A logarithmic involution · source reference

Printed question

(a) The curve $y=f(x)$ has horizontal asymptote $y=k$ and cuts the axes at $(a, 0)$ and $(0, b)$, where $a, b, k$ are non-zero constants. Given that $f^{-1}$ exists, state if possible the coordinates where the following curves cut the axes and their asymptotes: (i) $y=f(2x-3)$, (ii) $y=f^{-1}(x)$. [2]

(b) $g:x\mapsto\ln\left(\frac{e^x+5}{e^x-1}\right)$ for $x\in\mathbb R$, $x>0$.

(i) Find $g^{-1}(x)$ and state its domain. [3]

The function $h:x\mapsto1+\sqrt{9-(x-2)^2}$ for $x\in\mathbb R$, $-1\le x\le5$.

(ii) Find the exact solutions of $g^{2021}h(x)=\ln2$ in simplest form. [3]

[2021/NYJC/Prelim/I/5]

Printed working

(a)(i) $(a, 0)\mapsto((a+3)/2, 0)$; $(0, b)\mapsto(3/2, b)$. The source says “$((a+3)/2, 0)$ is the only point”; this is the only determinable axial intercept from the given data. Its $y$-intercept is $(0, f(-3))$ if $-3\in D_f$, and is not specified by $a, b, k$. Asymptote $y=k$.

(a)(ii) $(a, 0)\mapsto(0, a)$; $(0, b)\mapsto(b, 0)$; asymptote $x=k$.

(b)(i) $y=\ln((e^x+5)/(e^x-1))$ ⇒ $e^y=(e^x+5)/(e^x-1)$ ⇒ $(e^y-1)e^x=e^y+5$ ⇒ $e^x=(e^y+5)/(e^y-1)$ ⇒ $x=\ln((e^y+5)/(e^y-1))$.

Thus $g^{-1}(x)=\ln((e^x+5)/(e^x-1))$, $D_{g^{-1}}=R_g=(0, \infty)$.

(b)(ii) $g$ is self-inverse: $g^{-1}=g\Rightarrow gg(x)=x$, i. e. $g^2(x)=x$.

$$g^{2021}h(x)=\ln2\Rightarrow g^{1010\times2+1}h(x)=\ln2\Rightarrow gh(x)=\ln2.$$

$h(x)=g^{-1}(\ln2)=\ln7$.

$$1+\sqrt{9-(x-2)^2}=\ln7\Rightarrow9-(x-2)^2=(\ln7-1)^2,$$ $$(x-2)^2=9-(\ln7-1)^2=(2+\ln7)(4-\ln7),$$ $$x=2\pm\sqrt{(2+\ln7)(4-\ln7)}.$$

Both are in $[-1, 5]$.

Teacher notes

Transformation chain: $y=f(x)\xrightarrow{x\mapsto x-3}y=f(x-3)\xrightarrow{x\mapsto2x}y=f(2x-3)$. Effect: add3 to every original $x$ coordinate, then multiply the resulting value by 1/2.

If $(a, b)$ lies on $y=f(x)$, i. e. $b=f(a)$, then $(b, a)$ lies on $y=f^{-1}(x)$, i. e. $f^{-1}(b)=a$.

Iteration diagram: $g^{2021}h=g(g^2)\cdots(g^2)h$, 1010 identity pairs; after cancelling one there are1009, and continuing gives $gh$.

Answer key

(a)(i) $((a+3)/2, 0)$ “is the only point”; asymptote $y=k$, qualified above. (ii) $(a, 0)\to(0, a)$, $(0, b)\to(b, 0)$; asymptote $x=k$. (b)(i) inverse equals $g$, domain $(0, \infty)$. (ii) $x=2\pm\sqrt{(2+\ln7)(4-\ln7)}$.