2. Pure Mathematics AS 2

Syllabus
9709–2028–2029
Section
2
Level
AS

2.1 Algebra

Syllabus
9709–2028–2029
Topic
2.1
Level
AS

Use distance and sign regions for linear modulus problems

∣u∣|u| is the non-negative distance of uu from 00. Thus y=∣ax+b∣y=|ax+b| is a V-shaped graph with vertex where ax+b=0ax+b=0, slopes −∣a∣-|a| then ∣a∣|a|, and range y≥0y\ge0.

Form Equivalent relation
∣u∣=c|u|=c, c≥0c\ge0 u=cu=c or u=−cu=-c
∣u∣=∣v∣|u|=|v| u2=v2u^2=v^2, equivalently 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

If the other side contains xx, locate every zero of both expressions, split the number line into sign regions, remove the modulus with the correct sign in each region, solve and intersect with that region.

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

Check equality endpoints according to << or ≤\le. Non-linear graphs y=∣f(x)∣y=|f(x)| and y=f(∣x∣)y=f(|x|) are excluded from this objective.

Divide polynomials and control the remainder degree

For dividend $P(x)$ and non-zero divisor $D(x)$:P(x)=D(x)Q(x)+R(x),\qquad \deg R<\deg D.

Divisor Possible remainder
linear constant
quadratic linear mx+cmx+c (including zero/special cases)

Write every polynomial in descending powers and insert zero coefficients for missing powers. Divide leading terms, multiply the divisor, subtract the whole row, bring down the next term and repeat until the remainder degree is smaller than the divisor degree.

x^4+2x^2+3=(x^2+1)(x^2+1)+2.The quotient is $x^2+1$ and the remainder is $2$, whose degree is below $2$.

Verify divisor imesimes quotient ++ remainder equals the original dividend. Synthetic division is only a shortcut for suitable linear divisors, not a quadratic divisor.

Turn factors and remainders into evaluation equations

Division statement Evaluation statement
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, $a
e0∣|P(-b/a)=0$

Translate every factor or stated remainder into a substitution equation. If coefficients are unknown, solve the resulting simultaneous equations; then divide out confirmed factors and solve any lower-degree quotient when roots are required.

Let $P(x)=x^3+kx+6$. If $2x-1$ is a factor, $P(1/2)=0$: rac18+ rac{k}{2}+6=0\quad\Rightarrow\quad k=- rac{49}{4}.

For a non-zero remainder, keep the right side: if division by x+2x+2 leaves 55, then P(−2)=5P(-2)=5, not 00. Use polynomial division when the full quotient is needed.

The zero of ax+bax+b is −b/a-b/a. A factor statement and a remainder statement are different equations; no repeated-root derivative test is needed for this objective.

2.2 Logarithmic and exponential functions

Syllabus
9709–2028–2029
Topic
2.2
Level
AS

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.

2.3 Trigonometry

Syllabus
9709–2028–2029
Topic
2.3
Level
AS

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.

2.4 Differentiation

Syllabus
9709–2028–2029
Topic
2.4
Level
AS

Attach the inner derivative to each P2 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)

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.

rac d{dx}\ln(1+x^2)= rac{2x}{1+x^2},\qquad rac d{dx}[3e^{2x}-\cos(4x)]=6e^{2x}+4\sin(4x).

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.

The derivative of ln⁡g\ln g is g′/gg'/g, not 1/ln⁡g1/\ln g. Products and quotients are handled in the next objective rather than smuggled into a sum/chain shortcut.

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.

2.5 Integration

Syllabus
9709–2028–2029
Topic
2.5
Level
AS

Reverse five linear-inner derivatives with the $1/a$ factor

| 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

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.

General integration by substitution and integration by parts are not required in Pure Mathematics 2; use these direct reverse-derivative patterns only.

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.

Weight trapezium ordinates and use chord position to judge error

For $n$ equal strips, $h=(b-a)/n$ and ordinates $y_0,\ldots,y_n$:T= rac h2\left[y_0+y_n+2(y_1+\cdots+y_{n-1})
ight].

Confirm equal spacing, list all n+1n+1 ordinates in order, weight endpoints once and internal ordinates twice, then retain appropriate accuracy because this is an estimate.

Sketch on each strip Chord relative to curve Estimate
concave up chord above curve over-estimate
concave down chord below curve under-estimate

With $h=0.5$ at $x=0,0.5,1$, there are two strips and three ordinates:T=0.25(y_0+2y_1+y_2).

The number of ordinates is one more than the number of strips. If curvature changes, inspect or split the graph rather than claiming one global error direction from a single segment.

2.6 Numerical solution of equations

Syllabus
9709–2028–2029
Topic
2.6
Level
AS

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.

An iterative approximation is useful only when its sequence converges to the intended root

Iteration replaces x by x_{n+1}=g(x_n). A fixed point α satisfies g(α)=α, corresponding to a root of the rearranged equation.

Choose a starting value in the stated interval, compute enough figures during iteration, and stop using a tolerance on successive values or the residual. Different rearrangements can converge differently.

For x=cos x, starting x₀=0 gives a sequence approaching about 0.739; starting values should remain in a region where g behaves stably.

A few stable-looking digits do not prove convergence, and iteration can diverge or enter a cycle even when the equation has a root.

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.