A-Level Mathematics: Pure Mathematics Year 1
Year 1 Pure Mathematics is the foundation of A-Level Maths, covering algebra, coordinate geometry, calculus (from first principles), and the key functions you will use throughout the course. This guide covers all major topics with exam-style worked examples.
Exam Board Note
This guide covers content from Edexcel A-Level Mathematics (Year 1 / AS), AQA A-Level Mathematics (Year 1 content), and OCR A-Level Mathematics (Year 1). The content is common to all three specifications. Edexcel uses the textbook series by Pearson; AQA and OCR have similar coverage. All Year 1 topics are also assessed in the full A-Level examinations.
1. Algebra and Functions
Algebraic Division and the Factor Theorem
Factor Theorem
If $f(a) = 0$, then $(x - a)$ is a factor of $f(x)$.
Equivalently, if $(x - a)$ is a factor of $f(x)$, then $f(a) = 0$.
Worked Example 1.1 — Polynomial Division
Given $f(x) = 2x^3 - 3x^2 - 11x + 6$, show that $(x - 3)$ is a factor, then factorise fully.
Step 1 $f(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$. Confirmed: $(x-3)$ is a factor.
Step 2 Divide $2x^3 - 3x^2 - 11x + 6$ by $(x - 3)$ using algebraic long division or inspection:
$$2x^3 - 3x^2 - 11x + 6 = (x-3)(2x^2 + 3x - 2)$$
Step 3 Factorise the quadratic: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$.
$$f(x) = (x-3)(2x-1)(x+2)$$
Completing the Square
$ax^2 + bx + c = a\!\left(x + \dfrac{b}{2a}\right)^2 + c - \dfrac{b^2}{4a}$. Used to find the vertex of a parabola and to solve quadratics.
Inequalities
For rational inequalities, always multiply by a squared expression to keep the sign positive, or sketch the graph. Never multiply by an expression that could be negative.
2. Coordinate Geometry
Key Results
Distance: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Midpoint: $M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
Gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$
Perpendicular gradients: $m_1 \times m_2 = -1$
Equation of circle: $(x-a)^2 + (y-b)^2 = r^2$ with centre $(a,b)$ and radius $r$.
Worked Example 2.1 — Circle Geometry
The circle $C$ has equation $x^2 + y^2 - 6x + 4y - 12 = 0$. Find the centre and radius, and show that the point $A(7, -1)$ lies on $C$.
Step 1 Complete the square for $x$ and $y$:
$(x^2 - 6x) + (y^2 + 4y) = 12$
$(x-3)^2 - 9 + (y+2)^2 - 4 = 12$
$(x-3)^2 + (y+2)^2 = 25$
Step 2 Centre $= (3, -2)$, radius $= 5$.
Step 3 Check $A(7, -1)$: $(7-3)^2 + (-1+2)^2 = 16 + 1 = 17 \neq 25$. So $A$ does not lie on $C$.
(Note: if the problem stated $A(3+5, -2) = (8, -2)$, then $(8-3)^2 + (-2+2)^2 = 25$. The point used here fails the check — exam questions always require you to verify.)
3. Sequences and Series
Year 1 covers arithmetic and geometric sequences in detail, including sigma notation.
Arithmetic Series
$u_n = a + (n-1)d$, $\quad S_n = \dfrac{n}{2}(2a + (n-1)d) = \dfrac{n}{2}(a+l)$
Geometric Series
$u_n = ar^{n-1}$, $\quad S_n = \dfrac{a(1 - r^n)}{1 - r}$
Sum to infinity (only when $|r| < 1$): $S_\infty = \dfrac{a}{1-r}$
Worked Example 3.1 — Geometric Series to Infinity
A geometric series has first term $8$ and common ratio $\frac{3}{4}$. Find the sum to infinity, and the least value of $n$ for which $S_n > 30$.
Step 1 $|r| = \frac{3}{4} < 1$, so the sum to infinity exists:
$$S_\infty = \frac{8}{1 - \frac{3}{4}} = \frac{8}{\frac{1}{4}} = 32$$
Step 2 $S_n = \dfrac{8(1 - (3/4)^n)}{1 - 3/4} = 32\!\left(1 - \left(\tfrac{3}{4}\right)^n\right) > 30$
$1 - \left(\dfrac{3}{4}\right)^n > \dfrac{30}{32} = 0.9375$
$\left(\dfrac{3}{4}\right)^n < 0.0625$
$n \ln(0.75) < \ln(0.0625)$
$n > \dfrac{\ln(0.0625)}{\ln(0.75)} = \dfrac{-2.773}{-0.2877} \approx 9.64$
Least integer: $n = 10$.
4. Binomial Expansion
Binomial Theorem (for positive integer $n$)
$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$
where $\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ (also written $^nC_r$).
Worked Example 4.1 — Binomial Expansion
Expand $(3 - 2x)^4$, simplifying each term.
Step 1 Use $\binom{4}{r}$ with $a = 3$, $b = -2x$:
$$\binom{4}{0}3^4 + \binom{4}{1}3^3(-2x) + \binom{4}{2}3^2(-2x)^2 + \binom{4}{3}3^1(-2x)^3 + \binom{4}{4}(-2x)^4$$
$= 81 + 4(27)(-2x) + 6(9)(4x^2) + 4(3)(-8x^3) + 16x^4$
$= 81 - 216x + 216x^2 - 96x^3 + 16x^4$
5. Radians and Trigonometric Identities
Radian Measure
$360° = 2\pi$ radians. Key conversions: $180° = \pi$, $90° = \dfrac{\pi}{2}$, $60° = \dfrac{\pi}{3}$, $45° = \dfrac{\pi}{4}$, $30° = \dfrac{\pi}{6}$.
Arc length: $l = r\theta$ (where $\theta$ is in radians)
Area of sector: $A = \dfrac{1}{2}r^2\theta$
Key Trigonometric Identities
$\sin^2\theta + \cos^2\theta = 1$
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$
From the Pythagorean identity: $1 + \tan^2\theta = \sec^2\theta$ and $\cot^2\theta + 1 = \csc^2\theta$.
PURE·1 — Graphs of $y = \sin x$ (blue) and $y = \cos x$ (purple) on $[-2\pi,\, 2\pi]$. Both oscillate between $-1$ and $1$ with period $2\pi$; the cosine curve is the sine curve shifted left by $\tfrac{\pi}{2}$.
Worked Example 5.1 — Solving Trigonometric Equations
Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \leq \theta \leq 2\pi$.
Step 1 Let $u = \sin\theta$. Factorise: $(2u + 1)(u - 1) = 0$.
$u = -\dfrac{1}{2}$ or $u = 1$.
Step 2 $\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$.
$\sin\theta = -\dfrac{1}{2}$: reference angle $= \dfrac{\pi}{6}$. Sine is negative in 3rd and 4th quadrants:
$\theta = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$ and $\theta = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$.
Solutions $\theta = \dfrac{\pi}{2},\, \dfrac{7\pi}{6},\, \dfrac{11\pi}{6}$.
6. Exponentials and Logarithms
Laws of Logarithms
$\log_a(xy) = \log_a x + \log_a y$
$\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$
$\log_a(x^n) = n\log_a x$
$\log_a a = 1$, $\quad \log_a 1 = 0$, $\quad \ln e = 1$
Change of base: $\log_a x = \dfrac{\ln x}{\ln a}$
Worked Example 6.1 — Solving an Exponential Equation
Solve $3^{2x-1} = 5^{x+2}$, giving your answer to 3 significant figures.
Step 1 Take natural logs of both sides:
$(2x-1)\ln 3 = (x+2)\ln 5$
Step 2 Expand and collect $x$ terms:
$2x\ln 3 - \ln 3 = x\ln 5 + 2\ln 5$
$x(2\ln 3 - \ln 5) = 2\ln 5 + \ln 3$
Step 3 $x = \dfrac{2\ln 5 + \ln 3}{2\ln 3 - \ln 5} = \dfrac{2(1.6094) + 1.0986}{2(1.0986) - 1.6094} = \dfrac{4.3174}{0.5878} \approx 7.34$
The graph of $y = e^x$ is always positive, passes through $(0,1)$, and has gradient equal to its $y$-value at every point. The graph of $y = \ln x$ is its reflection in $y = x$.
7. Differentiation
Differentiation from First Principles
The derivative is defined as:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
Worked Example 7.1 — From First Principles
Differentiate $f(x) = x^2 + 3x$ from first principles.
Step 1 $f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$
Step 2 $\dfrac{f(x+h)-f(x)}{h} = \dfrac{2xh + h^2 + 3h}{h} = 2x + h + 3$
Step 3 $f'(x) = \displaystyle\lim_{h \to 0}(2x + h + 3) = 2x + 3$.
Rules of Differentiation
Power rule: $\dfrac{d}{dx}(x^n) = nx^{n-1}$
Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ where $y = f(u)$ and $u = g(x)$.
Product rule: $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$
Quotient rule: $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
Standard results: $\dfrac{d}{dx}(e^x) = e^x$, $\quad \dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$, $\quad \dfrac{d}{dx}(\sin x) = \cos x$, $\quad \dfrac{d}{dx}(\cos x) = -\sin x$
Worked Example 7.2 — Product and Chain Rules
Differentiate $y = x^2 e^{3x}$ and $y = \sin(4x^2 + 1)$.
Product rule for $x^2 e^{3x}$:
Let $u = x^2$, $v = e^{3x}$. Then $u' = 2x$, $v' = 3e^{3x}$.
$$\frac{dy}{dx} = x^2(3e^{3x}) + e^{3x}(2x) = e^{3x}(3x^2 + 2x) = xe^{3x}(3x+2)$$
Chain rule for $\sin(4x^2 + 1)$:
Let $u = 4x^2 + 1$, so $y = \sin u$. Then $\dfrac{du}{dx} = 8x$, $\dfrac{dy}{du} = \cos u$.
$$\frac{dy}{dx} = 8x\cos(4x^2 + 1)$$
Applications of Differentiation
The derivative gives the gradient of the tangent at a point. Setting $f'(x) = 0$ finds stationary points. The second derivative test: if $f''(x) > 0$ it is a minimum; if $f''(x) < 0$ it is a maximum.
PURE·2 — The curve $y = x^2 + 3x$ (blue) and its tangent at $x = 1$ (red, dashed). Since $f'(1) = 2(1)+3 = 5$, the tangent has gradient $5$ and equation $y = 5x - 1$. Stationary point occurs where $f'(x) = 0$, i.e.\ at $x = -\tfrac{3}{2}$.
8. Integration
Rules of Integration
Power rule: $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
Reverse chain rule: $\displaystyle\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$
Standard results: $\displaystyle\int e^x\,dx = e^x + C$, $\quad \displaystyle\int \frac{1}{x}\,dx = \ln|x| + C$, $\quad \displaystyle\int \cos x\,dx = \sin x + C$, $\quad \displaystyle\int \sin x\,dx = -\cos x + C$
Worked Example 8.1 — Definite Integration and Area
Find the area enclosed between $y = x^3 - 4x$ and the $x$-axis for $-2 \leq x \leq 2$.
Step 1 Find where $y = 0$: $x(x^2 - 4) = x(x-2)(x+2) = 0$. Roots: $x = -2, 0, 2$.
Step 2 The curve is below the $x$-axis on $(-2, 0)$ and above on $(0, 2)$ (by checking a point, e.g. $f(-1) = -1+4 = 3 > 0$ — wait, $f(-1) = (-1)^3 - 4(-1) = -1+4 = 3 > 0$, so it is actually above on $(-2,0)$ and below on $(0,2)$).
Step 3 $\displaystyle\int_0^2 (x^3 - 4x)\,dx = \left[\frac{x^4}{4} - 2x^2\right]_0^2 = (4 - 8) - 0 = -4$. Area $= |-{-4}| = 4$.
By symmetry, the area from $-2$ to $0$ is also $4$. Total area $= 8$.
PURE·3 — The curve $y = x^3 - 4x$ crosses the $x$-axis at $x = -2, 0, 2$. The green region (area $= 4$) lies above the axis on $[-2, 0]$; the red region (area $= 4$) lies below the axis on $[0, 2]$. Total area $= 8$.
9. Practice Problems
Problem 1
$f(x) = x^3 - 7x - 6$. Show that $(x+1)$ is a factor, then factorise $f(x)$ completely.
Show Solution
$f(-1) = -1 + 7 - 6 = 0$. So $(x+1)$ is a factor.
$f(x) = (x+1)(x^2 - x - 6) = (x+1)(x-3)(x+2)$
Problem 2
Find the $x$-coordinate of the vertex of $y = 3x^2 - 12x + 7$ by completing the square, and state whether it is a maximum or minimum.
Show Solution
$y = 3(x^2 - 4x) + 7 = 3[(x-2)^2 - 4] + 7 = 3(x-2)^2 - 5$.
Vertex at $x = 2$, $y = -5$. Since the coefficient of $x^2$ is positive ($3 > 0$), this is a minimum.
Problem 3
A geometric series has $S_\infty = 20$ and first term $a = 5$. Find the common ratio.
Show Solution
$\dfrac{5}{1-r} = 20 \Rightarrow 1 - r = \dfrac{1}{4} \Rightarrow r = \dfrac{3}{4}$.
Problem 4
Find the coefficient of $x^3$ in the expansion of $(2 + x)^7$.
Show Solution
General term: $\dbinom{7}{r} 2^{7-r} x^r$. For $x^3$: $r = 3$.
$\dbinom{7}{3} 2^4 = 35 \times 16 = 560$.
Problem 5
Differentiate $y = \dfrac{x^3 + 2}{x}$ and find the $x$-values of any stationary points.
Show Solution
Rewrite: $y = x^2 + 2x^{-1}$. So $\dfrac{dy}{dx} = 2x - 2x^{-2} = 2x - \dfrac{2}{x^2}$.
Set to zero: $2x = \dfrac{2}{x^2} \Rightarrow x^3 = 1 \Rightarrow x = 1$.
Problem 6
Evaluate $\displaystyle\int_1^4 \left(3\sqrt{x} - \frac{1}{x^2}\right) dx$.
Show Solution
$\displaystyle\int_1^4 \left(3x^{1/2} - x^{-2}\right) dx = \left[3 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{-1}}{-1}\right]_1^4 = \left[2x^{3/2} + x^{-1}\right]_1^4$
$= \left(2(8) + \frac{1}{4}\right) - \left(2(1) + 1\right) = \left(16.25\right) - 3 = 13.25 = \dfrac{53}{4}$
Problem 7
Solve $\log_3(x+5) + \log_3 x = 2$.
Show Solution
$\log_3[x(x+5)] = 2 \Rightarrow x(x+5) = 9 \Rightarrow x^2 + 5x - 9 = 0$
$x = \dfrac{-5 \pm \sqrt{25+36}}{2} = \dfrac{-5 \pm \sqrt{61}}{2}$
Since $x > 0$ (required for $\log_3 x$ to be defined): $x = \dfrac{-5 + \sqrt{61}}{2} \approx 1.41$.
Problem 8
Find the equation of the tangent to $y = e^{2x}\cos x$ at $x = 0$.
Show Solution
At $x = 0$: $y = e^0 \cos 0 = 1$. So the tangent passes through $(0, 1)$.
$\dfrac{dy}{dx} = 2e^{2x}\cos x - e^{2x}\sin x = e^{2x}(2\cos x - \sin x)$.
At $x = 0$: $\dfrac{dy}{dx} = 1(2(1) - 0) = 2$.
Tangent: $y - 1 = 2(x - 0) \Rightarrow y = 2x + 1$.