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CIE 9709 2024 November Paper 12 Q10

A Level / CIE / P1

CIE 9709 2024 Nov Paper 12 Paper · Question 10

题目

Problem

A function f\mathrm{f} with domain x>0x>0 is such that f(x)=8(2x3)1310x23\mathrm{f}'(x)=8(2x-3)^{\frac{1}{3}}-10x^{\frac{2}{3}}. It is given that the curve with equation y=f(x)y=\mathrm{f}(x) passes through the point (1,0)(1,0).

(a) Find the equation of the normal to the curve at the point (1,0)(1,0).

[3]

(b) Find f(x)\mathrm{f}(x).

[4]

It is given that the equation f(x)=0\mathrm{f}'(x)=0 can be expressed in the form

125x2128x+192=0.125x^2-128x+192=0.

(c) Determine, making your reasoning clear, whether f\mathrm{f} is an increasing function, a decreasing function or neither.

[3]
题目中文翻译

函数 f\mathrm{f} 的定义域为 x>0x>0,且 f(x)=8(2x3)1310x23\mathrm{f}'(x)=8(2x-3)^{\frac{1}{3}}-10x^{\frac{2}{3}}。已知方程为 y=f(x)y=\mathrm{f}(x) 的曲线经过点 (1,0)(1,0)

(a) 求曲线在点 (1,0)(1,0) 处的法线方程。

(b) 求 f(x)\mathrm{f}(x)

已知方程 f(x)=0\mathrm{f}'(x)=0 可以表示为

125x2128x+192=0.125x^2-128x+192=0.

(c) 判断 f\mathrm{f} 是增函数、减函数还是都不是,并清楚说明理由。

解答