题目
Problem
Let f ( x ) = x 3 + 2 x − 11 ( 3 + x ) ( 2 + x 2 ) \mathrm{f}(x)=\frac{x^3+2x-11}{(3+x)(2+x^2)} f ( x ) = ( 3 + x ) ( 2 + x 2 ) x 3 + 2 x − 11
(a) Express f ( x ) \mathrm{f}(x) f ( x )
(6)
(b) Hence obtain the expansion of f ( x ) \mathrm{f}(x) f ( x ) x x x x 2 x^2 x 2
(5)
题目中文翻译
设 f ( x ) = x 3 + 2 x − 11 ( 3 + x ) ( 2 + x 2 ) \mathrm{f}(x)=\frac{x^3+2x-11}{(3+x)(2+x^2)} f ( x ) = ( 3 + x ) ( 2 + x 2 ) x 3 + 2 x − 11
(a) 将 f ( x ) \mathrm{f}(x) f ( x )
(6)
(b) 由此求 f ( x ) \mathrm{f}(x) f ( x ) x x x x 2 x^2 x 2
(5)
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
展开
分子分母同次，先提取常数项。余下的部分分式在不可约二次因式上使用一次分子。
答题过程
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Since
( 3 + x ) ( 2 + x 2 ) = x 3 + 3 x 2 + 2 x + 6 , (3+x)(2+x^2)=x^3+3x^2+2x+6, ( 3 + x ) ( 2 + x 2 ) = x 3 + 3 x 2 + 2 x + 6 , polynomial division gives
f ( x ) = 1 + − 3 x 2 − 17 ( x + 3 ) ( x 2 + 2 ) . f(x)=1+\frac{-3x^2-17}{(x+3)(x^2+2)}. f ( x ) = 1 + ( x + 3 ) ( x 2 + 2 ) − 3 x 2 − 17 . Write
− 3 x 2 − 17 ( x + 3 ) ( x 2 + 2 ) = B x + 3 + C x + D x 2 + 2 . \frac{-3x^2-17}{(x+3)(x^2+2)}
=\frac{B}{x+3}+\frac{Cx+D}{x^2+2}. ( x + 3 ) ( x 2 + 2 ) − 3 x 2 − 17 = x + 3 B + x 2 + 2 C x + D . Multiplying through,
− 3 x 2 − 17 = B ( x 2 + 2 ) + ( C x + D ) ( x + 3 ) . -3x^2-17=B(x^2+2)+(Cx+D)(x+3). − 3 x 2 − 17 = B ( x 2 + 2 ) + ( C x + D ) ( x + 3 ) . Putting x = − 3 x=-3 x = − 3 − 44 = 11 B -44=11B − 44 = 11 B B = − 4 B=-4 B = − 4 x 2 x^2 x 2 x x x
B + C = − 3 ⟹ C = 1 , 3 C + D = 0 ⟹ D = − 3. \begin{align*}
B+C=&\,-3\quad\Longrightarrow\quad C=1,\\[4mm]
3C+D=&\,0\quad\Longrightarrow\quad D=-3.
\end{align*} B + C = 3 C + D = − 3 ⟹ C = 1 , 0 ⟹ D = − 3. The constant coefficient also checks: 2 B + 3 D = − 8 − 9 = − 17 2B+3D=-8-9=-17 2 B + 3 D = − 8 − 9 = − 17
f ( x ) = 1 − 4 x + 3 + x − 3 x 2 + 2 . \boxed{f(x)=1-\frac4{x+3}+\frac{x-3}{x^2+2}}. f ( x ) = 1 − x + 3 4 + x 2 + 2 x − 3 .
(b)
解法一
思路
展开
承接(a)，分别使用两个等比展开。乘一次分子时保留到二次项，最后按幂次合并。
答题过程
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Using the partial fractions from (a),
− 4 x + 3 = − 4 3 ( 1 + x 3 ) − 1 = − 4 3 ( 1 − x 3 + x 2 9 + ⋯ ) = − 4 3 + 4 9 x − 4 27 x 2 + ⋯ . \begin{align*}
-\frac4{x+3}
=&\,-\frac43\bigg(1+\frac x3\bigg)^{-1}\\[4mm]
=&\,-\frac43\bigg(1-\frac x3+\frac{x^2}9+\cdots\bigg)\\[4mm]
=&\,-\frac43+\frac49x-\frac4{27}x^2+\cdots.
\end{align*} − x + 3 4 = = = − 3 4 ( 1 + 3 x ) − 1 − 3 4 ( 1 − 3 x + 9 x 2 + ⋯ ) − 3 4 + 9 4 x − 27 4 x 2 + ⋯ . Also,
x − 3 x 2 + 2 = x − 3 2 ( 1 + x 2 2 ) − 1 = x − 3 2 ( 1 − x 2 2 + ⋯ ) = − 3 2 + 1 2 x + 3 4 x 2 + ⋯ . \begin{align*}
\frac{x-3}{x^2+2}
=&\,\frac{x-3}{2}\bigg(1+\frac{x^2}{2}\bigg)^{-1}\\
=&\,\frac{x-3}{2}\bigg(1-\frac{x^2}{2}+\cdots\bigg)\\
=&\,-\frac32+\frac12x+\frac34x^2+\cdots.
\end{align*} x 2 + 2 x − 3 = = = 2 x − 3 ( 1 + 2 x 2 ) − 1 2 x − 3 ( 1 − 2 x 2 + ⋯ ) − 2 3 + 2 1 x + 4 3 x 2 + ⋯ . Therefore
f ( x ) = 1 − 4 3 − 3 2 + ( 4 9 + 1 2 ) x + ( − 4 27 + 3 4 ) x 2 + ⋯ = − 11 6 + 17 18 x + 65 108 x 2 + ⋯ . \begin{align*}
f(x)=&\,1-\frac43-\frac32\\
&\,\hspace{2pt}+\bigg(\frac49+\frac12\bigg)x\\[4mm]
&\,\hspace{4pt}+\bigg(-\frac4{27}+\frac34\bigg)x^2+\cdots\\
=&\,\boxed{-\frac{11}{6}+\frac{17}{18}x+\frac{65}{108}x^2+\cdots}.
\end{align*} f ( x ) = = 1 − 3 4 − 2 3 + ( 9 4 + 2 1 ) x + ( − 27 4 + 4 3 ) x 2 + ⋯ − 6 11 + 18 17 x + 108 65 x 2 + ⋯ . Both expansions are valid for ∣ x ∣ < 2 |x|<\sqrt2 ∣ x ∣ < 2
