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CIE 9709 2025 Nov Paper 13 Q7

A Level / CIE / P1

CIE 9709 2025 Nov Paper 13 Paper · Question 7

题目

Problem

The function gg is defined by g(x)=2ax3+12g(x)=\frac{2}{ax-3}+\frac{1}{2} for x>3ax>\frac{3}{a}, where aa is a positive constant.

Find g1(x)g^{-1}(x) and hence verify that if a=6a=6 then g1(x)g(x)g^{-1}(x)\equiv g(x).

[4]
题目中文翻译

函数 gg 定义为 g(x)=2ax3+12g(x)=\frac{2}{ax-3}+\frac{1}{2},其中 x>3ax>\frac{3}{a}aa 是正的常数。

g1(x)g^{-1}(x),并由此验证当 a=6a=6 时,g1(x)g(x)g^{-1}(x)\equiv g(x)

解答