2019 H2 Mathematics Paper 1 Question 5

Functions

Answers

(i)

f−1(x)=12ln⁡(x+4){f^{-1}(x) = \frac{1}{2}\ln(x+4)}
Df−1=(−4,∞){D_{f^{-1}} = (-4, \infty)}

(ii)

x=ln⁡3−2{x=\ln 3 - 2}

Full solutions

(i)

y=e2x−4e2x=y+42x=ln⁡(y+4)x=12ln⁡(y+4)\begin{gather*} y = \mathrm{e}^{2x}- 4 \\ \mathrm{e}^{2x} = y + 4 \\ 2x = \ln(y+4) \\ x = \frac{1}{2} \ln (y+4) \end{gather*}
f−1(x)=12ln⁡(x+4)  ■f^{-1}(x) = \frac{1}{2}\ln(x+4) \; \blacksquare
Df−1=Rf=(−4,∞)  ■\begin{align*} D_{f^{-1}} &= R_f \\ &= (-4, \infty) \; \blacksquare \end{align*}

(ii)

fg(x)=5f(x+2)=5x+2=f−1(5)x+2=12ln⁡(5+4)x=12ln⁡9−2=ln⁡3−2  ■\begin{align*} fg(x) &= 5 \\ f(x+2) &= 5 \\ x+2 &= f^{-1}(5) \\ x + 2 &= \frac{1}{2} \ln(5+4) \\ x &= \frac{1}{2} \ln9 - 2 \\ &= \ln 3 - 2 \; \blacksquare \end{align*}