题目
Problem
The polynomial ax3−11x2+bx+4, where a and b are constants, is denoted by p(x). It is given that (x−2) is a factor of p(x), and that when p′(x) is divided by (x−2) the remainder is 24.
(a) Find the value of a and the value of b.
(5)
(b) (i) When a and b have these values, factorise p(x) completely.
(3)
(ii) Hence, solve the inequality p(x)≥0.
(2)
题目中文翻译
多项式 ax3−11x2+bx+4 记为 p(x),其中 a 和 b 为常数。已知 (x−2) 是 p(x) 的一个因式,且 p′(x) 除以 (x−2) 的余数为 24。
(a) 求 a 和 b 的值。
(5)
(b) (i) 当 a 和 b 取上述值时,将 p(x) 完全因式分解。
(3)
(ii) 由此解不等式 p(x)≥0。
(2)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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由因式定理得到 p(2)=0。对导函数应用余式定理,得到 p′(2)=24,联立求参数。
答题过程
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The factor condition gives
p(2)=4a+b=8a−44+2b+4=0,20.
Also p′(x)=3ax2−22x+b, so the remainder condition gives
p′(2)=12a+b=12a−44+b=24,68.
Subtracting the equations gives 8a=48, hence a=6 and b=−4.
(b)(i)
解法一
思路
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除去已知因式 x−2 后,继续分解得到的二次式。
答题过程
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With the values found in part (a), p(x)=6x3−11x2−4x+4. Since x−2 is a factor, write p(x)=(x−2)(6x2+Bx+C). Expanding gives
p(x)=6x3+(B−12)x2+(C−2B)x−2C.
Comparing coefficients gives B−12=−11, so B=1, and −2C=4, so C=−2. The x coefficient is then C−2B=−4, as required. Hence
p(x)===(x−2)(6x2+x−2)(x−2)(6x2+4x−3x−2)(x−2)(3x+2)(2x−1).
(b)(ii)
解法一
思路
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三个根都是单根,首项系数为正。按根从小到大分区间判断三个因子的乘积符号,并包含等号。
答题过程
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The roots are −2/3, 1/2 and 2. The sign of p(x) on the four successive intervals is negative, positive, negative, positive. Including the zeros, the solution is
−32≤x≤21orx≥2.