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CIE 9709 2024 June Paper 13 Q6

A Level / CIE / P1

CIE 9709 2024 June Paper 13 Paper · Question 6

题目

Problem

A curve passes through the point (45,−3)\big(\frac{4}{5}, -3\big) and is such that dydx=−20(5x−3)2\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-20}{(5x - 3)^2}.

(a) Find the equation of the curve.

[4]

(b) The curve is transformed by a stretch in the xx-direction with scale factor 12\frac{1}{2} followed by a translation of (210)\begin{pmatrix}2 \\ 10\end{pmatrix}.

Find the equation of the new curve.

[3]
题目中文翻译

一条曲线经过点 (45,−3)\big(\frac{4}{5}, -3\big),并且满足 dydx=−20(5x−3)2\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-20}{(5x - 3)^2}。

(a) 求这条曲线的方程。

[4]

(b) 这条曲线先经过 xx 方向、比例因子为 12\frac{1}{2} 的伸缩变换,然后再平移 (210)\begin{pmatrix}2 \\ 10\end{pmatrix}。

求新曲线的方程。

[3]

解答

(a)

解法一

思路

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把分母改成负指数积分,再用曲线经过的点确定积分常数。

答题过程

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On the branch containing x=45x=\frac45, integration gives

y=45x−3+C.y=\frac4{5x-3}+C.

At x=45x=\frac45, the fraction is 44, so −3=4+C-3=4+C and C=−7C=-7. The equation is y=45x−3−7y=\frac4{5x-3}-7.

(b)

解法一

思路

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横向伸缩到原来的一半时,先把式中的 xx 换成 2x2x。之后右移 22 时,把这一步得到的表达式中的 xx 换成 x−2x-2。

答题过程

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After the stretch the equation is y=4/(10x−3)−7y=4/(10x-3)-7. Translating by (210)\begin{pmatrix}2\\10\end{pmatrix} gives

y=410(x−2)−3−7+10=410x−23+3.y=\frac4{10(x-2)-3}-7+10=\frac4{10x-23}+3.