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

<p>The function $\mathrm{g}$, with domain the non-negative integers, is defined by</p>
<p>\[\mathrm{g}(n)=\begin{cases}1,&n=0,\\2\mathrm{g}\left(\dfrac n2\right),&n\text{ is positive and even},\\1+\mathrm{g}\left(\dfrac{n-1}{2}\right),&n\text{ is odd}.\end{cases}\]</p>
<ol class="abc-list"><li data-list-item-id="recursive-10" data-list-item-path="recursive-10"><div class="list-item-paragraph">Find $\mathrm{g}(7)$, $\mathrm{g}(10)$ and $\mathrm{g}(12)$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="recursive-11" data-list-item-path="recursive-11"><div class="list-item-paragraph">Find all values of $n$ for which $\mathrm{g}(n)=4$, showing your reasoning.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="recursive-12" data-list-item-path="recursive-12"><div class="list-item-paragraph">Hence explain why $\mathrm{g}$ does not have an inverse.<span class="inline-marks">[2]</span></div></li></ol>

The function $\mathrm{g}$, with domain the non-negative integers, is defined by

\[\mathrm{g}(n)=\begin{cases}1,&n=0,\\2\mathrm{g}\left(\dfrac n2\right),&n\text{ is positive and even},\\1+\mathrm{g}\left(\dfrac{n-1}{2}\right),&n\text{ is odd}.\end{cases}\]

Find $\mathrm{g}(7)$, $\mathrm{g}(10)$ and $\mathrm{g}(12)$.

Find all values of $n$ for which $\mathrm{g}(n)=4$, showing your reasoning.

Hence explain why $\mathrm{g}$ does not have an inverse.

One-to-One Functions and Inverse Existence

Working backwards through a recursive rule

Functions

<p>The function $\mathrm{f}$, with domain the non-negative integers, is defined by</p>
<p>\[\mathrm{f}(n)=\begin{cases}2,&n=0,\\3+\mathrm{f}\left(\dfrac n2\right),&n\text{ is positive and even},\\1+\mathrm{f}(n-1),&n\text{ is odd}.\end{cases}\]</p>
<ol class="abc-list"><li data-list-item-id="recursive-4" data-list-item-path="recursive-4"><div class="list-item-paragraph">Find $\mathrm{f}(6)$, $\mathrm{f}(11)$ and $\mathrm{f}(20)$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="recursive-5" data-list-item-path="recursive-5"><div class="list-item-paragraph">Does $\mathrm{f}$ have an inverse? Justify your answer.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="recursive-6" data-list-item-path="recursive-6"><div class="list-item-paragraph">Find an expression for&nbsp;$\mathrm{f}(2^m)$ in terms of $m$, where&nbsp;$m$ is a non-negative integer.<span class="inline-marks">[2]</span></div></li></ol>

The function $\mathrm{f}$, with domain the non-negative integers, is defined by

\[\mathrm{f}(n)=\begin{cases}2,&n=0,\\3+\mathrm{f}\left(\dfrac n2\right),&n\text{ is positive and even},\\1+\mathrm{f}(n-1),&n\text{ is odd}.\end{cases}\]

Find $\mathrm{f}(6)$, $\mathrm{f}(11)$ and $\mathrm{f}(20)$.

Does $\mathrm{f}$ have an inverse? Justify your answer.

Find an expression for $\mathrm{f}(2^m)$ in terms of $m$, where $m$ is a non-negative integer.

Evaluating a recursive rule and its powers-of-two inputs

<p>The functions $\mathrm{p}$ and $\mathrm{q}$ are continuous for&nbsp;$1\leq x\leq5$. The function $\mathrm{p}$ is strictly increasing, while $\mathrm{q}$ is strictly decreasing and $\mathrm{q}(x)&gt;0$. Selected values are shown below.</p>
<div class="content-block content-block--table"><div class="table-container"><table class="table--original-style" style="table-layout: fixed; width: 300px;"><colgroup><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"></colgroup><thead><tr><th style="text-align: center; --cell-justify: center">$x$</th><th style="text-align: center; --cell-justify: center">$1$</th><th style="text-align: center; --cell-justify: center">$2$</th><th style="text-align: center; --cell-justify: center">$3$</th><th style="text-align: center; --cell-justify: center">$4$</th><th style="text-align: center; --cell-justify: center">$5$</th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">$\mathrm{p}(x)$</td><td style="text-align: center; --cell-justify: center">$-10$</td><td style="text-align: center; --cell-justify: center">$-8$</td><td style="text-align: center; --cell-justify: center">$-6$</td><td style="text-align: center; --cell-justify: center">$-4$</td><td style="text-align: center; --cell-justify: center">$5$</td></tr><tr><td style="text-align: center; --cell-justify: center">$\mathrm{q}(x)$</td><td style="text-align: center; --cell-justify: center">$5$</td><td style="text-align: center; --cell-justify: center">$4$</td><td style="text-align: center; --cell-justify: center">$3$</td><td style="text-align: center; --cell-justify: center">$2$</td><td style="text-align: center; --cell-justify: center">$1$</td></tr></tbody></table></div></div>
<p>Using the information in the table,</p>
<ol class="abc-list"><li data-list-item-id="q1-a" data-list-item-path="q1-a"><div class="list-item-paragraph"><p>Find $\mathrm{p}\mathrm{q}(2)$.</p><span class="inline-marks">[1]</span></div></li><li data-list-item-id="q1-b" data-list-item-path="q1-b"><div class="list-item-paragraph"><p>Solve $\mathrm{p}^{-1}\mathrm{q}(x)=5$.</p><span class="inline-marks">[2]</span></div></li></ol>
<p>The function $\mathrm{r}$ is defined by $\mathrm{r}(x)=\frac{\mathrm{p}(x)}{\mathrm{q}(x)}$.</p>
<p>It is further given that&nbsp;$\mathrm{p}'(3)=3$ and $\mathrm{q}'(3)=-1$.</p>
<ol class="abc-list" start="3"><li data-list-item-id="q1-c" data-list-item-path="q1-c"><div class="list-item-paragraph">Explain why $\mathrm{r}^{-1}$ does not exist on the given domain.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="q1-d" data-list-item-path="q1-d"><div class="list-item-paragraph">Find the equation of the normal to $y=\mathrm{r}(x)$ at $x=3$.<span class="inline-marks">[3]</span></div></li></ol>

The functions $\mathrm{p}$ and $\mathrm{q}$ are continuous for $1\leq x\leq5$. The function $\mathrm{p}$ is strictly increasing, while $\mathrm{q}$ is strictly decreasing and $\mathrm{q}(x)>0$. Selected values are shown below.

The function $\mathrm{r}$ is defined by $\mathrm{r}(x)=\frac{\mathrm{p}(x)}{\mathrm{q}(x)}$.

It is further given that $\mathrm{p}'(3)=3$ and $\mathrm{q}'(3)=-1$.

Explain why $\mathrm{r}^{-1}$ does not exist on the given domain.

Find the equation of the normal to $y=\mathrm{r}(x)$ at $x=3$.

Composite Functions

Tangents and Normals at a Point

Quotient Rule

Inverse Functions

Normal from tabulated function values

Techniques of Differentiation

Applications of Differentiation

<p>The function $\mathrm{f}$ is defined by $\mathrm{f}:x\rightarrow x\left|x-3\right|,\ x\in\mathbb{R},\ x&lt;m$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}^{-1}$ does not exist if&nbsp;$m=4$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">State the largest possible value of $m$ for which $\mathrm{f}^{-1}$ exists.<span class="inline-marks">[1]</span></div></li></ol>
<p>Use the value of $m$ obtained in part <strong>(ii)</strong> for the following questions.</p>
<ol class="i-list" start="3"><li data-list-item-id="item-iii"><div class="list-item-paragraph">Find $\mathrm{f}^{-1}$, stating its domain.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-iv"><div class="list-item-paragraph">The composite function $\mathrm{fg}$ is defined by $\mathrm{fg}:x\rightarrow \frac{9}{4}-x^2,\ x\in\mathbb{R},\ x&lt;0$.</div><div class="list-item-paragraph">Find $\mathrm{g}(x)$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-v"><div class="list-item-paragraph">The function $\mathrm{h}$ is defined by $\mathrm{h}:x\rightarrow \frac{3}{x-a},\ x\in\mathbb{R},\ x&lt;a$.</div><ol class="abc-list"><li data-list-item-id="item-v-a"><div class="list-item-paragraph">The graph of $y=\mathrm{k}(x)$ is transformed by a stretch with scale factor 2 parallel to the $x$-axis, followed by a translation of 10 units in the negative $y$-direction. The equation of the resulting curve is $y=\mathrm{h}(x)$.</div><div class="list-item-paragraph">Find $\mathrm{k}(x)$.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\rightarrow x\left|x-3\right|,\ x\in\mathbb{R},\ x<m$.

Use the value of $m$ obtained in part **(ii)** for the following questions.

Modulus Functions

Restricting a Domain to Obtain an Inverse

Recovering a Function from a Composite

2014 HCI P2 Q4 (part)

<p>The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 - (x + 3)^2$, $x \in \mathbb{R}$, $x \ge -2$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Show that the inverse function of $\mathrm{f}$ exists. Find $\mathrm{f}^{-1}(x)$ and state its domain.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">On the same diagram, sketch the graphs of $y = \mathrm{f}(x)$, $y = \mathrm{f}^{-1}(x)$ and $y = \mathrm{f}^{-1}\mathrm{f}(x)$, showing clearly the relationship between the graphs.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Hence, find the exact solution of $\mathrm{f}^{-1}\mathrm{f}(x) \le \mathrm{f}(x)$.<span class="inline-marks">[3]</span></div></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 - (x + 3)^2$, $x \in \mathbb{R}$, $x \ge -2$.

Sketching a Function and Its Inverse

Sketching $y=f(f^{-1}(x))$ and $y=f^{-1}(f(x))$

2011 NJC P1 Q7

<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p>\[\begin{aligned} &\mathrm{f} : x \mapsto 5 - (x - 2)^2, \quad \text{for}\hspace{0.5em} x \in \mathbb{R}, \ x \ge 0, \\ &\mathrm{g} : x \mapsto \mathrm{e}^x + 3, \qquad \text{for}\hspace{0.5em} x \in \mathbb{R}. \end{aligned}\]</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Show that the composite function $\mathrm{fg}$ exists.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find an expression for&nbsp;$\mathrm{fg}(x)$ and state its domain.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the range of $\mathrm{fg}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}^{-1}$ does not exist.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-v" data-list-item-path="item-v"><div class="list-item-paragraph">If the domain of $\mathrm{f}$ is restricted to $[k, \infty)$, find the smallest value of $k$ such that&nbsp;$\mathrm{f}^{-1}$ exists.<span class="inline-marks">[1]</span></div></li></ol>
<p>For the rest of the question, the domain of $\mathrm{f}$ is now restricted to $[k, \infty)$, with the value of $k$ found in part <strong>(v)</strong>.</p>
<ol class="i-list" start="6"><li data-list-item-id="item-vi" data-list-item-path="item-vi"><div class="list-item-paragraph">Find $\mathrm{f}^{-1}(x)$ and state the domain of $\mathrm{f}^{-1}$.<span class="inline-marks">[3]</span></div></li></ol>

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

\[\begin{aligned} &\mathrm{f} : x \mapsto 5 - (x - 2)^2, \quad \text{for } x \in \mathbb{R}, \ x \ge 0, \\ &\mathrm{g} : x \mapsto \mathrm{e}^x + 3, \qquad \text{for } x \in \mathbb{R}. \end{aligned}\]

For the rest of the question, the domain of $\mathrm{f}$ is now restricted to $[k, \infty)$, with the value of $k$ found in part **(v)**.

Existence of Composite Functions

Range of a Composite Function

2024 MI PRU1 Q7

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">\[\mathrm{f} : x \mapsto \frac{-x - 2}{5x + 1}, \quad x \in \mathbb{R}, x \neq -\frac{1}{5}.\]</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Explain why both the function $\mathrm{f}^{-1}$ and composite function $\mathrm{f}^2$ exist.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find $\mathrm{f}^2(x)$ and state the range of $\mathrm{f}^2$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Determine the solution of the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>The function $g$ is defined by</p>
<p style="text-align: center">\[g : x \mapsto \frac{2x - 2}{5x + 4}, \quad x \in \mathbb{R}, x \neq -\frac{4}{5}, x \neq -\frac{1}{5}, x \neq \frac{2}{5}.\]</p>
<ol class="i-list" start="5"><li><div class="list-item-content"><span class="list-item-text">Verify that&nbsp;$g^2(x) = \mathrm{f}(x)$.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$k$ is an integer, find $g^{49}(k)$, giving your answer in terms of $k$.</span><span class="inline-marks">[3]</span></div></li></ol>

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

\[\mathrm{f} : x \mapsto \frac{-x - 2}{5x + 1}, \quad x \in \mathbb{R}, x \neq -\frac{1}{5}.\]

Explain why both the function $\mathrm{f}^{-1}$ and composite function $\mathrm{f}^2$ exist.

Find $\mathrm{f}^2(x)$ and state the range of $\mathrm{f}^2$.

Determine the solution of the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$.

\[g : x \mapsto \frac{2x - 2}{5x + 4}, \quad x \in \mathbb{R}, x \neq -\frac{4}{5}, x \neq -\frac{1}{5}, x \neq \frac{2}{5}.\]

Given that $k$ is an integer, find $g^{49}(k)$, giving your answer in terms of $k$.

Intersections of a Function and Its Inverse

Iterates $f^n$ for Large $n$

2014 VJC Promo Q8

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">\[\mathrm{f} : x \mapsto \frac{1}{x^2 + 2x}, \quad x \in \mathbb{R}, \quad x &lt; -1, \quad x \neq -2\]</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that&nbsp;$\mathrm{f}$ has an inverse.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Define $\mathrm{f}^{-1}$ in similar form.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The function $\mathrm{g}$ is such that&nbsp;$\mathrm{fg}(x) = x^2 + 1$. Find $\mathrm{g}(x)$.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>The function $\mathrm{h}$ is defined by</p>
<p style="text-align: center">\[\mathrm{h} : x \mapsto \frac{1}{x^2}, \quad x \in \mathbb{R}, \quad x \neq 0\]</p>
<ol class="abc-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Only one of the composite functions $\mathrm{fh}$ and $\mathrm{hf}$ exists. Give a definition (including the domain) of the composite that exists, and explain why the other composite does not exist.</span><span class="inline-marks">[3]</span></div></li></ol>

\[\mathrm{f} : x \mapsto \frac{1}{x^2 + 2x}, \quad x \in \mathbb{R}, \quad x < -1, \quad x \neq -2\]

Define $\mathrm{f}^{-1}$ in similar form.

The function $\mathrm{g}$ is such that $\mathrm{fg}(x) = x^2 + 1$. Find $\mathrm{g}(x)$.

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

\[\mathrm{h} : x \mapsto \frac{1}{x^2}, \quad x \in \mathbb{R}, \quad x \neq 0\]

Only one of the composite functions $\mathrm{fh}$ and $\mathrm{hf}$ exists. Give a definition (including the domain) of the composite that exists, and explain why the other composite does not exist.

2025 YIJC P1 Q11

<p>The function $\mathrm{f}$ is defined by $\mathrm{f}:x \mapsto \frac{1}{2}+e^{-x}$, $x \in \mathbb{R}$, $x \ge 0$.</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 $y=\mathrm{f}(x)$, $y=\mathrm{f}^{-1}(x)$ and $y=\mathrm{f}^{-1}\mathrm{f}(x)$ on the same diagram, showing clearly their relationship.<span class="inline-marks">[3]</span></div></li></ol>
<p>The function $\mathrm{g}$ is defined by $\mathrm{g}:x \mapsto 1+4x-x^2$, $x \le 2$.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Explain why $\mathrm{g}^{-1}$ exists.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Find $\mathrm{g}^{-1}(x)$ and state its domain.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Explain why the composite function $\mathrm{gf}$ exists.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-e" data-list-item-path="item-e"><div class="list-item-paragraph">Find the exact range of $\mathrm{gf}$.<span class="inline-marks">[2]</span></div></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f}:x \mapsto \frac{1}{2}+e^{-x}$, $x \in \mathbb{R}$, $x \ge 0$.

Sketch the graph of $y=\mathrm{f}(x)$, $y=\mathrm{f}^{-1}(x)$ and $y=\mathrm{f}^{-1}\mathrm{f}(x)$ on the same diagram, showing clearly their relationship.

The function $\mathrm{g}$ is defined by $\mathrm{g}:x \mapsto 1+4x-x^2$, $x \le 2$.

Find $\mathrm{g}^{-1}(x)$ and state its domain.

Explain why the composite function $\mathrm{gf}$ exists.

2025 TJC P2 Q3

<p>Sketch the graph of $\mathrm{f}\left( x \right)=\left\{ \begin{matrix}-1 \\x \\1\mathrm{ } \\\end{matrix}\right.\mathrm{}\begin{matrix}\mathrm{for} \\\mathrm{for}\\\mathrm{for}\\\end{matrix}\mathrm{   }\begin{matrix}x<-1 \\ -1\le x\le 1 \\x>1  \\\end{matrix}$ and state its range

</p>
<p>Determine whether $\mathrm{f}$ is a one-one function.</p>

Sketch the graph of $\mathrm{f}\left( x \right)=\left\{ \begin{matrix}-1 \\x \\1\mathrm{ } \\\end{matrix}\right.\mathrm{}\begin{matrix}\mathrm{for} \\\mathrm{for}\\\mathrm{for}\\\end{matrix}\mathrm{   }\begin{matrix}x<-1 \\ -1\le x\le 1 \\x>1  \\\end{matrix}$ and state its range



Determine whether $\mathrm{f}$ is a one-one function.

Piecewise-Defined Functions

Range of a Function

Worked Example 3

<p>Function $f$ is defined by
$f : x \mapsto x^{2} + 2x + 3,\ x \le -1$.

Show that $f^{-1}$ exists. Hence, find $f^{-1}$, stating its domain and range.</p>

Function $f$ is defined by
$f : x \mapsto x^{2} + 2x + 3,\ x \le -1$.

Show that $f^{-1}$ exists. Hence, find $f^{-1}$, stating its domain and range.

Worked Example 1

<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:

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

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

2009 HCI P1 Q9

<p>Sketch the graph of $\mathrm{f}\left( x \right)=\left\{ \begin{matrix}-1 \\x \\1\mathrm{ } \\\end{matrix}\right.\mathrm{}\begin{matrix}\mathrm{for} \\\mathrm{for}\\\mathrm{for}\\\end{matrix}\mathrm{   }\begin{matrix}x&lt;-1 \\ -1\le x\le 1 \\x&gt;1  \\\end{matrix}$ and state its range</p>
<p>Determine whether $f$ is a one-one function.</p>

Sketch the graph of $\mathrm{f}\left( x \right)=\left\{ \begin{matrix}-1 \\x \\1\mathrm{ } \\\end{matrix}\right.\mathrm{}\begin{matrix}\mathrm{for} \\\mathrm{for}\\\mathrm{for}\\\end{matrix}\mathrm{   }\begin{matrix}x<-1 \\ -1\le x\le 1 \\x>1  \\\end{matrix}$ and state its range

Determine whether $f$ is a one-one function.

<p>Function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto -{{x}^{2}}+4x+2,x&gt;2$. Show that&nbsp;${{\mathrm{f}}^{-1}}$ exists. Hence, find ${{\mathrm{f}}^{-1}}$, stating its domain and range. </p>

Function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto -{{x}^{2}}+4x+2,x>2$. Show that ${{\mathrm{f}}^{-1}}$ exists. Hence, find ${{\mathrm{f}}^{-1}}$, stating its domain and range. 

<p>Function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{x}^{2}}+2x+3,x\le -1$. Show that&nbsp;${{\mathrm{f}}^{-1}}$ exists. Hence, find ${{\mathrm{f}}^{-1}}$, stating its domain and range. </p>

Function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{x}^{2}}+2x+3,x\le -1$. Show that ${{\mathrm{f}}^{-1}}$ exists. Hence, find ${{\mathrm{f}}^{-1}}$, stating its domain and range. 

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \ln {{(x+1)}^{2}}+2$, $x\in \mathbb{R}$, $x\ge 0$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $\mathrm{f}$ has an inverse.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find ${{\mathrm{f}}^{-1}}(x)$ and state the domain of ${{\mathrm{f}}^{-1}}$.</div><p style="text-align: right;">[3]</p></li></ol>
<p>The function $\mathrm{g}$ is defined by</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto \frac{1-x}{x+1}$, $x\in \mathbb{R}$, $x\ne -1$.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">Given that ${{\mathrm{g}}^{-1}}$ exists, show that ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}$ exists, and find the range of ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find the exact value of $x$ such that ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}(x)=\frac{1}{2}$.</div><p style="text-align: right;">[3]</p></li></ol>

$\mathrm{f}:x\mapsto \ln {{(x+1)}^{2}}+2$, $x\in \mathbb{R}$, $x\ge 0$.

$\mathrm{g}:x\mapsto \frac{1-x}{x+1}$, $x\in \mathbb{R}$, $x\ne -1$.

2023 HCI P1 Q9

<ol class="abc-list" start="2"><li><div class="list-item-paragraph">The function $\mathrm{f}$ is defined by</div><div class="list-item-paragraph">\[\mathrm{f}(x) = \left\{ 
\begin{array}{ll}
2x + 1 & \text{for}\hspace{0.5em} 0 \le x &lt; 2 \\
(x - 4)^2 + 1 & \text{for}\hspace{0.5em} 2 \le x &lt; 4.
\end{array} 
\right.\]</div><div class="list-item-paragraph">It is further given that&nbsp;$\mathrm{f}\left( x \right)=\mathrm{f}\left( x+4 \right)$ for all real values of $x$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find the values of $\mathrm{f}\left( 1 \right)$ and $\mathrm{f}\left( 5 \right)$. Hence, explain why $\mathrm{f}$ is not one-one.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{f}\left( x \right)$ for&nbsp;$-4\le x&lt;8$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the range of $\mathrm{f}$ for&nbsp;$-4\le x&lt;8$.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

Periodic Functions

2014 SRJC Promo Q11

<p>Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \left| x+20 \right|$, $x\in \mathbb{R}$,</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto x+\frac{\alpha }{x}$, $x\in \mathbb{R}$, $x>0$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Explain why the composite function $\mathrm{gf}$ does not exist.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Determine the range of values of $\alpha $ such that ${{\mathrm{g}}^{-1}}$ exists.</div><p style="text-align: right;">[2]</p></li></ol>
<p>For the rest of the question, it is given that $\alpha =4$ and the domain of $\mathrm{g}$ is further restricted to $x\ge \beta $.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">Determine the least value of $\beta $ for which ${{\mathrm{g}}^{-1}}$ exists; using this value of $\beta $, hence find ${{\mathrm{g}}^{-1}}\left( x \right)$ and state its domain.</div><p style="text-align: right;">[4]</p></li></ol>

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

$\mathrm{f}:x\mapsto \left| x+20 \right|$, $x\in \mathbb{R}$,

$\mathrm{g}:x\mapsto x+\frac{\alpha }{x}$, $x\in \mathbb{R}$, $x>0$.

For the rest of the question, it is given that $\alpha =4$ and the domain of $\mathrm{g}$ is further restricted to $x\ge \beta $.

HCI Functions Tutorial Q9

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto 1-\frac{4}{{{x}^{2}}}$, for&nbsp;$x\in \mathbb{R}$, $x&lt;-1$.</p>
<ol class="i-list"><li data-list-item-id="item-1760869219899-3dmry2sxu" data-list-item-path="item-1760869219899-3dmry2sxu"><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{f}\left( x \right)$, stating the equation of the asymptote and the coordinates of the point where the curve crosses the axes.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760869220558-y6aku5zmc" data-list-item-path="item-1760869220558-y6aku5zmc"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}$ has an inverse. Hence, find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state its domain.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760869221556-oyo09zrqr" data-list-item-path="item-1760869221556-oyo09zrqr"><div class="list-item-paragraph">By considering the domains of $y=\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)$ and $y={{\mathrm{f}}^{-1}}\mathrm{f}\left( x \right)$, find the range of values of $x$ such that&nbsp;$\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)={{\mathrm{f}}^{-1}}\mathrm{f}\left( x \right)$.<span class="inline-marks">[2]</span></div></li></ol>
<p>Another function $\mathrm{g}$ is defined by</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto -{{\mathrm{e}}^{2x}}-2$, for&nbsp;$x\in \mathbb{R}$.</p>
<ol class="i-list" start="4"><li data-list-item-id="item-1760869259556-bm7s4t0qh" data-list-item-path="item-1760869259556-bm7s4t0qh"><div class="list-item-paragraph">Explain why $\mathrm{fg}$ exists. Find $\mathrm{fg}$ and its range.<span class="inline-marks">[3]</span></div></li></ol>

$\mathrm{f}:x\mapsto 1-\frac{4}{{{x}^{2}}}$, for $x\in \mathbb{R}$, $x<-1$.

Another function $\mathrm{g}$ is defined by

$\mathrm{g}:x\mapsto -{{\mathrm{e}}^{2x}}-2$, for $x\in \mathbb{R}$.

Sketching Rational Functions

Asymptotes and Discontinuities

2024 JPJC Promo Q13

Graphing Techniques

<ol class="abc-list"><li data-list-item-id="item-1760695372830-3qcqpg4zl" data-list-item-path="item-1760695372830-3qcqpg4zl"><div class="list-item-paragraph">The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</div><div class="list-item-paragraph" style="text-align: center">${\mathrm{f}:x\rightarrow\frac{x-2}{x+2},\quad x\in\mathbb{R},\; x\neq -2,}$</div><div class="list-item-paragraph" style="text-align: center">${\mathrm{g}:x\rightarrow -x^2,\quad x\in\mathbb{R},\; x &lt; -\sqrt{2},}$</div><ol class="i-list"><li data-list-item-id="subitem-1760695392997-5fvml0sqs" data-list-item-path="item-1760695372830-3qcqpg4zl/subitem-1760695392997-5fvml0sqs"><div class="list-item-paragraph">Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state its domain.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-1760695402716-fx2neqsjs" data-list-item-path="item-1760695372830-3qcqpg4zl/subitem-1760695402716-fx2neqsjs"><div class="list-item-paragraph">Find an expression for&nbsp;$\mathrm{fg}\left( x \right)$ and hence, or otherwise, find ${{\left( \mathrm{fg} \right)}^{-1}}\left( 3 \right)$.<span class="inline-marks">[4]</span></div></li></ol></li><li data-list-item-id="item-1760695373526-07tjvnbaj" data-list-item-path="item-1760695373526-07tjvnbaj"><div class="list-item-paragraph">It is given that</div><div class="list-item-paragraph" style="text-align: center">${\mathrm{h}(x)=\begin{cases} -4x-12 & \text{for}\hspace{0.5em} -4 \leq x \leq -2, \\ x\left|x\right| & \text{for}\hspace{0.5em} -2 &lt; x \leq -2 \end{cases}}$</div><div class="list-item-paragraph">and that&nbsp;$\mathrm{h}\left( x \right)=\mathrm{h}\left( x+6 \right)$ for all real values of $x$.</div><ol class="i-list"><li data-list-item-id="subitem-1760695466481-iguzm1f62" data-list-item-path="item-1760695373526-07tjvnbaj/subitem-1760695466481-iguzm1f62"><div class="list-item-paragraph">Evaluate $\mathrm{h}\left( -4 \right)$ and $\mathrm{h}\left( 12 \right)$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760695467066-0lyti35da" data-list-item-path="item-1760695373526-07tjvnbaj/subitem-1760695467066-0lyti35da"><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{h}\left( x \right)$ for&nbsp;$-4\le x\le 6$ and explain why $\mathrm{h}$ has no inverse.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

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