Pure Mathematics 3 (P3) – Functions and Core Extensions (Cambridge International AS & A Level Mathematics 9709)
1. Functions – Domain, Range & Restrictions
Learning outcomes (AO1 & AO2)
- Identify algebraic, radical, logarithmic and composite restrictions on a function.
- Express the domain in interval notation and describe the possible range.
- Apply the systematic procedure to new functions in exam questions.
Key ideas
For a real‑valued function \(f(x)\) the domain is the set of all real numbers for which the expression is defined; the range is the set of values that the function can actually take.
| Expression | Restriction | Resulting condition |
|---|
| \(\displaystyle\frac{1}{g(x)}\) | Denominator ≠ 0 | \(g(x)eq0\) |
| \(\sqrt[h]{g(x)}\) ( \(h\) even ) | Radicand ≥ 0 | \(g(x)\ge0\) |
| \(\log_b\bigl(g(x)\bigr)\) ( \(b>0,\;beq1\) ) | Argument > 0 | \(g(x)>0\) |
| Composite \(f\!\bigl(g(x)\bigr)\) | Domain of \(g\) ∩ pre‑image of domain of \(f\) | Combine the two sets of conditions |
Systematic procedure (AO2)
- List every restriction (denominator, even root, log, domain of inner function).
- Write each as an inequality or equation.
- Solve each condition and express the solution set.
- Take the intersection of all solution sets – this is the domain.
- If required, reason about the range (monotonicity, asymptotes, sign analysis, or by sketching).
Worked example
Find the domain of \(\displaystyle f(x)=\frac{\sqrt{2x-5}}{x^{2}-9}\).
- Radicand: \(2x-5\ge0\;\Rightarrow\;x\ge\frac52\).
- Denominator: \(x^{2}-9eq0\;\Rightarrow\;xeq\pm3\).
Intersection: \(\displaystyle x\in\Bigl[\frac52,3\Bigr)\cup(3,\infty)\).
Practice question
State the domain of \(\displaystyle g(x)=\frac{\log(x-1)}{\sqrt{4-x}}\).
Solution sketch: \(\log(x-1)\) requires \(x-1>0\Rightarrow x>1\); \(\sqrt{4-x}\) requires \(4-x\ge0\Rightarrow x\le4\). Intersection: \((1,4]\).
2. Modulus (Absolute‑Value) Function
Learning outcomes (AO1 & AO2)
- Apply the definition \(|x|=\begin{cases}x,&x\ge0\\-x,&x<0\end{cases}\).
- Solve linear equations and inequalities involving absolute values by case‑splitting.
- Interpret absolute‑value expressions geometrically on the number line.
Fundamental properties
- \(|x|\ge0\) for all real \(x\).
- \(|ab|=|a|\,|b|,\qquad |a/b|=|a|/|b|\;(beq0).\)
- \(|a+b|\le|a|+|b|\) (triangle inequality).
Solving strategy (AO2)
- Isolate the absolute‑value term.
- Replace \(|A|=B\) with the two equations \(A=B\) and \(A=-B\) (provided \(B\ge0\)).
- For \(|A|B\) rewrite as a double inequality:
\(\displaystyle -BB\) (with \(B>0\)).
Worked example – Equation
\(|2x-3|=5\)
- \(2x-3=5\;\Rightarrow\;x=4\).
- \(2x-3=-5\;\Rightarrow\;x=-1\).
Solution set \(\{-1,4\}\).
Worked example – Inequality
\(|x+2|<3\)
Interpret as \(-3
Solution interval \((-5,1)\).
Practice question
Solve \(|3t+7|\ge 2\).
Solution sketch: \(\;3t+7\le-2\) or \(3t+7\ge2\) → \(t\le-3\) or \(t\ge-\frac53\). Hence \((-\infty,-3]\cup[-\tfrac53,\infty)\).
3. Sketching Graphs of Functions
Learning outcomes (AO1 & AO2)
- Analyse a function using domain, intercepts, symmetry, asymptotes, critical points and points of inflection.
- Combine this information to produce a clear, accurate sketch suitable for exam marks.
Checklist for a complete sketch
- Domain & Range – from restrictions and end‑behaviour.
- Intercepts – solve \(f(x)=0\) (x‑intercepts) and evaluate \(f(0)\) (y‑intercept, if defined).
- Symmetry – test \(f(-x)=f(x)\) (even) or \(f(-x)=-f(x)\) (odd).
- Asymptotes
- Vertical: where the denominator = 0 or log/√ arguments become invalid.
- Horizontal: compare degrees of numerator/denominator or use limits.
- Oblique (slant): perform polynomial division when \(\deg(\text{numerator})=\deg(\text{denominator})+1\).
- Critical points – compute \(f'(x)\); solve \(f'(x)=0\) or where \(f'\) is undefined; classify using the first‑derivative sign test or the second‑derivative test.
- Points of inflection – solve \(f''(x)=0\) (where defined) and check for a sign change.
- Behaviour at domain boundaries – limits as \(x\) approaches vertical asymptotes or endpoints.
- Plot a few convenient points in each interval to confirm the shape.
Worked example – Sketch \(y=\displaystyle\frac{|x-1|}{x^{2}-4}\)
- Domain: \(x^{2}-4eq0\Rightarrow xeq\pm2\).
- Intercepts:
- y‑intercept: \(x=0\Rightarrow y=\frac{| -1|}{-4}=-\tfrac14\).
- x‑intercept: numerator = 0 → \(x=1\) (allowed).
- Symmetry: Not even nor odd (the shift in \(|x-1|\) destroys symmetry).
- Vertical asymptotes: \(x=-2,\;x=2\).
- Horizontal asymptote: \(\deg\) denominator (2) > \(\deg\) numerator (1) → \(y=0\).
- Derivative (quotient rule):
\[
f'(x)=\frac{(x^{2}-4)\,\operatorname{sgn}(x-1)-|x-1|\,(2x)}{(x^{2}-4)^{2}},
\qquad
\operatorname{sgn}(x-1)=\begin{cases}1,&x>1\\-1,&x<1\end{cases}.
\]
Solve \(f'(x)=0\) separately on \((-\infty,1)\) and \((1,\infty)\) to locate maxima/minima.
- Sign analysis: Use the sign of numerator and denominator to decide where the curve lies above or below the x‑axis.
- Key plotted points: \((-3,\tfrac45),\;(-1,\tfrac23),\;(0,-\tfrac14),\;(1,0),\;(3,\tfrac25).\)
- Connect the points respecting the asymptotes – the final sketch shows two separate branches with a hole at \(x=1\) and vertical asymptotes at \(\pm2\).
Practice question
Sketch \(y=\displaystyle\frac{x^{2}-4}{|x+2|}\). List domain, intercepts, asymptotes and one critical point.
Solution sketch:
- Domain: \(xeq-2\).
- x‑intercepts: \(x=\pm2\) (both allowed).
- y‑intercept: \(y=\frac{-4}{2}=-2\).
- Vertical asymptote: \(x=-2\) (function → ±∞ on either side).
- No finite horizontal asymptote (degree numerator = 2 > degree denominator = 1).
- Derivative \(y'=\frac{2x|x+2|-(x^{2}-4)\operatorname{sgn}(x+2)}{|x+2|^{2}}\). Setting \(y'=0\) gives a single critical point at \(x=0\) (a minimum, \(y=-2\)).
4. Algebra – Polynomials, Factorisation & Partial Fractions
Learning outcomes (AO1 & AO2)
- Perform polynomial long division and synthetic division.
- Apply the Factor Theorem and Remainder Theorem.
- Decompose proper rational functions into partial fractions.
- Use the Binomial Theorem for integer and fractional indices.
Key formulas
| Concept | Statement |
|---|
| Remainder Theorem | If \(f(x)\) is divided by \((x-a)\), the remainder is \(f(a)\). |
| Factor Theorem | \((x-a)\) is a factor of \(f(x)\) ⇔ \(f(a)=0\). |
| Binomial expansion (integer \(n\)) | \((a+b)^{n}= \displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k}\). |
| Binomial expansion (fractional \(n\)) | \((1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^{2}+\cdots,\;|x|<1.\) |
Polynomial division (synthetic)
To divide \(f(x)\) by \((x-a)\): write the coefficients of \(f(x)\), bring down the leading coefficient, multiply by \(a\), add down the column, repeat. The final row gives the coefficients of the quotient and the remainder.
Worked example – Partial fractions
Decompose \(\displaystyle \frac{2x+5}{(x-1)(x+2)}\).
- Assume \(\displaystyle\frac{2x+5}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}\).
- Multiply through: \(2x+5=A(x+2)+B(x-1)\).
- Set \(x=1\) → \(2(1)+5= A(3)\) ⇒ \(A=\frac{7}{3}\).
Set \(x=-2\) → \(-4+5= B(-3)\) ⇒ \(B=-\frac13\).
- Result: \(\displaystyle\frac{2x+5}{(x-1)(x+2)}=\frac{7/3}{x-1}-\frac{1/3}{x+2}\).
Practice question
Write \(\displaystyle\frac{x^{2}+4x+5}{(x+1)^{2}}\) in the form \(A+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}\).
Solution sketch: Perform polynomial division:
\[
\frac{x^{2}+4x+5}{(x+1)^{2}}=1+\frac{2x+4}{(x+1)^{2}}.
\]
Now set \(\displaystyle\frac{2x+4}{(x+1)^{2}}=\frac{B}{x+1}+\frac{C}{(x+1)^{2}}\) → \(2x+4=B(x+1)+C\).
Comparing coefficients gives \(B=2,\;C=2\). Hence
\[
1+\frac{2}{x+1}+\frac{2}{(x+1)^{2}}.
\]
5. Logarithmic & Exponential Functions
Learning outcomes (AO1 & AO2)
- Define exponential functions \(y=a^{x}\) (\(a>0,\;aeq1\)) and logarithmic functions \(y=\log_{a}x\).
- Apply the laws of logarithms and exponentials.
- Solve equations of the form \(a^{x}=b\) and \(\log_{a}x=c\).
- Linearise data using semi‑log and log‑log plots.
Logarithm laws
| Law | Expression |
|---|
| Product | \(\log_{a}(xy)=\log_{a}x+\log_{a}y\) |
| Quotient | \(\log_{a}\!\left(\dfrac{x}{y}\right)=\log_{a}x-\log_{a}y\) |
| Power | \(\log_{a}(x^{k})=k\log_{a}x\) |
| Change of base | \(\displaystyle\log_{a}x=\frac{\log_{b}x}{\log_{b}a}\) |
Exponential laws
- \(a^{m}a^{n}=a^{m+n}\)
- \(\dfrac{a^{m}}{a^{n}}=a^{m-n}\)
- \((a^{m})^{n}=a^{mn}\)
Solving example – Exponential equation
Solve \(3^{2x-1}=27\).
- Write \(27=3^{3}\) → \(3^{2x-1}=3^{3}\).
- Equate exponents: \(2x-1=3\) → \(2x=4\) → \(x=2\).
Solving example – Logarithmic equation
Solve \(\log_{2}(x+5)=3\).
- Rewrite as \(x+5=2^{3}=8\).
- Hence \(x=3\).
Practice question
Solve for \(x\): \(\displaystyle 5^{x}= \frac{1}{125}\).
Solution sketch: \(125=5^{3}\) so \(\frac{1}{125}=5^{-3}\). Equate exponents: \(x=-3\).
6. Quadratic Functions (Pure Mathematics 1)
Learning outcomes (AO1 & AO2)
- Complete the square and use it to find the vertex, axis of symmetry and minimum/maximum.
- Interpret the discriminant \(\Delta=b^{2}-4ac\) for real roots.
- Solve simultaneous linear‑quadratic systems.
- Apply quadratic functions to optimisation and area problems.
Key formulas
- Standard form: \(y=ax^{2}+bx+c\).
- Vertex form: \(y=a(x-h)^{2}+k\) where \(h=-\frac{b}{2a},\;k=c-\frac{b^{2}}{4a}\).
- Discriminant: \(\Delta=b^{2}-4ac\)
- \(\Delta>0\) – two distinct real roots.
- \(\Delta=0\) – one repeated real root (tangent to the x‑axis).
- \(\Delta<0\) – no real roots (graph lies wholly above or below the x‑axis).
Worked example – Vertex
Find the vertex of \(y=2x^{2}-8x+5\).
- \(h=-\dfrac{-8}{2\cdot2}=2\).
- \(k=5- \dfrac{(-8)^{2}}{4\cdot2}=5- \dfrac{64}{8}=5-8=-3\).
- Vertex \((2,-3)\); axis of symmetry \(x=2\).
Practice question
Determine the nature of the roots of \(3x^{2}+4x-7=0\) and solve the equation.
Solution sketch: \(\Delta=4^{2}-4\cdot3\cdot(-7)=16+84=100>0\) → two real roots. Using the quadratic formula,
\(x=\dfrac{-4\pm\sqrt{100}}{2\cdot3}=\dfrac{-4\pm10}{6}\) giving \(x=1\) or \(x=-\tfrac73\).
7. Coordinate Geometry (Pure Mathematics 1)
Learning outcomes (AO1 & AO2)
- Write equations of straight lines in slope‑intercept, point‑slope and general form.
- Find the distance between two points and the gradient of a line.
- Write the equation of a circle given centre and radius, or given three points.
- Use the perpendicular bisector method to find the centre of a circle.
Key formulas
- Distance: \(d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\).
- Mid‑point: \(\displaystyle\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)\).
- Line through \((x_{1},y_{1})\) and \((x_{2},y_{2})\): \(y-y_{1}=m(x-x_{1})\) where \(m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\).
- Circle with centre \((h,k)\) and radius \(r\): \((x-h)^{2}+(y-k)^{2}=r^{2}\).
Worked example – Circle from three points
Find the equation of the circle passing through \(A(1,2),\;B(4,6),\;C(5,2)\).
- Find the perpendicular bisectors of \(AB\) and \(BC\).
- Mid‑point of \(AB\): \((\tfrac{1+4}{2},\tfrac{2+6}{2})=(2.5,4)\). Slope of \(AB\): \(\frac{6-2}{4-1}= \frac{4}{3}\) → perpendicular slope \(-\tfrac34\). Equation: \(y-4=-\tfrac34(x-2.5)\).
- Mid‑point of \(BC\): \((\tfrac{4+5}{2},\tfrac{6+2}{2})=(4.5,4)\). Slope of \(BC\): \(\frac{2-6}{5-4}=-4\) → perpendicular slope \(\tfrac14\). Equation: \(y-4=\tfrac14(x-4.5)\).
- Solve the two linear equations → centre \((h,k)=(3,3)\).
- Radius \(r=\sqrt{(1-3)^{2}+(2-3)^{2}}=\sqrt{4+1}= \sqrt5\).
- Circle: \((x-3)^{2}+(y-3)^{2}=5.\)
Practice question
Find the equation of the line passing through \((-2,5)\) with gradient \(-3\).
Solution: \(y-5=-3(x+2)\) → \(y=-3x-1\).
8. Circular Measure (Pure Mathematics 1)
Learning outcomes (AO1 & AO2)
- Convert between degrees and radians.
- Use the definitions of arc length and sector area.
- Apply these formulas in geometry and trigonometry problems.
Key formulas
- Radians to degrees: \(\displaystyle 1\text{ rad}= \frac{180}{\pi}^{\circ}\).
- Degrees to radians: \(\displaystyle \theta^{\circ}= \frac{\pi}{180}\theta\) rad.
- Arc length: \(s=r\theta\) (θ in radians).
- Sector area: \(A=\frac12 r^{2}\theta\) (θ in radians).
Worked example
Find the length of an arc subtended by a central angle of \(45^{\circ}\) in a circle of radius \(8\) cm.
- Convert: \(\theta=45^{\circ}= \frac{\pi}{4}\) rad.
- Arc length \(s= r\theta = 8\cdot\frac{\pi}{4}=2\pi\) cm.
Practice question
A sector has radius \(10\) cm and area \(25\pi\) cm\(^2\). Find the central angle in degrees.
Solution: \(A=\frac12 r^{2}\theta\Rightarrow 25\pi=\frac12(10)^{2}\theta\Rightarrow 25\pi=50\theta\Rightarrow\theta=\frac{\pi}{2}\) rad = \(90^{\circ}\).
9. Basic Trigonometry (Pure Mathematics 1)
Learning outcomes (AO1 & AO2)
- Recall the six fundamental trig ratios for right‑angled triangles.
- Use the sine, cosine and tangent rules for any triangle.
- Solve simple trig equations and apply them to geometry problems.
Key identities
- \(\sin^{2}\theta+\cos^{2}\theta=1\).
- \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) (where \(\cos\thetaeq0\)).
- Co‑function: \(\sin(90^{\circ}-\theta)=\cos\theta\), etc.
Worked example – Solving a trig equation
Solve \(\sin\theta = \frac12\) for \(0^{\circ}\le\theta<360^{\circ}\).
Solutions: \(\theta=30^{\circ}\) or \(\theta=150^{\circ}\).
Practice question
In \(\triangle ABC\), \(AB=7\), \(AC=9\) and \(\angle BAC=60^{\circ}\). Find the length of \(BC\) using the cosine rule.
Solution: \(BC^{2}=7^{2}+9^{2}-2\cdot7\cdot9\cos60^{\circ}=49+81-126\cdot\frac12=130-63=67\). Hence \(BC=\sqrt{67}\) cm.
10. Series (Pure Mathematics 1)
Learning outcomes (AO1 & AO2)
- Write the nth term of an arithmetic progression (AP) and a geometric progression (GP).
- Find the sum of the first \(n\) terms of an AP and the sum to infinity of a convergent GP.
- Apply series formulae to word‑problem contexts.
Key formulas
- AP: \(a_{n}=a_{1}+(n-1)d\).
- Sum of first \(n\) terms: \(S_{n}= \dfrac{n}{2}\bigl(a_{1}+a_{n}\bigr)=\dfrac{n}{2}\bigl[2a_{1}+(n-1)d\bigr]\).
- GP: \(a_{n}=a_{1}r^{\,n-1}\).
- Sum of first \(n\) terms: \(S_{n}=a_{1}\dfrac{1-r^{n}}{1-r}\) ( \(req1\)).
- Infinite sum (|r|<1): \(S_{\infty}= \dfrac{a_{1}}{1-r}\).
Worked example – Sum of an AP
Find the sum of the first 20 terms of the AP: \(5, 9, 13,\dots\).
- Common difference \(d=4\); \(a_{1}=5\); \(a_{20}=5+19\cdot4=81\).
- \(S_{20}= \dfrac{20}{2}(5+81)=10\cdot86=860.\)
Practice question
A GP has first term \(3\) and common ratio \(\frac12\). Find the sum to infinity.
Solution: \(S_{\infty}= \dfrac{3}{1-\frac12}=6.\)
11. Differentiation & Integration (Basic – PM1)
Learning outcomes (AO1 & AO2)
- Differentiate polynomial functions using the power rule.
- Integrate simple powers of \(x\) (including constants) using the reverse power rule.
- Apply differentiation to find gradients, stationary points and to solve optimisation problems.
- Use integration to find areas under curves between two points.
Key rules
- Power rule (differentiation): \(\displaystyle\frac{d}{dx}x^{n}=nx^{\,n-1}\) (any real \(n\)).
- Power rule (integration): \(\displaystyle\int x^{n}\,dx=\frac{x^{\,n+1}}{n+1}+C\) (\(neq-1\)).
- Constant multiple: \(\displaystyle\frac{d}{dx}[k\,f(x)]=k\,f'(x)\), \(\displaystyle\int k\,f(x)\,dx=k\int f(x)\,dx\).
- Sum rule: differentiate or integrate term‑by‑term.
Worked example – Gradient of a curve
Find the gradient of \(y=4x^{3}-2x^{2}+5\) at \(x=1\).
- Derivative: \(y'=12x^{2}-4x\).
- At \(x=1\): \(y'(1)=12-4=8\). Gradient = 8.
Worked example – Area under a curve
Find the area between \(y=x^{2}\) and the x‑axis from \(x=0\) to \(x=3\).
- Integral: \(\displaystyle\int_{0}^{3} x^{2}\,dx=\Bigl[\frac{x^{3}}{3}\Bigr]_{0}^{3}= \frac{27}{3}=9.\)
Practice question
Find the stationary points of \(f(x)=x^{3}-6x^{2}+9x\) and classify them.
Solution: \(f'(x)=3x^{2}-12x+9=3(x^{2}-4x+3)=3(x-1)(x-3)\). Critical points at \(x=1,3\).
\(f''(x)=6x-12\). \(f''(1)=-6<0\) → local maximum at \((1,4)\).
\(f''(3)=6>0\) → local minimum at \((3,0)\).
12. Advanced Trigonometry (Pure Mathematics 2)
Learning outcomes (AO1 & AO2)
- Use the sine, cosine and tangent rules for any triangle.
- Apply the area formula \(\frac12ab\sin C\) and the law of cosines in problem solving.
- Solve trigonometric equations involving multiple angles and use identities to simplify.
Key identities
- \(\sin A=\sin(180^{\circ}-A)\), \(\cos A=-\cos(180^{\circ}-A)\).