题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
A curve C has equation y=f(x) where f(x) is a polynomial in x.
Given
f′(x)=2(3x−2)(x+5)
(a) deduce the range of values of x for which y is decreasing.
(2)
Given further that C cuts the x-axis at 27
(b) state a factor of f(x), giving the answer in the form (Ax+B) where A and B are integers.
(1)
(c) Hence find an expression for f(x) in a fully factorised form.
(6)
题目中文翻译
本题必须展示计算过程的所有步骤。
完全依赖计算器技术的解答不予接受。
曲线 C 的方程为 y=f(x),其中 f(x) 是关于 x 的多项式。
已知
f′(x)=2(3x−2)(x+5)
(a) 推出 y 递减时 x 的取值范围。
(2)
还已知 C 在 27 处与 x 轴相交。
(b) 写出 f(x) 的一个因式,答案写成 (Ax+B) 的形式,其中 A、B 是整数。
(1)
(c) 由此求 f(x) 的表达式,写成完全因式分解的形式。
(6)
解答
(a)
解法一
思路
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y decreasing 表示 f′(x)<0。先找导数的两个零点,再判断符号。
答题过程
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The critical values are found from
2(3x−2)(x+5)=0.
So
x=32orx=−5.
Since the quadratic 2(3x−2)(x+5) has positive leading coefficient, it is negative between its two roots.
Therefore, y is decreasing when
−5<x<32.
(b)
解法一
思路
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曲线在 x=27 处与 x 轴相交,所以 f(27)=0,对应因式是 2x−7。
答题过程
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Since C cuts the x-axis at x=27,
f(27)=0.
Therefore, a factor of f(x) is
2x−7.
(c)
解法一
思路
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先把 f′(x) 展开并积分,得到 f(x),但会有常数 c。再用 f(27)=0 求 c,最后因式分解。
答题过程
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First expand f′(x):
f′(x)===2(3x−2)(x+5)2(3x2+15x−2x−10)6x2+26x−20.
Integrate:
f(x)==∫(6x2+26x−20)dx2x3+13x2−20x+c.
Since f(27)=0,
=2(27)3+13(27)2−20(27)+c0.
Simplifying,
4343+637−70+c=245−70+c=c=00−175.
So
f(x)=2x3+13x2−20x−175.
Using the factor 2x−7, divide the polynomial. The first quotient term is x2:
=2x3+13x2−20x−175−(2x3−7x2)20x2−20x−175.
The next quotient term is 10x:
=20x2−20x−175−(20x2−70x)50x−175.
The final quotient term is 25:
50x−175−(50x−175)=0.
So
f(x)==(2x−7)(x2+10x+25)(2x−7)(x+5)2.
Therefore,
f(x)=(2x−7)(x+5)2.
解法二
思路
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先把导数展开并积分,保留积分常数 c。由(b)的因式设 f(x)=(2x−7)(Ax2+Bx+C),与积分结果比较系数,求出二次因式和常数。这条路线也明确展示了所需积分过程。
答题过程
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First expand and integrate:
f′(x)==f(x)==2(3x−2)(x+5)6x2+26x−20,∫(6x2+26x−20)dx2x3+13x2−20x+c.
From part (b), 2x−7 is a factor. Let
f(x)=(2x−7)(Ax2+Bx+C).
Expanding this gives
f(x)=2Ax3+(2B−7A)x2+(2C−7B)x−7C.
Compare with the integrated expression:
2A=A=2B−7A=2B−7=B=2C−7B=2C−70=C=21,131310,−20−2025.
The constant term is c=−7C=−175. Therefore,
f(x)==(2x−7)(x2+10x+25)(2x−7)(x+5)2.
Equivalently, dividing the integrated expression before finding c gives quotient x2+10x+25 and remainder c+175. Since 2x−7 is a factor, that remainder is zero, again giving c=−175.