2011 H2 Mathematics Paper 2 Question 3

Functions

Answers

(i)

f−1(x)=ex−3−12{f^{-1}(x) = \frac{\mathrm{e}^{x-3}-1}{2}}
Df−1=(−∞,∞){D_{f^{-1}}=(-\infty, \infty)}

(ii)

(iii)

−0.4847,5.482{-0.4847, 5.482}

Full solutions

(i)

y=ln⁡(2x+1)+3ln⁡(2x+1)=y−32x+1=ey−3x=ey−3−12\begin{gather*} y = \ln(2x+1)+3 \\ \ln(2x+1) = y-3 \\ 2x+1 = \mathrm{e}^{y-3} \\ x = \frac{\mathrm{e}^{y-3}-1}{2} \\ \end{gather*}
f−1(x)=ex−3−12  ■f^{-1}(x) = \frac{\mathrm{e}^{x-3}-1}{2} \; \blacksquare
Df−1=Rf=(−∞,∞)  ■\begin{align*} D_{f^{-1}} &= R_f \\ &= (-\infty, \infty) \; \blacksquare \end{align*}

(ii)

(iii)

The intersections of the curves is also the intersection between y=f(x){y=f(x)} and y=x{y=x} since y=x{y=x} is the line of symmetry between the two curves
ln⁡(2x+1)+3=xln⁡(2x+1)=x−3  ■\begin{gather*} \ln(2x+1)+3 = x \\ \ln(2x+1)=x-3 \; \blacksquare \end{gather*}
Using a GC, x-{x\textrm{-}}coordinates of the points of intersection:
x=−0.4847 (4 sf)  ■or x=5.482 (4 sf)  ■\begin{align*} x &= -0.4847 \textrm{ (4 sf)} \; \blacksquare \\ \textrm{or } \quad x &= 5.482 \textrm{ (4 sf)} \; \blacksquare \\ \end{align*}