A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<ol class="abc-list"><li data-list-item-id="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0" data-list-item-path="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0"><div class="list-item-paragraph">The function $\mathrm{f}$ is defined as follows.</div><div class="list-item-paragraph"><p style="text-align: center;">$\mathrm{f}:x\mapsto a(x-1)^2+b,\qquad x\in\mathbb{R},\ x&gt;0.$</p></div><ol class="i-list"><li data-list-item-id="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0-sub-0" data-list-item-path="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0/h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0-sub-0"><div class="list-item-paragraph">Given that&nbsp;$a&gt;0$ and $0&lt;b&lt;1$, show that&nbsp;$\mathrm{f}^2$ exists.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0-sub-1" data-list-item-path="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0/h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-0-sub-1"><div class="list-item-paragraph">Hence determine the range of $\mathrm{f}^2$, leaving your answer in terms of $b$.<span class="inline-marks">[1]</span></div></li></ol></li><li data-list-item-id="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1" data-list-item-path="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1"><div class="list-item-paragraph">The function $\mathrm{g}$ is defined as follows.</div><div class="list-item-paragraph"><p style="text-align: center;">$\mathrm{g}:x\mapsto \frac{x+4}{2x-1},\qquad x\in\mathbb{R},\ x \ne \frac{1}{2}.$</p></div><ol class="i-list"><li data-list-item-id="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1-sub-0" data-list-item-path="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1/h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1-sub-0"><div class="list-item-paragraph">Find $\mathrm{g}^{-1}(x)$ and state the domain of $\mathrm{g}^{-1}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1-sub-1" data-list-item-path="h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1/h2-2025-sajc-promo-q5-sajc-q5-item-block-0-item-1-sub-1"><div class="list-item-paragraph">Find the values of $c$ such that&nbsp;$\mathrm{g}^{2024}(c)=\mathrm{g}^{-1}(c)$. Explain your answers clearly.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Existence of Composite Functions

Range of a Composite Function

Iterates $f^n$ for Large $n$

Inverse Functions

Self-inverse Functions: $ \mathrm{f} = \mathrm{f}^{-1}$

2025 SAJC Promo Q5

Functions

<p>A function $\mathrm{f}$ is said to be self-inverse if&nbsp;$\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ for all $x$ in the domain of $\mathrm{f}$.</p>
<p>It is given that&nbsp;$\mathrm{g}$ is a self-inverse function and is defined by $\mathrm{g} : x \mapsto \frac{x+a}{3x+b}$, for&nbsp;$x \in \mathbb{R},\, x &gt; -\frac{b}{3}$,</p>
<p>where&nbsp;$a$ and $b$ are constants and $\mathrm{g}(1) = 5$.</p>
<ol class="i-list"><li data-list-item-id="imported-item-block-3-item-0" data-list-item-path="imported-item-block-3-item-0"><div class="list-item-paragraph">Show that&nbsp;$a = 9$ and $b = -1$.</div></li><li data-list-item-id="imported-item-block-3-item-1" data-list-item-path="imported-item-block-3-item-1"><div class="list-item-paragraph">Find $\mathrm{g}^{2021}(1)$.</div></li></ol>
<p>The function $\mathrm{h}$ is defined by</p>
<p style="text-align: center">\[\mathrm{h} : x \mapsto \left\vert{} 1-x \right\vert{}(x+5),\text{ for}\hspace{0.5em} x \in \mathbb{R},\, x \ge 2.\]</p>
<ol class="i-list" start="3"><li data-list-item-id="imported-item-block-6-item-0" data-list-item-path="imported-item-block-6-item-0"><div class="list-item-paragraph">Given that&nbsp;$\mathrm{h}^{-1}$ exists, find $\mathrm{h}^{-1}$ in similar form.</div></li><li data-list-item-id="imported-item-block-6-item-1" data-list-item-path="imported-item-block-6-item-1"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{gh}$ exists and find the exact range of $\mathrm{gh}$.</div></li></ol>

A function $\mathrm{f}$ is said to be self-inverse if $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ for all $x$ in the domain of $\mathrm{f}$.

It is given that $\mathrm{g}$ is a self-inverse function and is defined by $\mathrm{g} : x \mapsto \frac{x+a}{3x+b}$, for $x \in \mathbb{R},\, x > -\frac{b}{3}$,

where $a$ and $b$ are constants and $\mathrm{g}(1) = 5$.

The function $\mathrm{h}$ is defined by

\[\mathrm{h} : x \mapsto \left\vert{} 1-x \right\vert{}(x+5),\text{ for } x \in \mathbb{R},\, x \ge 2.\]

Given that $\mathrm{h}^{-1}$ exists, find $\mathrm{h}^{-1}$ in similar form.

Show that $\mathrm{gh}$ exists and find the exact range of $\mathrm{gh}$.

Modulus Functions

<p>A function $\mathrm{h}$ is self-inverse if&nbsp;$\mathrm{h}(x) = \mathrm{h}^{-1}(x)$ for all $x$ in the domain of $\mathrm{h}$.</p>
<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \[\mathrm{f}(x) = 2 + \frac{1}{x-2}, \quad \text{for}\hspace{0.5em} x \in \mathbb{R}, x &gt; 2\] \[\mathrm{g}(x) = \begin{cases} 3x - 1 & \text{for}\hspace{0.5em} x \le 1, \\ (x - 1)^2 + 2 & \text{for}\hspace{0.5em} x &gt; 1. \end{cases}\]</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Sketch the graph of $y = \mathrm{f}(x)$. With the aid of your graph, explain why $\mathrm{f}$ has an inverse.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}$ is self-inverse and find $\mathrm{f}^2(x)$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Hence, or otherwise, evaluate $\mathrm{f}^{2025}(4)$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Find an expression for&nbsp;$\mathrm{g}\mathrm{f}$ and state its domain.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-e" data-list-item-path="item-e"><div class="list-item-paragraph">Find the range of $\mathrm{g}\mathrm{f}$.<span class="inline-marks">[2]</span></div></li></ol>

A function $\mathrm{h}$ is self-inverse if $\mathrm{h}(x) = \mathrm{h}^{-1}(x)$ for all $x$ in the domain of $\mathrm{h}$.

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \[\mathrm{f}(x) = 2 + \frac{1}{x-2}, \quad \text{for } x \in \mathbb{R}, x > 2\] \[\mathrm{g}(x) = \begin{cases} 3x - 1 & \text{for } x \le 1, \\ (x - 1)^2 + 2 & \text{for } x > 1. \end{cases}\]

Composite Functions

Horizontal Line Test

2024 CJC Promo Q11

<p></p>
<ol class="abc-list"><li><div class="list-item-paragraph">A function $\mathrm{g}$ is said to be self-inverse if&nbsp;$\mathrm{g}(x) = \mathrm{g}^{-1}(x)$ for all $x$ in the domain of $\mathrm{g}$.<br>The function $\mathrm{g}$ is defined by\[\mathrm{g} : x \mapsto \frac{\alpha x + \beta}{x - 2}, \quad x \in \mathbb{R}, x \neq 2,\]where&nbsp;$\alpha$ and $\beta$ are constants.</div><ol class="i-list"><li><div class="list-item-paragraph">Find the values of $\alpha$ and $\beta$ for&nbsp;$\mathrm{g}$ to be a self-inverse function.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;$\mathrm{g}$ is self-inverse, find the value of $x$ where&nbsp;$\mathrm{g}^{2023}(x) = -\frac{\beta}{2}$.<span class="inline-marks">[2]</span></div></li></ol></li><li><div class="list-item-paragraph">The functions $\mathrm{f}$ and $\mathrm{h}$ are defined by \[\mathrm{f} : x \mapsto ex + 1, \quad x &gt; 0.\]\[\mathrm{h} : x \mapsto 3x^3 - 9x^2 - 12x + 36, \quad x \geq -2.\]</div><ol class="i-list"><li><div class="list-item-paragraph">Are $\mathrm{f}^2(x)$ and $[\mathrm{f}(x)]^2$ equivalent? Explain your answer.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">By considering $\mathrm{f}^2(x)$ and $\mathrm{f}^3(x)$, show that \[\mathrm{f}^n(x) = \frac{e^{n+1}x + e^n(1 - x) - 1}{e - 1}.\]<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Explain why the composite function $\mathrm{hf}$ exists.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the range of $\mathrm{hf}$.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

2023 HCI C1 BT Q12

<p>The function $\mathrm{f}$ is defined by\[\mathrm{f} : x \mapsto \frac{ax - 3}{5x - 4}, \text{ for}\hspace{0.5em} x \in \mathbb{R}, x \neq \frac{4}{5}, \text{ and}\hspace{0.5em} a \in \mathbb{R}\]</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find the range of $\mathrm{f}$ in terms of $a$.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>The function $\mathrm{f}$ is such that&nbsp;$\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ for all $x$ in the domain of $\mathrm{f}$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Find the value of $a$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find $\mathrm{f}^{2025}(2)$.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>Another function $\mathrm{g}$ is defined by\[\mathrm{g} : x \mapsto \left(x - \frac{1}{5}\right)^2 + \frac{2}{5}, \text{ for}\hspace{0.5em} x \in \mathbb{R}, 0 &lt; x &lt; \frac{3}{5}\]</p>
<ol class="abc-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Find the range of $\mathrm{fg}$.</span><span class="inline-marks">[2]</span></div></li></ol>

The function $\mathrm{f}$ is defined by\[\mathrm{f} : x \mapsto \frac{ax - 3}{5x - 4}, \text{ for } x \in \mathbb{R}, x \neq \frac{4}{5}, \text{ and } a \in \mathbb{R}\]

Find the range of $\mathrm{f}$ in terms of $a$.

The function $\mathrm{f}$ is such that $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ for all $x$ in the domain of $\mathrm{f}$.

Another function $\mathrm{g}$ is defined by\[\mathrm{g} : x \mapsto \left(x - \frac{1}{5}\right)^2 + \frac{2}{5}, \text{ for } x \in \mathbb{R}, 0 < x < \frac{3}{5}\]

Range of a Function

2025 HCI P1 Q8

<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined as follows:</p>
<p style="text-align: center">$\mathrm{f}: x \mapsto \dfrac{5 - x}{1 - x}, x \in \mathbb{R}, x \ne 1$</p>
<ol class="i-list"><li><div class="list-item-paragraph">Explain why $\mathrm{f}$ has an inverse, $\mathrm{f}^{-1}$, and show that&nbsp;$\mathrm{f}^{-1} = \mathrm{f}$.<br><span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Evaluate $\mathrm{f}^{51}(4)$.<span class="inline-marks">[2]</span></div></li></ol>

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined as follows:

$\mathrm{f}: x \mapsto \dfrac{5 - x}{1 - x}, x \in \mathbb{R}, x \ne 1$

One-to-One Functions and Inverse Existence

2009 HCI P1 Q9

<p>A function $g$ is said to be self-inverse if&nbsp;${{\mathrm{g}}^{-1}}\left( x \right)=\mathrm{g}\left( x \right)$ for all $x$ in the domain of $g$.</p>
<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">$\mathrm{f}\left( x \right)=\frac{ax+b}{x-2}$, $x\in \mathbb{R}$, $x\ne 2$,</p>
<p>where&nbsp;$a$ and $b$ are real constants.</p>
<p>It is given that&nbsp;$\mathrm{f}$ is self-inverse.</p>
<ol class="i-list"><li data-list-item-id="item-1760754661230-wjvgxti4y" data-list-item-path="item-1760754661230-wjvgxti4y"><div class="list-item-paragraph">Find $a$ and the value(s) that&nbsp;$b$ cannot take.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760754662182-gu7tautxi" data-list-item-path="item-1760754662182-gu7tautxi"><div class="list-item-paragraph">Explain why the composite function ${{\mathrm{f}}^{n}}$ always exists for all integers $n &gt; 1$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760754663285-l3fjvzny4" data-list-item-path="item-1760754663285-l3fjvzny4"><div class="list-item-paragraph">Hence find ${{\mathrm{f}}^{2025}}\left( 1 \right)$ in terms of $b$.<span class="inline-marks">[2]</span></div></li></ol>
<p>The function $\mathrm{h}$ is defined for all $x\in \mathbb{R}$ by</p>
<p style="text-align: center">$\mathrm{h}\left( x \right)=\left\{ \begin{matrix}
   3x+2\,\,\,\,\,\,\,\,x\ge 0,  \\
   2{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,x&lt;0.  \\
\end{matrix} \right.$</p>
<ol class="i-list" start="4"><li data-list-item-id="item-1760754729585-ngk7kpfvc" data-list-item-path="item-1760754729585-ngk7kpfvc"><div class="list-item-paragraph">Sketch the graph of $\mathrm{h}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760754730545-uain7zreu" data-list-item-path="item-1760754730545-uain7zreu"><div class="list-item-paragraph">Using the value of $a$ found in <strong>(i)</strong> and given that&nbsp;$b=1$, find $\mathrm{hf}\left( \frac{1}{2} \right)$.<span class="inline-marks">[1]</span></div></li></ol>

A function $g$ is said to be self-inverse if ${{\mathrm{g}}^{-1}}\left( x \right)=\mathrm{g}\left( x \right)$ for all $x$ in the domain of $g$.

$\mathrm{f}\left( x \right)=\frac{ax+b}{x-2}$, $x\in \mathbb{R}$, $x\ne 2$,

It is given that $\mathrm{f}$ is self-inverse.

The function $\mathrm{h}$ is defined for all $x\in \mathbb{R}$ by

$\mathrm{h}\left( x \right)=\left\{ \begin{matrix}
   3x+2\,\,\,\,\,\,\,\,x\ge 0,  \\
   2{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,x<0.  \\
\end{matrix} \right.$

Piecewise-Defined Functions

2024 SAJC P2 Q3

<ol class="abc-list"><li><div class="list-item-paragraph">Two functions are defined on the domain $0\le x\le \pi $ by $\mathrm{f}:x\mapsto \sin x$ and $\mathrm{g}:x\mapsto \cos x$.</div><ol class="i-list"><li><div class="list-item-paragraph">Explain why one of these functions has an inverse while the other does not.</div><p style="text-align: right;">[2]</p><div class="list-item-paragraph">If the domain is restricted to $0\le x\le k$, the inverse exists for both functions.</div></li><li><div class="list-item-paragraph">State the greatest possible value of $k$ in exact form.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Evaluate ${{\mathrm{f}}^{-1}}\mathrm{g}\left( \frac{\pi }{3} \right)$ in exact form.</div><p style="text-align: right;">[1]</p></li></ol></li><li><div class="list-item-paragraph">The function $\mathrm{h}$ is defined by:</div><div class="list-item-paragraph" style="text-align: center">$\mathrm{h}:x\mapsto \frac{x+a}{2x+b}$, for $x\in \mathbb{R}$, $x\ne k$.</div><ol class="i-list"><li><div class="list-item-paragraph">Give an expression for $k$ and explain why this value of $x$ has to be excluded from the domain of $\mathrm{h}$.</div><p style="text-align: right;">[2]</p><div class="list-item-paragraph">The function $\mathrm{h}$ is such that $\mathrm{h}\left( x \right)={{\mathrm{h}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{h}$.</div></li><li><div class="list-item-paragraph">Determine the possible values of $a$ and $b$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find an expression for ${{\mathrm{h}}^{-1}}\left( -4 \right)$.</div><p style="text-align: right;">[1]</p></li></ol></li></ol>

Inverse Trigonometric Functions

Restricting a Domain to Obtain an Inverse

Domain of a Function

2022 NYJC CT Q7

<ol class="abc-list"><li><div class="list-item-paragraph">The function $\mathrm{h}$ is defined by $\mathrm{h}:x\mapsto \frac{1}{2}{{x}^{2}}+3$, for $x\in \mathbb{R}$.</div><div class="list-item-paragraph">The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{x+1}{5x-1}$, for $x\in \mathbb{R}$, $x\ne 0.2$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find $\mathrm{gh}\left( 2 \right)$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the value of $x$ for which $\mathrm{g}\left( x \right)=1.4$.</div><p style="text-align: right;">[1]</p></li></ol></li><li><div class="list-item-paragraph">The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{x+a}{2x+b}$, for $x\in \mathbb{R}$, $x\ne k$.</div><ol class="i-list"><li><div class="list-item-paragraph">Give an expression for $k$ and explain why this value of $x$ has to be excluded from the domain of $\operatorname{f}$.</div><p style="text-align: right;">[2]</p><div class="list-item-paragraph">The function $\mathrm{f}$ is such that $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{f}$.</div></li><li><div class="list-item-paragraph">Determine the possible values of $a$ and of $b$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find an expression for ${{\mathrm{f}}^{-1}}\left( -4 \right)$.</div><p style="text-align: right;">[1]</p></li></ol></li></ol>

N2021 P2 Q3

<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined as follows</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \frac{3x-a}{x-b}$, $x\in \mathbb{R}$, $x\ne b$,</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto 3+8x-2{{x}^{2}}$, $x>-1$,</p>
<p>where $a$ and $b$ are positive constants and $a < b$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">By sketching the graph of $y=\mathrm{f}(x)$, explain why ${{\mathrm{f}}^{-1}}$ exists and find ${{\mathrm{f}}^{-1}}(x)$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Find the set of values of $b$ such that the composite function $\mathrm{fg}$ exists. Hence find the range of $\mathrm{fg}$ in terms of $a$ and $b$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">If ${{\mathrm{f}}^{-1}}(x)=\mathrm{f}(x)$ for all values of $x$, find the value of $b$. Hence find the value of ${{\mathrm{f}}^{(2017)}}(a)$, leaving your answer in terms of $a$.</div><p style="text-align: right;">[3]</p></li></ol>

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined as follows

$\mathrm{f}:x\mapsto \frac{3x-a}{x-b}$, $x\in \mathbb{R}$, $x\ne b$,

$\mathrm{g}:x\mapsto 3+8x-2{{x}^{2}}$, $x>-1$,

where $a$ and $b$ are positive constants and $a < b$.

2017 AJC MYCT Q7

<p>A function $\mathrm{f}$ is said to be self-inverse if $\mathrm{f}(x)={{\mathrm{f}}^{-1}}(x)$ for all $x$ in the domain of $\mathrm{f}$.</p>
<p>The function $\mathrm{g}$ is defined by</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto \sqrt{\frac{{{x}^{2}}+2}{{{x}^{2}}-1}},\mathrm{ }x>1$ .</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Sketch the curve $y=\mathrm{g}\left( x \right)$ stating the equations of the asymptotes clearly.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Define ${{\mathrm{g}}^{-1}}$ in a similar form and show that $\mathrm{g}$ is self-inverse.</span></div><p style="text-align: right;">[4]</p></li><li><div class="list-item-content"><span class="list-item-text">Show that ${{\mathrm{g}}^{2}}(x)=x$ and that ${{\mathrm{g}}^{3}}(x)=\mathrm{g}(x)$ . Hence find the values of $x$ for which $4-{{\mathrm{g}}^{50}}(x)={{\left[ {{\mathrm{g}}^{51}}(x) \right]}^{2}}$.</span></div><p style="text-align: right;">[4]</p></li></ol>

A function $\mathrm{f}$ is said to be self-inverse if $\mathrm{f}(x)={{\mathrm{f}}^{-1}}(x)$ for all $x$ in the domain of $\mathrm{f}$.

$\mathrm{g}:x\mapsto \sqrt{\frac{{{x}^{2}}+2}{{{x}^{2}}-1}},\mathrm{ }x>1$ .

Sketch the curve $y=\mathrm{g}\left( x \right)$ stating the equations of the asymptotes clearly.

Define ${{\mathrm{g}}^{-1}}$ in a similar form and show that $\mathrm{g}$ is self-inverse.

Show that ${{\mathrm{g}}^{2}}(x)=x$ and that ${{\mathrm{g}}^{3}}(x)=\mathrm{g}(x)$ . Hence find the values of $x$ for which $4-{{\mathrm{g}}^{50}}(x)={{\left[ {{\mathrm{g}}^{51}}(x) \right]}^{2}}$.

Asymptotes and Discontinuities

2016 JJC P1 Q6

Graphing Techniques

<p>A function $\mathrm{h}$ is said to self-inverse if $\mathrm{h}\left( x \right)={{\mathrm{h}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{h}$.</p>
<p>Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \frac{5x-3}{x-5}$, $x\in \mathbb{R}$, $x\ne a$, where $a$ is a constant,</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto \ln x$, $x\in \mathbb{R}$, $x\ge {{\mathrm{e}}^{10}}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">State the value of $a$ and explain why this value has to be excluded from the domain of $\mathrm{f}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Show that $\mathrm{f}$ is self-inverse.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Find the exact values of $b$ such that ${{\mathrm{f}}^{4}}\left( b \right)-2={{\mathrm{f}}^{-1}}\left( b \right)$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Find the exact range of $\mathrm{fg}$.</span></div><p style="text-align: right;">[2]</p></li></ol>

A function $\mathrm{h}$ is said to self-inverse if $\mathrm{h}\left( x \right)={{\mathrm{h}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{h}$.

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by

$\mathrm{f}:x\mapsto \frac{5x-3}{x-5}$, $x\in \mathbb{R}$, $x\ne a$, where $a$ is a constant,

$\mathrm{g}:x\mapsto \ln x$, $x\in \mathbb{R}$, $x\ge {{\mathrm{e}}^{10}}$.

State the value of $a$ and explain why this value has to be excluded from the domain of $\mathrm{f}$.

Find the exact values of $b$ such that ${{\mathrm{f}}^{4}}\left( b \right)-2={{\mathrm{f}}^{-1}}\left( b \right)$.

2018 ACJC P1 Q5

<p>Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \frac{x+a}{x+b}$ for $x\in \mathbb{R},$ $x\ne -b,$ $a\ne -1,$</p>
<p style="text-align: center">$g:x\mapsto x$ for $x\in \mathbb{R}$.</p>
<p>It is given that $\mathrm{ff}=g$.</p>
<p>Find the value of $b$.</p>
<p>Find ${{\mathrm{f}}^{-1}}(x)$ in terms of $x$ and $a$.</p><p style="text-align: right;">[5]</p>

$\mathrm{f}:x\mapsto \frac{x+a}{x+b}$ for $x\in \mathbb{R},$ $x\ne -b,$ $a\ne -1,$

Find ${{\mathrm{f}}^{-1}}(x)$ in terms of $x$ and $a$.

N2018 P1 Q5

<p>A function $\mathrm{h}$ is self-inverse if&nbsp;$\mathrm{h}\left( x \right)={{\mathrm{h}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{h}$.</p>
<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="text-align: center">$\mathrm{f}\left( x \right)=2+\frac{1}{x-2}$ $\,\,\,\,\,\,\,\,\,$ for&nbsp;$\,\,\,\,\,\,\,\,\,$ $x\in \mathbb{R},x&gt;2$</p>
<p style="text-align: center">$\mathrm{g}\left( x \right)=\left\{ \begin{matrix}3x-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for}\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\le 1 \\ {{\left( x-1 \right)}^{2}}+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for}\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&gt;1 \\\end{matrix} \right.$</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{f}(x)$. With the aid of your graph, explain why $\mathrm{f}$ has an inverse.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}$ is self-inverse and find ${{\mathrm{f}}^{2}}\left( x \right)$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Hence, or otherwise, evaluate ${{\mathrm{f}}^{2025}}\left( 4 \right)$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find an expression for&nbsp;$\mathrm{gf}$ and state its domain.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the range of $\mathrm{gf}$.<span class="inline-marks">[2]</span></div></li></ol>

A function $\mathrm{h}$ is self-inverse if $\mathrm{h}\left( x \right)={{\mathrm{h}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{h}$.

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by

$\mathrm{f}\left( x \right)=2+\frac{1}{x-2}$ $\,\,\,\,\,\,\,\,\,$ for $\,\,\,\,\,\,\,\,\,$ $x\in \mathbb{R},x>2$

$\mathrm{g}\left( x \right)=\left\{ \begin{matrix}3x-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for}\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\le 1 \\ {{\left( x-1 \right)}^{2}}+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for}\,\,\,\,\,\,\,\,\,\,\,\,\,\,x>1 \\\end{matrix} \right.$

Sketch the graph of $y=\mathrm{f}(x)$. With the aid of your graph, explain why $\mathrm{f}$ has an inverse.

Show that $\mathrm{f}$ is self-inverse and find ${{\mathrm{f}}^{2}}\left( x \right)$.

Hence, or otherwise, evaluate ${{\mathrm{f}}^{2025}}\left( 4 \right)$.

Find an expression for $\mathrm{gf}$ and state its domain.

<p>A function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{2x+3}{x-2}$, $x\in \mathbb{R}$, $x\ne 2$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the domain of ${{\mathrm{f}}^{-1}}$.<span class="inline-marks">[2]</span></div></li></ol>
<p>A function $\mathrm{h}$ is said to be self-inverse if&nbsp;$\mathrm{h}$ and ${{\mathrm{h}}^{-1}}$ are the same function.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}$ is self-inverse.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">State the value of ${{\mathrm{f}}^{2023}}(6)$.<span class="inline-marks">[1]</span></div></li></ol>
<p>Another function $\mathrm{g}$ has range $\left( k,k+10 \right]$, where&nbsp;$k$ is a constant.</p>
<ol class="i-list" start="4"><li><div class="list-item-paragraph">Find the range of values of $k$ such that the composite function $\mathrm{fg}$ exists.<span class="inline-marks">[3]</span></div></li></ol>
<p>Assume for the remainder of this question that&nbsp;$\mathrm{fg}$ exists and $k\ne 2$.</p>
<ol class="i-list" start="5"><li><div class="list-item-paragraph">Find the range of $\mathrm{fg}$ in terms of $k$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;$\mathrm{fg}\left( x \right)=\frac{2x+2}{1-6x}$, find $\mathrm{g}\left( x \right)$ in terms of $x$.<span class="inline-marks">[3]</span></div></li></ol>

A function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{2x+3}{x-2}$, $x\in \mathbb{R}$, $x\ne 2$.

Find the domain of ${{\mathrm{f}}^{-1}}$.

A function $\mathrm{h}$ is said to be self-inverse if $\mathrm{h}$ and ${{\mathrm{h}}^{-1}}$ are the same function.

State the value of ${{\mathrm{f}}^{2023}}(6)$.

Another function $\mathrm{g}$ has range $\left( k,k+10 \right]$, where $k$ is a constant.

Find the range of values of $k$ such that the composite function $\mathrm{fg}$ exists.

Assume for the remainder of this question that $\mathrm{fg}$ exists and $k\ne 2$.

Find the range of $\mathrm{fg}$ in terms of $k$.

Given that $\mathrm{fg}\left( x \right)=\frac{2x+2}{1-6x}$, find $\mathrm{g}\left( x \right)$ in terms of $x$.

Recovering a Function from a Composite

2023 NJC Promo Q7

<p>A function $\mathrm{f}$ is said to self-inverse if&nbsp;$\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{f}$.</p>
<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \frac{ax+7{{a}^{2}}}{x-a},x\in \mathbb{R},x\ne a,0 &lt; a &lt; 1$</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto \left( x-2 \right)\left( x+1 \right),x\in \mathbb{R},x&gt;3.$</p>
<ol class="i-list"><li><div class="list-item-paragraph">Explain why ${{\mathrm{f}}^{-1}}$ exists and show that&nbsp;$\mathrm{f}$ is self-inverse.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Hence, evaluate ${{\mathrm{f}}^{241}}\left( \frac{1}{a} \right)$ in terms of $a$, showing your working clearly.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Show that&nbsp;$\mathrm{fg}$ exists. Find the exact range of $\mathrm{fg}$ in terms of $a$.<span class="inline-marks">[3]</span></div></li></ol>

A function $\mathrm{f}$ is said to self-inverse if $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{f}$.

$\mathrm{f}:x\mapsto \frac{ax+7{{a}^{2}}}{x-a},x\in \mathbb{R},x\ne a,0 < a < 1$

$\mathrm{g}:x\mapsto \left( x-2 \right)\left( x+1 \right),x\in \mathbb{R},x>3.$

Explain why ${{\mathrm{f}}^{-1}}$ exists and show that $\mathrm{f}$ is self-inverse.

Hence, evaluate ${{\mathrm{f}}^{241}}\left( \frac{1}{a} \right)$ in terms of $a$, showing your working clearly.

Show that $\mathrm{fg}$ exists. Find the exact range of $\mathrm{fg}$ in terms of $a$.

Math Question Bank