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), where
f(x)=2x2−10xx∈R
(a) Solve the equation
f(∣x∣)=48
(3)
(b) Find the set of values of x for which
∣f(x)∣≥25x
(4)
题目中文翻译
图 2 给出了曲线 y=f(x) 的草图,其中
f(x)=2x2−10xx∈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∣)=48
(b) 求满足下式的 x 的取值集合
∣f(x)∣≥25x
解答
(a)
解法一
思路
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令 u=∣x∣,则 u≥0。先解关于 u 的二次方程,并利用 u 非负舍去无效根;最后由 ∣x∣=u 写出正、负两个 x 值。
答题过程
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Let u=∣x∣, where u≥0. Then
f(u)=48⟹2u2−10u=u2−5u−24=(u−8)(u+3)=4800.
Thus u=8 or u=−3. Since u=∣x∣≥0, reject u=−3.
Therefore
∣x∣=8,
and hence
x=−8, 8.
(b)
解法一
思路
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先注意当 x<0 时,右边 25x 为负数,而 ∣f(x)∣≥0,所以所有负数都满足不等式。对于 x≥0,再根据 f(x)=2x(x−5) 的正负分成 0≤x≤5 和 x≥5 两段求解。
答题过程
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If x<0, then 25x<0, while ∣f(x)∣≥0. Hence every x<0 satisfies the inequality.
For 0≤x≤5, f(x)≤0, so
∣f(x)∣=10x−2x2.
Therefore,
10x−2x2≥x(215−2x)≥25x0.
Within 0≤x≤5, this gives
0≤x≤415.
For x≥5, f(x)≥0, so
∣f(x)∣=2x2−10x.
Hence
2x2−10x≥x(2x−225)≥25x0.
Within x≥5, this gives
x≥425.
Combining all three regions,
x≤415orx≥425.
解法二
思路
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当 x<0 时不等式自动成立;当 x≥0 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
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As above, every x<0 satisfies the inequality.
For x≥0, both sides are non-negative, so squaring preserves the inequality:
(2x2−10x)2≥(25x)2.
Thus
x2((2x−10)2−425)≥41x2(4x−15)(4x−25)≥00.
Since x2≥0, a sign analysis gives
0≤x≤415orx≥425.
Including all x<0,
x≤415orx≥425.