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IAL 2021 Oct Q8

A Level / Edexcel / P1

IAL 2021 Oct Paper · Question 8

题目

Problem

Figure 4 shows a sketch of the curve CC with equation

y=4+12x−3x2.\begin{align*} y=4+12x-3x^2. \end{align*}

Figure 4

The point MM is the maximum turning point on CC.

(a)

(i) Write 4+12x−3x24+12x-3x^2 in the form

a+b(x+c)2\begin{align*} a+b(x+c)^2 \end{align*}

where aa, bb and cc are constants to be found.

(ii) Hence, or otherwise, state the coordinates of MM.

(5)

The line l1l_1 passes through OO and MM, as shown in Figure 4.

A line l2l_2 touches CC and is parallel to l1l_1.

(b) Find an equation for l2l_2.

(5)

解答

(a)

解法一

思路

展开

先提出 −3-3，再配方。配方形式会直接给出最大点，因为二次项系数为负。

答题过程

展开 4+12x−3x2= −3x2+12x+4= −3(x2−4x)+4= −3((x−2)2−4)+4= −3(x−2)2+12+4= 16−3(x−2)2.\begin{align*} 4+12x-3x^2 =&\,-3x^2+12x+4\\ =&\,-3(x^2-4x)+4\\ =&\,-3\bigl((x-2)^2-4\bigr)+4\\ =&\,-3(x-2)^2+12+4\\ =&\,16-3(x-2)^2. \end{align*}

Therefore

4+12x−3x2=16−3(x−2)2.\begin{align*} 4+12x-3x^2=16-3(x-2)^2. \end{align*}

The maximum value is 1616, when x=2x=2. Hence

M=(2,16).\begin{align*} M=(2,16). \end{align*}

(b)

解法一

思路

展开

直线 l1l_1 过 O(0,0)O(0,0) 和 M(2,16)M(2,16)，所以斜率是 88。l2l_2 与 l1l_1 平行，也有斜率 88。令切线为 y=8x+ky=8x+k，与曲线相切表示联立后只有一个交点，即判别式为 00。

答题过程

展开

The gradient of l1l_1 is

16−02−0=8.\begin{align*} \frac{16-0}{2-0}=8. \end{align*}

Since l2l_2 is parallel to l1l_1, let

l2:y=8x+k.\begin{align*} l_2:\quad y=8x+k. \end{align*}

At the point of contact,

4+12x−3x2= 8x+k3x2−4x+k−4= 0.\begin{align*} 4+12x-3x^2=&\,8x+k\\ 3x^2-4x+k-4=&\,0. \end{align*}

For a tangent, this quadratic has one repeated root, so its discriminant is 00:

(−4)2−4(3)(k−4)= 016−12k+48= 064−12k= 0k= 163.\begin{align*} (-4)^2-4(3)(k-4)=&\,0\\ 16-12k+48=&\,0\\ 64-12k=&\,0\\ k=&\,\frac{16}{3}. \end{align*}

Therefore

y=8x+163.\begin{align*} y=8x+\frac{16}{3}. \end{align*}

解法二

思路

展开

也可以直接用导数找切点：平行意味着切线斜率为 88，所以令曲线导数等于 88。

答题过程

展开 y= 4+12x−3x2,dydx= 12−6x.\begin{align*} y=&\,4+12x-3x^2,\\ \frac{dy}{dx}=&\,12-6x. \end{align*}

For l2l_2 to be parallel to l1l_1, its gradient must be 88:

12−6x= 8−6x= −4x= 23.\begin{align*} 12-6x=&\,8\\ -6x=&\,-4\\ x=&\,\frac23. \end{align*}

The corresponding yy coordinate is

y= 4+12(23)−3(23)2= 4+8−43= 323.\begin{align*} y =&\,4+12\left(\frac23\right)-3\left(\frac23\right)^2\\ =&\,4+8-\frac43\\ =&\,\frac{32}{3}. \end{align*}

So l2l_2 passes through (23,323)\left(\frac23,\frac{32}{3}\right) with gradient 88:

y−323= 8(x−23)y= 8x+163.\begin{align*} y-\frac{32}{3} =&\,8\left(x-\frac23\right)\\ y=&\,8x+\frac{16}{3}. \end{align*}

解法三

思路

展开

利用完全平方形式与待定系数法求解。 由直线 l1l_1 的斜率是 88，且 l2l_2 平行于 l1l_1，我们可以设 l2l_2 的方程为 y=8x+ky = 8x + k。 联立 l2l_2 与曲线 C:y=4+12x−3x2C: y = 4 + 12x - 3x^2 的方程，移项整理得到关于 xx 的一元二次方程：3x2−4x+(k−4)=03x^2 - 4x + (k - 4) = 0。 因为 l2l_2 是曲线的切线，说明此二次方程有重根，即必须可以写成完全平方式的形式：

3(x−c)2=3x2−6cx+3c2=0\begin{align*} 3(x - c)^2 = 3x^2 - 6cx + 3c^2 = 0 \end{align*}

对比 3x2−4x+(k−4)=03x^2 - 4x + (k - 4) = 0 与 3x2−6cx+3c2=03x^2 - 6cx + 3c^2 = 0 的系数：

−6c=  −4,3c2=  k−4.\begin{align*} -6c =&\,\, -4,\\[3mm] 3c^2 =&\,\, k - 4. \end{align*}

\begin{align*} 由此求得 c=23c = \frac{2}{3}。代入第二式可得： \end{align*}

3(23)2=k−4  ⟹  43=k−4  ⟹  k=1633\left(\frac{2}{3}\right)^2 = k - 4 \implies \frac{4}{3} = k - 4 \implies k = \frac{16}{3}

因此，切线 l2l_2 的方程为 y=8x+163y = 8x + \frac{16}{3}。该方法避开了判别式，思路十分简洁清爽。

答题过程

展开

The gradient of l1l_1 is:

16−02−0=8.\begin{align*} \frac{16 - 0}{2 - 0} = 8. \end{align*}

Since l2l_2 is parallel to l1l_1, let the equation of l2l_2 be:

y=8x+k.\begin{align*} y = 8x + k. \end{align*}

At the point of contact with the curve CC:

4+12x−3x2=  8x+k3x2−4x+(k−4)=  0.\begin{align*} 4 + 12x - 3x^2 =&\,\, 8x + k\\[3mm] 3x^2 - 4x + (k - 4) =&\,\, 0. \end{align*}

\begin{align*} Since l2l_2 is tangent to the curve, the quadratic equation must have a repeated root and can be written in the form: \end{align*}

3(x−c)2=  03x2−6cx+3c2=  0.\begin{align*} 3(x - c)^2 =&\,\, 0\\[3mm] 3x^2 - 6cx + 3c^2 =&\,\, 0. \end{align*}

\begin{align*} Compare the coefficients of 3x2−4x+(k−4)=03x^2 - 4x + (k - 4) = 0 and 3x2−6cx+3c2=03x^2 - 6cx + 3c^2 = 0: \end{align*}

−6c=  −43c2=  k−4.\begin{align*} -6c =&\,\, -4\\[3mm] 3c^2 =&\,\, k - 4. \end{align*}

From the first equation, we get:

c=23.c = \frac{2}{3}.

\begin{align*} Substitute c=23c = \frac{2}{3} into the second equation: \end{align*}

3(23)2=  k−443=  k−4k=  4+43k=  163.\begin{align*} 3\left(\frac{2}{3}\right)^2 =&\,\, k - 4\\[3mm] \frac{4}{3} =&\,\, k - 4\\[3mm] k =&\,\, 4 + \frac{4}{3}\\[3mm] k =&\,\, \frac{16}{3}. \end{align*}

\begin{align*} Therefore, the equation of l2l_2 is: \end{align*}

y=8x+163.y = 8x + \frac{16}{3}.