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P2 2024 May Q5

A Level / Edexcel / P2

IAL 2024 May Paper · Question 5

题目

Problem

In this question you must show detailed reasoning.

(a) Given that xx and yy are positive numbers such that

(x−y)3>x3−y3(x-y)^3>x^3-y^3

prove that

y>xy>x
(4)

(b) Using a counter example, show that the result in part (a) is not true for all real numbers.

(2)
题目中文翻译

本题必须给出详细推理。

(a) 已知 xx 和 yy 都是正数，且 (x−y)3>x3−y3(x-y)^3>x^3-y^3，证明 y>xy>x。（4分）

(b) 利用反例，说明第(a)问的结论并非对所有实数都成立。（2分）

解答

(a)

解法一

思路

展开

把左边展开，然后和右边比较。由于 x,yx,y 都是正数，所以可以安全地除以 3xy3xy，不改变不等号方向。

答题过程

展开

Expand the left hand side:

(x−y)3=x3−3x2y+3xy2−y3.\begin{align*} (x-y)^3=x^3-3x^2y+3xy^2-y^3. \end{align*}

Given

(x−y)3>x3−y3,\begin{align*} (x-y)^3>x^3-y^3, \end{align*}

we have

x3−3x2y+3xy2−y3> x3−y3−3x2y+3xy2> 03xy(y−x)> 0.\begin{align*} x^3-3x^2y+3xy^2-y^3 >&\, x^3-y^3\\ -3x^2y+3xy^2 >&\, 0\\ 3xy(y-x) >&\, 0. \end{align*}

Since xx and yy are positive,

3xy>0.\begin{align*} 3xy>0. \end{align*}

Therefore,

y−x>0,\begin{align*} y-x>0, \end{align*}

so

y>x.y>x.

解法二

思路

展开

因式分解法。已知立方差公式 x3−y3=(x−y)(x2+xy+y2)x^3 - y^3 = (x-y)(x^2 + xy + y^2)。将原不等式右侧展开后，将所有项移至左侧并提出公因式 (x−y)(x-y)。再利用 x,y>0x, y > 0 的条件化简即可。

答题过程

展开

Rewrite the right-hand side using the difference of cubes formula:

x3−y3=(x−y)(x2+xy+y2)\begin{align*} x^3 - y^3 = (x-y)(x^2 + xy + y^2) \end{align*}

The given inequality becomes:

(x−y)3>(x−y)(x2+xy+y2)\begin{align*} (x-y)^3 > (x-y)(x^2 + xy + y^2) \end{align*}

Rearrange to one side:

(x−y)3−(x−y)(x2+xy+y2)>0\begin{align*} (x-y)^3 - (x-y)(x^2 + xy + y^2) > 0 \end{align*}

Factor out (x−y)(x-y):

(x−y)A> 0,\begin{align*} (x-y)A>&\,0, \end{align*}

where

A= (x−y)2−(x2+xy+y2)= x2−2xy+y2 −x2−xy−y2= −3xy.\begin{align*} A=&\,(x-y)^2-(x^2+xy+y^2)\\ =&\,x^2-2xy+y^2\\ &\,\hspace{2pt}-x^2-xy-y^2\\ =&\,-3xy. \end{align*}

Hence

(x−y)(−3xy)> 03xy(y−x)> 0.\begin{align*} (x-y)(-3xy) >&\, 0\\ 3xy(y-x) >&\, 0. \end{align*}

Since xx and yy are positive numbers, we have 3xy>03xy > 0.

Therefore, we can divide both sides by 3xy3xy without changing the inequality sign:

y−x> 0y> x.\begin{align*} y-x >&\, 0\\ y >&\, x. \end{align*}

This completes the proof.

(b)

解法一

思路

展开

要推翻“对所有实数都成立”，只需找一组实数满足原不等式，但结论 y>xy>x 不成立。让 yy 为负数通常很容易构造。

答题过程

展开

Take

x=3,y=−1.\begin{align*} x=3,\qquad y=-1. \end{align*}

Then

(x−y)3= (3−(−1))3= 43= 64,\begin{align*} (x-y)^3=&\,(3-(-1))^3\\ =&\,4^3\\ =&\,64, \end{align*}

and

x3−y3= 33−(−1)3= 27+1= 28.\begin{align*} x^3-y^3=&\,3^3-(-1)^3\\ =&\,27+1\\ =&\,28. \end{align*}

So

(x−y)3>x3−y3.\begin{align*} (x-y)^3>x^3-y^3. \end{align*}

But

y=−1<3=x.\begin{align*} y=-1<3=x. \end{align*}

Therefore, the result is not true for all real numbers.