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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 PPxx 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 QQxx 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 MMxx 坐标。

解答

(a)

解法一

思路

展开

分别计算区间两端的函数值。函数在该区间连续,若两端异号,便可用介值定理说明区间内存在一个根。

答题过程

展开

Let

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

Then

f(1.25)=0.940<0f(-1.25)=-0.940\ldots<0

and

f(1.2)=0.214>0f(-1.2)=0.214\ldots>0

Since ff is continuous on [1.25,1.2][-1.25,-1.2], there is a root in this interval. Hence

1.25xP1.2\boxed{-1.25\leq x_P\leq-1.2}

(b)(i)

解法一

思路

展开

在给定迭代式中取 n=1n=1,直接代入 x1=6x_1=6

答题过程

展开 x2=12ln(2×6+3)+8=6.3637x_2=\sqrt{12\ln(2\times6+3)+8} =6.3637\ldots

Hence, to four decimal places,

x2=6.3637\boxed{x_2=6.3637}

(b)(ii)

解法一

思路

展开

继续使用同一迭代式,直到连续结果在四位小数下稳定。

答题过程

展开

Continuing the iteration,

nnxnx_n
16.00006.0000
26.36376.3637
36.40826.4082
46.41356.4135
56.41416.4141
66.41426.4142
76.41426.4142

Therefore the xx coordinate of QQ is

xQ=6.4142\boxed{x_Q=6.4142}

(c)

解法一

思路

展开

先求导并令导数为零,得到二次方程。两个代数根中,只有正根符合图中最大点 MM 的位置,也符合定义域 x>3/2x>-3/2

答题过程

展开

Differentiating,

dydx=122x+3x\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{12}{2x+3}-x

At MM,

122x+3x=0\frac{12}{2x+3}-x=0

Therefore

2x2+3x12=02x^2+3x-12=0

Using the quadratic formula,

x=3±1054x=\frac{-3\pm\sqrt{105}}{4}

The negative root is outside the domain x>3/2x>-3/2. Hence

xM=3+1054\boxed{x_M=\frac{-3+\sqrt{105}}{4}}