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IAL 2025 May A Q4

A Level / Edexcel / P1

IAL 2025 May A Paper · Question 4

题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

(i) Using the laws of indices, solve

24k−3=81−k42.\begin{align*} 2^{4k-3}=\frac{8^{1-k}}{4\sqrt2}. \end{align*}
(3)

(ii) Solve the equation

x3+23−1=x3−4,\begin{align*} \frac{x\sqrt3+2}{\sqrt3-1}=x\sqrt3-4, \end{align*}

giving your answer in the form a+b3a+b\sqrt3, where aa and bb are rational constants.

(4)

解答

(i)

解法一

思路

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把右边全部写成以 22 为底的幂。因为 8=238=2^3，4=224=2^2，2=21/2\sqrt2=2^{1/2}，所以分母 42=25/24\sqrt2=2^{5/2}。

答题过程

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Write every term as a power of 22:

81−k= (23)1−k= 23(1−k),\begin{align*} 8^{1-k}=&\,(2^3)^{1-k}\\ =&\,2^{3(1-k)}, \end{align*}

and

42= 22⋅21/2= 25/2.\begin{align*} 4\sqrt2=&\,2^2\cdot2^{1/2}\\ =&\,2^{5/2}. \end{align*}

Therefore

24k−3= 23(1−k)25/2= 23(1−k)−5/2.\begin{align*} 2^{4k-3} =&\,\frac{2^{3(1-k)}}{2^{5/2}}\\ =&\,2^{3(1-k)-5/2}. \end{align*}

Equating powers gives

4k−3= 3(1−k)−524k−3= 3−3k−524k−3= 12−3k7k= 72k= 12.\begin{align*} 4k-3=&\,3(1-k)-\frac52\\ 4k-3=&\,3-3k-\frac52\\ 4k-3=&\,\frac12-3k\\ 7k=&\,\frac72\\ k=&\,\frac12. \end{align*}

(ii)

解法一

思路

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先乘以 3−1\sqrt3-1 去分母，得到关于 xx 的一次方程。最后因为答案要求写成 a+b3a+b\sqrt3，所以要把分母有根号的形式有理化。

答题过程

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Starting with

x3+23−1=x3−4,\begin{align*} \frac{x\sqrt3+2}{\sqrt3-1}=x\sqrt3-4, \end{align*}

multiply both sides by 3−1\sqrt3-1:

x3+2= (x3−4)(3−1)= 3x−x3−43+4.\begin{align*} x\sqrt3+2=&\,(x\sqrt3-4)(\sqrt3-1)\\ =&\,3x-x\sqrt3-4\sqrt3+4. \end{align*}

Collect the xx terms:

x3+x3−3x= 2−43x(23−3)= 2−43.\begin{align*} x\sqrt3+x\sqrt3-3x=&\,2-4\sqrt3\\ x(2\sqrt3-3)=&\,2-4\sqrt3. \end{align*}

So

x= 2−4323−3= (2−43)(23+3)(23−3)(23+3)= 43+6−24−12312−9= −18−833= −6−833.\begin{align*} x=&\,\frac{2-4\sqrt3}{2\sqrt3-3}\\ =&\,\frac{(2-4\sqrt3)(2\sqrt3+3)} {(2\sqrt3-3)(2\sqrt3+3)}\\ =&\,\frac{4\sqrt3+6-24-12\sqrt3}{12-9}\\ =&\,\frac{-18-8\sqrt3}{3}\\ =&\,-6-\frac83\sqrt3. \end{align*}

解法二

思路

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也可以先把左边分母有理化，再整理成一次方程。这个方法一开始步骤较长，但后面方程的分母会消失。

答题过程

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Rationalise the left hand side first:

x3+23−1= (x3+2)(3+1)(3−1)(3+1)= 3x+x3+23+22.\begin{align*} \frac{x\sqrt3+2}{\sqrt3-1} =&\,\frac{(x\sqrt3+2)(\sqrt3+1)} {(\sqrt3-1)(\sqrt3+1)}\\ =&\,\frac{3x+x\sqrt3+2\sqrt3+2}{2}. \end{align*}

So the equation becomes

3x+x3+23+22= x3−4.\begin{align*} \frac{3x+x\sqrt3+2\sqrt3+2}{2}=&\,x\sqrt3-4. \end{align*}

Multiply by 22:

3x+x3+23+2= 2x3−8.\begin{align*} 3x+x\sqrt3+2\sqrt3+2=&\,2x\sqrt3-8. \end{align*}

Collect the xx terms:

3x−x3= −10−23x(3−3)= −10−23.\begin{align*} 3x-x\sqrt3=&\,-10-2\sqrt3\\ x(3-\sqrt3)=&\,-10-2\sqrt3. \end{align*}

So

x= −10−233−3= (−10−23)(3+3)(3−3)(3+3)= −30−103−63−69−3= −36−1636= −6−833.\begin{align*} x=&\,\frac{-10-2\sqrt3}{3-\sqrt3}\\ =&\,\frac{(-10-2\sqrt3)(3+\sqrt3)} {(3-\sqrt3)(3+\sqrt3)}\\ =&\,\frac{-30-10\sqrt3-6\sqrt3-6}{9-3}\\ =&\,\frac{-36-16\sqrt3}{6}\\ =&\,-6-\frac83\sqrt3. \end{align*}