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IAL 2024 Oct Q2

A Level / Edexcel / P1

IAL 2024 Oct Paper · Question 2

题目

Problem

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

Solutions relying on calculator technology are not acceptable.

(i) Simplify fully

3y3(2x4)34x2y4\begin{align*} \frac{3y^3(2x^4)^3}{4x^2y^4} \end{align*}
(3)

(ii) Find the exact value of aa such that

163+1=a27+4\begin{align*} \frac{16}{\sqrt3+1}=a\sqrt{27}+4 \end{align*}

Write your answer in the form p3+qp\sqrt3+q where pp and qq are fully simplified rational constants.

(4)

解答

(i)

解法一

思路

展开

先处理括号的三次方，再分别整理数字、xx 的幂和 yy 的幂。

答题过程

展开 3y3(2x4)34x2y4= 3y3⋅8x124x2y4= 24x12y34x2y4= 6x10y−1= 6x10y.\begin{align*} \frac{3y^3(2x^4)^3}{4x^2y^4} =&\,\frac{3y^3\cdot 8x^{12}}{4x^2y^4}\\ =&\,\frac{24x^{12}y^3}{4x^2y^4}\\ =&\,6x^{10}y^{-1}\\ =&\,\frac{6x^{10}}{y}. \end{align*}

(ii)

解法一

思路

展开

先把左边有理化，右边把 27\sqrt{27} 写成 333\sqrt3。这样两边都只含有 3\sqrt3，再把 aa 解出来。

答题过程

展开

Rationalise the denominator:

163+1= 163+1⋅3−13−1= 16(3−1)3−1= 8(3−1)= 83−8.\begin{align*} \frac{16}{\sqrt3+1} =&\,\frac{16}{\sqrt3+1}\cdot\frac{\sqrt3-1}{\sqrt3-1}\\ =&\,\frac{16(\sqrt3-1)}{3-1}\\ =&\,8(\sqrt3-1)\\ =&\,8\sqrt3-8. \end{align*}

Also,

27=33.\begin{align*} \sqrt{27}=3\sqrt3. \end{align*}

So

83−8= 3a3+43a3= 83−12a= 83−1233.\begin{align*} 8\sqrt3-8=&\,3a\sqrt3+4\\ 3a\sqrt3=&\,8\sqrt3-12\\ a=&\,\frac{8\sqrt3-12}{3\sqrt3}. \end{align*}

Now simplify:

a= 8333−1233= 83−43= 83−433= −433+83.\begin{align*} a =&\,\frac{8\sqrt3}{3\sqrt3}-\frac{12}{3\sqrt3}\\ =&\,\frac83-\frac{4}{\sqrt3}\\ =&\,\frac83-\frac{4\sqrt3}{3}\\ =&\,-\frac43\sqrt3+\frac83. \end{align*}

解法二

思路

展开

也可以先把等式两边乘以 3+1\sqrt3+1，避免一开始处理整个左边。最后会出现带根式的分母，再对 aa 的表达式有理化。

答题过程

展开

Since 27=33\sqrt{27}=3\sqrt3,

163+1=3a3+4.\begin{align*} \frac{16}{\sqrt3+1}=3a\sqrt3+4. \end{align*}

Multiply both sides by 3+1\sqrt3+1:

16= (3a3+4)(3+1)16= 9a+3a3+43+4.\begin{align*} 16=&\,(3a\sqrt3+4)(\sqrt3+1)\\ 16=&\,9a+3a\sqrt3+4\sqrt3+4. \end{align*}

So

12−43= a(9+33)a= 12−439+33.\begin{align*} 12-4\sqrt3=&\,a(9+3\sqrt3)\\ a=&\,\frac{12-4\sqrt3}{9+3\sqrt3}. \end{align*}

Rationalise the denominator:

a= 12−439+33⋅9−339−33= (12−43)(9−33)81−27= 108−363−363+3654= 144−72354= 83−433.\begin{align*} a =&\,\frac{12-4\sqrt3}{9+3\sqrt3}\cdot \frac{9-3\sqrt3}{9-3\sqrt3}\\ =&\,\frac{(12-4\sqrt3)(9-3\sqrt3)}{81-27}\\ =&\,\frac{108-36\sqrt3-36\sqrt3+36}{54}\\ =&\,\frac{144-72\sqrt3}{54}\\ =&\,\frac83-\frac43\sqrt3. \end{align*}

Therefore

a=−433+83.\begin{align*} a=-\frac43\sqrt3+\frac83. \end{align*}