2008 H2 Mathematics Paper 2 Question 4

Functions

Answers

(i)

(ii)

f−1(x)=4+x−1{f^{-1}(x) = 4 + \sqrt{x-1}}
Df−1=(1,∞){D_{f^{-1}} = (1, \infty) }

(iii)

(iv)

y=x{y=x}
x=92+1213{x=\frac{9}{2} + \frac{1}{2} \sqrt{13}}

Full solutions

(i)

(ii)

y=(x−4)2+1(x−4)2=y−1x−4=±y−1x=4±y−1\begin{gather*} y = (x-4)^2 + 1 \\ (x-4)^2 = y - 1 \\ x-4 = \pm \sqrt{y - 1} \\ x = 4 \pm \sqrt{y - 1} \\ \end{gather*}
Since x>4,{x>4, }
x=4+y−1x = 4 + \sqrt{y-1}
f−1(x)=4+x−1  ■f^{-1}(x) = 4 + \sqrt{x-1} \; \blacksquare
Df−1=Rf=(1,∞)  ■\begin{align*} D_{f^{-1}} &= R_f \\ &= (1, \infty) \; \blacksquare \end{align*}

(iii)

(iv)

To obtain the graph of y=f−1(x),{y=f^{-1}(x), } the graph of y=f(x){y=f(x)} must be reflected in the line y=x  ■{y=x \; \blacksquare}
The intersection of y=f(x){y=f(x)} and y=f−1(x){y=f^{-1}(x)} is also the intersection between y=f(x){y=f(x)} and y=x{y=x}
f(x)=f−1(x)f(x)=x(x−4)2+1=xx2−8x+16+1=xx2−9x+17=0\begin{gather*} f(x) = f^{-1}(x) \\ f(x) = x \\ (x-4)^2 + 1 = x \\ x^2 - 8x + 16 + 1 = x \\ x^2 - 9 x + 17 = 0 \end{gather*}
x=9±92−4(1)(17)2(1)=9±132\begin{align*} x &= \frac{9\pm\sqrt{9^2-4(1)(17)}}{2(1)} \\ &= \frac{9\pm\sqrt{13}}{2} \end{align*}
Since x>4,{x>4,}
x=92+1213  ■x=\frac{9}{2} + \frac{1}{2} \sqrt{13} \; \blacksquare