Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 March Paper 12 Q7

A Level / CIE / P1

CIE 9709 2026 March Paper 12 Paper · Question 7

题目

Problem

A curve, CC, is such that

dydx=(x−3)2x.\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x-3)^2}{\sqrt{x}}.

The curve passes through the point (4,7)(4,7).

(a) Find yy in terms of xx.

(5)

(b) The equation of a different curve, DD, is also such that dydx=(x−3)2x\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x-3)^2}{\sqrt{x}}. This curve instead passes through the point (4,k)(4,k).

Describe, in terms of kk, the transformation that maps curve CC onto curve DD.

(2)
题目中文翻译

曲线 CC 满足

dydx=(x−3)2x,\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x-3)^2}{\sqrt{x}},

且经过点 (4,7)(4,7)。

(a) 求用 xx 表示的 yy 的表达式。

(5)

(b) 另一条曲线 DD 的方程也满足 dydx=(x−3)2x\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x-3)^2}{\sqrt{x}},但该曲线经过点 (4,k)(4,k)。

用 kk 描述将曲线 CC 映射到曲线 DD 的变换。

(2)
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

展开分子后逐项改写成 xx 的幂,再积分并用已知点求常数。

答题过程

展开

For x>0x>0, the derivative is x3/2−6x1/2+9x−1/2x^{3/2}-6x^{1/2}+9x^{-1/2}. Integrating gives

y=25x5/2−4x3/2+18x1/2+C.y=\frac25x^{5/2}-4x^{3/2}+18x^{1/2}+C.

At x=4x=4, this is 64/5−32+36+C=84/5+C64/5-32+36+C=84/5+C. Setting it equal to 77 gives C=−49/5C=-49/5.

Therefore the required equation is

y=25x5/2−4x3/2+18x1/2−495,x>0.y=\frac25x^{5/2}-4x^{3/2}+18x^{1/2}-\frac{49}{5},\qquad x>0.

(b)

解法一

思路

展开

两个曲线的导数相同,所以函数值相差一个常数。比较 x=4x=4 时的纵坐标即可。

答题过程

展开

The difference between the two functions is constant on x>0x>0 and equals k−7k-7 at x=4x=4. Thus DD is obtained from CC by the translation with vector

(0k−7).\begin{pmatrix}0\\k-7\end{pmatrix}.