2018 H2 Mathematics Paper 1 Question 5

Functions

Answers

b=−1{b=-1}
f−1(x)=x+ax−1{f^{-1}(x)=\frac{x+a}{x-1}}

Full solutions

ff(x)=f(x+ax+b)=x+ax+b+ax+ax+b+b=x+a+a(x+b)x+a+b(x+b)=(a+1)x+(a+ab)(b+1)x+(a+b2)\begin{align*} ff(x) & = f \left( \frac{x+a}{x+b} \right) \\ &= \frac{\frac{x+a}{x+b}+a}{\frac{x+a}{x+b}+b} \\ &= \frac{x+a+a(x+b)}{x+a+b(x+b)} \\ &= \frac{(a+1)x + (a+ab)}{(b+1)x + (a+b^2)} \\ \end{align*}
ff(x)=g(x)(a+1)x+(a+ab)(b+1)x+(a+b2)=x(a+1)x+(a+ab)=(b+1)x2+(a+b2)x\begin{gather*} ff(x) = g(x) \\ \frac{(a+1)x + (a+ab)}{(b+1)x + (a+b^2)} = x \\ (a+1)x + (a+ab) = (b+1)x^2 + (a+b^2)x \end{gather*}
Comparing coefficients of x2,{x^2,}
b+1=0b=−1  ■\begin{align*} b+1 &= 0 \\ b &= -1 \; \blacksquare \end{align*}
ff(x)=g(x)ff(x)=xf(x)=f−1(x)f−1(x)=f(x)=x+ax+b=x+ax−1  ■\begin{align*} ff(x) &= g(x) \\ ff(x) &= x \\ f(x) &= f^{-1}(x) \\ f^{-1}(x) &= f(x) \\ &= \frac{x+a}{x+b} \\ &= \frac{x+a}{x-1} \; \blacksquare \end{align*}