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PSLE

PSLE Foundation Mathematics

<p>The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where&nbsp;$x,y&gt;0$. It is given that&nbsp;$y=2$ when&nbsp;$x=1$.</p>
<p>Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.</p>
<p>The estimated change in $y$ is therefore&nbsp;$h\times\text{current gradient}$. Let&nbsp;$y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is</p>
<p>\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]</p>
<p>At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.</p>
<ol class="abc-list"><li data-list-item-id="p2-tzc-q12-item01" data-list-item-path="p2-tzc-q12-item01"><div class="list-item-paragraph">Take $h=0.01$. First calculate the gradient at $(1,2)$. Use this gradient in the Euler formula to show that the first estimate, at $x_1=1.01$, is $y_1=2.06$.</div><div class="list-item-paragraph">Then calculate $y_2$, showing your substitution and giving its value to 5 decimal places. Continue the process to estimate $y$ when&nbsp;$x=1.05$, giving this estimate to 3 significant figures.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item02" data-list-item-path="p2-tzc-q12-item02"><div class="list-item-paragraph">Using the substitution $u=xy$, show that the differential equation can be written as&nbsp;$\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item03" data-list-item-path="p2-tzc-q12-item03"><div class="list-item-paragraph">Hence, by separating the variables and using the given initial condition, solve the differential equation, giving $y$ as a function of $x$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p2-tzc-q12-item04" data-list-item-path="p2-tzc-q12-item04"><div class="list-item-paragraph">State the largest possible domain of this solution containing $x=1$, and its corresponding range.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item05" data-list-item-path="p2-tzc-q12-item05"><div class="list-item-paragraph">By considering the shape of the solution curve for&nbsp;$1\leq x\leq1.05$ and the straight-line steps in Euler's method, state whether the approximation in part (a) is an overestimate or an underestimate. Give a reason.<span class="inline-marks">[1]</span></div></li></ol>

The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where $x,y>0$. It is given that $y=2$ when $x=1$.

Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.

The estimated change in $y$ is therefore $h\times\text{current gradient}$. Let $y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is

\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]

At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.

Euler's Method

Concavity of Graphs

Solving Differential Equations by Separation of Variables

Function Notation, Domain and Range

Solving Differential Equations by Substitution

Range of a Function

Product Rule

Domain of a Function

Sketching Solution Curves of Differential Equations

IB AA HL May 2026 P2 TZC Q12 [modified]

Functions

Differential Equations

Techniques of Differentiation

Applications of Differentiation

<p>A curve, $y=f(x)$, undergoes the transformations $A$, $B$, $C$ and $D$ in succession:</p>
<ol class="ABC-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">A translation by $5$ units in the negative $x$-direction.</div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">A scaling by a scale factor of $3$ parallel to the $x$-axis.</div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">A reflection about the $y$-axis.</div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">A translation by $2$ units in the negative $y$-direction.</div></li></ol>
<p>The equation of the resulting curve is $y=9-x+2\ln\left(5-\dfrac{x}{3}\right)$, where&nbsp;$x&lt;15$.</p>
<p>Find the equation of the original curve $y=f(x)$, and state its domain.<span class="inline-marks">[6]</span></p>

A curve, $y=f(x)$, undergoes the transformations $A$, $B$, $C$ and $D$ in succession:

The equation of the resulting curve is $y=9-x+2\ln\left(5-\dfrac{x}{3}\right)$, where $x<15$.

Find the equation of the original curve $y=f(x)$, and state its domain.

Recovering Original Graphs from Transformations

Recovering a logarithmic function after graph transformations

Transformations of Graphs

<ol class="abc-list"></ol>
<ol class="abc-list"><li data-list-item-id="item-1772957783014-v9o9bijfb"><div class="list-item-paragraph">Solve the equation $\log_{2}(x+2) - \log_{\sqrt{2}}(x-1) = 1$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b-i-14"><div class="list-item-paragraph">It is given that&nbsp;$\lg a - \lg b = \lg(a+b)$.</div><ol class="i-list"><li data-list-item-id="sub-item-i"><div class="list-item-content"><span class="list-item-text">Express $a$ in terms of $b$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="sub-item-ii"><div class="list-item-content"><span class="list-item-text">State the range of $b$.</span><span class="inline-marks">[1]</span></div></li></ol></li></ol>

Solve the equation $\log_2(x+2)-\log_{\sqrt{2}}(x-1)=1$.

It is given that $\lg a-\lg b=\lg(a+b)$.

Solving Logarithm Equations

Laws of Logarithms

2025 AHS PRELIMS P1 Q9

Exponential and Logarithmic Functions

<p>It is given that&nbsp;$\log_{3}(4 - x^{2}) - \log_{\sqrt{3}}(x - 1) = 1$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Explain clearly why $1 &lt; x &lt; 2$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence solve $\log_{3}(4 - x^{2}) - \log_{\sqrt{3}}(x - 1) = 1$ and show that it has only one solution.</span><span class="inline-marks">[4]</span></div></li></ol>

It is given that $\log_{3}(4 - x^{2}) - \log_{\sqrt{3}}(x - 1) = 1$.

Hence solve $\log_{3}(4 - x^{2}) - \log_{\sqrt{3}}(x - 1) = 1$ and show that it has only one solution.

IP3 Class Test (May 2025) Q8 (2023 ACSBR AM Prelim P2 Q8)

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">\[\mathrm{f}(x) = \frac{x + 1}{2x - 3} \text{ for}\hspace{0.5em} x \in \mathbb{R}, x \neq k.\]</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">State the value of $k$ and explain why this value has to be excluded from the domain of $\mathrm{f}$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Determine, with a reason, if&nbsp;$\mathrm{f}^2$ exists.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find $\mathrm{f}^{-1}(x)$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find $\mathrm{h}(x)$ for which $\mathrm{fh}(x) = \frac{3x - 7}{2x - 1}, x \neq \frac{1}{2}$.</span><span class="inline-marks">[3]</span></div></li></ol>

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

\[\mathrm{f}(x) = \frac{x + 1}{2x - 3} \text{ for } x \in \mathbb{R}, x \neq k.\]

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

Determine, with a reason, if $\mathrm{f}^2$ exists.

Hence find $\mathrm{h}(x)$ for which $\mathrm{fh}(x) = \frac{3x - 7}{2x - 1}, x \neq \frac{1}{2}$.

Existence of Composite Functions

Inverse Functions

Recovering a Function from a Composite

2025 NYJC P1 Q9

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Solve the equation $\log_2 (x + 3) - \log_4 (3x + 1) = 1$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">It is given that $\ln p - \ln(p + q) = \ln q$.</span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Express $p$ in terms of $q$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">State the range of values of $p$ and explain clearly why $0 &lt; q &lt; 1$.</span><span class="inline-marks">[2]</span></div></li></ol></li></ol>

Solve the equation $\log_2 (x + 3) - \log_4 (3x + 1) = 1$.

It is given that $\ln p - \ln(p + q) = \ln q$.

2023 CSS AM S3 EOY Q9 

<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>

Composite Functions

One-to-One Functions and Inverse Existence

Inverse Trigonometric Functions

Restricting a Domain to Obtain an Inverse

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

2022 NYJC CT Q7

<p>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \ln \left( {{x}^{2}}-x-2 \right)$, $c < x < d$, where $c < d < 0$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find the values of $c$ and $d$ such that the range of $\mathrm{f}$ consists of all real values which are less than $\mathrm{e}$.</div></li><li><div class="list-item-paragraph">Find an expression for ${{\mathrm{f}}^{-1}}\left( x \right)$ for all the values of $c$ and $d$ found in <strong>(a)</strong>.</div></li><li><div class="list-item-paragraph">Write down an expression for $\mathrm{h}\left( x \right)$ such that $\mathrm{hf}\left( x \right)={{x}^{2}}-x-2$.</div></li></ol>

$\mathrm{f}:x\mapsto \ln \left( {{x}^{2}}-x-2 \right)$, $c < x < d$, where $c < d < 0$.

2006 MJC Promo

<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 function $\mathrm{f}$ is defined by $\mathrm{f}:x\to \ln \left[ {{\left( x-1 \right)}^{2}}-k \right]$, where $x<-2$ and $k$ is a constant. Find the maximum value of $k$. Another function $\mathrm{g}$ is defined by $\mathrm{g}:x\to \tan x,-\frac{\pi }{2} < x < \frac{\pi }{2}.$ If $k=8$,</p>
<ol class="i-list"><li><div class="list-item-paragraph">define ${{\mathrm{f}}^{-1}}$, the inverse function of $\mathrm{f}$, in a similar form.</div></li><li><div class="list-item-paragraph">show that the composite function ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}$ exists and find the range of this composite function.</div></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\to \ln \left[ {{\left( x-1 \right)}^{2}}-k \right]$, where $x<-2$ and $k$ is a constant. Find the maximum value of $k$. Another function $\mathrm{g}$ is defined by $\mathrm{g}:x\to \tan x,-\frac{\pi }{2} < x < \frac{\pi }{2}.$ If $k=8$,

Range of a Composite Function

2008 AJC P1 Q11

<ol class="abc-list"><li><div class="list-item-paragraph">Given that $\lg x=a$ and $\lg y=b$, express $\lg \sqrt{\frac{1000y}{{{x}^{2}}}}$ in terms of $a$ and of $b$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Solve the equation ${{\log }_{4}}p-8{{\log }_{p}}4=2$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">State the range of values of $x$ for ${{\log }_{x-1}}\left( 6-x \right)$ to be defined.</div><p style="text-align: right;">[1]</p></li></ol>

2020 BGSS AM S4 P2 Q1

<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>The function $\mathrm{f}$ is defined by</p>
<p style="text-align: center">$\mathrm{f}:x\mapsto \frac{-x-3}{x+1}$, for $\mathrm{ }x\in \mathbb{R}$, $x\ne k$</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">State the value of $k$ 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">Find ${{\mathrm{f}}^{-1}}$. Hence find ${{\mathrm{f}}^{2}}\left( x \right)$.</span></div><p style="text-align: right;">[3]</p></li></ol>
<p>The function $\mathrm{g}$ is defined by</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto \sqrt{x+1}$, $x\in \mathbb{R}$, $x>-1$</p>
<ol class="i-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find the range of $\mathrm{fg}$.</span></div><p style="text-align: right;">[2]</p></li></ol>

$\mathrm{f}:x\mapsto \frac{-x-3}{x+1}$, for $\mathrm{ }x\in \mathbb{R}$, $x\ne k$

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

Find ${{\mathrm{f}}^{-1}}$. Hence find ${{\mathrm{f}}^{2}}\left( x \right)$.

$\mathrm{g}:x\mapsto \sqrt{x+1}$, $x\in \mathbb{R}$, $x>-1$

2023 SAJC Promo Q8

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">The curve $y=\mathrm{f}(x)$ cuts the axes at $(a,0)$ and $(0,b)$. It is given that ${{\mathrm{f}}^{-1}}(x)$ exists. State, if it is possible to do so, the coordinates of the points where the following curves cut the axes.</span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">$y=\mathrm{f}(2x)$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$y=\mathrm{f}(x-1)$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$y=\mathrm{f}(2x-1)$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$y={{\mathrm{f}}^{-1}}(x)$</span></div><p style="text-align: right;">[4]</p></li></ol></li><li><div class="list-item-content"><span class="list-item-text">The function $\mathrm{g}$ is defined by</span></div><div class="list-item-paragraph" style="text-align: center">$\mathrm{g}:x\mapsto 1-\frac{1}{1-x}$, where $x\in \mathbb{R}$, $x\ne a$.</div><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 $\mathrm{g}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Find ${{\mathrm{g}}^{2}}(x)$ and ${{\mathrm{g}}^{-1}}(x)$, giving your answers in simplified form.</span></div><p style="text-align: right;">[4]</p></li><li><div class="list-item-content"><span class="list-item-text">Find the values of $b$ such that ${{\mathrm{g}}^{2}}(b)={{\mathrm{g}}^{-1}}(b)$.</span></div><p style="text-align: right;">[2]</p></li></ol></li></ol>

The curve $y=\mathrm{f}(x)$ cuts the axes at $(a,0)$ and $(0,b)$. It is given that ${{\mathrm{f}}^{-1}}(x)$ exists. State, if it is possible to do so, the coordinates of the points where the following curves cut the axes.

Math Question Bank