4

x2 + x − 1 = 0 → x = − 1± 1+ 42

→ x = − 1

2+ 5

2, − 1

2− 5

2

∴ a8 − 8

1 a7b +

8

2 a6b2 − ... −

8

7 ab

7 + b8

= (a − b)8 = −12

+ 5

2

−12

− 5

2

8

= ( 5 )8 = 625

��� 72 ���� 231 ��� 1,�� 1 !�2���&����%/,-����! 2 ����� = !�2�

6

2 = 15

,�� 2 !�2�*�$���$�( ����ก�( !�2�7 × 6 × 5 × 4 = 840

,�� 3 *�ก x2 − 1x

15

→ Tr+ 1 = 15r

(x2)15− r ⋅

1x

r

*+�$�!�� ∴ r = 10x30−2r

xr= x0 → 30 − 3r = 0

∴ E*� ��&����� x ��� E*� ��& 11,�� 4

n

22 =

n

9 → n = 22 + 9 = 31

28 12 233 7

137

���กa: 50 b��&��9� 45

�%����� 307

30