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P3 2021 Jan Q3

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 3

题目

Problem

f(x)=3−x−2x+1+5x+262x2−3x−5x>4f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{2x^2-3x-5}\qquad x>4

(a) Show that

f(x)=ax+bcx+dx>4f(x)=\frac{ax+b}{cx+d}\qquad x>4

where aa, bb, cc and dd are integers to be found.

(4)

(b) Hence find f−1(x)f^{-1}(x).

(2)

(c) Find the domain of f−1f^{-1}.

(2)
题目中文翻译 f(x)=3−x−2x+1+5x+262x2−3x−5x>4f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{2x^2-3x-5}\qquad x>4

(a) 证明

f(x)=ax+bcx+dx>4f(x)=\frac{ax+b}{cx+d}\qquad x>4

其中 a,b,c,da,b,c,d 为待求整数。

(b) 由此求 f−1(x)f^{-1}(x)。

(c) 求 f−1f^{-1} 的定义域。

解答

(a)

解法一

思路

展开

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)(x+1),约去后即可读出 a,b,c,da,b,c,d。

答题过程

展开

We have

f(x)=3−x−2x+1+5x+262x2−3x−5f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{2x^2-3x-5}

First factorise

2x2−3x−52x^2-3x-5

as

2x2−3x−5=(2x−5)(x+1)2x^2-3x-5=(2x-5)(x+1)

So

f(x)=3−x−2x+1+5x+26(2x−5)(x+1)f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{(2x-5)(x+1)}

Use the common denominator

(x+1)(2x−5)(x+1)(2x-5)

Then

f(x)= 3(x+1)(2x−5)−(x−2)(2x−5)(x+1)(2x−5) +5x+26(x+1)(2x−5)\begin{align*} f(x) =&\,\frac{3(x+1)(2x-5)-(x-2)(2x-5)} {(x+1)(2x-5)}\\[4mm] &\,\hspace{2pt}+\frac{5x+26}{(x+1)(2x-5)} \end{align*}

Now expand the numerator:

3(x+1)(2x−5)=3(2x2−3x−5)=6x2−9x−153(x+1)(2x-5)=3(2x^2-3x-5)=6x^2-9x-15

Also,

(x−2)(2x−5)=2x2−9x+10(x-2)(2x-5)=2x^2-9x+10

So the numerator is

 6x2−9x−15−(2x2−9x+10)+5x+26= 4x2+5x+1= (4x+1)(x+1)\begin{align*} &\,6x^2-9x-15-(2x^2-9x+10)+5x+26\\[4mm] =&\,4x^2+5x+1\\[4mm] =&\,(4x+1)(x+1) \end{align*}

Therefore

f(x)= (4x+1)(x+1)(x+1)(2x−5)= 4x+12x−5\begin{align*} f(x) =&\,\frac{(4x+1)(x+1)}{(x+1)(2x-5)}\\[4mm] =&\,\frac{4x+1}{2x-5} \end{align*}

Hence

f(x)=4x+12x−5\boxed{f(x)=\frac{4x+1}{2x-5}}

so one possible set of constants is

a=4,b=1,c=2,d=−5\boxed{a=4,\quad b=1,\quad c=2,\quad d=-5}

(b)

解法一

思路

展开

令 y=f(x)y=f(x),把等式重新整理成以 xx 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

展开

Let

y=f(x)y=f(x)

Then

y=4x+12x−5y=\frac{4x+1}{2x-5}

Rearrange:

y(2x−5)=4x+1y(2x-5)=4x+1

So

2xy−5y=4x+12xy-5y=4x+1

Bring the xx terms to one side:

2xy−4x=5y+12xy-4x=5y+1

Factorise:

x(2y−4)=5y+1x(2y-4)=5y+1

Therefore

x=5y+12y−4x=\frac{5y+1}{2y-4}

Replace yy by xx:

f−1(x)=5x+12x−4\boxed{f^{-1}(x)=\frac{5x+1}{2x-4}}

(c)

解法一

思路

展开

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=4x=4,再结合 x→∞x\to\infty 时的水平渐近值确定开区间。

答题过程

展开

The domain of f−1f^{-1} is the range of ff.

For

f(x)=4x+12x−5,x>4f(x)=\frac{4x+1}{2x-5},\qquad x>4

the upper endpoint is found by using x=4x=4:

f(4)=173.f(4)=\frac{17}{3}.

As x→∞x\to\infty,

f(x)→42=2f(x)\to \frac{4}{2}=2

Therefore the range of ff is

2<f(x)<1732<f(x)<\frac{17}{3}

Hence the domain of f−1f^{-1} is

2<x<173\boxed{2<x<\frac{17}{3}}