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IAL 2019 Oct Q10

A Level / Edexcel / P1

IAL 2019 Oct Paper · Question 10

题目

Problem

Figure 6 shows a sketch of part of the curve with equation y=f(x)y=f(x), where

f(x)=(2x+5)(x3)2.\begin{align*} f(x)=(2x+5)(x-3)^2. \end{align*}

Figure 6

(a) Deduce the values of xx for which f(x)0f(x)\leq0.

(2)

The curve crosses the yy-axis at the point PP, as shown.

(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)

(c) Hence, or otherwise, find

(i) the coordinates of PP,

(ii) the gradient of the curve at PP.

(2)

The curve with equation y=f(x)y=f(x) is translated two units in the positive xx direction to a curve with equation y=g(x)y=g(x).

(d) (i) Find g(x)g(x), giving your answer in a simplified factorised form.

(ii) Hence state the yy intercept of the curve with equation y=g(x)y=g(x).

(3)

解答

(a)

解法一

思路

展开

由于 (x3)20(x-3)^2\geq0,函数符号主要由 2x+52x+5 决定;但 x=3x=3 时平方因式为 00,也要包含。

答题过程

展开 f(x)=(2x+5)(x3)2.\begin{align*} f(x)=(2x+5)(x-3)^2. \end{align*}

Since (x3)20(x-3)^2\geq0,

f(x)0\begin{align*} f(x)\leq0 \end{align*}

when

2x+50\begin{align*} 2x+5\leq0 \end{align*}

or when (x3)2=0(x-3)^2=0.

Therefore

x52orx=3.\begin{align*} x\leq-\frac52 \quad\text{or}\quad x=3. \end{align*}

(b)

解法一

思路

展开

先展开平方,再乘以 (2x+5)(2x+5),最后合并同类项。

答题过程

展开 f(x)=(2x+5)(x3)2=(2x+5)(x26x+9)=2x312x2+18x+5x230x+45=2x37x212x+45.\begin{align*} f(x) =&\,(2x+5)(x-3)^2\\ =&\,(2x+5)(x^2-6x+9)\\ =&\,2x^3-12x^2+18x+5x^2-30x+45\\ =&\,2x^3-7x^2-12x+45. \end{align*}

(c)

解法一

思路

展开

PPyy 轴截距,所以令 x=0x=0。斜率则用导函数在 x=0x=0 的值。

答题过程

展开

At x=0x=0,

f(0)=45.\begin{align*} f(0)=45. \end{align*}

Thus

P=(0,45).\begin{align*} P=(0,45). \end{align*}

Differentiate:

f(x)=6x214x12.\begin{align*} f'(x)=6x^2-14x-12. \end{align*}

At PP, x=0x=0, so the gradient is

f(0)=12.\begin{align*} f'(0)=-12. \end{align*}

(d)

解法一

思路

展开

向右平移 22 个单位,要把 f(x)f(x) 中的 xx 换成 x2x-2。然后令 x=0x=0 求新曲线的 yy 轴截距。

答题过程

展开 g(x)=f(x2)=(2(x2)+5)(x23)2=(2x+1)(x5)2.\begin{align*} g(x) =&\,f(x-2)\\ =&\,(2(x-2)+5)(x-2-3)^2\\ =&\,(2x+1)(x-5)^2. \end{align*}

At x=0x=0,

g(0)=(1)(5)2=25.\begin{align*} g(0)=(1)(-5)^2=25. \end{align*}

So the yy intercept is 2525, or (0,25)(0,25).