Re: 104新北市國中數學
發表於 : 2015年 6月 26日, 22:10
第 27 題
-b/a = c/a = a + b + c
b = -c
-b/a = c/a = a
第 28 題
a_1 = S_1 = p * a_1
若 a_1 ≠ 0 則 p = 1
S_n = n * a_n
a_1 + a_2 = S_2 = 2a_2
a_1 = a_2,不合題意
故 a_1 = 0
S_2 = a_2 = p * 2 * a_2
p = 1/2
第 32 題
x^2 + y^2 - x - 4y + k = 0
(x - 1/2)^2 + (y - 2)^2 = 17/4 - k
令 O(1/2,2),A(2,3),B(3,2)
AB 和圓 O 相交
OA < 半徑,OB > 半徑
OA^2 < 半徑^2,OB^2 > 半徑^2
-2 < k < 1
第 33 題
即比較 3a + 3,a^2 + 3a,√3(a + 3) 的大小
a = 1.732 < √3
故 a^2 < 3
a^2 + 3a < 3a + 3
√3(a + 3) - (3a + 3) = (3√3 - 3a) - (3 - √3a) = 3(√3 - a) - √3(√3 - a) = (√3 - a)(3 - √3) > 0
√3(a + 3) > 3a + 3 > a^2 + 3a
第 36 題
令 z = x + yi
√[(x - 4)^2 + y^2] = 3√[x^2 + (y - 4)^2]
(x - 4)^2 + y^2 = 9[x^2 + (y - 4)^2]
x^2 + y^2 + x - 9y = -16
(x + 1/2)^2 + (y - 9/2)^2 = -16 + 1/4 + 81/4 = 9/2
所求 = (9/2)π
-b/a = c/a = a + b + c
b = -c
-b/a = c/a = a
第 28 題
a_1 = S_1 = p * a_1
若 a_1 ≠ 0 則 p = 1
S_n = n * a_n
a_1 + a_2 = S_2 = 2a_2
a_1 = a_2,不合題意
故 a_1 = 0
S_2 = a_2 = p * 2 * a_2
p = 1/2
第 32 題
x^2 + y^2 - x - 4y + k = 0
(x - 1/2)^2 + (y - 2)^2 = 17/4 - k
令 O(1/2,2),A(2,3),B(3,2)
AB 和圓 O 相交
OA < 半徑,OB > 半徑
OA^2 < 半徑^2,OB^2 > 半徑^2
-2 < k < 1
第 33 題
即比較 3a + 3,a^2 + 3a,√3(a + 3) 的大小
a = 1.732 < √3
故 a^2 < 3
a^2 + 3a < 3a + 3
√3(a + 3) - (3a + 3) = (3√3 - 3a) - (3 - √3a) = 3(√3 - a) - √3(√3 - a) = (√3 - a)(3 - √3) > 0
√3(a + 3) > 3a + 3 > a^2 + 3a
第 36 題
令 z = x + yi
√[(x - 4)^2 + y^2] = 3√[x^2 + (y - 4)^2]
(x - 4)^2 + y^2 = 9[x^2 + (y - 4)^2]
x^2 + y^2 + x - 9y = -16
(x + 1/2)^2 + (y - 9/2)^2 = -16 + 1/4 + 81/4 = 9/2
所求 = (9/2)π