题目
Problem
f ( x ) = 5 x + 10 ( 1 − x ) ( 2 + 3 x ) \mathrm{f}(x)=\frac{5x+10}{(1-x)(2+3x)} f ( x ) = ( 1 − x ) ( 2 + 3 x ) 5 x + 10 (a) Write f ( x ) \mathrm{f}(x) f ( x )
(3)
(b) (i) Hence find, in ascending powers of x x x x 2 x^2 x 2 f ( x ) \mathrm{f}(x) f ( x )
(5)
(ii) Find the range of values of x x x
(1)
题目中文翻译
f ( x ) = 5 x + 10 ( 1 − x ) ( 2 + 3 x ) \mathrm{f}(x)=\frac{5x+10}{(1-x)(2+3x)} f ( x ) = ( 1 − x ) ( 2 + 3 x ) 5 x + 10 (a) 将 f ( x ) \mathrm{f}(x) f ( x )
(b) (i) 进而求 f ( x ) \mathrm{f}(x) f ( x ) x x x x 2 x^2 x 2
(ii) 求该展开式成立时 x x x
解答
(a)
解法一
思路
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设部分分式为 A 1 − x + B 2 + 3 x \frac{A}{1-x}+\frac{B}{2+3x} 1 − x A + 2 + 3 x B x x x A A A B B B
答题过程
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Let
5 x + 10 ( 1 − x ) ( 2 + 3 x ) = A 1 − x + B 2 + 3 x . \frac{5x+10}{(1-x)(2+3x)}
=\frac{A}{1-x}+\frac{B}{2+3x}. ( 1 − x ) ( 2 + 3 x ) 5 x + 10 = 1 − x A + 2 + 3 x B . Then
5 x + 10 = A ( 2 + 3 x ) + B ( 1 − x ) . 5x+10=A(2+3x)+B(1-x). 5 x + 10 = A ( 2 + 3 x ) + B ( 1 − x ) . Setting x = 1 x=1 x = 1
15 = 5 A , 15=5A, 15 = 5 A , so A = 3 A=3 A = 3 x = − 2 3 x=-\frac23 x = − 3 2
20 3 = 5 3 B , \frac{20}{3}=\frac{5}{3}B, 3 20 = 3 5 B , so B = 4 B=4 B = 4
f ( x ) = 3 1 − x + 4 2 + 3 x . \boxed{
\mathrm{f}(x)
=\frac{3}{1-x}+\frac{4}{2+3x}
}. f ( x ) = 1 − x 3 + 2 + 3 x 4 .
(b)(i)
解法一
思路
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承接 (a)，分别展开两个部分分式。第一项使用等比级数 ( 1 − x ) − 1 (1-x)^{-1} ( 1 − x ) − 1 2 2 2 ( 1 + 3 x 2 ) − 1 (1+\frac{3x}{2})^{-1} ( 1 + 2 3 x ) − 1 x 2 x^2 x 2
答题过程
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From part (a),
f ( x ) = 3 ( 1 − x ) − 1 + 2 ( 1 + 3 x 2 ) − 1 . \mathrm{f}(x)
=3(1-x)^{-1}
+2\bigg(1+\frac{3x}{2}\bigg)^{-1}. f ( x ) = 3 ( 1 − x ) − 1 + 2 ( 1 + 2 3 x ) − 1 . Using
( 1 + u ) − 1 = 1 − u + u 2 + ⋯ , (1+u)^{-1}=1-u+u^2+\cdots, ( 1 + u ) − 1 = 1 − u + u 2 + ⋯ , we obtain
3 ( 1 − x ) − 1 = 3 ( 1 + x + x 2 + ⋯ ) 3(1-x)^{-1}
=3(1+x+x^2+\cdots) 3 ( 1 − x ) − 1 = 3 ( 1 + x + x 2 + ⋯ ) and
2 ( 1 + 3 x 2 ) − 1 = 2 ( 1 − 3 x 2 + 9 x 2 4 + ⋯ ) = 2 − 3 x + 9 2 x 2 + ⋯ . \begin{align*}
2\bigg(1+\frac{3x}{2}\bigg)^{-1}
=&\,2\bigg(
1-\frac{3x}{2}+\frac{9x^2}{4}+\cdots
\bigg)\\
=&\,2-3x+\frac{9}{2}x^2+\cdots.
\end{align*} 2 ( 1 + 2 3 x ) − 1 = = 2 ( 1 − 2 3 x + 4 9 x 2 + ⋯ ) 2 − 3 x + 2 9 x 2 + ⋯ . Therefore,
f ( x ) = ( 3 + 3 x + 3 x 2 ) + ( 2 − 3 x + 9 2 x 2 ) + ⋯ = 5 + 15 2 x 2 + ⋯ . \begin{align*}
\mathrm{f}(x)
=&\,(3+3x+3x^2)\\
&\,\hspace{2pt}
+\bigg(2-3x+\frac{9}{2}x^2\bigg)
+\cdots\\
=&\,\boxed{5+\frac{15}{2}x^2+\cdots}.
\end{align*} f ( x ) = = ( 3 + 3 x + 3 x 2 ) + ( 2 − 3 x + 2 9 x 2 ) + ⋯ 5 + 2 15 x 2 + ⋯ .
(b)(ii)
解法一
思路
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两个级数必须同时收敛。分别写出 ∣ x ∣ < 1 \lvert x\rvert<1 ∣ x ∣ < 1 ∣ 3 x 2 ∣ < 1 \lvert\frac{3x}{2}\rvert<1 ∣ 2 3 x ∣ < 1
答题过程
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The expansion of ( 1 − x ) − 1 (1-x)^{-1} ( 1 − x ) − 1
∣ x ∣ < 1. |x|<1. ∣ x ∣ < 1. The expansion of
( 1 + 3 x 2 ) − 1 \big(1+\frac{3x}{2}\big)^{-1} ( 1 + 2 3 x ) − 1
∣ 3 x 2 ∣ < 1 , \bigg|\frac{3x}{2}\bigg|<1, 2 3 x < 1 , so
∣ x ∣ < 2 3 . |x|<\frac23. ∣ x ∣ < 3 2 . Both conditions must hold. Hence the required range is
− 2 3 < x < 2 3 . \boxed{-\frac23<x<\frac23}. − 3 2 < x < 3 2 . 