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IAL 2025 Jan Q4

A Level / Edexcel / P1

IAL 2025 Jan Paper · Question 4

题目

Problem

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

(i) Given that

y=ax\begin{align*} y=a^x \end{align*}

where aa is a positive constant, express, in simplest form, in terms of yy and aa

(a) a3x+1a^{3x+1}

(1)

(b) 5(3a1−x)−2\dfrac{5}{(3a^{1-x})^{-2}}

(3)

(ii) (a) Use the substitution p=9tp=9^t to show that the equation

3(34t+2+1)=82⋅9t\begin{align*} 3\left(3^{4t+2}+1\right)=82\cdot9^t \end{align*}

can be rewritten as

27p2−82p+3=0\begin{align*} 27p^2-82p+3=0 \end{align*}
(2)

(b) Hence solve

3(34t+2+1)=82⋅9t.\begin{align*} 3\left(3^{4t+2}+1\right)=82\cdot9^t. \end{align*}
(3)

解答

(i)(a)

解法一

思路

展开

由 y=axy=a^x，所以 a3x=(ax)3=y3a^{3x}=(a^x)^3=y^3。剩下的 a1a^1 留作 aa。

答题过程

展开 a3x+1= a3x⋅a= (ax)3a= ay3.\begin{align*} a^{3x+1} =&\,a^{3x}\cdot a\\ =&\,(a^x)^3a\\ =&\,ay^3. \end{align*}

(i)(b)

解法一

思路

展开

分母是负指数，除以 (3a1−x)−2(3a^{1-x})^{-2} 等于乘以 (3a1−x)2(3a^{1-x})^2。再把 a−xa^{-x} 写成 1y\frac1y。

答题过程

展开 5(3a1−x)−2= 5(3a1−x)2= 5⋅9a2−2x= 45a2a−2x.\begin{align*} \frac{5}{(3a^{1-x})^{-2}} =&\,5(3a^{1-x})^2\\ =&\,5\cdot9a^{2-2x}\\ =&\,45a^2a^{-2x}. \end{align*}

Since y=axy=a^x,

a−2x=(ax)−2=y−2.\begin{align*} a^{-2x}=(a^x)^{-2}=y^{-2}. \end{align*}

Therefore

5(3a1−x)−2= 45a2y−2= 45a2y2.\begin{align*} \frac{5}{(3a^{1-x})^{-2}} =&\,45a^2y^{-2}\\ =&\,\frac{45a^2}{y^2}. \end{align*}

(ii)(a)

解法一

思路

展开

关键是把 34t+23^{4t+2} 改写成含 9t9^t 的形式。因为

34t+2=32(32)2t=9(9t)2.\begin{align*} 3^{4t+2}=3^2(3^2)^{2t}=9(9^t)^2. \end{align*}

再代入 p=9tp=9^t。

答题过程

展开

Using p=9tp=9^t,

34t+2= 32⋅34t= 9(32)2t= 9(9t)2= 9p2.\begin{align*} 3^{4t+2} =&\,3^2\cdot3^{4t}\\ =&\,9(3^2)^{2t}\\ =&\,9(9^t)^2\\ =&\,9p^2. \end{align*}

So

3(34t+2+1)= 82⋅9t3(9p2+1)= 82p27p2+3= 82p27p2−82p+3= 0.\begin{align*} 3\left(3^{4t+2}+1\right)=&\,82\cdot9^t\\ 3(9p^2+1)=&\,82p\\ 27p^2+3=&\,82p\\ 27p^2-82p+3=&\,0. \end{align*}

This is the required equation.

(ii)(b)

解法一

思路

展开

先解 (ii)(a) 的二次方程得到 pp，再用 p=9tp=9^t 回到 tt。

答题过程

展开

From part (ii)(a),

27p2−82p+3= 0.\begin{align*} 27p^2-82p+3=&\,0. \end{align*}

Factorise:

27p2−82p+3= (27p−1)(p−3).\begin{align*} 27p^2-82p+3=&\,(27p-1)(p-3). \end{align*}

So

p=127orp=3.\begin{align*} p=\frac1{27} \quad\text{or}\quad p=3. \end{align*}

Since p=9tp=9^t,

9t= 3or9t=127.\begin{align*} 9^t=&\,3 \quad\text{or}\quad 9^t=\frac1{27}. \end{align*}

Write both sides as powers of 33:

(32)t= 31or(32)t=3−3.\begin{align*} (3^2)^t=&\,3^1 \quad\text{or}\quad (3^2)^t=3^{-3}. \end{align*}

Therefore

2t= 1or2t=−3.\begin{align*} 2t=&\,1 \quad\text{or}\quad 2t=-3. \end{align*}

Hence

t=12ort=−32.\begin{align*} t=\frac12 \quad\text{or}\quad t=-\frac32. \end{align*}