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P3 2020 Oct Q6

A Level / Edexcel / P3

IAL 2020 Oct Paper · Question 6

题目

Problem

Figure 3 shows a sketch of curve C1C_1 with equation y=5ex−1+3y=5e^{x-1}+3 and curve C2C_2 with equation y=10−x2y=10-x^2.

The point PP lies on C1C_1 and has yy coordinate 1818.

(a) Find the xx coordinate of PP, writing your answer in the form ln⁡k\ln k, where kk is a constant to be found.

(3)

The curve C1C_1 meets the curve C2C_2 at x=αx=\alpha and at x=βx=\beta, as shown in Figure 3.

(b) Using a suitable interval and a suitable function that should be stated, show that to 3 decimal places α=1.134\alpha=1.134.

(3)

The iterative equation

xn+1=−7−5exn−1x_{n+1}=-\sqrt{7-5e^{x_n-1}}

is used to find an approximation to β\beta.

Using this iterative formula with x1=−3x_1=-3

(c) find the value of x2x_2 and the value of β\beta, giving each answer to 6 decimal places.

(3)
题目中文翻译

图 3 给出了曲线 C1C_1：y=5ex−1+3y=5e^{x-1}+3 与曲线 C2C_2：y=10−x2y=10-x^2 的草图。

点 PP 在 C1C_1 上，且其 yy 坐标为 1818。

(a) 求点 PP 的 xx 坐标，并将答案写成 ln⁡k\ln k 的形式，其中 kk 为待求常数。

曲线 C1C_1 与曲线 C2C_2 分别在 x=αx=\alpha 与 x=βx=\beta 处相交，如图 3 所示。

(b) 选取合适的区间并写出合适的函数，证明 α=1.134\alpha=1.134（精确到小数点后 3 位）。

迭代公式

xn+1=−7−5exn−1x_{n+1}=-\sqrt{7-5e^{x_n-1}}

用于求 β\beta 的近似值。

当 x1=−3x_1=-3 时，

(c) 求 x2x_2 和 β\beta 的值，两者都精确到小数点后 6 位。

解答

(a)

Point PP lies on

y=5ex−1+3y=5e^{x-1}+3

and has yy coordinate 1818. Therefore

5ex−1+3=185e^{x-1}+3=18

So

5ex−1=155e^{x-1}=15

Hence

ex−1=3e^{x-1}=3

Taking natural logarithms,

x−1=ln⁡3x-1=\ln3

so

x=1+ln⁡3x=1+\ln3

Since

1=ln⁡e1=\ln e

we get

x=ln⁡e+ln⁡3=ln⁡(3e)x=\ln e+\ln3=\ln(3e)

Therefore

x=ln⁡(3e)\boxed{x=\ln(3e)}

(b)

The two curves meet when

5ex−1+3=10−x25e^{x-1}+3=10-x^2

Rearrange:

7−x2−5ex−1=07-x^2-5e^{x-1}=0

Let

F(x)=7−x2−5ex−1F(x)=7-x^2-5e^{x-1}

To show that α=1.134\alpha=1.134 to 3 decimal places, use the interval

1.1335<x<1.13451.1335<x<1.1345

Now

F(1.1335)=0.001071…>0F(1.1335)=0.001071\ldots>0

and

F(1.1345)=−0.006914…<0F(1.1345)=-0.006914\ldots<0

Since F(x)F(x) is continuous and changes sign between 1.13351.1335 and 1.13451.1345, there is a root in this interval.

Therefore

1.1335<α<1.13451.1335<\alpha<1.1345

Hence

α=1.134\boxed{\alpha=1.134}

to 3 decimal places.

(c)

The iteration formula is

xn+1=−7−5exn−1x_{n+1}=-\sqrt{7-5e^{x_n-1}}

with

x1=−3x_1=-3

For x2x_2:

x2=−7−5e−3−1=−7−5e−4=−2.628387681…\begin{aligned} x_2 &=-\sqrt{7-5e^{-3-1}}\\ &=-\sqrt{7-5e^{-4}}\\ &=-2.628387681\ldots \end{aligned}

So

x2=−2.628388\boxed{x_2=-2.628388}

to 6 decimal places.

Continuing the iteration gives

β=−2.620330\boxed{\beta=-2.620330}

to 6 decimal places.