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CIE 9709 2023 Nov Paper 33 Q1

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 1

题目

Problem

Find the set of values of xx satisfying the inequality ∣2x+1−2∣<0.5|2^{x+1}-2|<0.5, giving your answer to 33 significant figures.

(4)
题目中文翻译

求满足不等式 ∣2x+1−2∣<0.5|2^{x+1}-2|<0.5 的 xx 的取值范围,答案保留 33 位有效数字。

(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

解法一

思路

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把绝对值不等式化成双边不等式。底数 2>12>1,取对数后不等号方向不变。

答题过程

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The inequality is equivalent to

−12<2x+1−2<12,-\frac12<2^{x+1}-2<\frac12,

so

32<2x+1<52.\frac32<2^{x+1}<\frac52.

Taking logarithms and dividing by ln⁡2>0\ln2>0 gives

ln⁡(3/2)ln⁡2<x+1<ln⁡(5/2)ln⁡2.\frac{\ln(3/2)}{\ln2}<x+1< \frac{\ln(5/2)}{\ln2}.

Therefore

ln⁡(3/2)ln⁡2−1<x<ln⁡(5/2)ln⁡2−1.\frac{\ln(3/2)}{\ln2}-1<x< \frac{\ln(5/2)}{\ln2}-1.

The boundary values are −0.415037…-0.415037\ldots and 0.321928…0.321928\ldots. Thus, to 3 significant figures,

−0.415<x<0.322.\boxed{-0.415<x<0.322}.