math quota-cmu-g-455
TRANSCRIPT
������������ �������� ��.����������������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
��� 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
��� 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
��� 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/$"�)�� *5�� 2 �"9�5����9)��GH�/#I�
y = 5 - x2
(2, 1)m=(>
4
��� 35 ���� 138 ��� y = 46 /<�=(> m=(> = m8@��
dy
dx
(1, 4) m=(> = f (x) = f (1)
/�ก g(x) = f(x)x → g (x) = x f (x) − f(x)(1)
x2
� x = 1 ��� ∴ g (1) = f (1) − f(1) → −4 = f (1) − 4 f (1) = 0
∴ �>��>(*=(>=��� (1, 4) �$ @)�*&(� = 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
��� 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 /$"�)��*5�� 3 5����9)�@)�*&(� = −4
��� 40 ���� 141 ��� 2
/�ก f (x) = ∫ f (x)dx = ∫(2x + 1)dx = x2 + x + c
∴ f (2) = 2 → 22 + 2 + c = 2 c = −4
∴ f (x) = x2 + x − 4 → f (1) = −2
�(��(1� m=(> = −2 , m⊥ = 1
2
>*ก�#�>���#�@�� y − 3 = 1
2(x − 1)
∴ y = 1
2x + 5
2
��� 42 ���� 142 ��� 1y = 2x3 − x2 → dy
dx= 6x2 − 2x
6. /<�=(> m=(> = m8@��(a, b) 4 = 6x2 − 2x
4 = 6a2 − 2a
= 03a2 − a − 2
= 0(3a + 2)(a − 1)
∴ a = −23, 1
���������� m=(>L
m
y = xx + 1
(1, )12
�C�/<� (1, 3) �G check choice "�)��* 5�� 2 �GH�/#I��"9�5����9)
f (x)
(1, 3)m
m = f (1)=(>
m = 4(a,b)
x + 4y = 10
6
��� 43 ���� 142 ��� 2.46. /<�=(> m=(> = m8@��
dy
dx
(1, k) =k− 122
2k(x − 2)
=k− 122
−2k
m=(> = = k− 121− (−1) = k− 12
2k − 12 −4k
∴ k = 12
5= 2.4
��� 48 ���� 145 ��� 9m = @���C��><�5�� f(x) = x − 6 + 5
/�ก f(x) "�)�� ��! ∴ m = 5x − 6 ≥ 0
/�ก g(x) = −x4
4+ x3 − x2 + 4 → g (x) = −x3 + 3x2 − 2x = 0 → − x(x2 − 3x + 2) = 0
∴ −x(x − 2)(x − 1) = 0 x = 0, 1, 2
Max Min Max
∴ @��>F�><�>(*"(O: P#�� �$ ∴ 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
@)�*&(��>��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
/�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
��� 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
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
��� 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
������ ��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
��� 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
��� 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
��� 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
��� 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
��� 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
��� 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
��� 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
��� 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
������ ������� ������������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
��� 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
��� 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
��� 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
��� 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
��� 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 ����� ��$!��*+���� �����20
36(72) = 40
��� 52 ���� 221 ��� 3����%/ I �� ��!, ก��% ����%/ II �� ��!, ��! ����%/ III ��ก��%, ก��%*�กD*�% D%�� �!�$���! ��$�!�� ∴ S = {I, II}, E = {II} P(E) = 1
2
��� 53 ���� 222 ��� 3
A B C !�2�←4
→4←5
→5 n(S) = 4 × 5 × 5 × 4 = 400
E : A B C !�2�←3
→4←4
→5 n(E) = 4 × 5 × 4 × 3 = 240
∴ P(E) = 240
400= 0.6
+, ก���
27
��� 54 ���� 222 ��� 4n(S) =
14
3 = 364
∴ n(E) = 5
3 +
6
3 +
3
3 = 31 P(E) = 31
364
��� 55 ���� 223 ��� 2�� 4 �� ����%�(� ��$ 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
$� 3 ���% 1
∴n(E) = 23
3
1
1 P(E) =
233
244
= 1
6
28
��� 60 ���� 225 ��� 2��� Bก(�9ก ,��$ ���������� 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 : ����$��(( C � +D%�� �!�������������&���� �P(E2) = 24
64× 2
6= 2
16
∴ �!� 2 ก��� P(E) = 5
16+ 2
16= 7
16
��� 66 ���� 228 ��� 0.2��� P(A ∩C ) = P(A∪C) = b
*�ก = 1 a + b + c (1)
= 0.7a + b (2)
= 0.5b + c (3)
= 0.7 + 0.5(2) + (3) , (a + b + c) + b
= 1.2 ∴ b = 0.21 + b
a b
cC
A B
0
29
��� 67 ���� 228 ��� 4P(E) = 93
100= 0.93
��(,����&=ก ����%���&�$��� 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