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 $f$ and $g$ are defined by

f(x)=\ln(x^2+3)\qquad x\in\mathbb{R}

g(x)=\frac{3+5x}{x+2}\qquad x\in\mathbb{R},\ x>-2

(a) State the range of $f$ .

(1)

(b) Find $g^{-1}$ .

(3)

(2)

(d) Find the exact value of $a$ for which

g(e^{2a})=f\left(\sqrt{e^4-3}\right)

(3)

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 $f$ 与 $g$ 定义为

f(x)=\ln(x^2+3)\qquad x\in\mathbb{R}

g(x)=\frac{3+5x}{x+2}\qquad x\in\mathbb{R},\ x>-2

(a) 写出 $f$ 的值域。

(b) 求 $g^{-1}$ 。

(d) 求满足

g(e^{2a})=f\left(\sqrt{e^4-3}\right)

的 $a$ 的精确值。

解答

(a)

Since

x^2\ge 0

we have

x^2+3\ge 3

Therefore

\ln(x^2+3)\ge \ln3

So the range of $f$ is

\boxed{f(x)\ge \ln3}

Equivalently,

\boxed{[\ln3,\infty)}

(b)

Let

y=\frac{3+5x}{x+2}

Rearrange to make $x$ the subject:

\begin{aligned} y(x+2)&=3+5x \\ xy+2y&=3+5x \\ xy-5x&=3-2y \\ x(y-5)&=3-2y \\ x&=\frac{3-2y}{y-5} \end{aligned}

Therefore

\boxed{g^{-1}(x)=\frac{3-2x}{x-5}},\qquad x<5

The condition $x<5$ is the domain of $g^{-1}$ , because it is the range of $g$ .

(c)

Here $fg(0)$ means $f(g(0))$ .

First find $g(0)$ :

g(0)=\frac{3+5(0)}{0+2}=\frac32

Then

\begin{aligned} fg(0) &=f\left(\frac32\right) \\ &=\ln\left[\left(\frac32\right)^2+3\right] \\ &=\ln\left(\frac94+3\right) \\ &=\ln\left(\frac{21}{4}\right) \end{aligned}

\boxed{fg(0)=\ln\left(\frac{21}{4}\right)}

(d)

First simplify the right-hand side:

\begin{aligned} f\left(\sqrt{e^4-3}\right) &=\ln\left[\left(\sqrt{e^4-3}\right)^2+3\right] \\ &=\ln(e^4-3+3) \\ &=\ln(e^4) \\ &=4 \end{aligned}

g(e^{2a})=4

Using

g(x)=\frac{3+5x}{x+2}

we get

\frac{3+5e^{2a}}{e^{2a}+2}=4

Rearrange:

\begin{aligned} 3+5e^{2a}&=4(e^{2a}+2) \\ 3+5e^{2a}&=4e^{2a}+8 \\ e^{2a}&=5 \end{aligned}

Take natural logs:

2a=\ln5

Therefore

\boxed{a=\frac12\ln5}

IAL 2025 Oct Q1

题目

解答

(a)

(b)

(c)

(d)