Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2025 March Paper 12 Q2

A Level / CIE / P1

CIE 9709 2025 March Paper 12 Paper · Question 2

题目

Problem

The diagram shows the curve with equation y=2x2−5x+3y = 2x^2 - \frac{5}{x} + 3. The curve crosses the xx-axis at the point P(1,0)P(1, 0) and MM is a minimum point.

(a) Find the gradient of the curve at PP.

[2]

(b) Find the coordinates of MM. Give each coordinate correct to 3 significant figures.

[3]
题目中文翻译

图中显示了方程为 y=2x2−5x+3y = 2x^2 - \frac{5}{x} + 3 的曲线。该曲线在点 P(1,0)P(1, 0) 处与 xx 轴相交,且 MM 是一个极小点。

(a) 求曲线在 PP 点处的斜率。

[2]

(b) 求 MM 的坐标。每个坐标都精确到 3 位有效数字。

[3]

解答

(a)

解法一

思路

展开

把分式改写成负指数再求导。点 PP 的横坐标为 11。

答题过程

展开

For x≠0x\ne0, y′=4x+5/x2y'=4x+5/x^2. Hence the gradient at PP is 4+5=94+5=9.

(b)

解法一

思路

展开

驻点处导数为零。先求精确的横坐标,再用驻点关系简化纵坐标并舍入。

答题过程

展开

Setting 4x+5/x2=04x+5/x^2=0 gives x3=−5/4x^3=-5/4, so the real root is x=−5/43x=-\sqrt[3]{5/4}. At this root, 5=−4x35=-4x^3, so

y=2x2−5x+3=6x2+3.y=2x^2-\frac5x+3=6x^2+3.

The second derivative 4−10/x34-10/x^3 is positive at this negative root, confirming a minimum. To three significant figures, M=(−1.08,9.96)M=(-1.08,9.96).