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IAL 2021 Jan Q8

A Level / Edexcel / P1

IAL 2021 Jan Paper · Question 8

题目

Problem

Figure 4 shows a sketch of part of the curve CC with equation y=f(x)y=f(x), where

f(x)=(3x2)2(x4).\begin{align*} f(x)=(3x-2)^2(x-4). \end{align*}

Figure 4

(a) Deduce the values of xx for which f(x)>0f(x)>0.

(1)

(b) Expand f(x)f(x) to the form

ax3+bx2+cx+d\begin{align*} ax^3+bx^2+cx+d \end{align*}

where aa, bb, cc and dd are integers to be found.

(3)

The line ll, also shown in Figure 4, passes through the yy intercept of CC and is parallel to the xx-axis.

The line ll cuts CC again at points PP and QQ, also shown in Figure 4.

(c) Using algebra and showing your working, find the length of line PQPQ. Write your answer in the form k3k\sqrt3, where kk is a constant to be found.

(Solutions relying entirely on calculator technology are not acceptable.)

(5)

解答

(a)

解法一

思路

展开

(3x2)2(3x-2)^2 永远非负,且在 x=23x=\frac23 时为 00。函数符号主要由 (x4)(x-4) 决定,所以只有 x>4x>4 时为正。

答题过程

展开

Since (3x2)20(3x-2)^2\ge0, the sign of f(x)f(x) is determined by x4x-4 except at the repeated root x=23x=\frac23.

Therefore

f(x)>0forx>4.\begin{align*} f(x)>0\quad\text{for}\quad x>4. \end{align*}

(b)

解法一

思路

展开

先展开平方,再乘以 (x4)(x-4)

答题过程

展开 f(x)=(3x2)2(x4)=(9x212x+4)(x4)=9x336x212x2+48x+4x16=9x348x2+52x16.\begin{align*} f(x) =&\,(3x-2)^2(x-4)\\ =&\,(9x^2-12x+4)(x-4)\\ =&\,9x^3-36x^2-12x^2+48x+4x-16\\ =&\,9x^3-48x^2+52x-16. \end{align*}

(c)

解法一

思路

展开

直线 ll 经过 CCyy 轴截距,所以先找 f(0)f(0)。因为 ll 是水平线,P,QP,Qyy 坐标相同,长度 PQPQ 就是两个对应 xx 坐标之差。

答题过程

展开

The yy-intercept of CC is

f(0)=(3(0)2)2(04)=4(4)=16.\begin{align*} f(0)=&\,(3(0)-2)^2(0-4)\\ =&\,4(-4)\\ =&\,-16. \end{align*}

So line ll has equation

y=16.\begin{align*} y=-16. \end{align*}

Find where CC meets this line:

9x348x2+52x16=169x348x2+52x=0x(9x248x+52)=0.\begin{align*} 9x^3-48x^2+52x-16=&\,-16\\ 9x^3-48x^2+52x=&\,0\\ x(9x^2-48x+52)=&\,0. \end{align*}

The non-zero intersections satisfy

9x248x+52=0.\begin{align*} 9x^2-48x+52=0. \end{align*}

Using the quadratic formula,

x=48±4824(9)(52)18=48±43218=48±12318=8±233.\begin{align*} x =&\,\frac{48\pm\sqrt{48^2-4(9)(52)}}{18}\\ =&\,\frac{48\pm\sqrt{432}}{18}\\ =&\,\frac{48\pm12\sqrt3}{18}\\ =&\,\frac{8\pm2\sqrt3}{3}. \end{align*}

Therefore

PQ=8+2338233=433.\begin{align*} PQ =&\,\frac{8+2\sqrt3}{3}-\frac{8-2\sqrt3}{3}\\ =&\,\frac{4\sqrt3}{3}. \end{align*}