题目
Problem
f(x)=3−x+1x−2+2x2−3x−55x+26x>4
(a) Show that
f(x)=cx+dax+bx>4
where a, b, c and d are integers to be found.
(4)
(b) Hence find f−1(x).
(2)
(c) Find the domain of f−1.
(2)
题目中文翻译
f(x)=3−x+1x−2+2x2−3x−55x+26x>4
(a) 证明
f(x)=cx+dax+bx>4
其中 a,b,c,d 为待求整数。
(b) 由此求 f−1(x)。
(c) 求 f−1 的定义域。
解答
(a)
解法一
思路
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先因式分解二次式,再通分合并三个分式。所得分子含有因子 (x+1),约去后即可读出 a,b,c,d。
答题过程
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We have
f(x)=3−x+1x−2+2x2−3x−55x+26
First factorise
2x2−3x−5
as
2x2−3x−5=(2x−5)(x+1)
So
f(x)=3−x+1x−2+(2x−5)(x+1)5x+26
Use the common denominator
(x+1)(2x−5)
Then
f(x)=(x+1)(2x−5)3(x+1)(2x−5)−(x−2)(2x−5)+(x+1)(2x−5)5x+26
Now expand the numerator:
3(x+1)(2x−5)=3(2x2−3x−5)=6x2−9x−15
Also,
(x−2)(2x−5)=2x2−9x+10
So the numerator is
==6x2−9x−15−(2x2−9x+10)+5x+264x2+5x+1(4x+1)(x+1)
Therefore
f(x)==(x+1)(2x−5)(4x+1)(x+1)2x−54x+1
Hence
f(x)=2x−54x+1
so one possible set of constants is
a=4,b=1,c=2,d=−5
(b)
解法一
思路
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令 y=f(x),把等式重新整理成以 x 为主项的形式,最后交换自变量名称即可得到反函数。
答题过程
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Let
y=f(x)
Then
y=2x−54x+1
Rearrange:
y(2x−5)=4x+1
So
2xy−5y=4x+1
Bring the x terms to one side:
2xy−4x=5y+1
Factorise:
x(2y−4)=5y+1
Therefore
x=2y−45y+1
Replace y by x:
f−1(x)=2x−45x+1
(c)
解法一
思路
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反函数的定义域就是原函数的值域。按照 A Level 常用答题方式,代入端点 x=4,再结合 x→∞ 时的水平渐近值确定开区间。
答题过程
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The domain of f−1 is the range of f.
For
f(x)=2x−54x+1,x>4
the upper endpoint is found by using x=4:
f(4)=317.
As x→∞,
f(x)→24=2
Therefore the range of f is
2<f(x)<317
Hence the domain of f−1 is
2<x<317