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IAL 2024 May Q2

A Level / Edexcel / P1

IAL 2024 May Paper · Question 2

题目

Problem

(i) Given that m=2nm=2^n, express each of the following in simplest form in terms of mm.

(a) 2n+32^{n+3}

(1)

(b) 163n16^{3n}

(2)

(ii) In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Solve the equation

x3−3=x+3\begin{align*} x\sqrt3-3=x+\sqrt3 \end{align*}

giving your answer in the form p+q3p+q\sqrt3 where pp and qq are integers.

(3)

解答

(i)(a)

解法一

思路

展开

用指数法则 2n+3=2n⋅232^{n+3}=2^n\cdot2^3，再代入 2n=m2^n=m。

答题过程

展开 2n+3= 2n⋅23= 8m.\begin{align*} 2^{n+3} =&\,2^n\cdot2^3\\ =&\,8m. \end{align*}

(i)(b)

解法一

思路

展开

把 1616 写成 242^4，再把结果改写成 2n2^n 的幂。

答题过程

展开 163n= (24)3n= 212n= (2n)12= m12.\begin{align*} 16^{3n} =&\,(2^4)^{3n}\\ =&\,2^{12n}\\ =&\,(2^n)^{12}\\ =&\,m^{12}. \end{align*}

(ii)

解法一

思路

展开

先把含 xx 的项放在一边，提出 xx，再有理化分母。最后答案要整理成 p+q3p+q\sqrt3。

答题过程

展开 x3−3= x+3x3−x= 3+3x(3−1)= 3+3x= 3+33−1.\begin{align*} x\sqrt3-3=&\,x+\sqrt3\\ x\sqrt3-x=&\,3+\sqrt3\\ x(\sqrt3-1)=&\,3+\sqrt3\\ x=&\,\frac{3+\sqrt3}{\sqrt3-1}. \end{align*}

Rationalise the denominator:

x= 3+33−1⋅3+13+1= (3+3)(3+1)3−1= 33+3+3+32= 43+62= 3+23.\begin{align*} x =&\,\frac{3+\sqrt3}{\sqrt3-1}\cdot \frac{\sqrt3+1}{\sqrt3+1}\\ =&\,\frac{(3+\sqrt3)(\sqrt3+1)}{3-1}\\ =&\,\frac{3\sqrt3+3+3+\sqrt3}{2}\\ =&\,\frac{4\sqrt3+6}{2}\\ =&\,3+2\sqrt3. \end{align*}

解法二

思路

展开

题目要求答案形如 p+q3p+q\sqrt3，所以可以直接设 x=p+q3x=p+q\sqrt3，代回方程后比较有理部分和 3\sqrt3 部分。

答题过程

展开

Let

x=p+q3.\begin{align*} x=p+q\sqrt3. \end{align*}

Substitute into x3−3=x+3x\sqrt3-3=x+\sqrt3:

(p+q3)3−3= p+q3+3p3+3q−3= p+(q+1)3.\begin{align*} (p+q\sqrt3)\sqrt3-3=&\,p+q\sqrt3+\sqrt3\\ p\sqrt3+3q-3=&\,p+(q+1)\sqrt3. \end{align*}

Compare the rational parts and the 3\sqrt3 parts:

3q−3= p,p= q+1.\begin{align*} 3q-3=&\,p,\\ p=&\,q+1. \end{align*}

Substitute p=q+1p=q+1 into 3q−3=p3q-3=p:

3q−3= q+12q= 4q= 2.\begin{align*} 3q-3=&\,q+1\\ 2q=&\,4\\ q=&\,2. \end{align*}

Then

p=q+1=3.\begin{align*} p=q+1=3. \end{align*}

Therefore

x=3+23.\begin{align*} x=3+2\sqrt3. \end{align*}