题目
Problem
The diagram shows the curve y = ( x + 5 ) 3 − 2 x y=(x+5)\sqrt{3-2x} y = ( x + 5 ) 3 − 2 x M M M
(a) Find the exact coordinates of M M M
(5)
(b) Using the substitution u = 3 − 2 x u=3-2x u = 3 − 2 x x x x a 13 a\sqrt{13} a 13 a a a
(5)
题目中文翻译
图中显示了曲线 y = ( x + 5 ) 3 − 2 x y=(x+5)\sqrt{3-2x} y = ( x + 5 ) 3 − 2 x M M M
(a) 求 M M M
(5)
(b) 使用代换 u = 3 − 2 x u=3-2x u = 3 − 2 x x x x a 13 a\sqrt{13} a 13 a a a
(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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用乘积法则求导，将根式分母通分后，分子为一次式。代回原函数求精确纵坐标。
答题过程
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For x < 3 / 2 x<3/2 x < 3/2
d y d x = 3 − 2 x − x + 5 3 − 2 x = 3 − 2 x − ( x + 5 ) 3 − 2 x = − 3 x − 2 3 − 2 x . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\sqrt{3-2x}
-\frac{x+5}{\sqrt{3-2x}}\\[4mm]
=&\,\frac{3-2x-(x+5)}{\sqrt{3-2x}}\\[4mm]
=&\,\frac{-3x-2}{\sqrt{3-2x}}.
\end{align*} d x d y = = = 3 − 2 x − 3 − 2 x x + 5 3 − 2 x 3 − 2 x − ( x + 5 ) 3 − 2 x − 3 x − 2 . At a stationary point, − 3 x − 2 = 0 -3x-2=0 − 3 x − 2 = 0 x = − 2 / 3 x=-2/3 x = − 2/3
The derivative changes from positive to negative there. Substituting into the curve,
y = 13 3 13 3 = 13 39 9 . y=\frac{13}{3}\sqrt{\frac{13}{3}}
=\frac{13\sqrt{39}}9. y = 3 13 3 13 = 9 13 39 . Thus
M = ( − 2 3 , 13 39 9 ) . \boxed{M=\bigg(-\frac23,\frac{13\sqrt{39}}9\bigg)}. M = ( − 3 2 , 9 13 39 ) .
解法二
思路
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也可先平方消去根式，再对 y 2 y^2 y 2 y = 0 y=0 y = 0
答题过程
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Squaring the curve equation,
y 2 = ( x + 5 ) 2 ( 3 − 2 x ) = − 2 x 3 − 17 x 2 − 20 x + 75. y^2=(x+5)^2(3-2x)=-2x^3-17x^2-20x+75. y 2 = ( x + 5 ) 2 ( 3 − 2 x ) = − 2 x 3 − 17 x 2 − 20 x + 75. Differentiating,
2 y d y d x = − 6 x 2 − 34 x − 20 = − 2 ( 3 x + 2 ) ( x + 5 ) . 2y\frac{\mathrm{d}y}{\mathrm{d}x}
=-6x^2-34x-20=-2(3x+2)(x+5). 2 y d x d y = − 6 x 2 − 34 x − 20 = − 2 ( 3 x + 2 ) ( x + 5 ) . A stationary point must therefore have x = − 2 / 3 x=-2/3 x = − 2/3 x = − 5 x=-5 x = − 5
At x = − 5 x=-5 x = − 5 y = 0 y=0 y = 0 13 ≠ 0 \sqrt{13}\ne0 13 = 0 x = − 2 / 3 x=-2/3 x = − 2/3 y ≠ 0 y\ne0 y = 0
y = 13 3 13 3 = 13 39 9 . y=\frac{13}{3}\sqrt{\frac{13}{3}}=\frac{13\sqrt{39}}9. y = 3 13 3 13 = 9 13 39 . Thus the required maximum is
M = ( − 2 3 , 13 39 9 ) . \boxed{M=\bigg(-\frac23,\frac{13\sqrt{39}}9\bigg)}. M = ( − 3 2 , 9 13 39 ) .
(b)
解法一
思路
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先找阴影的两个横轴交点，再使用指定的 u = 3 − 2 x u=3-2x u = 3 − 2 x
答题过程
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The intercepts are x = − 5 x=-5 x = − 5 x = 3 / 2 x=3/2 x = 3/2
A = ∫ − 5 3 / 2 ( x + 5 ) 3 − 2 x d x . A=\int_{-5}^{3/2}(x+5)\sqrt{3-2x}\,\mathrm{d}x. A = ∫ − 5 3/2 ( x + 5 ) 3 − 2 x d x . Put u = 3 − 2 x u=3-2x u = 3 − 2 x
x = 3 − u 2 , x + 5 = 13 − u 2 , d x = − 1 2 d u . \begin{align*}
x=&\,\frac{3-u}{2},\\[4mm]
x+5=&\,\frac{13-u}{2},\\[4mm]
\mathrm{d}x=&\,-\frac12\,\mathrm{d}u.
\end{align*} x = x + 5 = d x = 2 3 − u , 2 13 − u , − 2 1 d u . The limits become u = 13 u=13 u = 13 u = 0 u=0 u = 0
A = − 1 4 ∫ 13 0 ( 13 − u ) u 1 / 2 d u = 1 4 ∫ 0 13 ( 13 u 1 / 2 − u 3 / 2 ) d u = 1 4 [ 26 3 u 3 / 2 − 2 5 u 5 / 2 ] 0 13 = 1 4 ( 338 3 − 338 5 ) 13 = 169 15 13 . \begin{align*}
A
=&\,-\frac14\int_{13}^{0}(13-u)u^{1/2}\,\mathrm{d}u\\[4mm]
=&\,\frac14\int_0^{13}(13u^{1/2}-u^{3/2})\,\mathrm{d}u\\[4mm]
=&\,\frac14\bigg[\frac{26}{3}u^{3/2}-\frac25u^{5/2}\bigg]_0^{13}\\[4mm]
=&\,\frac14\bigg(\frac{338}{3}-\frac{338}{5}\bigg)\sqrt{13}\\[4mm]
=&\,\boxed{\frac{169}{15}\sqrt{13}}.
\end{align*} A = = = = = − 4 1 ∫ 13 0 ( 13 − u ) u 1/2 d u 4 1 ∫ 0 13 ( 13 u 1/2 − u 3/2 ) d u 4 1 [ 3 26 u 3/2 − 5 2 u 5/2 ] 0 13 4 1 ( 3 338 − 5 338 ) 13 15 169 13 . 
