3. Pure Mathematics A2 3

Syllabus
9709–2028–2029
Section
3
Level
A2

3.1 Algebra

Syllabus
9709–2028–2029
Topic
3.1
Level
A2

Use the linear modulus graph and sign-consistent cases

y=∣ax+b∣y=|ax+b| is a V-graph with vertex (−b/a,0)(-b/a,0) for $a
e0,range, rangey\ge0,andbranchgradients, and branch gradients-|a|thenthen|a|.Itrepresentsdistancefromthezeroof. It represents distance from the zero ofax+b$.

Form Equivalent condition
∣u∣=∣v∣|u|=|v| u=vu=v or u=−vu=-v
∣x−a∣<b|x-a|<b, b>0b>0 a−b<x<a+ba-b<x<a+b
∣x−a∣>b|x-a|>b, b>0b>0 x<a−bx<a-b or x>a+bx>a+b

For ∣f(x)∣=g(x)|f(x)|=g(x) require g(x)≥0g(x)\ge0, then solve both sign branches and check the original. For variable-side inequalities, split at zeros of both sides and intersect each result with its sign interval.

$|3x-2|=|2x+7|$ gives $3x-2=2x+7$ or $3x-2=-(2x+7)$, so $x=9$ or $x=-1$.

Endpoint inclusion follows << versus ≤\le. Non-linear graphs y=∣f(x)∣y=|f(x)| and y=f(∣x∣)y=f(|x|) are excluded.

Control polynomial division by the remainder degree

P(x)=D(x)Q(x)+R(x),\qquad \deg R<\deg D.A linear divisor leaves a constant remainder; a quadratic divisor may leave $mx+c$.

Write descending powers with zero coefficients for gaps. Repeatedly divide leading terms, multiply the whole divisor, subtract, and stop only when the remaining degree is lower than the divisor degree.

x^4+2x^2+3=(x^2+1)(x^2+1)+2.Thus quotient $x^2+1$ and remainder $2$.

Handle dividends of degree at most 44 and either linear or quadratic divisors. Synthetic division is only a shortcut for suitable linear divisors.

Always report both quotient and remainder, including zero, and verify DQ+R=PDQ+R=P.

Convert divisor statements into polynomial evaluations

Statement Evaluation
remainder on division by x−cx-c is kk P(c)=kP(c)=k
x−cx-c is a factor P(c)=0P(c)=0
ax+bax+b is a factor P(−b/a)=0P(-b/a)=0

Turn every factor or remainder statement into an equation. Solve simultaneous equations for unknown coefficients where needed; divide out confirmed factors before solving the lower-degree polynomial.

For $P(x)=x^3+kx+6$, if $2x-1$ is a factor then $P(1/2)=0$, giving $1/8+k/2+6=0$ and $k=-49/4$.

If division by x+2x+2 leaves remainder 55, use P(−2)=5P(-2)=5, not zero. Use full polynomial division when the quotient is also required.

The zero of ax+bax+b is −b/a-b/a; keep the sign and non-zero remainder exactly as stated.

Write one numerator term for every permitted denominator factor

Denominator structure Partial-fraction form
(ax+b)(cx+d)(ex+f)(ax+b)(cx+d)(ex+f) A/(ax+b)+B/(cx+d)+C/(ex+f)A/(ax+b)+B/(cx+d)+C/(ex+f)
(ax+b)(cx+d)2(ax+b)(cx+d)^2 A/(ax+b)+B/(cx+d)+C/(cx+d)2A/(ax+b)+B/(cx+d)+C/(cx+d)^2
(ax+b)(cx2+d)(ax+b)(cx^2+d) A/(ax+b)+(Bx+C)/(cx2+d)A/(ax+b)+(Bx+C)/(cx^2+d)

Factor the denominator, write the complete matching template, multiply through by the original denominator, then substitute convenient roots and/or equate coefficients to solve all constants.

rac1{x(x+1)}= rac1x- rac1{x+1}.Multiplication by $x(x+1)$ verifies $1=(x+1)-x$.

Substitute the solved coefficients back and recombine as a check before using the decomposition in later algebra or integration.

A repeated factor needs every power, and a quadratic factor needs a linear numerator. Cases where numerator degree exceeds denominator degree are excluded here; do not add an improper-division method.

Expand rational powers and transform the validity condition

For rational $n$ and $|x|<1$:(1+x)^n=1+nx+ rac{n(n-1)}{2!}x^2+ rac{n(n-1)(n-2)}{3!}x^3+\cdots.Useonlyasmanyinitialtermsasrequested.Use only as many initial terms as requested.

Rewrite the expression as a constant multiple of (1+u)n(1+u)^n, substitute uu into the displayed initial terms, expand and collect powers. Do not seek a general term; it is excluded.

(1-2x)^{-1/2}=1+x+ rac32x^2+ rac52x^3+\cdots,obtained with $n=-1/2$ and $u=-2x$.

Transform the convergence condition with the same substitution: ∣u∣<1|u|<1. In the example, ∣−2x∣<1|-2x|<1, so ∣x∣<1/2|x|<1/2. For a shifted/scaled u(x)u(x), solve the resulting inequality and report the full set.

For non-integer rational nn the expansion is generally infinite, not a finite P1 binomial. The standard ∣x∣<1|x|<1 condition applies to the inner series variable uu, not automatically to the original xx.

3.2 Logarithmic and exponential functions

Syllabus
9709–2028–2029
Topic
3.2
Level
A2

Translate index statements with the three logarithm laws

For $a>0$, $a e1$ and $y>0$:a^x=y\quad\Longleftrightarrow\quad \log_a y=x.Hence $\log_a1=0$ and $\log_a a=1$.

Structure Logarithm law (positive arguments)
product log⁡a(MN)=log⁡aM+log⁡aN\log_a(MN)=\log_aM+\log_aN
quotient log⁡a(M/N)=log⁡aM−log⁡aN\log_a(M/N)=\log_aM-\log_aN
power log⁡a(Mp)=plog⁡aM\log_a(M^p)=p\log_aM

Record the positivity conditions before combining or expanding logs. Move coefficients into powers when useful, combine to one logarithm, then translate back to index form.

For $x>0$,\log_2(8x)-\log_2x=\log_2(8)=3.

log⁡(a+b)\log(a+b) does not split. Change-of-base formulae are explicitly excluded from this objective.

Read exponential and natural-log graphs as inverse shapes

y=e^x\quad\Longleftrightarrow\quad x=\ln y.Thus $\ln(e^x)=x$ for real $x$, while $e^{\ln x}=x$ requires $x>0$. Their graphs reflect in $y=x$.

Graph Domain Range Intercept/asymptote
y=ekxy=e^{kx}, $k
e0∣allreal| all realx∣|y>0∣|(0,1);horizontalasymptote; horizontal asymptotey=0$
y=ln⁡xy=\ln x x>0x>0 all real yy (1,0)(1,0); vertical asymptote x=0x=0

For k>0k>0, ekxe^{kx} increases; for k<0k<0, it decreases. Both stay positive and approach, but never cross, the xx-axis in one direction.

$3e^{2x}=12$ gives $e^{2x}=4$, so $x= frac12\ln4$.

An exponential graph has no xx-intercept, and ln⁡x\ln x is undefined for x≤0x\le0. Do not treat ln⁡(x2)=2ln⁡x\ln(x^2)=2\ln x as valid when x<0x<0.

Bring an unknown exponent down with logarithms

Rearrange until each exponential expression is positive, take ln⁡\ln of both sides, use ln⁡(ag(x))=g(x)ln⁡a\ln(a^{g(x)})=g(x)\ln a, then solve the resulting algebraic equation and check in the original.

5^{2x-1}=12\quad\Rightarrow\quad(2x-1)\ln5=\ln12\quad\Rightarrow\quad x= rac{1+\ln12/\ln5}{2}.

Base form Monotonic direction
au<ava^{u}<a^{v} with a>1a>1 u<vu<v
au<ava^{u}<a^{v} with 0<a<10<a<1 u>vu>v

If terms such as a2xa^{2x} and axa^x occur together, set t=axt=a^x with t>0t>0, solve the polynomial in tt, reject non-positive roots, then take logs to recover xx.

Logs can only be taken after both sides are positive. An inequality reverses for a decreasing base 0<a<10<a<1, not merely because logarithms were used.

Linearising a relationship makes a model testable with a straight-line graph

Transform variables so the model becomes Y=mX+c. For y=ab^x, plotting ln y against x gives gradient ln b and intercept ln a; for y=ax^n, plotting ln y against ln x gives gradient n.

Transform measured uncertainties and units consistently, then interpret the gradient and intercept in the original parameters. A straight plot supports the model but does not prove causation.

If ln y=0.7x+1.2, then a=e^{1.2} and b=e^{0.7} for y=ab^x.

The intercept is not always the original constant; it may be ln a or another transformed quantity.

3.3 Trigonometry

Syllabus
9709–2028–2029
Topic
3.3
Level
A2

Build reciprocal trig graphs from denominator behaviour

\sec x= rac1{\cos x},\qquad \cosec x= rac1{\sin x},\qquad \cot x= rac1{ an x}= rac{\cos x}{\sin x},whereverthedenominatorisnon−zero.wherever the denominator is non-zero.

Function Period Range Vertical asymptotes
sec⁡x\sec x 2π2\pi y≤−1y\le-1 or y≥1y\ge1 cos⁡x=0\cos x=0
cosec⁡x\cosec x 2π2\pi y≤−1y\le-1 or y≥1y\ge1 sin⁡x=0\sin x=0
cot⁡x\cot x π\pi all real yy sin⁡x=0\sin x=0

Start with the corresponding cosine, sine or tangent graph over the required angles. Its denominator zeros become reciprocal asymptotes; where the denominator is ±1\pm1, the reciprocal is also ±1\pm1; use the denominator sign between asymptotes to choose each branch.

sec⁡(π/3)=2\sec(\pi/3)=2 because cos⁡(π/3)=1/2\cos(\pi/3)=1/2. At x=π/2x=\pi/2, cosine is zero, so secant is undefined and has a vertical asymptote.

Reciprocal functions have no zeros: 1/f(x)1/f(x) cannot equal 00. cosec⁡x\cosec x is not sin⁡−1x\sin^{-1}x; inverse notation means a principal angle, not a reciprocal.

Select the identity family that exposes the required form

1+ an^2A=\sec^2A,\qquad 1+\cot^2A=\cosec^2A.Usethesetoexchangeareciprocalsquareforatangent/cotangentsquare.Use these to exchange a reciprocal square for a tangent/cotangent square.

\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B, an(A\pm B)= rac{ an A\pm an B}{1\mp an A an B}.

\sin2A=2\sin A\cos A,\quad \cos2A=\cos^2A-\sin^2A=1-2\sin^2A=2\cos^2A-1, an2A= rac{2 an A}{1- an^2A}.

Write asin⁡heta+bcos⁡heta=Rsin⁡(heta+α)a\sin heta+b\cos heta=R\sin( heta+\alpha) by matching Rcos⁡α=aR\cos\alpha=a and Rsin⁡α=bR\sin\alpha=b. Thus R=a2+b2R=\sqrt{a^2+b^2} and choose the quadrant of α\alpha from both signs. A cosine form is equally valid when coefficient matching is adjusted.

For exact values, expand a compound angle built from known angles. For equations, rewrite to one function or RR-form, use its range to decide existence, find all solutions in the stated interval, and check any denominator restrictions.

The sign in the cosine compound formula is opposite the sign between the angles. Identity manipulation does not remove excluded values introduced by dividing by sine, cosine or another expression.

3.4 Differentiation

Syllabus
9709–2028–2029
Topic
3.4
Level
A2

Attach the inner derivative to each A2 base derivative

f(x)f(x) f′(x)f'(x)
exe^x exe^x
ln⁡x\ln x 1/x1/x (x>0x>0)
sin⁡x\sin x cos⁡x\cos x
cos⁡x\cos x −sin⁡x-\sin x
anxan x sec⁡2x\sec^2x (where defined)
tan⁡−1x\tan^{-1}x 1/(1+x2)1/(1+x^2)

For differentiable $g$: rac d{dx}e^{g}=g'e^g,\quad rac d{dx}\ln g= rac{g'}g,\quad rac d{dx}\sin g=g'\cos g, rac d{dx}\cos g=-g'\sin g,\quad rac d{dx} an g=g'\sec^2g.Forinversetangent:For inverse tangent:\frac d{dx}\tan^{-1}(g)=\frac{g'}{1+g^2}.

\frac d{dx}\ln(1+x^2)=\frac{2x}{1+x^2},\qquad \frac d{dx}\tan^{-1}(3x)=\frac{3}{1+9x^2}.

Identify the outer function, write its base derivative with the inner expression unchanged, multiply by the inner derivative, then combine constant multiples, sums and differences.

Do not confuse tan⁡−1x\tan^{-1}x with 1/tan⁡x1/\tan x. Derivatives of sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x are explicitly not required.

Differentiate each changing factor in a product or quotient

If $y=u(x)v(x)$,y'=u'v+uv'.If $y=u(x)/v(x)$ and $v e0$,y'= rac{u'v-uv'}{v^2}.

Label uu and vv, compute u′u' and v′v' separately using chain rules where needed, substitute without changing the quotient numerator order, then factor or simplify.

rac d{dx}(x^2e^x)=2xe^x+x^2e^x=e^x(x^2+2x).

rac d{dx}\left( rac{\sin x}{x}
ight)= rac{x\cos x-\sin x}{x^2},\qquad x
e0.

The product derivative is not u′v′u'v', and the quotient derivative is not u′/v′u'/v'. Logarithmic differentiation is not required by this objective.

Extract gradients from parametric and implicit definitions

Definition Route to dy/dxdy/dx
x=x(t), y=y(t)x=x(t),\ y=y(t) dy/dx=(dy/dt)/(dx/dt)dy/dx=(dy/dt)/(dx/dt) when $dx/dt
e0$
F(x,y)=0F(x,y)=0 differentiate both sides in xx, attach dy/dxdy/dx to every yy derivative, then collect

If $x=t-e^{2t}$ and $y=t+e^{2t}$, rac{dy}{dx}= rac{1+2e^{2t}}{1-2e^{2t}}wherever $dx/dt e0$.

For $x^2+y^2=xy+7$:2x+2y rac{dy}{dx}=y+x rac{dy}{dx},soso rac{dy}{dx}= rac{y-2x}{2y-x}when $2y-x e0$.

Find the parameter or point coordinates first, evaluate the tangent gradient mm, then use y−y0=m(x−x0)y-y_0=m(x-x_0). A non-vertical normal has gradient −1/m-1/m; handle horizontal/vertical tangent cases geometrically.

Parametric gradient is dy/dtdy/dt divided by dx/dtdx/dt, not the reverse. In implicit differentiation, d(y2)/dx=2y dy/dxd(y^2)/dx=2y\,dy/dx, not 2y2y.

3.5 Integration

Syllabus
9709–2028–2029
Topic
3.5
Level
A2

Match one of the six A2 standard antiderivatives

| Integrand ($a

e0$) Antiderivative
eax+be^{ax+b} eax+b/a+Ce^{ax+b}/a+C
1/(ax+b)1/(ax+b) (1/a)ln⁡∣ax+b∣+C(1/a)\ln|ax+b|+C
sin⁡(ax+b)\sin(ax+b) −cos⁡(ax+b)/a+C-\cos(ax+b)/a+C
cos⁡(ax+b)\cos(ax+b) sin⁡(ax+b)/a+C\sin(ax+b)/a+C
sec⁡2(ax+b)\sec^2(ax+b) an(ax+b)/a+Can(ax+b)/a+C
1/(x2+a2)1/(x^2+a^2), a>0a>0 (1/a)tan⁡−1(x/a)+C(1/a)\tan^{-1}(x/a)+C

Match the whole integrand to one row, keep the linear inner expression unchanged, divide by its gradient aa, carry any outside constant, and include +C+C for an indefinite integral.

\int 3e^{2x-1},dx= rac32e^{2x-1}+C,\int rac{5}{3x+4},dx= rac53\ln|3x+4|+C.

Differentiate the answer: the chain factor aa must cancel the inserted 1/a1/a. For definite integrals, use an interval that does not cross a point where the integrand is undefined.

These are direct reverse-derivative forms. Do not introduce substitution, integration by parts or partial fractions unless a later objective explicitly calls for them.

Reduce squared trig functions before integrating

\sin^2u= rac{1-\cos2u}{2},\qquad \cos^2u= rac{1+\cos2u}{2}.These follow from the two useful forms of $\cos2u$.

First rewrite the squared sine or cosine as a constant plus/minus a double-angle cosine. Then integrate term by term using the linear-inner rule, including the factor created by the doubled angle.

\int\sin^2x,dx=\int rac{1-\cos2x}{2},dx= rac x2- rac{\sin2x}{4}+C.

\int\cos^2(2x),dx=\int rac{1+\cos4x}{2},dx= rac x2+ rac{\sin4x}{8}+C.

The square is on the trig value, so the ordinary power integration rule does not apply. This objective uses trig identities with the direct P2 antiderivatives, not a general substitution method.

Integrate each permitted partial-fraction term by its own pattern

First decompose using only the three denominator structures approved in 3.1. Then integrate every resulting linear, repeated-linear or linear-over-quadratic term separately.

Term Antiderivative pattern
A/(ax+b)A/(ax+b) (A/a)ln⁡∣ax+b∣(A/a)\ln|ax+b|
A/(ax+b)2A/(ax+b)^2 −A/[a(ax+b)]-A/[a(ax+b)]
(Bx+C)/(ax2+c)(Bx+C)/(ax^2+c) split numerator into a multiple of 2ax2ax plus a constant; log plus possible inverse tangent

\int\frac1{x(x+1)},dx=\int\left(\frac1x-\frac1{x+1}\right)dx=\ln|x|-\ln|x+1|+C.

Recombine the decomposition before integration, then differentiate the final antiderivative. Preserve intervals that do not cross denominator zeros.

Do not assign a constant numerator to an irreducible quadratic when a linear numerator is required, and do not integrate the original quotient by dividing numerator and denominator termwise.

Recognise a scaled logarithmic derivative

Where $f(x)\ne0$:\int k\frac{f'(x)}{f(x)},dx=k\ln|f(x)|+C.

Differentiate the denominator or inner function, compare it with the numerator, factor out the required constant, then apply the rule. If a leftover remains, split it and use another approved standard form.

\int\frac{x}{x^2+1},dx=\frac12\int\frac{2x}{x^2+1},dx=\frac12\ln(x^2+1)+C.

\int\tan x,dx=\int\frac{\sin x}{\cos x},dx=-\ln|\cos x|+C.

The numerator need only be proportional to f′f', not identical. Absolute values are required unless ff is known positive on the interval.

Reverse the product rule with integration by parts

\int u,dv=uv-\int v,du.Choose $u$ to simplify on differentiation and $dv$ to have a known antiderivative.

Identify the product (write ln⁡x=1⋅ln⁡x\ln x=1\cdot\ln x if needed), state u,dv,du,vu,dv,du,v, substitute into the formula, evaluate the remaining simpler integral and add CC.

\int xe^x,dx=xe^x-\int e^x,dx=e^x(x-1)+C.

\int\ln x,dx=x\ln x-x+C,\qquad x>0.

The tree short label says inverse tangent, but the official objective requires integration by parts, including products such as xtan⁡−1xx\tan^{-1}x. Do not swap dudu with vv.

Apply the given substitution to every part of the integral

Use the substitution supplied by the question. Find dudu, rewrite every factor and dxdx in uu, and do not mix variables. For a definite integral, convert both limits immediately or substitute back before using the original limits.

Given $u=\sin x$:\int\sin^2x\cos x,dx=\int u^2,du=\frac{u^3}{3}+C=\frac{\sin^3x}{3}+C.

For definite limits x=a,bx=a,b, replace them by u(a),u(b)u(a),u(b) and finish entirely in uu. If the substitution is not one-to-one over the interval, follow the question structure carefully and verify the transformed bounds.

Differentiate an indefinite answer or compare a definite result numerically/sign-wise with the original integrand.

This objective asks for use of a given substitution, not invention of a general substitution strategy. The differential factor and all limits are part of the substitution.

3.6 Numerical solution of equations

Syllabus
9709–2028–2029
Topic
3.6
Level
A2

Locate a root by a graph crossing or a sign-change bracket

Rewrite the equation as f(x)=0f(x)=0 and locate where the graph y=f(x)y=f(x) crosses the xx-axis, or plot the two sides separately and locate their intersection. The graph supplies an approximate root or a search interval.

If $f$ is continuous on $[a,b]$ and $f(a)f(b)<0$, then at least one root lies in $(a,b)$. Evaluate consecutive integers or progressively closer endpoints when requested.

For a continuous ff, f(1)<0f(1)<0 and f(2)>0f(2)>0 locates at least one root between 11 and 22. State the function values or their signs, not just the interval.

A graph gives visual approximate evidence; a sign-change bracket gives endpoint evidence. A narrower bracket gives a tighter location but remains an interval, not the exact root.

A sign change guarantees at least one root under continuity, not uniqueness. A repeated/touching root may have no sign change, so graphical evidence can still matter.

Fixed-point iteration refines a root only when the sequence behaves well

With x_{n+1}=g(x_n), a fixed point α satisfies g(α)=α. Starting from x₀ creates a sequence intended to approach α.

Choose a valid starting value, retain guard digits, and stop when successive values or the residual meet the tolerance. Verify the final value in f(x)=0.

For x=cosx, x₀=1 gives iterates that approach approximately 0.739; reporting the last iterate without a tolerance is incomplete.

Iteration can oscillate or diverge even when a root exists; apparent agreement of early digits is not a proof.

Relate, run and verify a fixed-point iteration

For xn+1=F(xn)x_{n+1}=F(x_n), a convergent limit α\alpha must satisfy α=F(α)\alpha=F(\alpha). Rearrange that fixed-point equation to confirm it is the original equation whose root is required, including any domain restrictions.

Use the stated starting value, keep guard digits, tabulate nn and xnx_n, apply the same formula repeatedly, watch for settling/divergence/cycling, and continue until successive values justify the prescribed rounded answer.

To solve $x^3+x-1=0$, the given rearrangement $x_{n+1}=(1-x_n)^{1/3}$ has fixed-point equation $x^3=1-x$, hence $x^3+x-1=0$. Run it only from the given/appropriate start.

For a requested number of decimal places, obtain successive values that round consistently at that precision and substitute the reported approximation into the original equation as a residual sense-check when practical.

An algebraically related iteration may fail to converge or may approach a different root. The derivative condition for convergence is explicitly not required in this syllabus; judge only from the given task and observed sequence behaviour.

3.7 Vectors

Syllabus
9709–2028–2029
Topic
3.7
Level
A2

Recognise the standard notations for one vector

Meaning 2D 3D
column components (xy)\begin{pmatrix}x\\y\end{pmatrix} (xyz)\begin{pmatrix}x\\y\\z\end{pmatrix}
basis-vector form xi+yjx\mathbf i+y\mathbf j xi+yj+zkx\mathbf i+y\mathbf j+z\mathbf k

AB→\overrightarrow{AB} is the directed vector from AA to BB; a bold/lowercase symbol such as a\mathbf a can name the same free vector. Equality means equal corresponding components.

A point P(x,y,z)P(x,y,z) names a location. Its position vector OP→\overrightarrow{OP} has the same numerical entries only because the origin and direction from OO are specified.

$\begin{pmatrix}2\\-1\\4\end{pmatrix}=2\mathbf i-\mathbf j+4\mathbf k$.

Use arrow/bold/column notation consistently. Magnitude and unit-vector calculations belong to objective 3, not to the notation definition.

Interpret component operations as geometric vector equations

Add/subtractcomponentwiseandscaleeverycomponent:Add/subtract componentwise and scale every component:(a_1,a_2,a_3)\pm(b_1,b_2,b_3),\qquad k\mathbf a.A negative $k$ reverses direction.

Directed displacements chain head-to-tail: AB→+BC→=AC→\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}. Opposite direction gives BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}.

Geometry Position-vector equation
OABCOABC parallelogram OB→=OA→+OC→\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC} (with the stated vertex order)
midpoint MM of ABAB OM→=12(OA→+OB→)\overrightarrow{OM}=\tfrac12(\overrightarrow{OA}+\overrightarrow{OB})

If OA→=(2,1)\overrightarrow{OA}=(2,1) and OB→=(8,5)\overrightarrow{OB}=(8,5), midpoint MM has position (5,3)(5,3). Check by equal displacements AM→=MB→=(3,2)\overrightarrow{AM}=\overrightarrow{MB}=(3,2).

The general ratio theorem and scalar product are not included in this objective. Use only component operations, geometric chaining, parallelogram and midpoint.

Move between positions, displacements, magnitudes and unit directions

Quantity Formula
displacement AA to BB AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}
magnitude of v=(v1,v2,v3)\mathbf v=(v_1,v_2,v_3) ∣v∣=v12+v22+v32|\mathbf v|=\sqrt{v_1^2+v_2^2+v_3^2}
unit vector along non-zero v\mathbf v v^=v/∣v∣\hat{\mathbf v}=\mathbf v/|\mathbf v|

Keep one origin for position vectors, subtract final minus initial for displacement, calculate its magnitude, then divide every component by that magnitude if a unit direction is required. The same workflow works in 2D or 3D.

From $A(1,-2,3)$ to $B(3,1,9)$:\overrightarrow{AB}=(2,3,6),\quad |\overrightarrow{AB}|=7,\quad \hat{\mathbf{AB}}=\tfrac17(2,3,6).

A unit vector must have magnitude 11. Reversing direction negates the unit vector but keeps the same magnitude.

AB→\overrightarrow{AB} is final minus initial; dividing by ∣v∣2|\mathbf v|^2 does not normalise the vector.

Construct a vector line from a point and a non-zero direction

Symbol in r=a+tb\mathbf r=\mathbf a+t\mathbf b Meaning
r\mathbf r position vector of a general point on the line
a\mathbf a position vector of one fixed point on the line
t∈Rt\in\mathbb R scalar parameter
b≠0\mathbf b\ne\mathbf0 direction vector

Given point AA and direction b\mathbf b, use r=OA→+tb\mathbf r=\overrightarrow{OA}+t\mathbf b. Given points A,BA,B, use direction AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}.

Through $A(1,2,-1)$ and $B(4,0,5)$:\mathbf r=\begin{pmatrix}1\2\-1\end{pmatrix}+t\begin{pmatrix}3\-2\6\end{pmatrix}.

A different point on the same line or any non-zero scalar multiple of the direction produces an equivalent equation.

A point vector fixes location; a direction vector controls movement. Do not use A+BA+B as the direction between two points.

Classify two 3D lines by direction and component consistency

Check Conclusion
directions proportional parallel; test a point to distinguish coincident/distinct
directions not proportional and one parameter pair satisfies every component intersecting
directions not proportional and component equations are inconsistent skew

Use different parameters for the two lines. Equate position components, solve two equations, then verify the same parameter pair in the remaining component. If consistent, substitute to find the intersection point.

For $\mathbf r=\mathbf a+t\mathbf b$ and $\mathbf r=\mathbf c+s\mathbf d$, solve\mathbf a+t\mathbf b=\mathbf c+s\mathbf d.Avalid3Dintersectionrequiresallthreescalarequationssimultaneously.A valid 3D intersection requires all three scalar equations simultaneously.

If d=kb\mathbf d=k\mathbf b, test whether c−a\mathbf c-\mathbf a is also parallel to b\mathbf b. If yes the equations describe the same line; otherwise they are distinct parallel lines.

Two non-parallel 3D lines need not intersect. Shortest distance between skew lines and the equation of their common perpendicular are explicitly not required.

Use scalar products for line angles and perpendicular feet

\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta.Non-zero vectors are perpendicular exactly when their dot product is $0$.

For line directions b,d\mathbf b,\mathbf d, compute cos⁡θ=(b⋅d)/(∣b∣∣d∣)\cos\theta=(\mathbf b\cdot\mathbf d)/(|\mathbf b||\mathbf d|). If the smaller angle between lines is requested, use the acute value, equivalently ∣b⋅d∣|\mathbf b\cdot\mathbf d| in the numerator.

For line $\mathbf r=\mathbf a+t\mathbf b$ and point position $\mathbf p$, the foot is $\mathbf q=\mathbf a+t\mathbf b$ where(\mathbf p-\mathbf q)\cdot\mathbf b=0.Solve for $t$, then state $\mathbf q$.

This foot method works in 2D or 3D and in cuboid/tetrahedron contexts: the joining vector from the foot to the point must be perpendicular to the line direction.

The dot product is a scalar. Knowledge of the vector (cross) product is explicitly not required.

3.8 Differential equations

Syllabus
9709–2028–2029
Topic
3.8
Level
A2

Translate rate language into a signed differential equation

Statement about y(t)y(t) Differential equation
rate proportional to yy dy/dt=kydy/dt=ky
rate of decrease proportional to yy dy/dt=−kydy/dt=-ky, k>0k>0
rate proportional to A−yA-y dy/dt=k(A−y)dy/dt=k(A-y)

Name dependent/independent variables, convert “rate of change” to a derivative, make the right side depend only on stated quantities, introduce kk with sign/units, then use any given value and rate at the same instant to evaluate kk.

If $M$ decreases at a rate proportional to $M$ and when $M=50$ kg its rate is $-3$ kg h$^{-1}$, then\frac{dM}{dt}=-kM,\quad -3=-50k,\quad k=0.06\ \text{h}^{-1}.

Check dimensions: in dy/dt=kydy/dt=ky, kk has units inverse to the independent variable. State an initial relation separately if supplied.

This objective is formulation, not solution. Integrating factors are not part of the syllabus; do not introduce them.

Separate, integrate and state the general solution

ForFor\frac{dy}{dx}=f(x)g(y),movevariablestomove variables to\frac{1}{g(y)},dy=f(x),dxwheredivisionislegal,thenintegratebothsidesandcombineconstants.where division is legal, then integrate both sides and combine constants.

Check constant solutions from g(y)=0g(y)=0 before dividing. Separate completely, use any required technique from 3.5, include one arbitrary constant, then rearrange explicitly or leave a valid implicit relation as appropriate.

\frac{dy}{dx}=xy\Rightarrow\frac{dy}{y}=x,dx\Rightarrow\ln|y|=\frac{x^2}{2}+C,so $y=Ae^{x^2/2}$; $A=0$ includes the constant solution $y=0$.

Differentiate the general solution and substitute it into the original equation. Record intervals/domain restrictions created by logarithms or division.

Only separable first-order equations are required. Do not use a first-order linear integrating factor method.

Use one initial point to select a particular solution

A general first-order solution contains one arbitrary constant. Substitute the stated condition y(x0)=y0y(x_0)=y_0 into that integrated family to determine it, then state the resulting particular solution.

Solve generally first, apply the actual initial point (not automatically x=0x=0), solve the constant exactly, and check both the differential equation and the initial condition.

If $dy/dx=3x^2-2$ and $y(1)=4$, theny=x^3-2x+C,\quad4=1-2+C,so $C=5$ and $y=x^3-2x+5$.

Choose the branch/interval containing the initial point when logs, square roots or implicit relations create multiple possibilities.

This syllabus objective concerns first-order equations; do not add second-order condition counting or unrelated initial-velocity theory.

Translate a solution back into model meaning and limits

Feature of solution Context question
variable and derivative what quantity/rate, with what units?
sign/monotonicity increasing or decreasing when?
constants/initial value what do they represent?
limiting value or growth what happens as time increases?
domain when are time and predicted values physically meaningful?

If M(t)=50e−0.06tM(t)=50e^{-0.06t} kg for t≥0t\ge0, then M(0)=50M(0)=50 kg, the model predicts continuous decay, M(t)>0M(t)>0, and M(t)→0M(t)\to0 as t→∞t\to\infty. The constant 0.060.06 has units h−1^{-1} if tt is hours.

To find when $M$ reaches $20$ kg, solve50e^{-0.06t}=20\Rightarrow t=-\frac{\ln(0.4)}{0.06}.Interpretthepositiveresultinhours,notasanewmodelconstant.Interpret the positive result in hours, not as a new model constant.

Also verify the formula satisfies the ODE and initial data, but algebraic validity is only the start: state the contextual conclusion in words and sensible accuracy.

A mathematically valid formula may be unrealistic outside the stated time/value range. Do not claim negative time, negative populations/masses or indefinite model validity without support.

3.9 Complex numbers

Syllabus
9709–2028–2029
Topic
3.9
Level
A2

Name every part of a complex number and fix an argument branch

For z=x+iyz=x+iy with real x,yx,y:

Notation Meaning
Re⁡z=x\operatorname{Re}z=x real part
Im⁡z=y\operatorname{Im}z=y imaginary part (not iyiy)
∣z∣=x2+y2|z|=\sqrt{x^2+y^2} modulus
arg⁡z=θ\arg z=\theta directed angle from positive real axis
z∗=x−iyz^*=x-iy conjugate

A non-zero complex number has arguments differing by 2π2\pi. Unless specified, use a consistent principal interval such as −π<θ≤π-\pi<\theta\le\pi; 0≤θ<2π0\le\theta<2\pi may also be convenient. arg⁡0\arg0 is undefined.

$x+iy=u+iv$ if and only if $x=u$ and $y=v$. Alsozz^*=x^2+y^2=|z|^2.

For z=−1+i3z=-1+i\sqrt3, ∣z∣=2|z|=2 and a principal argument is 2π/32\pi/3; its conjugate is −1−i3-1-i\sqrt3.

Equality requires both components, not merely equal moduli. Quadrant determines argument; tan⁡−1(y/x)\tan^{-1}(y/x) alone is insufficient.

Complex arithmetic is checked by matching real and imaginary parts

Add complex numbers componentwise, multiply by expanding and replace i² with −1, and divide by multiplying numerator and denominator by the denominator’s conjugate.

After simplifying, write the answer in a+bi form. Equality of complex numbers requires equality of both components.

(1+2i)/(3−i) =[(1+2i)(3+i)]/10=(1+7i)/10.

A complex fraction is not simplified by dividing real parts and imaginary parts separately.

Complete real polynomial roots with the conjugate pair

If a polynomial has real coefficients and a+iba+ib (b≠0b\ne0) is a root, then a−iba-ib is also a root. The result is conditional on real coefficients.

ThepairgivestherealquadraticfactorThe pair gives the real quadratic factor[x-(a+ib)][x-(a-ib)]=(x-a)^2+b^2.

Write the conjugate root immediately, multiply the pair to form a real factor, divide the cubic/quartic polynomial by it, then solve the remaining lower-degree factor and verify the total degree.

If a real cubic has roots $1+i$ and $2$, it also has $1-i$, so a monic polynomial is(x-2)[(x-1)^2+1].

Conjugate pairing is not guaranteed for non-real coefficients. The broad fundamental theorem/root count is not the learning target here.

Place $x+iy$ at $(x,y)$ on an Argand diagram

An Argand diagram has horizontal real axis and vertical imaginary axis. The complex number z=x+iyz=x+iy is represented by point/vector (x,y)(x,y) from the origin.

Algebra Argand meaning
Re⁡z=x\operatorname{Re}z=x horizontal coordinate
Im⁡z=y\operatorname{Im}z=y vertical coordinate
∣z∣|z| distance from origin
arg⁡z\arg z directed angle from positive real axis

−2+3i-2+3i is plotted at (−2,3)(-2,3) in quadrant II; its modulus is 13\sqrt{13} and its argument must lie in the chosen quadrant-II branch.

Label axes, scale and reference points. Read a plotted point back as real coordinate plus ii times imaginary coordinate.

This is a coordinate representation, not a graph y=f(x)y=f(x). Polar multiplication belongs to the next objective.

Multiply and divide complex numbers with a polar ledger

WriteWritez=r(\cos\theta+i\sin\theta)=re^{i\theta},\qquad r>0.

Operation Modulus Argument
z1z2z_1z_2 r1r2r_1r_2 θ1+θ2\theta_1+\theta_2
z1/z2z_1/z_2 r1/r2r_1/r_2 θ1−θ2\theta_1-\theta_2

[2e^{i\pi/3}][3e^{-i\pi/6}]=6e^{i\pi/6},\qquad \frac{2e^{i\pi/3}}{3e^{-i\pi/6}}=\frac23e^{i\pi/2}.

Operate moduli and arguments separately, reduce the final argument to the requested interval, and convert to Cartesian form only if asked.

Do not add moduli during multiplication. De Moivre powers and general nnth roots are not objectives in this syllabus section and must not be imported.

Find both exact square roots in Cartesian form

To solve $z^2=p+iq$, set $z=a+ib$ with real $a,b$:(a+ib)^2=(a^2-b^2)+2abi.Hence $a^2-b^2=p$ and $2ab=q$.

Also $a^2+b^2=|p+iq|=\sqrt{p^2+q^2}$. Add/subtract this with $a^2-b^2=p$ to find $a^2,b^2$, then use $2ab=q$ for signs.

For $z^2=5+12i$:a^2-b^2=5,\quad2ab=12,\quad a^2+b^2=13,so $(a,b)=(3,2)$ or $(-3,-2)$ andz=\pm(3+2i).

Square both answers and show full Cartesian working. The two roots are always opposites for a non-zero complex number.

Do not take square roots of real and imaginary parts separately, and do not report only a principal root.

Read complex operations as transformations in the Argand plane

Operation Geometrical effect
z↦z∗z\mapsto z^* reflect in real axis
z↦z+wz\mapsto z+w translate by vector ww
z1−z2z_1-z_2 displacement from point z2z_2 to z1z_1
multiply by reiθre^{i\theta} scale distances from origin by rr, rotate by θ\theta
divide by reiθre^{i\theta} scale by 1/r1/r, rotate by −θ-\theta

Multiplication by i=eiπ/2i=e^{i\pi/2} rotates every point 90∘90^\circ anticlockwise about the origin without changing modulus. Multiplication by −2-2 scales by 22 and rotates by π\pi.

Addition/subtraction use parallelogram/displacement geometry; multiplication/division use origin-centred scale and rotation. Conjugates keep modulus and negate the argument within branch conventions.

Predict the geometric result, then confirm with Cartesian or polar arithmetic.

A transformation effect is not a locus condition. Powers/general roots are not needed for this objective.

Translate complex conditions into Argand loci and boundary rules

Condition Locus
∣z−a∣=r|z-a|=r circle centre aa, radius rr
∣z−a∣<r|z-a|<r / ≤r\le r interior, boundary excluded/included
∣z−a∣>∣z−b∣|z-a|>|z-b| points closer to bb than aa: one side of perpendicular bisector
∣z−a∣=∣z−b∣|z-a|=|z-b| perpendicular bisector of segment abab
arg⁡(z−a)=α\arg(z-a)=\alpha ray from aa at angle α\alpha, endpoint aa excluded

Plot reference points first, draw each boundary, decide included/excluded style from equality, shade the correct side/interior/exterior, then intersect all regions.

∣z−(2+i)∣<3|z-(2+i)|<3 is the open disc centred at (2,1)(2,1) with radius 33. Adding arg⁡(z−(2+i))=π/4\arg(z-(2+i))=\pi/4 restricts to points on the corresponding ray that also lie inside the disc.

For equal distances, do not draw a circle: the locus is a straight perpendicular bisector. Test one simple point to choose a side for an inequality.

arg⁡(z−a)\arg(z-a) is undefined at z=az=a, and is a direction condition rather than distance. De Moivre powers/roots are unrelated and not required.