题目

Problem

f(x)=(8-3x)^{\frac43}\qquad 0<x<\frac83

(a) Show that the binomial expansion of $f(x)$ in ascending powers of $x$ up to and including the term in $x^3$ is

A-8x+\frac{x^2}{2}+Bx^3+\ldots

where $A$ and $B$ are constants to be found.

(4)

(b) Use proof by contradiction to prove that the curve with equation

y=8+8x-\frac{15}{2}x^2

does not intersect the curve with equation

y=A-8x+\frac{x^2}{2}+Bx^3\qquad 0<x<\frac83

where $A$ and $B$ are the constants found in part (a).

(Solutions relying on calculator technology are not acceptable.)

(4)

题目中文翻译

f(x)=(8-3x)^{\frac43}\qquad 0<x<\frac83

(a) 证明 $f(x)$ 按 $x$ 的升幂展开，直到并包括 $x^3$ 项，为

A-8x+\frac{x^2}{2}+Bx^3+\ldots

其中 $A,B$ 为待求常数。

(b) 用反证法证明曲线

y=8+8x-\frac{15}{2}x^2

与曲线

y=A-8x+\frac{x^2}{2}+Bx^3\qquad 0<x<\frac83

不相交，其中 $A,B$ 是 (a) 中求得的常数。

不接受依赖计算器技术的解法。

解答

(a)

解法一

思路

展开

先把 $(8-3x)^{\frac43}$ 写成适合二项式展开的形式：

16\left(1-\frac{3x}{8}\right)^{\frac43}

然后使用标准展开式，保留到 $x^3$ 项。最后和题目给出的形式比较，就能读出 $A$ 和 $B$ 。

答题过程

展开

Rewrite

(8-3x)^{\frac43} =8^{\frac43}\left(1-\frac{3x}{8}\right)^{\frac43}

Since

8^{\frac43}=16

we have

f(x)=16\left(1-\frac{3x}{8}\right)^{\frac43}

Use

(1+u)^n =1+nu+\frac{n(n-1)}{2}u^2 +\frac{n(n-1)(n-2)}{6}u^3+\cdots

Here

n=\frac43,\qquad u=-\frac{3x}{8}

\begin{aligned} \left(1-\frac{3x}{8}\right)^{\frac43} &=1+\frac43\left(-\frac{3x}{8}\right) +\frac{\frac43\cdot\frac13}{2}\left(-\frac{3x}{8}\right)^2 \\ &\quad +\frac{\frac43\cdot\frac13\cdot\left(-\frac23\right)}{6} \left(-\frac{3x}{8}\right)^3+\cdots \end{aligned}

This becomes

\left(1-\frac{3x}{8}\right)^{\frac43} =1-\frac{x}{2}+\frac{x^2}{32}+\frac{x^3}{384}+\cdots

Therefore

\begin{aligned} f(x) &=16\left(1-\frac{x}{2}+\frac{x^2}{32}+\frac{x^3}{384}+\cdots\right) \\ &=16-8x+\frac{x^2}{2}+\frac{x^3}{24}+\cdots \end{aligned}

Hence

\boxed{A=16,\qquad B=\frac1{24}}

(b)

解法一

思路

展开

反证法的核心是：假设两条曲线有交点，那么同一个 $x$ 必须同时满足两个方程。把 (a) 中得到的 $A=16$ 、 $B=\dfrac{1}{24}$ 代入后整理，会得到

0=8(x-1)^2+\frac{x^3}{24}

但在 $0<x<\dfrac83$ 中，右边严格大于 $0$ ，所以不可能等于 $0$ ，产生矛盾。

答题过程

展开

Assume, for contradiction, that the two curves intersect for some $x$ in the interval

0<x<\frac83

Then

8+8x-\frac{15}{2}x^2 =16-8x+\frac{x^2}{2}+\frac{x^3}{24}

Move all terms to one side:

0=8-16x+8x^2+\frac{x^3}{24}

The quadratic part can be written as

8-16x+8x^2=8(x-1)^2

So the equation becomes

0=8(x-1)^2+\frac{x^3}{24}

But in the interval

0<x<\frac83

we have

\frac{x^3}{24}>0

and

8(x-1)^2\ge0

Even when $8(x-1)^2=0$ , we still have $\dfrac{x^3}{24}>0$ because $x>0$ .

Therefore

8(x-1)^2+\frac{x^3}{24}>0

This cannot equal $0$ , which is a contradiction.

Hence the two curves do not intersect.

IAL 2024 May Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程