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CIE 9709 2026 March Paper 12 Q4

A Level / CIE / P1

CIE 9709 2026 March Paper 12 Paper · Question 4

题目

Problem

The function f\mathrm{f} is defined by

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

for x<3x<3.

(a) Find an expression for f−1(x)\mathrm{f}^{-1}(x).

(3)

(b) Given that

ff(x)≡px+qr+sx,\mathrm{ff}(x)\equiv\frac{px+q}{r+sx},

find the possible values of the integers pp, qq, rr and ss.

(3)
题目中文翻译

函数 f\mathrm{f} 定义为

f(x)=2x+1x−3,\mathrm{f}(x)=\frac{2x+1}{x-3},

其中 x<3x<3。

(a) 求 f−1(x)\mathrm{f}^{-1}(x) 的表达式。

(3)

(b) 已知

ff(x)≡px+qr+sx,\mathrm{ff}(x)\equiv\frac{px+q}{r+sx},

求整数 pp、qq、rr 和 ss 的可能值。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

令原函数等于 yy,通分后集中含 xx 的项,再交换输入和输出。

答题过程

展开

From y(x−3)=2x+1y(x-3)=2x+1 we get x(y−2)=3y+1x(y-2)=3y+1. Thus

f−1(x)=3x+1x−2,x<2.\mathrm f^{-1}(x)=\frac{3x+1}{x-2},\qquad x<2.

Indeed, f(x)=2+7/(x−3)\mathrm f(x)=2+7/(x-3) has range (−∞,2)(-\infty,2) on x<3x<3.

(b)

解法一

思路

展开

把 f(x)\mathrm f(x) 当作外层函数的输入,再用同一个分母化简分子和分母。

答题过程

展开

Because f(x)<2<3\mathrm f(x)<2<3 for x<3x<3, the composition is defined throughout x<3x<3. We have

f(f(x))= 2(2x+1)+(x−3)(2x+1)−3(x−3)= 5x−110−x.\begin{align*} \mathrm f(\mathrm f(x))=&\,\frac{2(2x+1)+(x-3)}{(2x+1)-3(x-3)}\\ =&\,\frac{5x-1}{10-x}. \end{align*}

One possible set is p=5p=5, q=−1q=-1, r=10r=10 and s=−1s=-1. Multiplying all four by the same nonzero integer gives further representations.