2008年6月27日 星期五

高考題一則

今年高考純數科出現了一些「不難但很陌生」的題目,難倒不少考生。試題多樣化是大趨勢,讓我也炮製一道「高考程度但在高考很少出現」的題目:
證明方程 x4 - 8x3 + 22x2 - 24x + 10 = 0 沒有實根。

7 則留言:

Pop 提到...

Let f(x) be the given function

check
f(0) = 10, f(1) = 1, f(2) = 2
so, f(0),f(1),f(2) <> 0 (mode 3)
Thus, there is no integral roots for f(x).

However, the question asks for proving no real roots.
========================

If x < 0, f(x) > 0. thus no real roots for x <0.

f'(x) = 4x^3 - 24x^2 + 44x - 24
set f'(x) = 0
=> x^3 - 6x^2 + 11x - 6 = 0
=> (x-1)(x-2)(x-3) = 0
so the three turning points are at x=1, 2, 3

we have
f(0) = 10, f(1) = 1, f(2) = 2

check f(3) = 1
and f(x>3) > 0 as f'(x>3) > 0

It is not difficult to sketch the graph of f(x)

Thus f(x) > 0 for all real x.

koopa koo 提到...

Here is a simpler proof:

f(x) = (x^2 - 4x + 3)^2 + 1. Done.

Marco_Dick 提到...

哈,f(x) = (x^2 - 4x + 3)^2 + 1 是 Kahoo 的「立題原意」嗎?

人见人爱的小西 提到...

路过~~~

Kahoo 提到...

不,「立題原意」是用微分的,不過 Koopa 的做法令我想到一些出題的好主意……

Pop 提到...

ha, actually I tried to completing the square...but not get it done...so give up. :D

邪劉丸 提到...

我的做法和POP差不多:
differentiate f(x)=0,
=> 4(x-1)(x-2)(x-3)=0
=> x=1,2,3
sub into f(x),
f(1)=1
f(2)=2
f(3)=1
but f(x)=0
so it is in consistent
so f(x)have no real solution