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IAL 2022 Jan Q7

A Level / Edexcel / P1

IAL 2022 Jan Paper · Question 7

题目

Problem

Figure 3 shows a sketch of part of the curve with equation y=f(x)y=f(x), where

f(x)=(x+4)(x−2)(2x−9).\begin{align*} f(x)=(x+4)(x-2)(2x-9). \end{align*}

Figure 3

Given that the curve with equation y=f(x)−py=f(x)-p passes through the point with coordinates (0,50)(0,50),

(a) find the value of the constant pp.

(2)

Given that the curve with equation y=f(x+q)y=f(x+q) passes through the origin,

(b) write down the possible values of the constant qq.

(2)

(c) Find f′(x)f'(x).

(4)

(d) Hence find the range of values of xx for which the gradient of the curve with equation y=f(x)y=f(x) is less than −18-18.

(3)

解答

(a)

解法一

思路

展开

曲线 y=f(x)−py=f(x)-p 经过 (0,50)(0,50)，所以把 x=0,y=50x=0,y=50 代入即可。先求 f(0)f(0)。

答题过程

展开 f(0)= (0+4)(0−2)(2(0)−9)= 4(−2)(−9)= 72.\begin{align*} f(0) =&\,(0+4)(0-2)(2(0)-9)\\ =&\,4(-2)(-9)\\ =&\,72. \end{align*}

Since y=f(x)−py=f(x)-p passes through (0,50)(0,50),

50= 72−pp= 22.\begin{align*} 50=&\,72-p\\ p=&\,22. \end{align*}

(b)

解法一

思路

展开

曲线 y=f(x+q)y=f(x+q) 经过原点，代表当 x=0x=0 时，f(q)=0f(q)=0。所以 qq 必须是原函数 f(x)f(x) 的根。

答题过程

展开

The curve y=f(x+q)y=f(x+q) passes through the origin, so

f(0+q)=0.\begin{align*} f(0+q)=0. \end{align*}

Therefore

f(q)= 0(q+4)(q−2)(2q−9)= 0.\begin{align*} f(q)=&\,0\\ (q+4)(q-2)(2q-9)=&\,0. \end{align*}

Hence

q=−4,q=2,q=92.\begin{align*} q=-4,\quad q=2,\quad q=\frac92. \end{align*}

(c)

解法一

思路

展开

先把 f(x)f(x) 展开成三次多项式，再逐项求导。这样比直接用乘积法则更适合 P1。

答题过程

展开

First expand f(x)f(x):

f(x)= (x+4)(x−2)(2x−9)= (x2+2x−8)(2x−9)= 2x3−9x2+4x2−18x−16x+72= 2x3−5x2−34x+72.\begin{align*} f(x) =&\,(x+4)(x-2)(2x-9)\\ =&\,(x^2+2x-8)(2x-9)\\ =&\,2x^3-9x^2+4x^2-18x-16x+72\\ =&\,2x^3-5x^2-34x+72. \end{align*}

Therefore

f′(x)= 6x2−10x−34.\begin{align*} f'(x) =&\,6x^2-10x-34. \end{align*}

(d)

解法一

思路

展开

梯度就是 f′(x)f'(x)，所以要解不等式 f′(x)<−18f'(x)<-18。先把边界方程 f′(x)=−18f'(x)=-18 解出来，再判断二次不等式在哪个区间成立。

答题过程

展开

We need

f′(x)<−18.\begin{align*} f'(x)&<-18. \end{align*}

Using f′(x)=6x2−10x−34f'(x)=6x^2-10x-34,

6x2−10x−34<−186x2−10x−16<03x2−5x−8<0(3x−8)(x+1)<0.\begin{align*} 6x^2-10x-34&<-18\\ 6x^2-10x-16&<0\\ 3x^2-5x-8&<0\\ (3x-8)(x+1)&<0. \end{align*}

The roots are

x=−1,x=83.\begin{align*} x=-1,\quad x=\frac83. \end{align*}

Since the quadratic opens upwards, it is negative between the roots. Therefore

−1<x<83.\begin{align*} -1<x<\frac83. \end{align*}