题目
Problem
f ( x ) = 4 x 3 + p x 2 + 8 x + q \begin{align*}
\text{f}(x)=&\,4x^3+px^2+8x+q
\end{align*} f ( x ) = 4 x 3 + p x 2 + 8 x + q where p p p q q q
Given that
( 2 x + 3 ) (2x+3) ( 2 x + 3 ) f ( x ) \text{f}(x) f ( x ) f ( x ) \text{f}(x) f ( x ) − 5 -5 − 5 ( x + 2 ) (x+2) ( x + 2 )
(a) (i) show that p = 10 p=10 p = 10
(ii) find the value of q q q
(5)
(b) Hence find the range of values of x x x f ( x ) \text{f}(x) f ( x )
(Solutions based entirely on calculator technology are not acceptable.)
(4)
题目中文翻译
f ( x ) = 4 x 3 + p x 2 + 8 x + q \text{f}(x)=4x^3+px^2+8x+q f ( x ) = 4 x 3 + p x 2 + 8 x + q 其中 p p p q q q
( 2 x + 3 ) (2x+3) ( 2 x + 3 ) f ( x ) \text{f}(x) f ( x ) f ( x ) \text{f}(x) f ( x ) ( x + 2 ) (x+2) ( x + 2 ) − 5 -5 − 5
(a) (i) 证明 p = 10 p=10 p = 10 q q q
(b) 由此求 f ( x ) \text{f}(x) f ( x ) x x x
（完全依赖计算器技术的解答不予接受。）
解答
(a)
解法一
思路
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用因式定理得出 f ( − 3 / 2 ) = 0 \text{f}(-3/2)=0 f ( − 3/2 ) = 0 f ( − 2 ) = − 5 \text{f}(-2)=-5 f ( − 2 ) = − 5 p p p q q q
答题过程
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Since 2 x + 3 2x+3 2 x + 3
f ( − 3 2 ) = 0. \text{f}\bigg(-\frac32\bigg)=0. f ( − 2 3 ) = 0. Substituting gives
− 27 2 + 9 p 4 − 12 + q = 0 9 p + 4 q = 102. \begin{align*}
-\frac{27}{2}+\frac{9p}{4}-12+q=&\,0\\
9p+4q=&\,102. \tag{1}
\end{align*} − 2 27 + 4 9 p − 12 + q = 9 p + 4 q = 0 102. ( 1 ) The remainder on division by x + 2 x+2 x + 2 − 5 -5 − 5
f ( − 2 ) = − 5 − 32 + 4 p − 16 + q = − 5 4 p + q = 43. \begin{align*}
\text{f}(-2)=&\,-5\\
-32+4p-16+q=&\,-5\\
4p+q=&\,43. \tag{2}
\end{align*} f ( − 2 ) = − 32 + 4 p − 16 + q = 4 p + q = − 5 − 5 43. ( 2 ) Four times equation (2) minus equation (1) gives
16 p + 4 q − ( 9 p + 4 q ) = 172 − 102 7 p = 70 p = 10. (shown) \begin{align*}
16p+4q-(9p+4q)=&\,172-102\\
7p=&\,70\\
p=&\,10.\quad\text{(shown)}
\end{align*} 16 p + 4 q − ( 9 p + 4 q ) = 7 p = p = 172 − 102 70 10. (shown) Then
q = 43 − 4 ( 10 ) = 3. q=43-4(10)=3. q = 43 − 4 ( 10 ) = 3.
解法二
思路
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直接作两次多项式除法。除以 2 x + 3 2x+3 2 x + 3 x + 2 x+2 x + 2 − 5 -5 − 5
答题过程
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First divide f ( x ) \text{f}(x) f ( x ) 2 x + 3 2x+3 2 x + 3
f ( x ) − ( 2 x + 3 ) ( 2 x 2 ) = ( p − 6 ) x 2 + 8 x + q . \begin{align*}
&\,\text{f}(x)-(2x+3)(2x^2)\\
=&\,(p-6)x^2+8x+q.
\end{align*} = f ( x ) − ( 2 x + 3 ) ( 2 x 2 ) ( p − 6 ) x 2 + 8 x + q . ( p − 6 ) x 2 + 8 x + q − ( 2 x + 3 ) p − 6 2 x = 34 − 3 p 2 x + q . \begin{align*}
&\,(p-6)x^2+8x+q\\
&\,\hspace{2pt}-(2x+3)\frac{p-6}{2}x\\
=&\,\frac{34-3p}{2}x+q.
\end{align*} = ( p − 6 ) x 2 + 8 x + q − ( 2 x + 3 ) 2 p − 6 x 2 34 − 3 p x + q . 34 − 3 p 2 x + q − ( 2 x + 3 ) 34 − 3 p 4 = 9 p + 4 q − 102 4 . \begin{align*}
&\,\frac{34-3p}{2}x+q\\
&\,\hspace{2pt}-(2x+3)\frac{34-3p}{4}\\
=&\,\frac{9p+4q-102}{4}.
\end{align*} = 2 34 − 3 p x + q − ( 2 x + 3 ) 4 34 − 3 p 4 9 p + 4 q − 102 . Since 2 x + 3 2x+3 2 x + 3
9 p + 4 q = 102. (1) 9p+4q=102. \tag{1} 9 p + 4 q = 102. ( 1 ) Next divide by x + 2 x+2 x + 2
f ( x ) − ( x + 2 ) ( 4 x 2 ) = ( p − 8 ) x 2 + 8 x + q . \begin{align*}
&\,\text{f}(x)-(x+2)(4x^2)\\
=&\,(p-8)x^2+8x+q.
\end{align*} = f ( x ) − ( x + 2 ) ( 4 x 2 ) ( p − 8 ) x 2 + 8 x + q . ( p − 8 ) x 2 + 8 x + q − ( x + 2 ) ( p − 8 ) x = ( 24 − 2 p ) x + q . \begin{align*}
&\,(p-8)x^2+8x+q\\
&\,\hspace{2pt}-(x+2)(p-8)x\\
=&\,(24-2p)x+q.
\end{align*} = ( p − 8 ) x 2 + 8 x + q − ( x + 2 ) ( p − 8 ) x ( 24 − 2 p ) x + q . ( 24 − 2 p ) x + q − ( x + 2 ) ( 24 − 2 p ) = 4 p + q − 48. \begin{align*}
&\,(24-2p)x+q-(x+2)(24-2p)\\
=&\,4p+q-48.
\end{align*} = ( 24 − 2 p ) x + q − ( x + 2 ) ( 24 − 2 p ) 4 p + q − 48. Setting this remainder equal to − 5 -5 − 5
4 p + q = 43. (2) 4p+q=43. \tag{2} 4 p + q = 43. ( 2 ) Subtracting (1) from four times (2),
7 p = 70 p = 10. (shown) q = 43 − 4 ( 10 ) = 3. \begin{align*}
7p=&\,70\\
p=&\,10.\quad\text{(shown)}\\
q=&\,43-4(10)=3.
\end{align*} 7 p = p = q = 70 10. (shown) 43 − 4 ( 10 ) = 3.
(b)
解法一
思路
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承接 (a) 的 p = 10 p=10 p = 10 q = 3 q=3 q = 3
答题过程
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Using p = 10 p=10 p = 10 q = 3 q=3 q = 3
f ( x ) = 4 x 3 + 10 x 2 + 8 x + 3 f ′ ( x ) = 12 x 2 + 20 x + 8 = 4 ( 3 x + 2 ) ( x + 1 ) . \begin{align*}
\text{f}(x)=&\,4x^3+10x^2+8x+3\\
\text{f}'(x)=&\,12x^2+20x+8\\
=&\,4(3x+2)(x+1).
\end{align*} f ( x ) = f ′ ( x ) = = 4 x 3 + 10 x 2 + 8 x + 3 12 x 2 + 20 x + 8 4 ( 3 x + 2 ) ( x + 1 ) . The derivative is zero at x = − 1 x=-1 x = − 1 x = − 2 3 x=-\dfrac23 x = − 3 2 f ( x ) \text{f}(x) f ( x )
− 1 < x < − 2 3 . -1<x<-\frac23. − 1 < x < − 3 2 . 求根公式也给出相同的边界：
x = − 5 ± 25 − 24 6 = − 5 ± 1 6 = − 1 or − 2 3 . \begin{align*}
x=&\,\frac{-5\pm\sqrt{25-24}}6\\
=&\,\frac{-5\pm1}{6}\\
=&\,-1\text{ or }-\frac23.
\end{align*} x = = = 6 − 5 ± 25 − 24 6 − 5 ± 1 − 1 or − 3 2 . 