2. Pure Mathematics AS 2
- Syllabus
- 9709–2028–2029
- Section
- 2
- Level
- AS

∣u∣ is the non-negative distance of u from 0. Thus y=∣ax+b∣ is a V-shaped graph with vertex where ax+b=0, slopes −∣a∣ then ∣a∣, and range y≥0.
| Form | Equivalent relation |
|---|---|
| ∣u∣=c, c≥0 | u=c or u=−c |
| ∣u∣=∣v∣ | u2=v2, equivalently 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 |
If the other side contains x, 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 ≤. Non-linear graphs y=∣f(x)∣ and y=f(∣x∣) are excluded from this objective.
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+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 imes quotient + remainder equals the original dividend. Synthetic division is only a shortcut for suitable linear divisors, not a quadratic divisor.
| Division statement | Evaluation statement |
|---|---|
| remainder on division by x−c is k | P(c)=k |
| x−c is a factor | P(c)=0 |
| ax+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+2 leaves 5, then P(−2)=5, not 0. Use polynomial division when the full quotient is needed.
The zero of ax+b is −b/a. A factor statement and a remainder statement are different equations; no repeated-root derivative test is needed for this objective.
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) |
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 lng is g′/g, not 1/lng. Products and quotients are handled in the next objective rather than smuggled into a sum/chain shortcut.
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 |
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.
General integration by substitution and integration by parts are not required in Pure Mathematics 2; use these direct reverse-derivative patterns only.
\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.
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+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.
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.
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.
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.