Cambridge Syllabus Notes

IGCSE Additional Mathematics (0606) – Comprehensive Summary Notes

1. Functions

1.1 Key terminology

Term	Definition / What to check
Function	A rule that assigns to each element $x$ in a domain a unique element $y$ in a range.
One‑to‑one (injective)	If $f(x_1)=f(x_2)$ implies $x_1=x_2$. Graphically, it passes the horizontal‑line test.
Many‑to‑one	Two different $x$‑values give the same $y$ (fails the horizontal‑line test). Such a function has no inverse.
Inverse function $f^{-1}$	Exists only for one‑to‑one functions. Obtain by swapping $x$ and $y$ in $y=f(x)$ and solving for $y$.
Composition $f\circ g$	$(f\circ g)(x)=f\bigl(g(x)\bigr)$. Domain = $\{x\mid x\in\text{dom}(g)\text{ and }g(x)\in\text{dom}(f)\}$.

1.2 How to decide if a function is invertible

Identify the domain.
Apply the horizontal‑line test (or prove $f'(x)>0$ or $f'(x)<0$ on the domain).
If the test fails, restrict the domain to a monotonic interval before finding an inverse.

Example 1 – Invertibility

Determine whether $f(x)=x^{2}$ is invertible on the following domains and write the inverse where it exists.

Domain $x\ge0$: Passes the horizontal‑line test. $y=x^{2}\;\Rightarrow\;x=\sqrt{y}\;$ ⇒ $f^{-1}(x)=\sqrt{x}$, $x\ge0$.
Domain $x\in\mathbb{R}$: Fails the test (e.g. $f(-2)=f(2)=4$). No inverse.

1.3 Sketching a function

Find intercepts (set $x=0$ for $y$‑intercept, $y=0$ for $x$‑intercepts).
Identify symmetry (even, odd, or neither).
Determine monotonicity using $f'(x)$ (if known) or by testing intervals.
Locate asymptotes (vertical, horizontal, slant) where applicable.
Plot a few key points and sketch the curve.

2. Quadratic Functions

2.1 Standard and vertex forms

Standard: $y=ax^{2}+bx+c\;(aeq0)$.
Vertex (completed‑square) form: $y=a(x-h)^{2}+k$, where $h=-\dfrac{b}{2a}$ and $k=f(h)$.

2.2 Discriminant

For $ax^{2}+bx+c=0$, $D=b^{2}-4ac$.

D	Nature of roots
$>0$	Two distinct real roots.
$=0$	One repeated real root (tangent to the $x$‑axis).
$<0$	No real roots (parabola does not cross the $x$‑axis).

2.3 Solving quadratic inequalities

Find the roots (or vertex) of the corresponding quadratic equation.
Draw a sign chart or test a point in each interval defined by the roots.
Include the equality sign if the inequality is “$\le$” or “$\ge$”.

Example 2 – Solving a quadratic inequality

Solve $2x^{2}-8x+6\le0$.

Divide by 2: $x^{2}-4x+3\le0$.
Factor: $(x-1)(x-3)\le0$.
Sign chart: negative between the roots. $\displaystyle 1\le x\le3$.

3. Polynomials – Factor & Remainder Theorems

3.1 Statements

Remainder theorem: When $P(x)$ is divided by $(x-a)$, the remainder is $P(a)$.
Factor theorem: $(x-a)$ is a factor of $P(x)$ iff $P(a)=0$.

3.2 Synthetic division (quick method for linear divisors)

Write the coefficients of $P(x)$.
Bring down the leading coefficient.
Multiply by $a$, add to the next coefficient, repeat.
The final number is the remainder; the row above gives the coefficients of the reduced polynomial.

Example 3 – Factorising a cubic

Factor $P(x)=x^{3}-6x^{2}+11x-6$.

Possible rational roots: $\pm1,\pm2,\pm3,\pm6$.
Test $x=1$: $P(1)=0$ ⇒ $(x-1)$ is a factor.
Synthetic division by $x-1$ yields $x^{2}-5x+6$.
Factor the quadratic: $(x-2)(x-3)$.
Result: $P(x)=(x-1)(x-2)(x-3)$.

4. Equations & Inequalities

4.1 Linear equations

$ax+b=0\;\Rightarrow\;x=-\dfrac{b}{a}$ (provided $aeq0$).

4.2 Quadratic equations

Factorising (when possible).
Completing the square.
Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

4.3 Absolute‑value equations & inequalities

Equation $|ax+b|=c$ ($c\ge0$) ⇒ $ax+b=c$ or $ax+b=-c$.
Inequality $|ax+b|
Inequality $|ax+b|>c$ ⇒ $ax+b-c$.

4.4 Modulus (two absolute values)

Solve each absolute value separately, then combine the resulting conditions using intersection (for “and”) or union (for “or”).

Example 4 – Solving an absolute‑value inequality

Solve $|2x-5|\le3$.

$-3\le2x-5\le3$.
Add 5: $2\le2x\le8$.
Divide by 2: $1\le x\le4$.

5. Simultaneous Equations

5.1 Linear pairs

Use substitution or elimination. Choose the method that gives the simplest arithmetic.

5.2 Non‑linear pairs

Substitution: express one variable from one equation and substitute.
Elimination: manipulate equations to remove a term (often after squaring or factoring).
Factorisation after substitution is common (e.g., quadratic in one variable).

Example 5 – Non‑linear simultaneous equations

Solve $\displaystyle\begin{cases} xy=6 \\ x+y=5 \end{cases}$.

From $x+y=5$, $y=5-x$.
Substitute: $x(5-x)=6\;\Rightarrow\;-x^{2}+5x-6=0$.
Multiply by $-1$: $x^{2}-5x+6=0\;\Rightarrow\;(x-2)(x-3)=0$.
Solutions: $(x,y)=(2,3)$ or $(3,2)$.

6. Logarithmic & Exponential Functions

6.1 Forms and basic properties

Exponential: $y=a^{x}$, $a>0$, $aeq1$.
Logarithmic: $x=\log_{a}y\;\Longleftrightarrow\;y=a^{x}$.

6.2 Logarithm laws

$\log_{a}(mn)$	= $\log_{a}m+\log_{a}n$
$\log_{a}\!\left(\dfrac{m}{n}\right)$	= $\log_{a}m-\log_{a}n$
$\log_{a}(m^{k})$	= $k\log_{a}m$
Change of base	$\displaystyle\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$ (any convenient base $c$, often $10$ or $e$)

6.3 Natural logarithm

$\ln x=\log_{e}x$, where $e\approx2.71828$.

6.4 Solving exponential equations

Take logarithms of both sides (any base) and isolate the exponent.

Example 6 – Solving an exponential equation

Solve $5^{2x-1}=125$.

Write $125=5^{3}$.
Equate exponents: $2x-1=3\;\Rightarrow\;2x=4\;\Rightarrow\;x=2$.

7. Straight‑Line Graphs (Linearisation)

7.1 General straight‑line equation

$y=mx+c$, where $m$ = gradient, $c$ = $y$‑intercept.

7.2 Transforming common non‑linear relationships to a straight line

Original form	Transformation	Resulting straight‑line equation
$y=Ax^{n}$ (power law)	Take common (or natural) logs	$\log y=n\log x+\log A$
$y=Ae^{kx}$ (exponential)	Take natural log	$\ln y=kx+\ln A$
$y=\dfrac{A}{x}$ (inverse)	Plot $y$ against $\dfrac1x$	$y=\!A\left(\dfrac1x\right)$ (gradient $A$)
$y=A\log x$ (logarithmic)	Plot $y$ against $\log x$	$y=A\log x$ (gradient $A$)

Example 7 – Linearising a power relationship

Data are believed to follow $y=2.5\,x^{1.8}$. Show how to obtain a straight line.

Take common logs: $\log y=\log2.5+1.8\log x$.
Plot $\log y$ (vertical) against $\log x$ (horizontal).
Gradient should be $1.8$, intercept $\log2.5\approx0.398$.

8. Coordinate Geometry of the Circle

8.1 Standard equation

$(x-h)^{2}+(y-k)^{2}=r^{2}$, centre $C(h,k)$, radius $r$.

8.2 Finding points of intersection

Substitute the line equation $y=mx+c$ (or $x=ky+d$) into the circle.
Solve the resulting quadratic for the remaining variable.
Use the discriminant to decide: $D>0$ (two points), $D=0$ (tangent), $D<0$ (no real intersection).

8.3 Tangent at a known point $(x_{1},y_{1})$ on the circle

Using the point‑gradient form derived from the circle:

\[ (x_{1}-h)(x-h)+(y_{1}-k)(y-k)=r^{2} \]

which simplifies to the linear equation

\[ (x_{1}-h)x+(y_{1}-k)y = r^{2}+hx_{1}+ky_{1}-h^{2}-k^{2}. \]

8.4 Two‑circle problems

Subtract the two circle equations to obtain the line joining their centres (or the radical line).
Combine this line with either circle equation to solve for the intersection points.

Example 8 – Tangent to a circle

Find the equation of the tangent to $(x-2)^{2}+(y+1)^{2}=9$ at $P(5,-1)$.

Centre $C(2,-1)$, radius $r=3$.
Radius $CP$ is horizontal ⇒ slope $0$ ⇒ tangent is vertical: $x=5$.
Using the formula: $(5-2)(x-2)+( -1+1)(y+1)=9\;\Rightarrow\;3(x-2)=9\;\Rightarrow\;x=5$.

Calculus – Differentiation in Connected Rates, Small Increments & Approximations

1. Core ideas

The derivative $\displaystyle\frac{dy}{dx}$ gives the instantaneous rate of change of $y$ with respect to $x$.
In the IGCSE syllabus three applications are emphasised:
1. Related (connected) rates of change.
2. Small (infinitesimal) increments $\Delta x,\;\Delta y$.
3. Linear (tangent‑line) approximations.

2. Connected (Related) Rates of Change

2.1 General procedure

Write the relationship linking the variables (often a geometric formula).
Differentiate **implicitly** with respect to the independent variable (usually time $t$) using the chain rule: \[ \frac{d}{dt}\bigl[f(x)\bigr]=f'(x)\,\frac{dx}{dt}. \]
Insert the known values of the variables and any given rates.
Solve algebraically for the required unknown rate.

2.2 Common patterns

Relationship	Typical differentiation
Area of a circle $A=\pi r^{2}$	$\displaystyle\frac{dA}{dt}=2\pi r\,\frac{dr}{dt}$
Volume of a sphere $V=\frac{4}{3}\pi r^{3}$	$\displaystyle\frac{dV}{dt}=4\pi r^{2}\,\frac{dr}{dt}$
Perimeter of a rectangle $P=2(l+w)$	$\displaystyle\frac{dP}{dt}=2\!\left(\frac{dl}{dt}+\frac{dw}{dt}\right)$
Right‑triangle $x^{2}+y^{2}=z^{2}$	$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$

Example 9 – Expanding circular oil spill (revisited)

Given $dr/dt=0.5\;\text{m min}^{-1}$ at $r=10\,$m, find $dA/dt$.

\[ \frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi(10)(0.5)=10\pi\;\text{m}^{2}\!\text{min}^{-1}. \]

Example 10 – Ladder sliding down a wall

A 5 m ladder leans against a vertical wall. The foot slides away at $1.2\;\text{m s}^{-1}$. Find the speed of the top when the foot is $3\;$m from the wall.

Relation (Pythagoras): $x^{2}+y^{2}=5^{2}$, where $x$ = distance of foot, $y$ = height of top.
Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\;\Rightarrow\;x\frac{dx}{dt}+y\frac{dy}{dt}=0$.
When $x=3$, $y=\sqrt{5^{2}-3^{2}}=4$ m. Substitute: $3(1.2)+4\frac{dy}{dt}=0\;\Rightarrow\;\frac{dy}{dt}=-\dfrac{3.6}{4}=-0.9\;\text{m s}^{-1}$.
Interpretation: the top is descending at $0.9\;\text{m s}^{-1}$.

3. Small Increments ($\Delta x$, $\Delta y$)

3.1 Concept

If $x$ changes by a small amount $\Delta x$, the corresponding change in $y=f(x)$ is approximated by

\[ \Delta y\;\approx\;f'(x)\,\Delta x. \]

This is the linearisation of $f$ at the point $x$.

3.2 When to use

When the exact change is difficult to compute but $\Delta x$ is known to be small.
In error‑propagation problems (e.g., measurement uncertainties).

Example 11 – Approximate change in a square’s area

A square has side $s=10.0\,$cm. The side length is measured to the nearest $0.1\,$cm. Estimate the maximum possible error in the area $A=s^{2}$.

Derivative: $\displaystyle\frac{dA}{ds}=2s$.
Maximum $\Delta s=0.05\,$cm (half the unit of the last decimal place).
Approximate error: $\Delta A\approx2s\,\Delta s=2(10.0)(0.05)=1.0\;\text{cm}^{2}$.

4. Linear Approximations (Tangent‑Line Estimates)

4.1 Formula

The tangent line to $y=f(x)$ at $x=a$ is

\[ y\approx f(a)+f'(a)(x-a). \]

This provides a quick estimate of $f(x)$ for $x$ close to $a$.

4.2 Commonly used approximations

Function	Linearisation at $x=a$
$\sqrt{x}$	$\displaystyle\sqrt{a}+\frac{1}{2\sqrt{a}}(x-a)$
$\ln x$	$\displaystyle\ln a+\frac{1}{a}(x-a)$
$e^{x}$	$\displaystyle e^{a}+e^{a}(x-a)$
$\sin x$ (radians)	$\displaystyle\sin a+\cos a\,(x-a)$
$\cos x$ (radians)	$\displaystyle\cos a-\sin a\,(x-a)$

Example 12 – Estimate $\sqrt{4.1}$

Choose $a=4$ (easy square root). $f(x)=\sqrt{x}$, $f'(x)=\dfrac{1}{2\sqrt{x}}$.
Linearisation: $\sqrt{4.1}\approx\sqrt{4}+\frac{1}{2\sqrt{4}}(4.1-4)=2+\frac{1}{4}(0.1)=2+0.025=2.025$.
Actual value $=2.0249\ldots$, confirming the accuracy.

Example 13 – Approximate $\ln(1.05)$

Take $a=1$, $f(x)=\ln x$, $f'(x)=1/x$.
Linearisation: $\ln(1.05)\approx\ln1+\frac{1}{1}(1.05-1)=0+0.05=0.05$.
Actual $\ln(1.05)=0.04879\ldots$ – good for a quick estimate.

5. Summary of When to Use Each Technique

Situation	Best tool
Two quantities linked by a known equation, and you know one rate of change.	Related‑rates differentiation.
Exact change is hard to compute but the increment $\Delta x$ is tiny.	Small‑increment approximation $\Delta y\approx f'(x)\Delta x$.
Need a quick numerical estimate of $f(x)$ near a point where $f(a)$ and $f'(a)$ are simple.	Linear (tangent‑line) approximation $f(a)+f'(a)(x-a)$.

6. Practice Questions (selected)

When a spherical balloon is being inflated, its radius increases at $0.3\;\text{cm s}^{-1}$. Find the rate at which its volume is increasing when the radius is $5\,$cm.
A car travels along a straight road. Its position is $s(t)=4t^{2}+3t$ (metres, $t$ in seconds). Approximate the change in position between $t=2\,$s and $t=2.1\,$s using the small‑increment method.
Use a linear approximation to estimate $\cos(0.1)$ radians.
A conical tank has radius $r$ and height $h$ related by $r=0.5h$. Water is being pumped in so that the height rises at $0.02\;$m s$^{-1}$. Find the rate at which the volume of water is increasing when $h=4\;$m.

Answers (for teacher’s reference):

$\displaystyle\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}=4\pi(5)^{2}(0.3)=30\pi\;\text{cm}^{3}\!\text{s}^{-1}$.
$\displaystyle ds/dt=8t+3$, so at $t=2$, $ds/dt=19\;\text{m s}^{-1}$. Approximate $\Delta s\approx19(0.1)=1.9\;$m.
$\displaystyle\cos x\approx1-\frac{x^{2}}{2}$ near $0$. For $x=0.1$, $\cos(0.1)\approx1-\frac{0.01}{2}=0.995$ (actual $0.995004\ldots$).
Volume of cone $V=\frac13\pi r^{2}h=\frac13\pi(0.5h)^{2}h=\frac{\pi}{12}h^{3}$. $dV/dt=\frac{\pi}{4}h^{2}\,dh/dt=\frac{\pi}{4}(4)^{2}(0.02)=0.32\pi\;\text{m}^{3}\!\text{s}^{-1}$.

Apply differentiation to connected rates of change, small increments and approximations