CalcGospel 國際數學圖譜
返回

IAL 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)

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\\[4mm] -3x^2y+3xy^2 >&\,\, 0\\[4mm] 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)[(x−y)2−(x2+xy+y2)]>  0(x−y)[(x2−2xy+y2)−(x2+xy+y2)]>  0(x−y)(−3xy)>  03xy(y−x)>  0.\begin{align*} (x-y)\left[ (x-y)^2 - (x^2 + xy + y^2) \right] >&\,\, 0\\[4mm] (x-y)\left[ (x^2 - 2xy + y^2) - (x^2 + xy + y^2) \right] >&\,\, 0\\[4mm] (x-y)(-3xy) >&\,\, 0\\[4mm] 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\\[4mm] 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.