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CIE 9709 2026 March Paper 32 Q1

A Level / CIE / P3

CIE 9709 2026 March Paper 32 Paper · Question 1

题目

Problem

Solve the equation 4×2x+2−5×22−x=34\times2^{x+2}-5\times2^{2-x}=3. Give your answer correct to 33 significant figures.

(4)
题目中文翻译

解方程 4×2x+2−5×22−x=34\times2^{x+2}-5\times2^{2-x}=3。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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令正数变量 t=2xt=2^x,把负指数写成倒数,化为二次方程。舍去负根后再取对数。

答题过程

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Let t=2x>0t=2^x>0. Since 2x+2=4t2^{x+2}=4t and 22−x=4/t2^{2-x}=4/t,

16t−20t= 3,16t2−3t−20= 0,t= 3±128932.\begin{align*} 16t-\frac{20}{t}=&\,3,\\ 16t^2-3t-20=&\,0,\\ t=&\,\frac{3\pm\sqrt{1289}}{32}. \end{align*}

The negative root is impossible because t>0t>0. Hence

x= ln⁡((3+1289)/32)ln⁡2= 0.281796…=0.282.\begin{align*} x=&\,\frac{\ln\big((3+\sqrt{1289})/32\big)}{\ln2}\\[4mm] =&\,0.281796\ldots=\boxed{0.282}. \end{align*}