Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 March Paper 12 Q9

A Level / CIE / P1

CIE 9709 2022 March Paper 12 Paper · Question 9

题目

Problem

Functions f\mathrm{f}, g\mathrm{g} and h\mathrm{h} are defined as follows:

f:x↦ x−4x12+1for x≥0,g:x↦ mx2+nfor x≥−2, where m and n are constants,h:x↦ x12−2for x≥0.\begin{align*} \mathrm{f}:x\mapsto&\,x-4x^{\frac12}+1\quad\text{for }x\ge0,\\ \mathrm{g}:x\mapsto&\,mx^2+n\quad\text{for }x\ge-2,\text{ where }m\text{ and }n\text{ are constants},\\ \mathrm{h}:x\mapsto&\,x^{\frac12}-2\quad\text{for }x\ge0. \end{align*}

(a) Solve the equation f(x)=0\mathrm{f}(x)=0, giving your solutions in the form x=a+bcx=a+b\sqrt c, where aa, bb and cc are integers.

(4)

(b) Given that f(x)≡gh(x)\mathrm{f}(x)\equiv\mathrm{gh}(x), find the values of mm and nn.

(4)
题目中文翻译

函数 f\mathrm{f}、g\mathrm{g} 和 h\mathrm{h} 定义如下:

f:x↦ x−4x12+1,x≥0,g:x↦ mx2+n,x≥−2,h:x↦ x12−2,x≥0,\begin{align*} \mathrm{f}:x\mapsto&\,x-4x^{\frac12}+1,\quad x\ge0,\\ \mathrm{g}:x\mapsto&\,mx^2+n,\quad x\ge-2,\\ \mathrm{h}:x\mapsto&\,x^{\frac12}-2,\quad x\ge0, \end{align*}

其中 mm 和 nn 为常数。

(a) 解方程 f(x)=0\mathrm{f}(x)=0,答案写成 x=a+bcx=a+b\sqrt c 的形式,其中 aa、bb 和 cc 为整数。

(4)

(b) 已知 f(x)≡gh(x)\mathrm{f}(x)\equiv\mathrm{gh}(x),求 mm 和 nn 的值。

(4)
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 u=xu=\sqrt x,先解关于 uu 的二次方程。筛选非负根后再平方回到 xx。

答题过程

展开

The equation is u2−4u+1=0u^2-4u+1=0 for u=x≥0u=\sqrt x\ge0. By the quadratic formula,

u=4±16−42=2±3.u=\frac{4\pm\sqrt{16-4}}2=2\pm\sqrt3.

Both roots are positive. Squaring gives

x=(2±3)2=7±43.x=(2\pm\sqrt3)^2=7\pm4\sqrt3.

解法二

思路

展开

也可以先孤立根式,再两边平方。候选根必须满足平方前的方程。

答题过程

展开

From x+1=4xx+1=4\sqrt x, squaring gives x2−14x+1=0x^2-14x+1=0. By the quadratic formula, x=7±43x=7\pm4\sqrt3. Both are nonnegative since 49>4849>48. For either root, x+1x+1 and 4x4\sqrt x are nonnegative and have equal squares, so both satisfy the original equation.

(b)

解法一

思路

展开

将 h(x)\mathrm h(x) 代入 g\mathrm g,展开并比较 xx 项和常数项。

答题过程

展开

For x≥0x\ge0, h(x)=x−2≥−2\mathrm h(x)=\sqrt x-2\ge-2, so the composition is defined. It is

gh(x)=m(x−2)2+n=mx−4mx+4m+n.\mathrm{gh}(x)=m(\sqrt x-2)^2+n=mx-4m\sqrt x+4m+n.

Matching the coefficient of xx gives m=1m=1. Matching the constant term then gives 4+n=14+n=1, so n=−3n=-3.