我要問高一數學的問題~(有3題喔！)

Question

第一題：
f(x)= x四次 + ax三次 + bx二次 + cx + d
若 f(1)=10 f(2)=20 f(3)=30
求 f(7)+ f(-3) =？

第二題：
x、y、z 屬於 R
x+y+z=3
xy+yz+zx= -9
求 x 的 最大值和最小值？

第三題：
f(x)= 2x四次 + x三次 - 2x -a
g(x)=2x三次 - x二次 - x -b
H.C.F為二次
求(a,b)=？


麻煩大家幫忙囉～
這三題實在讓我想了很久！

Answer 1

1. f(x) = x4 + ax3 + bx2 + cx + d
x=1 => 1 + a + b + c + d = 10 ...(1)
x=2 => 16 + 8a + 4b + 2c + d = 20 ...(2)
x=3 => 81 + 27a + 9b + 3c + d = 30 ...(3)
(1) + (3) - 2 * (2) => 6a + b = -25
f(7)+ f(-3) = (2401 + 343a + 49b + 7c + d) + (81 - 27a + 9b - 3c + d)
= 2482 + 316a + 58b + 4c + 2d
= 2482 + 300a + 50b + 2(8a + 4b + 2c + d)
= 2482 + 50(6a + b) + 2(4)
= 2482 - 1250 + 8
= 1240
2. x+y+z=3; xy+yz+zx= -9
(x+y+z)2 = x2 + y2 + z2 + 2(xy+yz+zx)
x2 + y2 + z2 = 9 - 2(-9) = 27
歌西不等式: (y*1 + z*1)2 <= (y2 + z2)(12 + 12)
=> (3 - x)2 <= (27 - x2)2
=> x2 - 6x + 9 <= 54 - 2x2
=> 3x2 - 6x - 45 <= 0
=> x2 - 2x - 15 <= 0
=> (x+3)(x-5) <= 0
=> -3 <= x <= 5
3. f(x) = 2x4 + x3 - 2x - a, g(x) = 2x3 - x2 - x - b
f(x) - 2g(x) = 2x3 + x2 + (b-2)x - a ... (1)
(1) - g(x) = 2x2 + (b-1)x + (b-a) <= f(x), g(x) H.C.F
g(x) = (x - b/2)*(H.C.F.) + (b2/2 - 3b/2 + a -1)x + (b2/2 - b - ab/2)
=> b2/2 - 3b/2 + a -1 = 0 ... (1)
and b2/2 - b - ab/2 = 0 ... (2)
(2) => b(b - 2 - a) = 0
如果 b = 0 => g(x) = x(2x2 - x - 1)
由 H.C.F. = 2x2 + (b-1)x + (b-a) = 2x2 - x - 1 => a = 1
(a, b) = (1, 0)
如果 b ≠ 0 => b - 2 = a 代入 (1)
b2/2 - 3b/2 + b - 2 -1 = 0
=> b2 - b - 6 = 0
=> b = 3 or -2
=> a = 1 or -4
(a, b) = (1, 0) or (3, 1) or (-2, -4)
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.

Answer 2

第一題
f(1)=10 f(2)=20 f(3)=30
你有沒有發現它们都有共通點
f(1)=10 表示 1X10=10 那麼 f(2)=20 也是
　總而言之任意數的１０次方所以
　f(7)+ f(-3) =？
　就是　f(7)＝７０　f(-3)＝－３０（　-3X10= -30　）
那麼　f（７０）＋f（－３０）＝ f(４０)　（正負得負）你懂吧
其它兩題　我忘了　　不好意思喔！