题目

Problem

The function $f$ is defined by

f(x)=\frac{5x-3}{x-4}\qquad x>4

(a) Show, by using calculus, that $f$ is a decreasing function.

(3)

(b) Find $f^{-1}$

(3)

ff(x)=\frac{ax+b}{x+c}

where $a$ , $b$ and $c$ are constants to be found.

(ii) Deduce the range of $ff$ .

(5)

题目中文翻译

函数 $f$ 定义为

f(x)=\frac{5x-3}{x-4}\qquad x>4

(a) 用微积分证明 $f$ 是单调递减函数。

(b) 求 $f^{-1}$ 。

ff(x)=\frac{ax+b}{x+c}

其中 $a$ 、 $b$ 和 $c$ 是需要求出的常数。

(ii) 由此写出 $ff$ 的值域。

解答

(a)

We have

f(x)=\frac{5x-3}{x-4}

Using the quotient rule,

f'(x)=\frac{5(x-4)-(5x-3)}{(x-4)^2}

Simplifying,

\begin{aligned} f'(x) &=\frac{5x-20-5x+3}{(x-4)^2} \\ &=\frac{-17}{(x-4)^2} \end{aligned}

For $x>4$ ,

(x-4)^2>0

f'(x)<0

Therefore $f$ is a decreasing function.

(b)

Let

y=\frac{5x-3}{x-4}

Then

y(x-4)=5x-3

xy-4y=5x-3

Collect the $x$ terms:

xy-5x=4y-3

Thus

x(y-5)=4y-3

and hence

x=\frac{4y-3}{y-5}

Therefore

f^{-1}(x)=\frac{4x-3}{x-5}

Since $x>4$ for $f$ , the range of $f$ is $f(x)>5$ .

So the domain of $f^{-1}$ is

x>5

Hence

\boxed{f^{-1}(x)=\frac{4x-3}{x-5},\qquad x>5}

(c)(i)

We need

ff(x)=f(f(x))

Using

f(x)=\frac{5x-3}{x-4}

we get

ff(x) =\frac{5\left(\frac{5x-3}{x-4}\right)-3} {\left(\frac{5x-3}{x-4}\right)-4}

Simplify the numerator:

5\left(\frac{5x-3}{x-4}\right)-3 =\frac{25x-15-3(x-4)}{x-4} =\frac{22x-3}{x-4}

Simplify the denominator:

\left(\frac{5x-3}{x-4}\right)-4 =\frac{5x-3-4(x-4)}{x-4} =\frac{x+13}{x-4}

Therefore

ff(x) =\frac{\frac{22x-3}{x-4}}{\frac{x+13}{x-4}} =\frac{22x-3}{x+13}

\boxed{ff(x)=\frac{22x-3}{x+13}}

Hence

\boxed{a=22,\qquad b=-3,\qquad c=13}

(c)(ii)

From part (b), the range of $f$ is

f(x)>5

So in $ff(x)=f(f(x))$ , the input to the second $f$ is greater than $5$ .

For

u>5

we have

f(u)=\frac{5u-3}{u-4}=5+\frac{17}{u-4}

Since

u>5

we have

u-4>1

0<\frac{17}{u-4}<17

Therefore

5<f(u)<22

Hence the range of $ff$ is

\boxed{5<ff(x)<22}

IAL 2022 Jan Q6

题目

解答

(a)

(b)

(c)(i)

(c)(ii)