题目
Problem
Let f ( x ) = 17 x 2 − 7 x + 16 ( 2 + 3 x 2 ) ( 2 − x ) \mathrm{f}(x)=\dfrac{17x^2-7x+16}{(2+3x^2)(2-x)} f ( x ) = ( 2 + 3 x 2 ) ( 2 − x ) 17 x 2 − 7 x + 16
(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 3 x^3 x 3
(5)
(c) State the set of values of x x x (b) is valid. Give your answer in an exact form.
(1)
题目中文翻译
设 f ( x ) = 17 x 2 − 7 x + 16 ( 2 + 3 x 2 ) ( 2 − x ) \mathrm{f}(x)=\dfrac{17x^2-7x+16}{(2+3x^2)(2-x)} f ( x ) = ( 2 + 3 x 2 ) ( 2 − x ) 17 x 2 − 7 x + 16
(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 3 x^3 x 3
(5)
(c) 写出使 (b) 问中的展开式成立的 x x x
(1)
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
不可约二次因子的分子须设为一次式。清分母后先用特殊值求一个系数，再比较其余系数。
答题过程
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Write
f ( x ) = A x + B 2 + 3 x 2 + C 2 − x . \mathrm f(x)=\frac{Ax+B}{2+3x^2}+\frac{C}{2-x}. f ( x ) = 2 + 3 x 2 A x + B + 2 − x C . Clearing denominators,
17 x 2 − 7 x + 16 = ( A x + B ) ( 2 − x ) + C ( 2 + 3 x 2 ) . \begin{align*}
17x^2-7x+16=&\,(Ax+B)(2-x)\\[4mm]
&\,\hspace{2pt}+C(2+3x^2).
\end{align*} 17 x 2 − 7 x + 16 = ( A x + B ) ( 2 − x ) + C ( 2 + 3 x 2 ) . Putting x = 2 x=2 x = 2 70 = 14 C 70=14C 70 = 14 C C = 5 C=5 C = 5
Comparing constant terms,
16 = 2 B + 2 C , B = 3. 16=2B+2C,\qquad B=3. 16 = 2 B + 2 C , B = 3. Comparing coefficients of x x x
− 7 = 2 A − B , A = − 2. -7=2A-B,\qquad A=-2. − 7 = 2 A − B , A = − 2. The x 2 x^2 x 2 − A + 3 C = 2 + 15 = 17 -A+3C=2+15=17 − A + 3 C = 2 + 15 = 17
f ( x ) = 3 − 2 x 2 + 3 x 2 + 5 2 − x . \boxed{\mathrm f(x)=\frac{3-2x}{2+3x^2}+\frac5{2-x}}. f ( x ) = 2 + 3 x 2 3 − 2 x + 2 − x 5 .
(b)
解法一
思路
展开
承接(a)，二次因子只含 x 2 x^2 x 2 x x x x 3 x^3 x 3
答题过程
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Using part (a),
3 − 2 x 2 + 3 x 2 = 3 − 2 x 2 ( 1 + 3 2 x 2 ) − 1 = 3 − 2 x 2 ( 1 − 3 2 x 2 + ⋯ ) = 3 2 − x − 9 4 x 2 + 3 2 x 3 + ⋯ . \begin{align*}
\frac{3-2x}{2+3x^2}
=&\,\frac{3-2x}{2}
\bigg(1+\frac32x^2\bigg)^{-1}\\[4mm]
=&\,\frac{3-2x}{2}\bigg(1-\frac32x^2+\cdots\bigg)\\[4mm]
=&\,\frac32-x-\frac94x^2+\frac32x^3+\cdots.
\end{align*} 2 + 3 x 2 3 − 2 x = = = 2 3 − 2 x ( 1 + 2 3 x 2 ) − 1 2 3 − 2 x ( 1 − 2 3 x 2 + ⋯ ) 2 3 − x − 4 9 x 2 + 2 3 x 3 + ⋯ . Also,
5 2 − x = 5 2 ( 1 − x 2 ) − 1 = 5 2 ( 1 + x 2 + x 2 4 + x 3 8 + ⋯ ) = 5 2 + 5 4 x + 5 8 x 2 + 5 16 x 3 + ⋯ . \begin{align*}
\frac5{2-x}
=&\,\frac52\bigg(1-\frac x2\bigg)^{-1}\\[4mm]
=&\,\frac52\bigg(1+\frac x2+\frac{x^2}{4}
+\frac{x^3}{8}+\cdots\bigg)\\[4mm]
=&\,\frac52+\frac54x+\frac58x^2+\frac5{16}x^3+\cdots.
\end{align*} 2 − x 5 = = = 2 5 ( 1 − 2 x ) − 1 2 5 ( 1 + 2 x + 4 x 2 + 8 x 3 + ⋯ ) 2 5 + 4 5 x + 8 5 x 2 + 16 5 x 3 + ⋯ . Adding,
f ( x ) = 4 + ( − 1 + 5 4 ) x + ( − 9 4 + 5 8 ) x 2 + ( 3 2 + 5 16 ) x 3 + ⋯ = 4 + 1 4 x − 13 8 x 2 + 29 16 x 3 + ⋯ . \begin{align*}
\mathrm f(x)
=&\,4+\bigg(-1+\frac54\bigg)x
+\bigg(-\frac94+\frac58\bigg)x^2\\[4mm]
&\,\hspace{2pt}+\bigg(\frac32+\frac5{16}\bigg)x^3+\cdots\\[4mm]
=&\,\boxed{4+\frac14x-\frac{13}{8}x^2+\frac{29}{16}x^3+\cdots}.
\end{align*} f ( x ) = = 4 + ( − 1 + 4 5 ) x + ( − 4 9 + 8 5 ) x 2 + ( 2 3 + 16 5 ) x 3 + ⋯ 4 + 4 1 x − 8 13 x 2 + 16 29 x 3 + ⋯ .
(c)
解法一
思路
展开
两个二项展开的条件都要满足，取较严格的条件。
答题过程
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For the expansion involving 1 + 3 2 x 2 1+\tfrac32x^2 1 + 2 3 x 2
∣ 3 2 x 2 ∣ < 1 ⟹ ∣ x ∣ < 2 3 . \bigg|\frac32x^2\bigg|<1
\quad\Longrightarrow\quad
|x|<\sqrt{\frac23}. 2 3 x 2 < 1 ⟹ ∣ x ∣ < 3 2 . For the expansion involving 1 − x / 2 1-x/2 1 − x /2 ∣ x / 2 ∣ < 1 |x/2|<1 ∣ x /2∣ < 1 ∣ x ∣ < 2 |x|<2 ∣ x ∣ < 2
Their common range is
− 2 3 < x < 2 3 . \boxed{-\sqrt{\frac23}<x<\sqrt{\frac23}}. − 3 2 < x < 3 2 . 