A quarter ellipse becomes a self-inverse circle · source reference

Printed question

Sketch the curve $y^2+ax^2=5$ for $x\ge0$, $y\ge0$, where $a$ is positive. [1]

$f:x\mapsto\sqrt{5-ax^2}$, $x\in\mathbb R$, $0\le x\le\sqrt{5/a}$; $g:x\mapsto1+e^{-x}$, $x\in\mathbb R$, $x\ge0$.

Show that $f^{-1}$ exists and define $f^{-1}$ in a similar form. [4]

Given $f^2(x)=x$ for all $x\in\mathbb R$, $0\le x\le\sqrt{5/a}$,

(i) show $a=1$ without evaluating $f^2(x)$, [2]

(ii) using (i), show $fg$ exists and find its corresponding range. [3]

[2011/ACJC/Prelim/I/9]

Printed working

$y^2+ax^2=5\Rightarrow\frac{y^2}{(\sqrt5)^2}+\frac{x^2}{(\sqrt{5/a})^2}=1$. Native first-quadrant ellipse has intercepts $\sqrt5$ and $\sqrt{5/a}$.

Any horizontal line $y=k$, $k\in\mathbb R$, cuts $f$ at most once ⇒ $f$ one-one ⇒ $f^{-1}$ exists (shown).

Let $y=\sqrt{5-ax^2}$, $0\le x\le\sqrt{5/a}$. Then $y^2=5-ax^2$ ⇒ $x^2=(5-y^2)/a$ ⇒ $x=\sqrt{(5-y^2)/a}$ since $x\ge0$.

$f^{-1}:x\mapsto\sqrt{(5-x^2)/a}$, $x\in\mathbb R$, $0\le x\le\sqrt5$.

(i) $f^2(x)=x\Rightarrow f(x)=f^{-1}(x)$ ⇒ $\sqrt{5-ax^2}=\sqrt{5/a-x^2/a}$ ⇒ $5=5/a$ and $a=1/a$ ⇒ $a=1$ (shown).

Alternative method: $f=f^{-1}$ ⇒ graph symmetric about $y=x$ ⇒ $f$ is part of a circle centered $(0, 0)$ ⇒ $\sqrt{5/a}=\sqrt5$ ⇒ $a=1$.

(ii) $R_g=(1, 2]$, $D_f=[0, \sqrt5]$. Since $R_g\subseteq D_f$, $fg$ exists. Domain/range flow: $D_g=[0, \infty)\xrightarrow{g}R_g=(1, 2]\xrightarrow{f}R_{fg}=[1, 2)$. Thus $R_{fg}=[1, 2)$.

Teacher notes

Why the self-inverse inference works: $f^2(x)=x\Rightarrow f^{-1}f^2(x)=f^{-1}(x)\Rightarrow f^{-1}f(f(x))=f^{-1}(x)\Rightarrow f(x)=f^{-1}(x)$ since $f^{-1}f(x)=x$.

The graph of $g$ starts at closed $(0, 2)$, tends to the excluded asymptote $y=1$; the highlighted ellipse arc maps open input 1 to open output 2 and closed input 2 to closed output 1.

Answer key

Inverse as above, $0\le x\le\sqrt5$; native ellipse sketch. (ii) $R_{fg}=[1, 2)$.

Native diagrams