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IAL 2022 Jan Q5

A Level / Edexcel / P3

IAL 2022 Jan Paper · Question 5

题目

Problem

Figure 1 shows a sketch of part of the curve with equation

y=6ln⁡(2x+3)−12x2+4x>−32y=6\ln(2x+3)-\frac12x^2+4\qquad x>-\frac32

The curve cuts the negative xx-axis at the point PP, as shown in Figure 1.

(a) Show that the xx coordinate of PP lies in the interval [−1.25,−1.2][-1.25,-1.2]

(2)

The curve cuts the positive xx-axis at the point QQ, also shown in Figure 1.

Using the iterative formula

xn+1=12ln⁡(2xn+3)+8 with x1=6x_{n+1}=\sqrt{12\ln(2x_n+3)+8}\text{ with }x_1=6

(b) (i) find, to 4 decimal places, the value of x2x_2

(ii) find, by continued iteration, the xx coordinate of QQ. Give your answer to 4 decimal places.

(3)

The curve has a maximum turning point at MM, as shown in Figure 1.

(c) Using calculus and showing each stage of your working, find the xx coordinate of MM.

(4)
题目中文翻译

图 1 给出了部分曲线的示意图，其方程为

y=6ln⁡(2x+3)−12x2+4x>−32y=6\ln(2x+3)-\frac12x^2+4\qquad x>-\frac32

该曲线与负 xx 轴交于点 PP，如图所示。

(a) 证明点 PP 的 xx 坐标位于区间 [−1.25,−1.2][-1.25,-1.2] 内。

该曲线还与正 xx 轴交于点 QQ，如图所示。

使用迭代公式

xn+1=12ln⁡(2xn+3)+8，其中 x1=6x_{n+1}=\sqrt{12\ln(2x_n+3)+8}\text{，其中 }x_1=6

(b) (i) 求 x2x_2 的值，精确到小数点后 4 位。

(ii) 用继续迭代的方法求点 QQ 的 xx 坐标，答案精确到小数点后 4 位。

曲线在点 MM 处有一个极大转折点，如图所示。

(c) 用微积分并写出每一步过程，求点 MM 的 xx 坐标。

解答

(a)

Let

y=6ln⁡(2x+3)−12x2+4y=6\ln(2x+3)-\frac12x^2+4

At x=−1.25x=-1.25,

y=6ln⁡(0.5)−12(−1.25)2+4y=6\ln(0.5)-\frac12(-1.25)^2+4

So

y=−0.940…y=-0.940\ldots

At x=−1.2x=-1.2,

y=6ln⁡(0.6)−12(−1.2)2+4y=6\ln(0.6)-\frac12(-1.2)^2+4

So

y=0.214…y=0.214\ldots

Therefore

y(−1.25)<0andy(−1.2)>0y(-1.25)<0 \qquad\text{and}\qquad y(-1.2)>0

Since the function is continuous on [−1.25,−1.2][-1.25,-1.2], there is a root in this interval.

Hence the xx coordinate of PP lies in

[−1.25,−1.2]\boxed{[-1.25,-1.2]}

(b)(i)

Using

xn+1=12ln⁡(2xn+3)+8x_{n+1}=\sqrt{12\ln(2x_n+3)+8}

with x1=6x_1=6,

x2=12ln⁡(15)+8x_2=\sqrt{12\ln(15)+8}

Therefore

x2=6.3637\boxed{x_2=6.3637}

to 4 decimal places.

(b)(ii)

Continuing the iteration gives

x1=6,x2=6.3637,x3=6.4082,x4=6.4135,x5=6.4141,x6=6.4142\begin{aligned} x_1&=6,\\ x_2&=6.3637,\\ x_3&=6.4082,\\ x_4&=6.4135,\\ x_5&=6.4141,\\ x_6&=6.4142 \end{aligned}

Hence the xx coordinate of QQ is

6.4142\boxed{6.4142}

(c)

For

y=6ln⁡(2x+3)−12x2+4y=6\ln(2x+3)-\frac12x^2+4

求导得

dydx=6⋅22x+3−x\frac{dy}{dx} =6\cdot\frac{2}{2x+3}-x

So

dydx=122x+3−x\frac{dy}{dx} =\frac{12}{2x+3}-x

At a stationary point,

dydx=0\frac{dy}{dx}=0

Hence

122x+3−x=0\frac{12}{2x+3}-x=0

So

122x+3=x\frac{12}{2x+3}=x

Therefore

12=x(2x+3)12=x(2x+3)

and

2x2+3x−12=02x^2+3x-12=0

Using the quadratic formula,

x=−3±32−4(2)(−12)2(2)x=\frac{-3\pm\sqrt{3^2-4(2)(-12)}}{2(2)}

So

x=−3±1054x=\frac{-3\pm\sqrt{105}}{4}

Since the maximum point MM is on the positive side of the graph,

x=−3+1054\boxed{x=\frac{-3+\sqrt{105}}{4}}