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>Given&nbsp;$h(x)=1-|x-2|$, where&nbsp;$0\le x\le4$, and $H(x)=\dfrac{1}{h(x)}$, the range of $H(x)$ is the set&nbsp;$R$ such that&nbsp;$R=\{y:y\ge a\text{ or}\hspace{0.5em}y\le b\}$ where&nbsp;$a,b\in\mathbb Z$.</p>
<p>Determine the values of $a$ and $b$.</p>

Given $h(x)=1-|x-2|$, where $0\le x\le4$, and $H(x)=\dfrac{1}{h(x)}$, the range of $H(x)$ is the set $R$ such that $R=\{y:y\ge a\text{ or }y\le b\}$ where $a,b\in\mathbb Z$.

Modulus Functions

Range of a Function

Reciprocal Graph, $y=\frac{1}{f(x)}$

Graphs of Modulus Function

Two branches in a reciprocal modulus range

Functions

Modulus Function

Transformations of Graphs

<p>For&nbsp;$-2\le x\le3$, define $p(x)=|2x-1|-3$ and $q(x)=\dfrac{1}{p(x)}$ wherever $p(x)\ne0$. The range of $q$ is $\{y:y\le a\text{ or}\hspace{0.5em}y\ge b\}$. Find the exact values of $a$ and $b$.<span class="inline-marks">[4]</span></p>

For $-2\le x\le3$, define $p(x)=|2x-1|-3$ and $q(x)=\dfrac{1}{p(x)}$ wherever $p(x)\ne0$. The range of $q$ is $\{y:y\le a\text{ or }y\ge b\}$. Find the exact values of $a$ and $b$.

Reciprocal range of a stretched modulus trough

<ol class="abc-list"><li data-list-item-id="h2-2025-acjc-promo-q4-acjc-q4-a" data-list-item-path="h2-2025-acjc-promo-q4-acjc-q4-a"><div class="list-item-paragraph">The diagram shows the graphs of $y=|a-bx|$ and $y=|mx-6|$, where&nbsp;$a$, $b$ and $m$ are positive constants. The graph of $y=|a-bx|$ meets the $x$-axis at the point $\left(\dfrac94,0\right)$.</div><div class="graph-block" data-graph-config="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" data-width="360" data-height="250" style="width:360px;height:250px;max-width:100%;display:block;margin:0 auto;"></div><div class="list-item-paragraph">The solution set of the inequality $|mx-6|&lt;|a-bx|$ is given as $\{x:x\in\mathbb{R},\ x&lt;1\text{ or}\hspace{0.5em}x&gt;3\}$.</div><div class="list-item-paragraph">Find the values of $a$, $b$ and $m$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1787798435359-r5rp7s74v" data-list-item-path="item-1787798435359-r5rp7s74v"><div class="list-item-paragraph"><p>Given that&nbsp;$k$ is a constant and $k&gt;1$, solve the inequality $\dfrac{kx^2+1}{x^2+(k+1)x+k}&gt;-1$, giving your answer in terms of $k$.</p><span class="inline-marks">[4]</span></div></li></ol>

Graphical Method

Inequalities involving Rational Functions

Inequalities involving Modulus Functions

2025 ACJC Promo Q4

Inequalities

<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 $\mathrm{gh}$.

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}$

<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.

One-to-One Functions and Inverse Existence

Restricting a Domain to Obtain an Inverse

Recovering a Function from a Composite

2014 HCI P2 Q4 (part)

<p>It is given that</p>
<p style="text-align: center">\[\mathrm{f}(x) = \begin{cases} \quad\;\;\; \frac{1}{2}x^2, & \text{for}\hspace{0.5em} 1 \le x &lt; 2, \\ \frac{1}{4}(-3x + 14), & \text{for}\hspace{0.5em} 2 \le x &lt; 4, \end{cases}\]</p>
<p>and that&nbsp;$\mathrm{f}(x) = \mathrm{f}(x + 3)$ for all real values of $x$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">State the value of $\mathrm{f}(0)$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Sketch the graph of $y = \mathrm{f}(x)$ for&nbsp;$0 \le x \le 5$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">The function $\mathrm{g}$ is given by $\mathrm{g}(x) = \frac{4}{3}|x - 2|$ for&nbsp;$x \in \mathbb{R}$. By sketching the graph of $y = \mathrm{g}(x)$ on the same diagram as in <strong>part (b)</strong>, solve the inequality $\mathrm{f}(x) &gt; \mathrm{g}(x)$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">The function $\mathrm{h}$ is given by $\mathrm{h}(x) = \sqrt{3} \sin(\pi x) + \cos(\pi x) + 1$ for&nbsp;$0 \le x \le 2$. Explain why the composite function $\mathrm{hf}$ exists and find its range in exact form.<span class="inline-marks">[4]</span></div></li></ol>

\[\mathrm{f}(x) = \begin{cases} \quad\;\;\; \frac{1}{2}x^2, & \text{for } 1 \le x < 2, \\ \frac{1}{4}(-3x + 14), & \text{for } 2 \le x < 4, \end{cases}\]

and that $\mathrm{f}(x) = \mathrm{f}(x + 3)$ for all real values of $x$.

Piecewise-Defined Functions

Periodic Functions

2024 RI Promo Q10

<p>The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \sqrt{5 + 2x}$, for&nbsp;$x \in \mathbb{R}$, $x &gt; -\frac{5}{2}$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find the value of $a$ such that&nbsp;$\mathrm{f}^2(a) = 5$.<span class="inline-marks">[2]</span></div></li></ol>
<p>Another function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto \left|\frac{x}{1 - 2x}\right|$, for&nbsp;$x \in \mathbb{R}$, $x \le k$.</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">State the largest value of $k$ for&nbsp;$\mathrm{g}^{-1}$ to exist.<span class="inline-marks">[1]</span></div></li></ol>
<p>It is now given that&nbsp;$k = -1$.</p>
<ol class="abc-list" start="3"><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 $\mathrm{fg}$ exists. State the exact range of $\mathrm{fg}$.<span class="inline-marks">[2]</span></div></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \sqrt{5 + 2x}$, for $x \in \mathbb{R}$, $x > -\frac{5}{2}$.

Another function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto \left|\frac{x}{1 - 2x}\right|$, for $x \in \mathbb{R}$, $x \le k$.

Composite Functions

2024 ASRJC Promo Q7

<p>The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="margin-left: 4rem">$\mathrm{f}(x)=1+3\mathrm{e}^{-x}, \quad\quad\quad\, x \in \mathbb{R}, x&gt;0,$ <br>$\mathrm{g}(x)=|x-1|(x-3), \quad x \in \mathbb{R}, x&lt;c \text{ where}\hspace{0.5em} c \text{ is a real constant.}$</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$c=4$, determine if the composite functions $\mathrm{fg}$ and $\mathrm{gf}$ exist, justifying your answers. Find the range of the composite function that exists.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$\mathrm{g}^{-1}$ exists, state the largest possible value of $c$. Using this value of $c$ find $\mathrm{g}^{-1}(x)$.</span><span class="inline-marks">[4]</span></div></li></ol>

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

$\mathrm{f}(x)=1+3\mathrm{e}^{-x}, \quad\quad\quad\, x \in \mathbb{R}, x>0,$ 
$\mathrm{g}(x)=|x-1|(x-3), \quad x \in \mathbb{R}, x<c \text{ where } c \text{ is a real constant.}$

Given that $c=4$, determine if the composite functions $\mathrm{fg}$ and $\mathrm{gf}$ exist, justifying your answers. Find the range of the composite function that exists.

Given that $\mathrm{g}^{-1}$ exists, state the largest possible value of $c$. Using this value of $c$ find $\mathrm{g}^{-1}(x)$.

2020 RI P2 Q3

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">\[\mathrm{f} : x \rightarrow \left|\frac{2x + 4}{3 - x}\right|, \quad x \in \mathbb{R}, \quad x \neq 3.\]</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $y = \mathrm{f}(x)$, giving the equations of any asymptotes and the coordinates of the points where the curve meets the axes.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence state the range of $\mathrm{f}$.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Explain why the function $\mathrm{f}^2$ does not exist.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>The domain of $\mathrm{f}$ is further restricted to $x \leqslant a$, where&nbsp;$a$ is a constant.</p>
<ol class="abc-list" start="4"><li><div class="list-item-content"><span class="list-item-text">State the greatest value of $a$ such that the function $\mathrm{f}^{-1}$ exists.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find $\mathrm{f}^{-1}(x)$ and state its domain.</span><span class="inline-marks">[4]</span></div></li></ol>

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

\[\mathrm{f} : x \rightarrow \left|\frac{2x + 4}{3 - x}\right|, \quad x \in \mathbb{R}, \quad x \neq 3.\]

Sketch the graph of $y = \mathrm{f}(x)$, giving the equations of any asymptotes and the coordinates of the points where the curve meets the axes.

Explain why the function $\mathrm{f}^2$ does not exist.

The domain of $\mathrm{f}$ is further restricted to $x \leqslant a$, where $a$ is a constant.

State the greatest value of $a$ such that the function $\mathrm{f}^{-1}$ exists.

Hence find $\mathrm{f}^{-1}(x)$ and state its domain.

Sketching Rational Functions

Asymptotes and Discontinuities

N2023 P1 Q7

Graphing Techniques

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">\[\mathrm{f} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le k\]</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that the largest exact value of $k$ such that&nbsp;$\mathrm{f}^{-1}$ exists is $\frac{1}{3}\pi$. Find $\mathrm{f}^{-1}(x)$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">It is given that&nbsp;$k = \frac{1}{3}\pi$. In a single diagram, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$, labelling your graphs clearly.<span class="inline-marks">[3]</span></div></li></ol>
<p>The function $\mathrm{h}$ is defined by $\mathrm{h} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le \frac{1}{3}\pi, \, x \neq \frac{1}{12}\pi$.</p>
<p>Another function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto \left| 3 - \frac{1}{x} \right|, \, -3 \le x \le 3, \, x \neq 0$.</p>
<ol class="abc-list" start="3"><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Explain clearly why $\mathrm{gh}$ exists. Find $\mathrm{gh}(x)$ and its range.<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">Supposing $(\mathrm{gh})^{-1}$ exists for a restricted domain. Find the exact value of $x$ for which $(\mathrm{gh})^{-1}(x) = 0$. Show your working clearly.<span class="inline-marks">[2]</span></div></li></ol>

\[\mathrm{f} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le k\]

Show that the largest exact value of $k$ such that $\mathrm{f}^{-1}$ exists is $\frac{1}{3}\pi$. Find $\mathrm{f}^{-1}(x)$.

It is given that $k = \frac{1}{3}\pi$. In a single diagram, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$, labelling your graphs clearly.

The function $\mathrm{h}$ is defined by $\mathrm{h} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le \frac{1}{3}\pi, \, x \neq \frac{1}{12}\pi$.

Another function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto \left| 3 - \frac{1}{x} \right|, \, -3 \le x \le 3, \, x \neq 0$.

Explain clearly why $\mathrm{gh}$ exists. Find $\mathrm{gh}(x)$ and its range.

Supposing $(\mathrm{gh})^{-1}$ exists for a restricted domain. Find the exact value of $x$ for which $(\mathrm{gh})^{-1}(x) = 0$. Show your working clearly.

Sketching a Function and Its Inverse

Inverse Trigonometric Functions

2024 DHS Promo Q10

<p>The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 + \frac{3}{\mathrm{e}^x - 2}$ for&nbsp;$x \in \mathbb{R}$, $x &gt; \ln 2$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find $\mathrm{f}^{-1}(x)$ and state its domain.</span><span class="inline-marks">[3]</span></div></li></ol>
<p>Another function $\mathrm{g}$ is defined by</p>
<p>\[\mathrm{g} : x \mapsto \begin{cases} 2(x - 1)^2 + 2 & \text{for}\hspace{0.5em} x \le 0, \\ 2 + |2 - x| & \text{for}\hspace{0.5em} 0 &lt; x \le 3. \end{cases}\]</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $y = \mathrm{g}(x)$ and explain why the composite function $\mathrm{f}^{-1}\mathrm{g}$ exists.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find the exact value of $k$ such that&nbsp;$\mathrm{f}^{-1}\mathrm{g}(k) = \ln \frac{5}{2}$.</span><span class="inline-marks">[3]</span></div></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 + \frac{3}{\mathrm{e}^x - 2}$ for $x \in \mathbb{R}$, $x > \ln 2$.

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

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

\[\mathrm{g} : x \mapsto \begin{cases} 2(x - 1)^2 + 2 & \text{for } x \le 0, \\ 2 + |2 - x| & \text{for } 0 < x \le 3. \end{cases}\]

Sketch the graph of $y = \mathrm{g}(x)$ and explain why the composite function $\mathrm{f}^{-1}\mathrm{g}$ exists.

Hence find the exact value of $k$ such that $\mathrm{f}^{-1}\mathrm{g}(k) = \ln \frac{5}{2}$.

2025 RI P1 Q7

<ol class="abc-list"><li><div class="list-item-paragraph">Solve the following integral.</div><ol class="i-list"><li><div class="list-item-paragraph">$\displaystyle \displaystyle\int \frac{1-2x}{\sqrt{3+2x-x^2}} \, \mathrm{d}x$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">$\displaystyle \displaystyle\int x \ln(2-x^2) \, \mathrm{d}x$, where&nbsp;$-\sqrt{2} &lt; x &lt; \ \sqrt{2}$.<span class="inline-marks">[3]</span></div></li></ol></li><li><div class="list-item-paragraph">A function $\mathrm{f}$ is defined by $\mathrm{f}(x) = \mathrm{e}^x - 2$.</div><ol class="i-list"><li><div class="list-item-paragraph">Sketch the graph of $y = |\mathrm{f}(x)|$, indicating clearly the equation(s) of asymptote(s), if any.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Hence find $\displaystyle \displaystyle\int_0^{\ln 3} |\mathrm{f}(x)| \, \mathrm{d}x$, giving your answer in exact form.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Sketching Exponential Functions and Solving Equations Graphically

Integration by Splitting the Numerator

Integration by Parts

Area Bounded by Curves

Integration by Completing the Square

Standard Integrals with Quadratic Denominators

2025 ASRJC P2 Q5

Techniques of Integration

Applications of Integration

<p style="text-align: center">$\mathrm{f}:x\mapsto -{{e}^{\left| 3-x \right|}},x\in \mathbb{R}$ and $\mathrm{g}:x\mapsto -1-{{\left( x-1 \right)}^{2}},x\in \mathbb{R}$.</p>
<p>Determine, with reasons, whether $\mathrm{fg}$ and $\mathrm{gf}$ exist. If the composite function exists, give the rule, domain and range in exact form.</p>

$\mathrm{f}:x\mapsto -{{e}^{\left| 3-x \right|}},x\in \mathbb{R}$ and $\mathrm{g}:x\mapsto -1-{{\left( x-1 \right)}^{2}},x\in \mathbb{R}$.

Determine, with reasons, whether $\mathrm{fg}$ and $\mathrm{gf}$ exist. If the composite function exists, give the rule, domain and range in exact form.

<p>Functions $\mathrm{f}$ and $\mathrm{g}$ are defined as:</p>
<p style="text-align: center">${\mathrm{f}(x) = 4a^2 - x^2,\quad x\in\mathbb{R},}$<br>${\mathrm{g}(x) = \left\vert{}4x - a\right\vert{},\quad x\in\mathbb{R},}$</p>
<p>where&nbsp;$a$ is a constant such that&nbsp;$a\ge 1$.</p>
<ol class="abc-list"><li data-list-item-id="item-1761381825391-laso2kvsk" data-list-item-path="item-1761381825391-laso2kvsk"><div class="list-item-paragraph">Sketch on the same diagram the graph of $\mathrm{f}\left( x \right)$ and the graph of $\mathrm{g}\left( x \right)$. The sketch must include the coordinates of any points where each graph meets the coordinate axes and the exact $x$ coordinates of intersections in terms of $a$.</div></li><li data-list-item-id="item-1761381826051-7bvnu9kpd" data-list-item-path="item-1761381826051-7bvnu9kpd"><div class="list-item-paragraph">Find, in exact form where appropriate, the solutions of the equation:</div><div class="list-item-paragraph" style="text-align: center">$4-{{x}^{2}}=\left| 4x-1 \right|$.</div></li></ol>

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined as:

${\mathrm{f}(x) = 4a^2 - x^2,\quad x\in\mathbb{R},}$
${\mathrm{g}(x) = \left\vert{}4x - a\right\vert{},\quad x\in\mathbb{R},}$

where $a$ is a constant such that $a\ge 1$.

Modulus Graphs and Inequalities Q1

<ol class="abc-list"><li><div class="list-item-paragraph">Without using a calculator, solve the inequality</div><div class="list-item-paragraph" style="text-align: center">$\frac{2{{x}^{3}}+{{x}^{2}}-5x+4}{x+1}>x$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Show that there are no solutions to the equation</div><div class="list-item-paragraph" style="text-align: center">$b\left| x-1 \right|=x-b$, where $b>1$ is a fixed constant.</div><p style="text-align: right;">[4]</p></li></ol>

2022 RI J1 CT Q5

<ol class="abc-list"><li data-list-item-id="item-1761043308513-c4swo9cg0" data-list-item-path="item-1761043308513-c4swo9cg0"><div class="list-item-paragraph">Sketch, on the same axes, the graphs of $y=\left| {{x}^{2}}-5 \right|$ and $y=2x-1$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1761043309105-l0q6x1cfl" data-list-item-path="item-1761043309105-l0q6x1cfl"><div class="list-item-paragraph">Find the exact solutions of $\left| {{x}^{2}}-5 \right|=2x-1$.<span class="inline-marks">[4]</span></div></li></ol>

N2021 P1 Q2

<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

<ol class="abc-list"><li data-list-item-id="item-1760868194386-z58j953ij" data-list-item-path="item-1760868194386-z58j953ij"><div class="list-item-paragraph">Sketch the curve with equation $y=\left| \frac{2x-q}{x-1} \right|$, where&nbsp;$q&gt;2$, stating the equations of the asymptotes and the coordinates of the points where the curve meets the axes.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760868195061-9l7iiy1l5" data-list-item-path="item-1760868195061-9l7iiy1l5"><div class="list-item-paragraph">Hence, by adding a suitable line on the same diagram in part <strong>(a)</strong>, solve the inequality</div><div class="list-item-paragraph">$\left| \frac{2x-q}{x-1} \right|-2x&gt;0$, giving your answer in terms of $q$.<span class="inline-marks">[3]</span></div></li></ol>

2024 HCI Promo Q4

<ol class="abc-list"><li data-list-item-id="item-1760868105916-qaij0z4q7" data-list-item-path="item-1760868105916-qaij0z4q7"><div class="list-item-paragraph">Sketch the graph of $y=\frac{4-x}{2+3x-2{{x}^{2}}}$. Give the equations of the asymptotes and the coordinates of the point where the curve intersects the $x$-axis.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760868106726-lon3wsd9x" data-list-item-path="item-1760868106726-lon3wsd9x"><div class="list-item-paragraph">Solve the inequality $\frac{4-x}{2+3x-2{{x}^{2}}}&gt;0$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760868107251-ftvwa4m4s" data-list-item-path="item-1760868107251-ftvwa4m4s"><div class="list-item-paragraph">Hence solve the inequality $\frac{4-\left| x \right|}{2+3\left| x \right|-2{{x}^{2}}}&gt;0$.<span class="inline-marks">[2]</span></div></li></ol>

Inequalities involving Substitution

2024 ASRJC Promo Q4

<p>Solve exactly the inequality $\frac{3x-2}{x-1}\le \frac{4}{x+2}$.<span class="inline-marks">[3]</span></p>
<p>Hence solve $\frac{3\left| x \right|-2}{\left| x \right|-1}\le \frac{4}{\left| x \right|+2}$.<span class="inline-marks">[2]</span></p>

Solve exactly the inequality $\frac{3x-2}{x-1}\le \frac{4}{x+2}$.

Hence solve $\frac{3\left| x \right|-2}{\left| x \right|-1}\le \frac{4}{\left| x \right|+2}$.

Math Question Bank