2013 H2 Mathematics Paper 2 Question 1

Functions

Answers

(i)

The composite function fg{fg} does not exist since Rg⊈Df{R_g \not \subseteq D_f}

(ii)

gf(x)=−3−3x1−x{gf(x)=\frac{-3-3x}{1-x}}
(gf)−1(5)=4{(gf)^{-1}(5)=4}

Full solutions

(i)

Rg=(−∞,∞)Df=(−∞,1)∪(1,∞)\begin{align*} R_g &= (-\infty, \infty) \\ D_f &= (-\infty, 1) \cup (1, \infty) \\ \end{align*}
Hence the composite function fg{fg} does not exist because Rg⊈Df  ■{R_g \not \subseteq D_f \; \blacksquare}

(ii)

gf(x)=g(2+x1−x)=1−2(2+x1−x)=1−4+2x1−x=1−x−4−2x1−x=−3−3x1−x  ■\begin{align*} gf(x) &= g\left( \frac{2+x}{1-x} \right) \\ &= 1 - 2\left( \frac{2+x}{1-x} \right) \\ &= 1 - \frac{4+2x}{1-x} \\ &= \frac{1-x-4-2x}{1-x}\\ &= \frac{-3-3x}{1-x} \; \blacksquare \end{align*}
Let x=(gf)−1(5)gf(x)=5−3−3x1−x=5−3−3x=5−5x2x=8x=4\begin{align*} \textrm{Let } \qquad x &= (gf)^{-1}(5) \\ gf(x) &= 5 \\ \frac{-3-3x}{1-x} &= 5 \\ -3-3x &= 5 - 5x \\ 2x &= 8 \\ x = 4 \end{align*}
(gf)−1(5)=4  ■(gf)^{-1}(5)=4 \; \blacksquare