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CIE 9709 2022 November Paper 11 Q8

A Level / CIE / P1

CIE 9709 2022 Nov Paper 11 Paper · Question 8

题目

Problem

The function f\mathrm{f} is defined by f(x)=2−34x−p\mathrm{f}(x)=2-\dfrac{3}{4x-p} for x>p4x>\dfrac{p}{4}, where pp is a constant.

(a) Find f′(x)\mathrm{f}'(x) and hence determine whether f\mathrm{f} is an increasing function, a decreasing function or neither.

(3)

(b) Express f−1(x)\mathrm{f}^{-1}(x) in the form pa−bcx−d\dfrac{p}{a}-\dfrac{b}{cx-d}, where aa, bb, cc and dd are integers.

(4)

(c) Hence state the value of pp for which f−1(x)≡f(x)\mathrm{f}^{-1}(x)\equiv\mathrm{f}(x).

(1)
题目中文翻译

函数 f\mathrm{f} 定义为 f(x)=2−34x−p\mathrm{f}(x)=2-\dfrac{3}{4x-p},其中 x>p4x>\dfrac{p}{4},pp 为常数。

(a) 求 f′(x)\mathrm{f}'(x),并由此判断 f\mathrm{f} 是递增函数、递减函数,还是两者都不是。

(3)

(b) 将 f−1(x)\mathrm{f}^{-1}(x) 写成 pa−bcx−d\dfrac{p}{a}-\dfrac{b}{cx-d} 的形式,其中 aa、bb、cc 和 dd 为整数。

(4)

(c) 由此写出使 f−1(x)≡f(x)\mathrm{f}^{-1}(x)\equiv\mathrm{f}(x) 成立的 pp 值。

(1)
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

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把分母改写为负一次幂并求导。导数的分母为正的平方。

答题过程

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The chain rule gives

f′(x)=12(4x−p)2>0\mathrm f'(x)=\frac{12}{(4x-p)^2}>0

for every x>p/4x>p/4. Thus f\mathrm f is increasing on its domain.

(b)

解法一

思路

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令函数值为 yy,把含 xx 的分母解出来。反函数的输入范围由原函数的值域决定。

答题过程

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From y=2−3/(4x−p)y=2-3/(4x-p) we get (y−2)(4x−p)=−3(y-2)(4x-p)=-3. Therefore

f−1(x)=p4−34x−8,x<2.\mathrm f^{-1}(x)=\frac p4-\frac3{4x-8},\qquad x<2.

Indeed, on x>p/4x>p/4, the term 3/(4x−p)3/(4x-p) runs through all positive values, giving the original range (−∞,2)(-\infty,2).