题目
Problem
Figure 1 shows a sketch of part of the curve with equation
y=6ln(2x+3)−21x2+4x>−23
The curve cuts the negative x-axis at the point P, as shown in Figure 1.
(a) Show that the x coordinate of P lies in the interval [−1.25,−1.2]
(2)
The curve cuts the positive x-axis at the point Q, also shown in Figure 1.
Using the iterative formula
xn+1=12ln(2xn+3)+8 with x1=6
(b) (i) find, to 4 decimal places, the value of x2
(ii) find, by continued iteration, the x coordinate of Q. Give your answer to 4 decimal places.
(3)
The curve has a maximum turning point at M, as shown in Figure 1.
(c) Using calculus and showing each stage of your working, find the x coordinate of M.
(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)−21x2+4x>−23
该曲线与负 x 轴交于点 P,如图所示。
(a) 证明点 P 的 x 坐标位于区间 [−1.25,−1.2] 内。
该曲线还与正 x 轴交于点 Q,如图所示。
使用迭代公式
xn+1=12ln(2xn+3)+8,其中 x1=6
(b) (i) 求 x2 的值,精确到小数点后 4 位。
(ii) 用继续迭代的方法求点 Q 的 x 坐标,答案精确到小数点后 4 位。
曲线在点 M 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 M 的 x 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)−21x2+4
Then
f(−1.25)=−0.940…<0
and
f(−1.2)=0.214…>0
Since f is continuous on [−1.25,−1.2], there is a root in this interval. Hence
−1.25≤xP≤−1.2
(b)(i)
解法一
思路
展开
在给定迭代式中取 n=1,直接代入 x1=6。
答题过程
展开
x2=12ln(2×6+3)+8=6.3637…
Hence, to four decimal places,
x2=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,
| n | xn |
|---|
| 1 | 6.0000 |
| 2 | 6.3637 |
| 3 | 6.4082 |
| 4 | 6.4135 |
| 5 | 6.4141 |
| 6 | 6.4142 |
| 7 | 6.4142 |
Therefore the x coordinate of Q is
xQ=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 M 的位置,也符合定义域 x>−3/2。
答题过程
展开
Differentiating,
dxdy=2x+312−x
At M,
2x+312−x=0
Therefore
2x2+3x−12=0
Using the quadratic formula,
x=4−3±105
The negative root is outside the domain x>−3/2. Hence
xM=4−3+105