3. Pure Mathematics A2 3
- Syllabus
- 9709–2028–2029
- Section
- 3
- Level
- A2
y=∣ax+b∣ is a V-graph with vertex (−b/a,0) for $a
e0,rangey\ge0,andbranchgradients-|a|then|a|.Itrepresentsdistancefromthezeroofax+b$.
| Form | Equivalent condition |
|---|---|
| ∣u∣=∣v∣ | u=v or u=−v |
| ∣x−a∣<b, b>0 | a−b<x<a+b |
| ∣x−a∣>b, b>0 | x<a−b or x>a+b |
For ∣f(x)∣=g(x) require g(x)≥0, 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 ≤. Non-linear graphs y=∣f(x)∣ and y=f(∣x∣) are excluded.
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 4 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=P.
| Statement | Evaluation |
|---|---|
| remainder on division by x−c is k | P(c)=k |
| x−c is a factor | P(c)=0 |
| ax+b is a factor | P(−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+2 leaves remainder 5, use P(−2)=5, not zero. Use full polynomial division when the quotient is also required.
The zero of ax+b is −b/a; keep the sign and non-zero remainder exactly as stated.
| Denominator structure | Partial-fraction form |
|---|---|
| (ax+b)(cx+d)(ex+f) | A/(ax+b)+B/(cx+d)+C/(ex+f) |
| (ax+b)(cx+d)2 | A/(ax+b)+B/(cx+d)+C/(cx+d)2 |
| (ax+b)(cx2+d) | A/(ax+b)+(Bx+C)/(cx2+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.
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.
Rewrite the expression as a constant multiple of (1+u)n, substitute u 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. In the example, ∣−2x∣<1, so ∣x∣<1/2. For a shifted/scaled u(x), solve the resulting inequality and report the full set.
For non-integer rational n the expansion is generally infinite, not a finite P1 binomial. The standard ∣x∣<1 condition applies to the inner series variable u, not automatically to the original x.
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 | loga(MN)=logaM+logaN |
| quotient | loga(M/N)=logaM−logaN |
| power | loga(Mp)=plogaM |
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) does not split. Change-of-base formulae are explicitly excluded from this objective.
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=ekx, $k | |||
| e0∣allrealx∣y>0∣(0,1);horizontalasymptotey=0$ | |||
| y=lnx | x>0 | all real y | (1,0); vertical asymptote x=0 |
For k>0, ekx increases; for k<0, it decreases. Both stay positive and approach, but never cross, the x-axis in one direction.
$3e^{2x}=12$ gives $e^{2x}=4$, so $x= frac12\ln4$.
An exponential graph has no x-intercept, and lnx is undefined for x≤0. Do not treat ln(x2)=2lnx as valid when x<0.
Rearrange until each exponential expression is positive, take ln of both sides, use ln(ag(x))=g(x)lna, 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<av with a>1 | u<v |
| au<av with 0<a<1 | u>v |
If terms such as a2x and ax occur together, set t=ax with t>0, solve the polynomial in t, reject non-positive roots, then take logs to recover x.
Logs can only be taken after both sides are positive. An inequality reverses for a decreasing base 0<a<1, not merely because logarithms were used.
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.
\sec x=rac1{\cos x},\qquad \cosec x=rac1{\sin x},\qquad \cot x=rac1{ an x}=rac{\cos x}{\sin x},whereverthedenominatorisnon−zero.
| Function | Period | Range | Vertical asymptotes |
|---|---|---|---|
| secx | 2π | y≤−1 or y≥1 | cosx=0 |
| cosecx | 2π | y≤−1 or y≥1 | sinx=0 |
| cotx | π | all real y | sinx=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, the reciprocal is also ±1; use the denominator sign between asymptotes to choose each branch.
sec(π/3)=2 because cos(π/3)=1/2. At x=π/2, cosine is zero, so secant is undefined and has a vertical asymptote.
Reciprocal functions have no zeros: 1/f(x) cannot equal 0. cosecx is not sin−1x; inverse notation means a principal angle, not a reciprocal.
1+ an^2A=\sec^2A,\qquad 1+\cot^2A=\cosec^2A.Usethesetoexchangeareciprocalsquareforatangent/cotangentsquare.
\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 asinheta+bcosheta=Rsin(heta+α) by matching Rcosα=a and Rsinα=b. Thus R=a2+b2 and choose the quadrant of α 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 R-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.
| f(x) | f′(x) |
|---|---|
| ex | ex |
| lnx | 1/x (x>0) |
| sinx | cosx |
| cosx | −sinx |
| anx | sec2x (where defined) |
| tan−1x | 1/(1+x2) |
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:\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 with 1/tanx. Derivatives of sin−1x and cos−1x are explicitly not required.
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 u and v, compute u′ and 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′, and the quotient derivative is not u′/v′. Logarithmic differentiation is not required by this objective.
| Definition | Route to dy/dx |
|---|---|
| x=x(t), y=y(t) | dy/dx=(dy/dt)/(dx/dt) when $dx/dt |
| e0$ | |
| F(x,y)=0 | differentiate both sides in x, attach dy/dx to every y 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+2yrac{dy}{dx}=y+xrac{dy}{dx},sorac{dy}{dx}=rac{y-2x}{2y-x}when $2y-x e0$.
Find the parameter or point coordinates first, evaluate the tangent gradient m, then use y−y0=m(x−x0). A non-vertical normal has gradient −1/m; handle horizontal/vertical tangent cases geometrically.
Parametric gradient is dy/dt divided by dx/dt, not the reverse. In implicit differentiation, d(y2)/dx=2ydy/dx, not 2y.
| Integrand ($a
| e0$) | Antiderivative |
|---|---|
| eax+b | eax+b/a+C |
| 1/(ax+b) | (1/a)ln∣ax+b∣+C |
| sin(ax+b) | −cos(ax+b)/a+C |
| cos(ax+b) | sin(ax+b)/a+C |
| sec2(ax+b) | an(ax+b)/a+C |
| 1/(x2+a2), a>0 | (1/a)tan−1(x/a)+C |
Match the whole integrand to one row, keep the linear inner expression unchanged, divide by its gradient a, carry any outside constant, and include +C for an indefinite integral.
\int 3e^{2x-1},dx=rac32e^{2x-1}+C,\intrac{5}{3x+4},dx=rac53\ln|3x+4|+C.
Differentiate the answer: the chain factor a must cancel the inserted 1/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.
\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=\intrac{1-\cos2x}{2},dx=rac x2-rac{\sin2x}{4}+C.
\int\cos^2(2x),dx=\intrac{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.
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/a)ln∣ax+b∣ |
| A/(ax+b)2 | −A/[a(ax+b)] |
| (Bx+C)/(ax2+c) | split numerator into a multiple of 2ax 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.
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′, not identical. Absolute values are required unless f is known positive on the interval.
\int u,dv=uv-\int v,du.Choose $u$ to simplify on differentiation and $dv$ to have a known antiderivative.
Identify the product (write lnx=1⋅lnx if needed), state u,dv,du,v, substitute into the formula, evaluate the remaining simpler integral and add C.
\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−1x. Do not swap du with v.
Use the substitution supplied by the question. Find du, rewrite every factor and dx in u, 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,b, replace them by u(a),u(b) and finish entirely in u. 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.
Rewrite the equation as f(x)=0 and locate where the graph y=f(x) crosses the x-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 f, f(1)<0 and f(2)>0 locates at least one root between 1 and 2. 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.
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.
For xn+1=F(xn), a convergent limit α must satisfy α=F(α). 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 n and xn, 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.
| Meaning | 2D | 3D |
|---|---|---|
| column components | (xy) | xyz |
| basis-vector form | xi+yj | xi+yj+zk |
AB is the directed vector from A to B; a bold/lowercase symbol such as a can name the same free vector. Equality means equal corresponding components.
A point P(x,y,z) names a location. Its position vector OP has the same numerical entries only because the origin and direction from O 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.
Add/subtractcomponentwiseandscaleeverycomponent:(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. Opposite direction gives BA=−AB.
| Geometry | Position-vector equation |
|---|---|
| OABC parallelogram | OB=OA+OC (with the stated vertex order) |
| midpoint M of AB | OM=21(OA+OB) |
If OA=(2,1) and OB=(8,5), midpoint M has position (5,3). Check by equal displacements AM=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.
| Quantity | Formula |
|---|---|
| displacement A to B | AB=OB−OA |
| magnitude of v=(v1,v2,v3) | ∣v∣=v12+v22+v32 |
| unit vector along non-zero v | v^=v/∣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 1. Reversing direction negates the unit vector but keeps the same magnitude.
AB is final minus initial; dividing by ∣v∣2 does not normalise the vector.
| Symbol in r=a+tb | Meaning |
|---|---|
| r | position vector of a general point on the line |
| a | position vector of one fixed point on the line |
| t∈R | scalar parameter |
| b=0 | direction vector |
Given point A and direction b, use r=OA+tb. Given points A,B, use direction AB=OB−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+B as the direction between two points.
| 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.
If d=kb, test whether c−a is also parallel to 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.
\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, compute cosθ=(b⋅d)/(∣b∣∣d∣). If the smaller angle between lines is requested, use the acute value, equivalently ∣b⋅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.
| Statement about y(t) | Differential equation |
|---|---|
| rate proportional to y | dy/dt=ky |
| rate of decrease proportional to y | dy/dt=−ky, k>0 |
| rate proportional to 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 k with sign/units, then use any given value and rate at the same instant to evaluate k.
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=ky, k 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.
For\frac{dy}{dx}=f(x)g(y),movevariablesto\frac{1}{g(y)},dy=f(x),dxwheredivisionislegal,thenintegratebothsidesandcombineconstants.
Check constant solutions from g(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.
A general first-order solution contains one arbitrary constant. Substitute the stated condition y(x0)=y0 into that integrated family to determine it, then state the resulting particular solution.
Solve generally first, apply the actual initial point (not automatically x=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.
| 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−0.06t kg for t≥0, then M(0)=50 kg, the model predicts continuous decay, M(t)>0, and M(t)→0 as t→∞. The constant 0.06 has units h−1 if t is hours.
To find when $M$ reaches $20$ kg, solve50e^{-0.06t}=20\Rightarrow t=-\frac{\ln(0.4)}{0.06}.Interpretthepositiveresultinhours,notasanewmodelconstant.
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.
For z=x+iy with real x,y:
| Notation | Meaning |
|---|---|
| Rez=x | real part |
| Imz=y | imaginary part (not iy) |
| ∣z∣=x2+y2 | modulus |
| argz=θ | directed angle from positive real axis |
| z∗=x−iy | conjugate |
A non-zero complex number has arguments differing by 2π. Unless specified, use a consistent principal interval such as −π<θ≤π; 0≤θ<2π may also be convenient. 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+i3, ∣z∣=2 and a principal argument is 2π/3; its conjugate is −1−i3.
Equality requires both components, not merely equal moduli. Quadrant determines argument; tan−1(y/x) alone is insufficient.
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.
If a polynomial has real coefficients and a+ib (b=0) is a root, then a−ib is also a root. The result is conditional on real coefficients.
Thepairgivestherealquadraticfactor[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.
An Argand diagram has horizontal real axis and vertical imaginary axis. The complex number z=x+iy is represented by point/vector (x,y) from the origin.
| Algebra | Argand meaning |
|---|---|
| Rez=x | horizontal coordinate |
| Imz=y | vertical coordinate |
| ∣z∣ | distance from origin |
| argz | directed angle from positive real axis |
−2+3i is plotted at (−2,3) in quadrant II; its modulus is 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 i times imaginary coordinate.
This is a coordinate representation, not a graph y=f(x). Polar multiplication belongs to the next objective.
Writez=r(\cos\theta+i\sin\theta)=re^{i\theta},\qquad r>0.
| Operation | Modulus | Argument |
|---|---|---|
| z1z2 | r1r2 | θ1+θ2 |
| z1/z2 | r1/r2 | θ1−θ2 |
[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 nth roots are not objectives in this syllabus section and must not be imported.
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.
| Operation | Geometrical effect |
|---|---|
| z↦z∗ | reflect in real axis |
| z↦z+w | translate by vector w |
| z1−z2 | displacement from point z2 to z1 |
| multiply by reiθ | scale distances from origin by r, rotate by θ |
| divide by reiθ | scale by 1/r, rotate by −θ |
Multiplication by i=eiπ/2 rotates every point 90∘ anticlockwise about the origin without changing modulus. Multiplication by −2 scales by 2 and rotates by π.
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.
| Condition | Locus |
|---|---|
| ∣z−a∣=r | circle centre a, radius r |
| ∣z−a∣<r / ≤r | interior, boundary excluded/included |
| ∣z−a∣>∣z−b∣ | points closer to b than a: one side of perpendicular bisector |
| ∣z−a∣=∣z−b∣ | perpendicular bisector of segment ab |
| arg(z−a)=α | ray from a at angle α, endpoint a 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 is the open disc centred at (2,1) with radius 3. Adding arg(z−(2+i))=π/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) is undefined at z=a, and is a direction condition rather than distance. De Moivre powers/roots are unrelated and not required.