Skip to content
CalcGospel 國際數學圖譜
返回

P3 2024 Oct Q3

A Level / Edexcel / P3

IAL 2024 Oct Paper · Question 3

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 2 shows a sketch of the curve with equation y=f(x)y=f(x), where

f(x)=2x2−10xx∈Rf(x)=2x^2-10x\qquad x\in\mathbb{R}

(a) Solve the equation

f(∣x∣)=48f(|x|)=48
(3)

(b) Find the set of values of xx for which

∣f(x)∣≥52x|f(x)|\geq \frac{5}{2}x
(4)
题目中文翻译

图 2 给出了曲线 y=f(x)y=f(x) 的草图,其中

f(x)=2x2−10xx∈Rf(x)=2x^2-10x\qquad x\in\mathbb{R}

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

(a) 解方程

f(∣x∣)=48f(|x|)=48

(b) 求满足下式的 xx 的取值集合

∣f(x)∣≥52x|f(x)|\geq \frac{5}{2}x

解答

(a)

解法一

思路

展开

令 u=∣x∣u=|x|,则 u≥0u\geq0。先解关于 uu 的二次方程,并利用 uu 非负舍去无效根;最后由 ∣x∣=u|x|=u 写出正、负两个 xx 值。

答题过程

展开

Let u=∣x∣u=|x|, where u≥0u\geq0. Then

f(u)=48⟹2u2−10u= 48u2−5u−24= 0(u−8)(u+3)= 0.\begin{align*} f(u)=48 \quad\Longrightarrow\quad 2u^2-10u=&\,48\\ u^2-5u-24=&\,0\\ (u-8)(u+3)=&\,0. \end{align*}

Thus u=8u=8 or u=−3u=-3. Since u=∣x∣≥0u=|x|\geq0, reject u=−3u=-3.

Therefore

∣x∣=8,|x|=8,

and hence

x=−8, 8.\boxed{x=-8,\ 8}.

(b)

解法一

思路

展开

先注意当 x<0x<0 时,右边 52x\dfrac52x 为负数,而 ∣f(x)∣≥0|f(x)|\geq0,所以所有负数都满足不等式。对于 x≥0x\geq0,再根据 f(x)=2x(x−5)f(x)=2x(x-5) 的正负分成 0≤x≤50\leq x\leq5 和 x≥5x\geq5 两段求解。

答题过程

展开

If x<0x<0, then 52x<0\dfrac52x<0, while ∣f(x)∣≥0|f(x)|\geq0. Hence every x<0x<0 satisfies the inequality.

For 0≤x≤50\leq x\leq5, f(x)≤0f(x)\leq0, so

∣f(x)∣=10x−2x2.|f(x)|=10x-2x^2.

Therefore,

10x−2x2≥ 52xx(152−2x)≥ 0.\begin{align*} 10x-2x^2\geq&\,\frac52x\\ x\bigg(\frac{15}{2}-2x\bigg)\geq&\,0. \end{align*}

Within 0≤x≤50\leq x\leq5, this gives

0≤x≤154.0\leq x\leq\frac{15}{4}.

For x≥5x\geq5, f(x)≥0f(x)\geq0, so

∣f(x)∣=2x2−10x.|f(x)|=2x^2-10x.

Hence

2x2−10x≥ 52xx(2x−252)≥ 0.\begin{align*} 2x^2-10x\geq&\,\frac52x\\ x\bigg(2x-\frac{25}{2}\bigg)\geq&\,0. \end{align*}

Within x≥5x\geq5, this gives

x≥254.x\geq\frac{25}{4}.

Combining all three regions,

x≤154orx≥254.\boxed{x\leq\frac{15}{4} \quad\text{or}\quad x\geq\frac{25}{4}}.

解法二

思路

展开

当 x<0x<0 时不等式自动成立;当 x≥0x\geq0 时,两边都非负,因此可以安全地平方。整理后得到三个临界值,再用符号分析确定区间。

答题过程

展开

As above, every x<0x<0 satisfies the inequality.

For x≥0x\geq0, both sides are non-negative, so squaring preserves the inequality:

(2x2−10x)2≥(52x)2.\big(2x^2-10x\big)^2 \geq\bigg(\frac52x\bigg)^2.

Thus

x2((2x−10)2−254)≥ 014x2(4x−15)(4x−25)≥ 0.\begin{align*} x^2\bigg((2x-10)^2-\frac{25}{4}\bigg)\geq&\,0\\ \frac14x^2(4x-15)(4x-25)\geq&\,0. \end{align*}

Since x2≥0x^2\geq0, a sign analysis gives

0≤x≤154orx≥254.0\leq x\leq\frac{15}{4} \quad\text{or}\quad x\geq\frac{25}{4}.

Including all x<0x<0,

x≤154orx≥254.\boxed{x\leq\frac{15}{4} \quad\text{or}\quad x\geq\frac{25}{4}}.