2015 H2 Mathematics Paper 2 Question 3

Functions

Answers

(ai)
All horizontal lines y=k,{y=k, } k∈R{k \in \mathbb{R}} cuts the graph of y=f(x){y=f(x)} at most once. Hence f{f} is a one-one function and has an inverse
(aii)
f−1(x)=x−1x{f^{-1}(x) = \sqrt{\frac{x-1}{x}}}
Df−1=(−∞,0){D_{f^{-1}} = (-\infty, 0)}

(b)

(−∞,1−123]∪[1+123,∞){\left(-\infty, 1 - \frac{1}{2} \sqrt{3} \right]}\allowbreak {\cup \left[ 1 + \frac{1}{2} \sqrt{3}, \infty \right)}

Full solutions

(ai)
All horizontal lines y=k,{y=k, } k∈R{k \in \mathbb{R}} cuts the graph of y=f(x){y=f(x)} at most once. Hence f{f} is a one-one function and has an inverse ■{\blacksquare}
(aii)
y=11−x2y−x2y=1x2y=y−1x=±y−1y\begin{gather*} y = \frac{1}{1-x^2} \\ y - x^2y = 1 \\ x^2 y = y-1 \\ x = \pm \sqrt{\frac{y-1}{y}} \end{gather*}
Since x>1,{x>1,}
x=y−1yf−1(x)=x−1x  ■\begin{align*} x &= \sqrt{\frac{y-1}{y}} \\ f^{-1}(x) &= \sqrt{\frac{x-1}{x}} \; \blacksquare \end{align*}
Df−1=Rf=(−∞,0)  ■\begin{align*} D_{f^{-1}} &= R_f \\ &= (-\infty, 0) \; \blacksquare \end{align*}

(b)

y=2+x1−x2y−yx2=2+xyx2+x+2−y=0\begin{gather*} y = \frac{2+x}{1-x^2} \\ y - yx^2 = 2 + x \\ yx^2 + x + 2-y = 0 \end{gather*}
For the range of g,{g,}
b2−4ac≥012−4(y)(2−y)≥01−8y+4y2≥0\begin{gather*} b^2 - 4ac \geq 0 \\ 1^2 - 4(y)(2-y) \geq 0 \\ 1 - 8y + 4y^2 \geq 0 \\ \end{gather*}
Roots of 4y2−8y+1=0:{4y^2-8y+1=0:}
y=8±82−4(4)2(4)=8±488=8±438=1±123\begin{align*} y &= \frac{8\pm \sqrt{8^2-4(4)}}{2(4)} \\ &= \frac{8 \pm \sqrt{48}}{8} \\ &= \frac{8 \pm 4 \sqrt{3}}{8} \\ &= 1 \pm \frac{1}{2} \sqrt{3} \end{align*}
x≤1−123   or   x≥1+123x \leq 1 - \frac{1}{2} \sqrt{3} \; \textrm{ or } \; x \geq 1 + \frac{1}{2} \sqrt{3}
Rg=(−∞,1−123]∪[1+123,∞)  ■R_g = \left(-\infty, 1 - \frac{1}{2} \sqrt{3} \right] \cup \left[ 1 + \frac{1}{2} \sqrt{3}, \infty \right) \; \blacksquare