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IAL 2022 May Q10

A Level / Edexcel / P1

IAL 2022 May Paper · Question 10

题目

Problem

Figure 5 shows a sketch of the curve CC with equation

y=27x3+17x2−52x+k\begin{align*} y=\frac27x^3+\frac17x^2-\frac52x+k \end{align*}

Figure 5

where kk is a constant.

(a) Find dydx\dfrac{dy}{dx}.

(2)

The line ll, shown in Figure 5, is the normal to CC at the point AA with xx coordinate −72-\dfrac72.

Given that ll is also a tangent to CC at the point BB,

(b) show that the xx coordinate of the point BB is a solution of the equation

12x2+4x−33=0.\begin{align*} 12x^2+4x-33=0. \end{align*}
(4)

(c) Hence find the xx coordinate of BB, justifying your answer.

(2)

Given that the yy intercept of ll is −1-1,

(d) find the value of kk.

(4)

解答

(a)

解法一

思路

展开

逐项求导即可，常数 kk 求导后为 00。

答题过程

展开 dydx= 27(3x2)+17(2x)−52= 67x2+27x−52.\begin{align*} \frac{dy}{dx} =&\,\frac27(3x^2)+\frac17(2x)-\frac52\\ =&\,\frac67x^2+\frac27x-\frac52. \end{align*}

(b)

解法一

思路

展开

先求曲线在 AA 处的切线斜率，再取负倒数得到法线 ll 的斜率。因为同一条直线 ll 又是 BB 处的切线，所以 BB 处的导数等于这个斜率。

答题过程

展开

At AA, x=−72x=-\frac72. The gradient of the tangent to CC is

dydx= 67(−72)2+27(−72)−52= 67⋅494−1−52= 212−72= 7.\begin{align*} \frac{dy}{dx} =&\,\frac67\left(-\frac72\right)^2 +\frac27\left(-\frac72\right)-\frac52\\ =&\,\frac67\cdot\frac{49}{4}-1-\frac52\\ =&\,\frac{21}{2}-\frac72\\ =&\,7. \end{align*}

So the gradient of the normal ll is

−17.\begin{align*} -\frac17. \end{align*}

At BB, the same line ll is a tangent to CC, so

67x2+27x−52= −17.\begin{align*} \frac67x^2+\frac27x-\frac52=&\,-\frac17. \end{align*}

Multiply by 1414:

12x2+4x−35= −212x2+4x−33= 0.\begin{align*} 12x^2+4x-35=&\,-2\\ 12x^2+4x-33=&\,0. \end{align*}

This is the required equation.

(c)

解法一

思路

展开

解上一小题的二次方程。图中点 BB 在正 xx 方向，所以舍去负根。

答题过程

展开

Solve

12x2+4x−33=0.\begin{align*} 12x^2+4x-33=0. \end{align*}

Factorise:

12x2+4x−33= (2x−3)(6x+11).\begin{align*} 12x^2+4x-33 =&\,(2x-3)(6x+11). \end{align*}

So

x=32orx=−116.\begin{align*} x=\frac32 \quad\text{or}\quad x=-\frac{11}{6}. \end{align*}

From the sketch, BB has a positive xx coordinate, so

xB=32.\begin{align*} x_B=\frac32. \end{align*}

(d)

解法一

思路

展开

已知 ll 的斜率是 −17-\frac17，且 yy 轴截距是 −1-1，所以可写出 ll。

然后用 AA 的 xx 坐标在直线上求出 AA 的 yy 坐标，再代入曲线方程求 kk。

答题过程

展开

The line ll has gradient −17-\frac17 and yy-intercept −1-1.

So

y=−17x−1.\begin{align*} y=-\frac17x-1. \end{align*}

At AA, x=−72x=-\frac72, so

y= −17(−72)−1= 12−1= −12.\begin{align*} y =&\,-\frac17\left(-\frac72\right)-1\\ =&\,\frac12-1\\ =&\,-\frac12. \end{align*}

Thus A=(−72,−12)A=\left(-\frac72,-\frac12\right). Substitute into the curve:

−12= 27(−72)3+17(−72)2−52(−72)+k.\begin{align*} -\frac12 =&\,\frac27\left(-\frac72\right)^3 +\frac17\left(-\frac72\right)^2\\ &\quad -\frac52\left(-\frac72\right)+k. \end{align*}

Simplify:

−12= −494+74+354+k= −74+k.\begin{align*} -\frac12 =&\,-\frac{49}{4}+\frac74+\frac{35}{4}+k\\ =&\,-\frac74+k. \end{align*}

Therefore

k=54.\begin{align*} k=\frac54. \end{align*}