题目
Problem
Let
f ( x ) = 5 x 2 + 8 x − 3 ( x − 2 ) ( 2 x 2 + 3 ) . \mathrm{f}(x)=\frac{5x^2+8x-3}{(x-2)(2x^2+3)}. f ( x ) = ( x − 2 ) ( 2 x 2 + 3 ) 5 x 2 + 8 x − 3 . (a) Express f ( x ) \mathrm{f}(x) f ( x )
(5)
(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 ) = 5 x 2 + 8 x − 3 ( x − 2 ) ( 2 x 2 + 3 ) . \mathrm{f}(x)=\frac{5x^2+8x-3}{(x-2)(2x^2+3)}. f ( x ) = ( x − 2 ) ( 2 x 2 + 3 ) 5 x 2 + 8 x − 3 . (a) 将 f ( x ) \mathrm{f}(x) f ( x )
(5)
(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
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对不可约二次因子设一次分子。清分母后先代入 x = 2 x=2 x = 2
答题过程
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Write
f ( x ) = A x − 2 + B x + C 2 x 2 + 3 . \mathrm f(x)=\frac{A}{x-2}+\frac{Bx+C}{2x^2+3}. f ( x ) = x − 2 A + 2 x 2 + 3 B x + C . Then
5 x 2 + 8 x − 3 = A ( 2 x 2 + 3 ) + ( B x + C ) ( x − 2 ) . \begin{align*}
5x^2+8x-3=&\,A(2x^2+3)\\[4mm]
&\,\hspace{2pt}+(Bx+C)(x-2).
\end{align*} 5 x 2 + 8 x − 3 = A ( 2 x 2 + 3 ) + ( B x + C ) ( x − 2 ) . At x = 2 x=2 x = 2 33 = 11 A 33=11A 33 = 11 A A = 3 A=3 A = 3
Comparing coefficients of x 2 x^2 x 2 5 = 2 A + B 5=2A+B 5 = 2 A + B B = − 1 B=-1 B = − 1 − 3 = 3 A − 2 C -3=3A-2C − 3 = 3 A − 2 C C = 6 C=6 C = 6
Hence
f ( x ) = 3 x − 2 + 6 − x 2 x 2 + 3 . \boxed{\mathrm f(x)=\frac3{x-2}+\frac{6-x}{2x^2+3}}. f ( x ) = x − 2 3 + 2 x 2 + 3 6 − x .
(b)
解法一
思路
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承接(a)，第一个分式展开成几何级数；第二个分母只含 x 2 x^2 x 2
答题过程
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Using ( 1 + u ) − 1 = 1 − u + u 2 + ⋯ (1+u)^{-1}=1-u+u^2+\cdots ( 1 + u ) − 1 = 1 − u + u 2 + ⋯ ∣ u ∣ < 1 |u|<1 ∣ u ∣ < 1
3 x − 2 = − 3 2 ( 1 − x 2 ) − 1 = − 3 2 − 3 4 x − 3 8 x 2 + ⋯ . \begin{align*}
\frac3{x-2}
=&\,-\frac32\bigg(1-\frac x2\bigg)^{-1}\\[4mm]
=&\,-\frac32-\frac34x-\frac38x^2+\cdots.
\end{align*} x − 2 3 = = − 2 3 ( 1 − 2 x ) − 1 − 2 3 − 4 3 x − 8 3 x 2 + ⋯ . Also,
6 − x 2 x 2 + 3 = 6 − x 3 ( 1 + 2 3 x 2 ) − 1 = ( 2 − x 3 ) ( 1 − 2 3 x 2 + ⋯ ) = 2 − x 3 − 4 3 x 2 + ⋯ . \begin{align*}
\frac{6-x}{2x^2+3}
=&\,\frac{6-x}{3}\bigg(1+\frac23x^2\bigg)^{-1}\\[4mm]
=&\,\bigg(2-\frac x3\bigg)\bigg(1-\frac23x^2+\cdots\bigg)\\[4mm]
=&\,2-\frac x3-\frac43x^2+\cdots.
\end{align*} 2 x 2 + 3 6 − x = = = 3 6 − x ( 1 + 3 2 x 2 ) − 1 ( 2 − 3 x ) ( 1 − 3 2 x 2 + ⋯ ) 2 − 3 x − 3 4 x 2 + ⋯ . Adding,
f ( x ) = 1 2 − ( 3 4 + 1 3 ) x − ( 3 8 + 4 3 ) x 2 + ⋯ = 1 2 − 13 12 x − 41 24 x 2 + ⋯ . \begin{align*}
\mathrm f(x)
=&\,\frac12-\bigg(\frac34+\frac13\bigg)x\\[4mm]
&\,\hspace{2pt}-\bigg(\frac38+\frac43\bigg)x^2+\cdots\\[4mm]
=&\,\boxed{\frac12-\frac{13}{12}x-\frac{41}{24}x^2+\cdots}.
\end{align*} f ( x ) = = 2 1 − ( 4 3 + 3 1 ) x − ( 8 3 + 3 4 ) x 2 + ⋯ 2 1 − 12 13 x − 24 41 x 2 + ⋯ . The common range of validity is ∣ x ∣ < 3 / 2 |x|<\sqrt{3/2} ∣ x ∣ < 3/2
