CalcGospel 國際數學圖譜
返回

IAL 2025 May Q1

A Level / Edexcel / P3

IAL 2025 May Paper · Question 1

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The functions ff and gg are defined by

f(x)=2x3x+1x∈R, x≥0f(x)=\frac{2x}{3x+1}\qquad x\in\mathbb{R},\ x\ge 0 g(x)=4−x2x∈R, x≥0g(x)=4-x^2\qquad x\in\mathbb{R},\ x\ge 0

(a) Find the value of gf(1)gf(1).

(2)

(b) Find the range of ff.

(2)

(c) Find f−1(x)f^{-1}(x).

(2)

(d) Solve f−1(x)=f(x)f^{-1}(x)=f(x).

(2)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

函数 ff 与 gg 定义为

f(x)=2x3x+1x∈R, x≥0f(x)=\frac{2x}{3x+1}\qquad x\in\mathbb{R},\ x\ge 0 g(x)=4−x2x∈R, x≥0g(x)=4-x^2\qquad x\in\mathbb{R},\ x\ge 0

(a) 求 gf(1)gf(1) 的值；

(b) 求 ff 的值域；

(c) 求 f−1(x)f^{-1}(x)；

(d) 解方程 f−1(x)=f(x)f^{-1}(x)=f(x)。

解答

(a)

Here gf(1)gf(1) means g(f(1))g(f(1)).

First,

f(1)=2(1)3(1)+1=24=12f(1)=\frac{2(1)}{3(1)+1}=\frac{2}{4}=\frac12

So

gf(1)=g(12)gf(1)=g\left(\frac12\right)

Using

g(x)=4−x2g(x)=4-x^2

we get

g(12)=4−(12)2=4−14=154g\left(\frac12\right)=4-\left(\frac12\right)^2 =4-\frac14 =\frac{15}{4}

Therefore

gf(1)=154\boxed{gf(1)=\frac{15}{4}}

(b)

For

f(x)=2x3x+1,x≥0f(x)=\frac{2x}{3x+1},\qquad x\ge 0

we have

f(0)=0f(0)=0

Also, as xx increases,

2x3x+1\frac{2x}{3x+1}

approaches

23\frac23

but never reaches it.

So the range of ff is

0≤f(x)<23\boxed{0\le f(x)<\frac23}

(c)

Let

y=2x3x+1y=\frac{2x}{3x+1}

Rearrange to make xx the subject:

y(3x+1)=2x3xy+y=2x3xy−2x=−yx(3y−2)=−yx=−y3y−2\begin{aligned} y(3x+1)&=2x \\ 3xy+y&=2x \\ 3xy-2x&=-y \\ x(3y-2)&=-y \\ x&=\frac{-y}{3y-2} \end{aligned}

Thus

x=y2−3yx=\frac{y}{2-3y}

Therefore

f−1(x)=x2−3x\boxed{f^{-1}(x)=\frac{x}{2-3x}}

(d)

Solve

f−1(x)=f(x)f^{-1}(x)=f(x)

So

x2−3x=2x3x+1\frac{x}{2-3x}=\frac{2x}{3x+1}

Cross-multiply:

x(3x+1)=2x(2−3x)x(3x+1)=2x(2-3x)

Expand:

3x2+x=4x−6x23x^2+x=4x-6x^2

So

9x2−3x=09x^2-3x=0

Factorise:

3x(3x−1)=03x(3x-1)=0

Therefore

x=0orx=13\boxed{x=0\quad\text{or}\quad x=\frac13}