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IAL 2021 June Q2

A Level / Edexcel / P1

IAL 2021 June 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.

f(x)=ax3+(6a+8)x2−a2x\begin{align*} f(x)=ax^3+(6a+8)x^2-a^2x \end{align*}

where aa is a positive constant.

Given f(−1)=32f(-1)=32,

(a)

(i) show that the only possible value for aa is 33,

(ii) Using a=3a=3 solve the equation

f(x)=0.\begin{align*} f(x)=0. \end{align*}
(5)

(b) Hence find all real solutions of

(i)

3y+26y2/3−9y1/3=0,\begin{align*} 3y+26y^{2/3}-9y^{1/3}=0, \end{align*}

(ii)

3(93z)+26(92z)−9(9z)=0.\begin{align*} 3(9^{3z})+26(9^{2z})-9(9^z)=0. \end{align*}
(5)

解答

(a)

解法一

思路

展开

先把 x=−1x=-1 代入 f(x)f(x)，用 f(−1)=32f(-1)=32 得到关于 aa 的二次方程。因为 aa 是正数，所以负根要舍去。

答题过程

展开 f(−1)= a(−1)3+(6a+8)(−1)2−a2(−1)= −a+6a+8+a2= a2+5a+8.\begin{align*} f(-1) =&\,a(-1)^3+(6a+8)(-1)^2-a^2(-1)\\ =&\,-a+6a+8+a^2\\ =&\,a^2+5a+8. \end{align*}

Given f(−1)=32f(-1)=32,

a2+5a+8= 32a2+5a−24= 0(a+8)(a−3)= 0.\begin{align*} a^2+5a+8=&\,32\\ a^2+5a-24=&\,0\\ (a+8)(a-3)=&\,0. \end{align*}

So

a=−8ora=3.\begin{align*} a=-8\quad\text{or}\quad a=3. \end{align*}

Since aa is positive,

a=3.\begin{align*} a=3. \end{align*}

Using a=3a=3,

f(x)= 3x3+26x2−9x= x(3x2+26x−9)= x(3x−1)(x+9).\begin{align*} f(x) =&\,3x^3+26x^2-9x\\ =&\,x(3x^2+26x-9)\\ =&\,x(3x-1)(x+9). \end{align*}

Therefore

x=0,x=13,x=−9.\begin{align*} x=0,\quad x=\frac13,\quad x=-9. \end{align*}

(b)(i)

解法一

思路

展开

把 y1/3y^{1/3} 看成上一小题里的 xx。设 u=y1/3u=y^{1/3}，则 y2/3=u2y^{2/3}=u^2，而 y=u3y=u^3。不过这个方程正好等于 u(3u2+26u−9)=0u(3u^2+26u-9)=0。

答题过程

展开

Let

u=y1/3.\begin{align*} u=y^{1/3}. \end{align*}

Then

y2/3=u2,y=u3.\begin{align*} y^{2/3}=u^2,\qquad y=u^3. \end{align*}

The equation becomes

3u3+26u2−9u= 0.\begin{align*} 3u^3+26u^2-9u=&\,0. \end{align*}

Using part (a),

u=0,u=13,u=−9.\begin{align*} u=0,\quad u=\frac13,\quad u=-9. \end{align*}

Since u=y1/3u=y^{1/3},

y= 03,(13)3,(−9)3= 0,127,−729.\begin{align*} y=&\,0^3,\quad \left(\frac13\right)^3,\quad (-9)^3\\ =&\,0,\quad \frac1{27},\quad -729. \end{align*}

(b)(ii)

解法一

思路

展开

设 u=9zu=9^z，则 92z=u29^{2z}=u^2，93z=u39^{3z}=u^3。因为 9z9^z 一定为正，所以只能使用上一小题中正的解。

答题过程

展开

Let

u=9z.\begin{align*} u=9^z. \end{align*}

Then the equation becomes

3u3+26u2−9u=0.\begin{align*} 3u^3+26u^2-9u=0. \end{align*}

From part (a),

u=0,u=13,u=−9.\begin{align*} u=0,\quad u=\frac13,\quad u=-9. \end{align*}

But 9z>09^z>0, so the only possible value is

9z=13.\begin{align*} 9^z=\frac13. \end{align*}

Since 9=329=3^2,

(32)z= 3−132z= 3−12z= −1z= −12.\begin{align*} (3^2)^z=&\,3^{-1}\\ 3^{2z}=&\,3^{-1}\\ 2z=&\,-1\\ z=&\,-\frac12. \end{align*}