题目
Problem
The equation of a curve is y = f ( x ) y = f(x) y = f ( x ) , where f ( x ) = ( 2 x − 1 ) 3 x − 2 − 2 f(x) = (2x - 1)\sqrt{3x - 2} - 2 f ( x ) = ( 2 x − 1 ) 3 x − 2 − 2 . The following points lie on the curve. Non-exact values have been given correct to 5 5 5 decimal places.
A ( 2 , 4 ) , B ( 2.0001 , k ) , C ( 2.001 , 4.00625 ) , D ( 2.01 , 4.06261 ) , E ( 2.1 , 4.63566 ) , F ( 3 , 11.22876 ) A(2, 4),\ B(2.0001, k),\ C(2.001, 4.00625),\ D(2.01, 4.06261),\ E(2.1, 4.63566),\ F(3, 11.22876) A ( 2 , 4 ) , B ( 2.0001 , k ) , C ( 2.001 , 4.00625 ) , D ( 2.01 , 4.06261 ) , E ( 2.1 , 4.63566 ) , F ( 3 , 11.22876 )
(a) Find the value of k k k . Give your answer correct to 5 5 5 decimal places.
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The table shows the gradients of the chords A B AB A B , A C AC A C , A D AD A D and A F AF A F .
Chord A B AB A B A C AC A C A D AD A D A E AE A E A F AF A F Gradient of chord 6.2501 6.2501 6.2501 6.2511 6.2511 6.2511 6.2608 6.2608 6.2608 7.2288 7.2288 7.2288
(b) Find the gradient of the chord A E AE A E . Give your answer correct to 4 4 4 decimal places.
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(c) Deduce the value of f ′ ( 2 ) f'(2) f ′ ( 2 ) using the values in the table.
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题目中文翻译
一条曲线的方程为 y = f ( x ) y = f(x) y = f ( x ) ,其中 f ( x ) = ( 2 x − 1 ) 3 x − 2 − 2 f(x) = (2x - 1)\sqrt{3x - 2} - 2 f ( x ) = ( 2 x − 1 ) 3 x − 2 − 2 。以下点在该曲线上。非精确值均已给出到 5 5 5 位小数。
A ( 2 , 4 ) , B ( 2.0001 , k ) , C ( 2.001 , 4.00625 ) , D ( 2.01 , 4.06261 ) , E ( 2.1 , 4.63566 ) , F ( 3 , 11.22876 ) A(2, 4),\ B(2.0001, k),\ C(2.001, 4.00625),\ D(2.01, 4.06261),\ E(2.1, 4.63566),\ F(3, 11.22876) A ( 2 , 4 ) , B ( 2.0001 , k ) , C ( 2.001 , 4.00625 ) , D ( 2.01 , 4.06261 ) , E ( 2.1 , 4.63566 ) , F ( 3 , 11.22876 )
(a) 求 k k k 的值。答案精确到 5 5 5 位小数。
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表格显示弦 A B AB A B 、A C AC A C 、A D AD A D 和 A F AF A F 的斜率。
弦 A B AB A B A C AC A C A D AD A D A E AE A E A F AF A F 弦的斜率 6.2501 6.2501 6.2501 6.2511 6.2511 6.2511 6.2608 6.2608 6.2608 7.2288 7.2288 7.2288
(b) 求弦 A E AE A E 的斜率。答案精确到 4 4 4 位小数。
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(c) 利用表中的数值推断 f ′ ( 2 ) f'(2) f ′ ( 2 ) 的值。
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解答
(a)
解法一
思路
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把 B B B 的横坐标代入函数,并在最后保留五位小数。
答题过程
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Substitution gives
k = ( 2 ( 2.0001 ) − 1 ) 3 ( 2.0001 ) − 2 − 2 = 4.00063 \begin{align*}
k=&\,(2(2.0001)-1)\sqrt{3(2.0001)-2}-2\\
=&\,4.00063
\end{align*} k = = ( 2 ( 2.0001 ) − 1 ) 3 ( 2.0001 ) − 2 − 2 4.00063
to five decimal places.
(b)
解法一
思路
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用纵坐标差除以横坐标差。
答题过程
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The gradient of A E AE A E is
4.63566 − 4 2.1 − 2 = 6.3566 \frac{4.63566-4}{2.1-2}=6.3566 2.1 − 2 4.63566 − 4 = 6.3566
to four decimal places.
(c)
解法一
思路
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第二个点越靠近 A A A ,割线斜率越接近切线斜率。从表中离 A A A 最近的几项看极限。
答题过程
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As the second point approaches A A A , the gradients 6.2608 6.2608 6.2608 , 6.2511 6.2511 6.2511 and 6.2501 6.2501 6.2501 tend towards 6.25 6.25 6.25 . This suggests f ′ ( 2 ) = 6.25 f'(2)=6.25 f ′ ( 2 ) = 6.25 .