Objective
Provide a concise, systematic set of Cambridge IGCSE Additional Mathematics (0606) notes that cover all 13 content areas of the syllabus, with a focus on clear procedures, key formulae, and worked examples.
1. Numbers and Algebra
1.1 Indices and Surds
Index laws (valid for non‑zero bases):
$a^{m}\,a^{n}=a^{m+n}$
$\displaystyle\frac{a^{m}}{a^{n}}=a^{\,m-n}$
$(a^{m})^{n}=a^{\,mn}$
$a^{0}=1,\qquad a^{-n}= \dfrac{1}{a^{n}}$
Surds – irrational roots written in radical form.
Product rule: $\sqrt{a}\,\sqrt{b}= \sqrt{ab}$ (for $a,b\ge0$).
Rationalising denominator: $\displaystyle\frac{1}{\sqrt{a}}=\frac{\sqrt{a}}{a}$.
1.2 Factorisation
Technique Typical Form Result
Common factor $ab+ac$ $a(b+c)$ Difference of squares $a^{2}-b^{2}$ $(a-b)(a+b)$ Perfect square $a^{2}\pm2ab+b^{2}$ $(a\pm b)^{2}$ Quadratic $ax^{2}+bx+c$ $a(x-r_{1})(x-r_{2})$ (if roots $r_{1},r_{2}$ exist) Cubic (simple) $x^{3}+px^{2}+qx+r$ Factor out a linear term, then use quadratic factorisation.
1.3 Partial Fractions (Simple Cases)
For $ \displaystyle\frac{A}{x-a} + \frac{B}{x-b}$ when $\displaystyle\frac{P(x)}{(x-a)(x-b)}$ is proper.
2. Functions
2.1 Definition, Domain & Range
A function $y=f(x)$ assigns exactly one $y$ to each permissible $x$.
Domain: set of all $x$ for which $f(x)$ is defined.
Range: set of all possible $y$ values.
2.2 Composite Functions
If $u=g(x)$ and $y=f(u)$ then the composite is $y=f(g(x))$.
Domain of $f\!\circ\!g$ = $\{x\in\text{Dom}(g)\mid g(x)\in\text{Dom}(f)\}$.
2.3 Inverse Functions
Swap $x$ and $y$ and solve for $y$.
Exist only if $f$ is one‑to‑one (horizontal‑line test).
Graph of $f^{-1}$ is a reflection of $f$ in the line $y=x$.
2.4 Sketching Transformations
Transformation Effect on $y=f(x)$
$y=f(x)+k$ Shift up $k$ units. $y=f(x-k)$ Shift right $k$ units. $y=af(x)$ Vertical stretch by $|a|$ (reflection if $a<0$). $y=f(bx)$ Horizontal compression by factor $1/|b|$ (reflection if $b<0$).
Worked Example 1 – Inverse Function
Find $f^{-1}(x)$ for $f(x)=\dfrac{2x-3}{5}$.
Write $y=\dfrac{2x-3}{5}$.
Swap: $x=\dfrac{2y-3}{5}$.
Multiply by $5$: $5x=2y-3$.
Solve for $y$: $2y=5x+3\;\Rightarrow\;y=\dfrac{5x+3}{2}$.
Thus $f^{-1}(x)=\dfrac{5x+3}{2}$.
3. Quadratic Functions
3.1 Standard and Vertex Forms
Standard: $y=ax^{2}+bx+c$.
Vertex (completed square): $y=a(x-h)^{2}+k$, where $h=-\dfrac{b}{2a}$ and $k=f(h)$.
3.2 Discriminant
For $ax^{2}+bx+c=0$, the discriminant $D=b^{2}-4ac$ determines the nature of roots:
$D>0$ – two distinct real roots (graph cuts $x$‑axis at two points).
$D=0$ – one repeated real root (tangent to $x$‑axis).
$D<0$ – no real roots (graph does not intersect $x$‑axis).
3.3 Solving Quadratics
Factorisation (when possible).
Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
Completing the square.
3.4 Quadratic Inequalities
Use sign‑chart or graph: determine intervals where the quadratic expression is $>0$ or $<0$.
Worked Example 2 – Vertex and Roots
Find the vertex and the roots of $y=2x^{2}-8x+5$.
Vertex: $h=-\dfrac{b}{2a}= -\dfrac{-8}{4}=2$, $k=2(2)^{2}-8(2)+5= -3$. Vertex $(2,-3)$.
Discriminant: $D=(-8)^{2}-4\cdot2\cdot5=64-40=24>0$.
Roots: $x=\dfrac{8\pm\sqrt{24}}{4}= \dfrac{8\pm2\sqrt{6}}{4}=2\pm\frac{\sqrt{6}}{2}$.
4. Polynomials
4.1 Degree, Leading Coefficient & End Behaviour
Degree = highest exponent of $x$.
Sign of leading coefficient determines whether the graph rises or falls as $|x|\to\infty$.
4.2 Remainder and Factor Theorems
Remainder theorem: If $P(x)$ is divided by $(x-a)$, remainder $=P(a)$.Factor theorem: $(x-a)$ is a factor of $P(x)$ iff $P(a)=0$.
4.3 Polynomial Division (Long Division)
Used to obtain quotient $Q(x)$ and remainder $R(x)$ such that $P(x)=D(x)Q(x)+R(x)$.
Worked Example 3 – Factor Theorem
Show that $x-2$ is a factor of $P(x)=x^{3}-6x^{2}+11x-6$ and find the remaining quadratic factor.
Evaluate $P(2)=8-24+22-6=0\;\Rightarrow\;x-2$ is a factor.
Divide $P(x)$ by $x-2$ (synthetic division):
2 | 1 -6 11 -6
2 -8 6
----------------
1 -4 3 0
Quotient $=x^{2}-4x+3$.
Thus $P(x)=(x-2)(x^{2}-4x+3)=(x-2)(x-1)(x-3)$.
5. Equations and Inequalities
5.1 Linear Equations
Standard form $ax+by=c$. Solve by substitution or elimination.
5.2 Absolute‑Value Equations
$|ax+b|=c\;(c\ge0)$ → $ax+b=c$ or $ax+b=-c$.
$|ax+b|=dx+e$ – solve by considering cases where $dx+e\ge0$ and $dx+e<0$.
5.3 Modulus (Piece‑wise) Equations
Write $|f(x)|$ as a piece‑wise definition and solve each branch.
Worked Example 4 – Absolute Value
Solve $|2x-5|=x+1$.
Case 1: $2x-5\ge0\;(x\ge2.5)$ → $2x-5=x+1\;\Rightarrow\;x=6$ (satisfies $x\ge2.5$).
Case 2: $2x-5<0\;(x<2.5)$ → $-(2x-5)=x+1\;\Rightarrow\;-2x+5=x+1\;\Rightarrow\;3x=4\;\Rightarrow\;x=\frac{4}{3}$ (satisfies $x<2.5$).
Solutions: $x=6$ and $x=\dfrac{4}{3}$.
6. Simultaneous Equations
6.1 Linear Systems (Two Variables)
Elimination or substitution.
Check for parallel (no solution) or coincident (infinitely many solutions) lines.
6.2 Non‑Linear Systems
Typical approach: solve one equation for a variable, substitute into the other, then solve the resulting quadratic or cubic.
Worked Example 5 – Non‑Linear System
Solve $\begin{cases} y= x^{2}+1 \\ x+y=5 \end{cases}$.
Substitute $y$ from first into second: $x+(x^{2}+1)=5\;\Rightarrow\;x^{2}+x-4=0$.
Quadratic formula: $x=\dfrac{-1\pm\sqrt{1+16}}{2}= \dfrac{-1\pm\sqrt{17}}{2}$.
Corresponding $y$: $y=x^{2}+1$ → two points $\bigl(\frac{-1+\sqrt{17}}{2},\;\frac{9+\sqrt{17}}{2}\bigr)$ and $\bigl(\frac{-1-\sqrt{17}}{2},\;\frac{9-\sqrt{17}}{2}\bigr)$.
7. Straight‑Line Graphs
7.1 Equation Forms
Slope‑intercept: $y=mx+c$.
Point‑slope: $y-y_{1}=m(x-x_{1})$.
General: $ax+by=c$.
7.2 Gradient, Parallel & Perpendicular
Gradient $m=\dfrac{\Delta y}{\Delta x}$.
Parallel lines: equal gradients.
Perpendicular lines: $m_{1}m_{2}=-1$ (provided neither is vertical).
7.3 Transforming to Straight‑Line Form
Many relationships become linear after algebraic manipulation (e.g., $y=ab^{x}\;\Rightarrow\;\ln y = x\ln b+\ln a$).
Worked Example 6 – Straight‑Line Form
Show that the data $x$ versus $y$ for $y=5e^{0.3x}$ can be plotted as a straight line.
Take natural logs: $\ln y = \ln 5 +0.3x$.
Set $Y=\ln y$, $X=x$, then $Y=0.3X+\ln5$ – a straight line with gradient $0.3$ and intercept $\ln5$.
8. Circles and Other Conics
8.1 Circle – Standard Form
$(x-h)^{2}+(y-k)^{2}=r^{2}$ where $(h,k)$ is the centre and $r$ the radius.
8.2 Tangent to a Circle
At point $(x_{0},y_{0})$ on the circle, the radius $OP$ is perpendicular to the tangent.
Equation of tangent: $(x_{0}-h)(x-h)+(y_{0}-k)(y-k)=r^{2}$ simplifies to $ (x_{0}-h)x+(y_{0}-k)y = r^{2}$.
8.3 General Conics (Brief)
Ellipse: $\displaystyle\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$.
Hyperbola: $\displaystyle\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$ (or vice‑versa).
Standard forms are used for identifying centre, vertices and asymptotes.
Worked Example 7 – Tangent to a Circle
Find the equation of the tangent to $x^{2}+y^{2}=25$ at the point $(3,4)$.
Centre $(0,0)$, radius $5$.
Using $x_{0}x+y_{0}y=r^{2}$: $3x+4y=25$.
9. Trigonometry
9.1 Basic Ratios
Ratio Definition (right‑angled triangle)
$\sin\theta$ Opposite / Hypotenuse $\cos\theta$ Adjacent / Hypotenuse $\tan\theta$ Opposite / Adjacent
9.2 Fundamental Identities
$\sin^{2}\theta+\cos^{2}\theta=1$.
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$.
Co‑function: $\sin(90^{\circ}-\theta)=\cos\theta$, etc.
9.3 Solving Triangles (SAS, ASA, SSS)
Use the law of sines and cosines when necessary:
\[
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C},\qquad
c^{2}=a^{2}+b^{2}-2ab\cos C.
\]
9.4 Trigonometric Graphs
Period $2\pi$, amplitude $1$ (for $\sin$ and $\cos$).
Key points at $0,\;\frac{\pi}{2},\;\pi,\;\frac{3\pi}{2},\;2\pi$.
Transformations: $y=a\sin(bx+c)+d$.
Worked Example 8 – Solving a Triangle
In $\triangle ABC$, $a=7$, $b=9$, and $\angle C=60^{\circ}$. Find side $c$.
\[
c^{2}=a^{2}+b^{2}-2ab\cos C=7^{2}+9^{2}-2\cdot7\cdot9\cdot\frac12=49+81-63=67.
\]
\[
c=\sqrt{67}\approx8.19.
\]
10. Differentiation – Gradients, Tangents, Normals & Stationary Points
10.1 Differentiation Rules (Recap)
Power, constant, root, product, quotient, chain (as in the original notes).
Trigonometric: $\dfrac{d}{dx}\sin x=\cos x$, $\dfrac{d}{dx}\cos x=-\sin x$, $\dfrac{d}{dx}\tan x=\sec^{2}x$.
Exponential/Logarithmic: $\dfrac{d}{dx}e^{x}=e^{x}$, $\dfrac{d}{dx}\ln x=\dfrac1x$.
10.2 Procedure for Explicit Functions
Find $f'(x)$.
Gradient at $x_{0}$: $m_{t}=f'(x_{0})$ (if defined).
Point on curve: $(x_{0},f(x_{0}))$.
Tangent: $y-f(x_{0})=m_{t}(x-x_{0})$.
Normal gradient $m_{n}=-\dfrac{1}{m_{t}}$ (if $m_{t}eq0$); otherwise normal is vertical $x=x_{0}$.
Special cases: vertical tangent (undefined $f'$) → $x=x_{0}$; horizontal tangent ($m_{t}=0$) → $y=f(x_{0})$.
10.3 Implicit Differentiation
Given $F(x,y)=0$, differentiate w.r.t. $x$: $\displaystyle F_{x}+F_{y}\frac{dy}{dx}=0\;\Rightarrow\;\frac{dy}{dx}= -\frac{F_{x}}{F_{y}}$ (provided $F_{y}eq0$).
10.4 Stationary Points
Set $f'(x)=0$ (or $\dfrac{dy}{dx}=0$ for implicit curves) and solve for $x$.
Classify using:
First‑derivative test: sign change of $f'$ around the point.Second‑derivative test: if $f''(x_{0})>0$ → minimum; $<0$ → maximum; $=0$ → inconclusive.
10.5 Related‑Rates
Express the geometric relationship between the quantities.
Differentiate implicitly with respect to time $t$ (or other independent variable).
Substitute known values and solve for the required rate.
Worked Example 9 – Tangent & Normal (Explicit)
Find the tangent and normal to $y=2x^{3}-3x^{2}+x-5$ at $x=2$.
$f'(x)=6x^{2}-6x+1$ → $m_{t}=13$.
Point: $(2,1)$.
Tangent: $y-1=13(x-2)\;\Rightarrow\;y=13x-25$.
Normal: $m_{n}=-\dfrac{1}{13}$ → $y-1=-\dfrac{1}{13}(x-2)\;\Rightarrow\;y=-\dfrac{1}{13}x+\dfrac{15}{13}$.
Worked Example 10 – Implicit Differentiation (Circle)
For $x^{2}+y^{2}=25$, find tangent and normal at $(3,4)$.
Differentiate: $2x+2y\,\dfrac{dy}{dx}=0\;\Rightarrow\;\dfrac{dy}{dx}= -\dfrac{x}{y}$.
At $(3,4)$: $m_{t}= -\dfrac34$.
Tangent: $y-4=-\dfrac34(x-3)\;\Rightarrow\;3x+4y=25$.
Normal: $m_{n}= \dfrac43$ → $y-4=\dfrac43(x-3)\;\Rightarrow\;4x-3y=0$.
Worked Example 11 – Stationary Points Classification
For $y=x^{3}-6x^{2}+9x+1$, locate and classify stationary points.
$y'=3x^{2}-12x+9=3(x-1)(x-3)$ → $x=1,3$.
$y(1)=5$, $y(3)=1$ → points $(1,5)$ and $(3,1)$.
$y''=6x-12$ → $y''(1)=-6<0$ (max), $y''(3)=6>0$ (min).
Worked Example 12 – Related‑Rates (Cone Filling)
A conical tank (radius 5 cm, height 12 cm) is filled at $4\;\text{cm}^{3}\!/\text{s}$. Find $\dfrac{dh}{dt}$ when $h=6\,$cm.
Volume $V=\dfrac13\pi r^{2}h$, with $r=\dfrac{5}{12}h$ → $V=\dfrac{25\pi}{432}h^{3}$.
Differentiate: $\displaystyle\frac{dV}{dt}= \frac{25\pi}{144}h^{2}\frac{dh}{dt}$.
Insert $dV/dt=4$ and $h=6$: $4= \frac{25\pi}{144}(36)\frac{dh}{dt}\;\Rightarrow\;\frac{dh}{dt}= \frac{4}{25\pi}\cdot\frac{144}{36}= \frac{16}{25\pi}\;\text{cm/s}.$
11. Integration – Indefinite and Definite
11.1 Basic Antiderivatives
Integrand Integral + C
$\displaystyle\int x^{n}\,dx$ $\displaystyle\frac{x^{n+1}}{n+1}+C\;(neq-1)$ $\displaystyle\int \frac{1}{x}\,dx$ $\ln|x|+C$ $\displaystyle\int e^{x}\,dx$ $e^{x}+C$ $\displaystyle\int a^{x}\,dx$ $\dfrac{a^{x}}{\ln a}+C\;(a>0,\;aeq1)$ $\displaystyle\int \sin x\,dx$ $-\cos x+C$ $\displaystyle\int \cos x\,dx$ $\sin x+C$ $\displaystyle\int \sec^{2}x\,dx$ $\tan x+C$
11.2 Integration by Substitution (Reverse Chain Rule)
If $u=g(x)$ then $\displaystyle\int f(g(x))g'(x)\,dx=\int f(u)\,du$.
11.3 Definite Integrals
For continuous $f$ on $[a,b]$:
\[
\int_{a}^{b} f(x)\,dx = F(b)-F(a),\qquad F'(x)=f(x).
\]
Interpretation: signed area between the curve and the $x$‑axis.
11.4 Area Between Two Curves
If $f(x)\ge g(x)$ on $[a,b]$, area $A=\displaystyle\int_{a}^{b}[f(x)-g(x)]\,dx$.
Worked Example 13 – Definite Integral (Area)
Find the area enclosed by $y=x^{2}$ and $y=4x-x^{2}$.
Set $x^{2}=4x-x^{2}\;\Rightarrow\;2x^{2}-4x=0\;\Rightarrow\;x=0$ or $x=2$.
Upper curve $=4x-x^{2}$, lower $=x^{2}$ on $[0,2]$.
Area $A=\displaystyle\int_{0}^{2}\bigl[(4x-x^{2})-x^{2}\bigr]dx=\int_{0}^{2}(4x-2x^{2})dx$.
$A=\Bigl[2x^{2}-\frac{2}{3}x^{3}\Bigr]_{0}^{2}= \bigl(8-\frac{16}{3}\bigr)-0=\frac{8}{3}\;$ square units.
12. Kinematics Applications
12.1 Core Relationships
\[
v(t)=\frac{ds}{dt},\qquad a(t)=\frac{dv}{dt}= \frac{d^{2}s}{dt^{2}}.
\]
12.2 From Acceleration to Motion
Integrate $a(t)$ to obtain $v(t)=\int a(t)\,dt + C_{1}$ (constant $C_{1}$ = initial velocity).
Integrate $v(t)$ to obtain $s(t)=\int v(t)\,dt + C_{2}$ (constant $C_{2}$ = initial displacement).
12.3 From Velocity to Displacement
Displacement over $[t_{1},t_{2}]$ is $\displaystyle\int_{t_{1}}^{t_{2}} v(t)\,dt$.
Worked Example 14 – Motion from Acceleration
A particle has $a(t)=6t$ (m s$^{-2}$). At $t=0$ its velocity is $2\,$m s$^{-1}$ and its displacement is $5\,$m. Find $v(t)$ and $s(t)$.
Integrate $a$: $v(t)=\int6t\,dt=3t^{2}+C_{1}$. Using $v(0)=2$, $C_{1}=2$ → $v(t)=3t^{2}+2$.
Integrate $v$: $s(t)=\int(3t^{2}+2)\,dt= t^{3}+2t+C_{2}$. Using $s(0)=5$, $C_{2}=5$ → $s(t)=t^{3}+2t+5$.
13. Vectors
13.1 Representation
A vector $\mathbf{u}$ in the plane can be written in component form $\mathbf{u}= \langle u_{x},u_{y}\rangle$ or as magnitude & direction $|\mathbf{u}|,\;\theta$.
13.2 Operations
Addition: $\mathbf{u}+\mathbf{v}= \langle u_{x}+v_{x},\;u_{y}+v_{y}\rangle$.
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