2010 H2 Mathematics Paper 2 Question 4

Functions

Answers

(i)

(ii)

Least value of k=0{k=0}

(iii)

fg(x)=(x−3)2(4−x)(x−2){fg(x)=\frac{(x-3)^2}{(4-x)(x-2)}}

(iv)

2<x<3{2<x<3} or 3<x<4{3<x<4}

(v)

Rfg=(−∞,−1)∪(0,∞)R_{fg} \allowbreak {= \left( -\infty, -1 \right) \cup \left( 0, \infty \right)}

Full solutions

(i)

(ii)

Since the graph of y=f(x){y=f(x)} has a maximum point at x=0,{x=0, } the least value of k{k} for which f−1{f^{-1}} exists is k=0  ■{k=0 \; \blacksquare}
This is because if the domain of f{f} is {x∈R,x≥0,x≠1},{\{x \in \mathbb{R}, x \geq 0, x \neq 1\},} then any horizontal line y=k,k∈R{y=k, k \in \mathbb{R}} would cut the graph of y=f(x){y=f(x)} at most once so f{f} is one-one and f−1{f^{-1}} exists ■{\blacksquare}

(iii)

fg(x)=f(1x−3)=1(1x−3)2−1=11−(x−3)2(x−3)2=(x−3)21−(x2−6x+9)=(x−3)2−x2+6x−8=(x−3)2(4−x)(x−2)  ■\begin{align*} fg(x) &= f \left( \frac{1}{x-3} \right) \\ &= \frac{1}{\left( \frac{1}{x-3} \right)^2 - 1} \\ &= \frac{1}{\frac{1-(x-3)^2}{(x-3)^2}} \\ &= \frac{(x-3)^2}{1-(x^2-6x+9)} \\ &= \frac{(x-3)^2}{-x^2+6x-8} \\ &= \frac{(x-3)^2}{(4-x)(x-2)} \; \blacksquare \\ \end{align*}

(iv)

(x−3)2(4−x)(x−2)>0 \frac{(x-3)^2}{(4-x)(x-2)} > 0
2<x<3   or   3<x<4  ■2 < x < 3 \; \textrm{ or } \; 3 < x < 4 \; \blacksquare

(v)

Rg=(−∞,−1)∪(−1,0)∪(0,1)∪(1,∞)R_g = \left( -\infty, -1 \right) \cup \left( -1, 0 \right) \cup \left( 0, 1 \right) \cup \left( 1, \infty \right)
Rfg=(−∞,−1)∪(0,∞)R_{fg} = \left( -\infty, -1 \right) \cup \left( 0, \infty \right)