A-Level Mathematics: Pure Mathematics Year 2
Year 2 Pure Mathematics extends every topic from Year 1 and introduces powerful new techniques, particularly in calculus. This guide covers proof, partial fractions, advanced trigonometry, parametric and implicit differentiation, integration by parts and by substitution, differential equations, and numerical methods.
Exam Board Note
This guide covers Edexcel A-Level Mathematics Year 2 (Pearson), AQA A-Level Mathematics Year 2, and OCR A-Level Mathematics (H240) Year 2. All topics appear in the full A-Level Papers 1 and 2. Numerical methods (Newton-Raphson and iteration) appear on all three specifications.
1. Proof
A-Level requires you to construct and evaluate mathematical proofs. Common methods:
Methods of Proof
Proof by deduction: Use known results and logical steps to reach the conclusion.
Proof by exhaustion: Check all possible cases.
Proof by counter-example: Find one example that disproves a statement.
Proof by contradiction: Assume the opposite is true, then derive a contradiction.
Worked Example 1.1 — Proof by Contradiction
Prove that $\sqrt{2}$ is irrational.
Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$ where $p, q \in \mathbb{Z}$ and $\gcd(p,q) = 1$ (fraction in lowest terms).
Step 1 Squaring: $2 = \dfrac{p^2}{q^2}$, so $p^2 = 2q^2$. Thus $p^2$ is even, which means $p$ must be even. Write $p = 2k$.
Step 2 $(2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2$. So $q^2$ is even, meaning $q$ is even.
Contradiction Both $p$ and $q$ are even, so $\gcd(p,q) \geq 2$. This contradicts $\gcd(p,q) = 1$. Therefore $\sqrt{2}$ is irrational.
2. Partial Fractions
Partial fractions decompose a rational expression into a sum of simpler fractions. This is essential for integration.
Worked Example 2.1 — Partial Fractions with Repeated Factor
Express $\dfrac{5x + 3}{(x-1)^2(x+2)}$ in partial fractions.
Step 1 The form is $\dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{x+2}$.
Multiply through by $(x-1)^2(x+2)$:
$5x + 3 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$
Step 2 Substitute $x = 1$: $8 = 3B$, so $B = \dfrac{8}{3}$.
Substitute $x = -2$: $-7 = 9C$, so $C = -\dfrac{7}{9}$.
Expand and compare $x^2$ coefficients: $0 = A + C$, so $A = \dfrac{7}{9}$.
Result $\dfrac{7/9}{x-1} + \dfrac{8/3}{(x-1)^2} - \dfrac{7/9}{x+2}$
3. Functions and Modelling
Year 2 extends knowledge of functions to include modulus functions, composite inverse functions with restricted domains, and modelling with exponential functions.
Modulus Function
$|f(x)|$ reflects any part of the graph below the $x$-axis to above it.
To solve $|f(x)| = k$ (for $k > 0$): solve $f(x) = k$ and $f(x) = -k$.
To solve $|f(x)| < k$: solve $-k < f(x) < k$.
Worked Example 3.1 — Modulus Equation
Solve $|2x - 3| = |x + 1|$.
Method Square both sides (valid since both sides are non-negative):
$(2x-3)^2 = (x+1)^2$
$4x^2 - 12x + 9 = x^2 + 2x + 1$
$3x^2 - 14x + 8 = 0$
$(3x-2)(x-4) = 0$
$x = \dfrac{2}{3}$ or $x = 4$.
Check both in the original: both valid.
4. Binomial Expansion for Any $n$
Generalised Binomial Theorem
For $|x| < 1$ and any rational $n$:
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$
This is an infinite series when $n$ is not a positive integer.
For $(a + bx)^n$, write as $a^n\!\left(1 + \dfrac{b}{a}x\right)^n$ and apply the formula with $\dfrac{b}{a}x$ in place of $x$. Valid for $\left|\dfrac{b}{a}x\right| < 1$.
Worked Example 4.1 — Binomial with Negative Index
Find the first four terms of $(1 + 2x)^{-2}$ and state the values of $x$ for which it is valid.
Step 1 Use the generalised formula with $n = -2$, $x \to 2x$:
$(1+2x)^{-2} = 1 + (-2)(2x) + \dfrac{(-2)(-3)}{2!}(2x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(2x)^3 + \cdots$
Step 2 Simplify each term:
$= 1 - 4x + \dfrac{6}{2}(4x^2) + \dfrac{-24}{6}(8x^3) + \cdots$
$= 1 - 4x + 12x^2 - 32x^3 + \cdots$
Validity: $|2x| < 1 \Rightarrow |x| < \dfrac{1}{2}$.
5. Radians and Further Trigonometry
Reciprocal Trig Functions
$\sec\theta = \dfrac{1}{\cos\theta}$, $\quad \csc\theta = \dfrac{1}{\sin\theta}$, $\quad \cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$
Double Angle Formulae
$\sin 2A = 2\sin A\cos A$
$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
$\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$
Compound Angle Formulae
$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$
Worked Example 5.1 — $R\sin(\theta + \alpha)$ Form
Write $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Hence find the maximum value and the smallest positive $\theta$ at which it occurs.
Step 1 $R\sin(\theta+\alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta$.
Matching: $R\cos\alpha = 3$ and $R\sin\alpha = 4$.
Step 2 $R = \sqrt{3^2 + 4^2} = 5$. $\tan\alpha = \dfrac{4}{3}$, so $\alpha = \arctan\!\left(\dfrac{4}{3}\right) \approx 0.927$ rad.
Step 3 Maximum value $= 5$ (when $\sin(\theta + \alpha) = 1$, i.e. $\theta + \alpha = \dfrac{\pi}{2}$).
$\theta = \dfrac{\pi}{2} - \alpha \approx \dfrac{\pi}{2} - 0.927 \approx 0.644$ rad.
6. Parametric Equations
A curve defined parametrically is given as $x = f(t)$ and $y = g(t)$ for parameter $t$. To find the Cartesian equation, eliminate $t$.
Worked Example 6.1 — Parametric to Cartesian
A curve has parametric equations $x = 3\cos t$, $y = 2\sin t$, $0 \leq t \leq 2\pi$. Find the Cartesian equation and identify the curve.
Step 1 $\dfrac{x}{3} = \cos t$ and $\dfrac{y}{2} = \sin t$. Use $\cos^2 t + \sin^2 t = 1$:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
This is an ellipse with semi-major axis $3$ along the $x$-axis and semi-minor axis $2$ along the $y$-axis.
PURE2·1 — The parametric ellipse $x = 3\cos t,\; y = 2\sin t$. The Cartesian form is $\tfrac{x^2}{9}+\tfrac{y^2}{4}=1$. Semi-major axis $a = 3$ (horizontal) and semi-minor axis $b = 2$ (vertical).
7. Differentiation (Implicit and Parametric)
Implicit Differentiation
For an equation not solved for $y$, differentiate every term with respect to $x$, using the chain rule: $\dfrac{d}{dx}[f(y)] = f'(y)\dfrac{dy}{dx}$.
Worked Example 7.1 — Implicit Differentiation
Find $\dfrac{dy}{dx}$ for $x^3 + y^3 - 3xy = 0$ (the folium of Descartes).
Step 1 Differentiate each term w.r.t. $x$:
$3x^2 + 3y^2\dfrac{dy}{dx} - 3\!\left(y + x\dfrac{dy}{dx}\right) = 0$
Step 2 Collect $\dfrac{dy}{dx}$ terms:
$3y^2\dfrac{dy}{dx} - 3x\dfrac{dy}{dx} = 3y - 3x^2$
$\dfrac{dy}{dx} = \dfrac{3y - 3x^2}{3y^2 - 3x} = \dfrac{y - x^2}{y^2 - x}$
Parametric Differentiation
Parametric Derivative
If $x = f(t)$ and $y = g(t)$, then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.
Second derivative: $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right) = \dfrac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{dx/dt}$
8. Integration Techniques
Integration by Parts
Integration by Parts Formula
$$\int u \frac{dv}{dx}\,dx = uv - \int v \frac{du}{dx}\,dx$$
Choose $u$ to be the function that simplifies when differentiated. The LIATE rule helps: Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential — choose $u$ from earlier in this list.
Worked Example 8.1 — Integration by Parts
Find $\displaystyle\int x^2 e^x\,dx$.
Step 1 Let $u = x^2$, $\dfrac{dv}{dx} = e^x$. Then $\dfrac{du}{dx} = 2x$, $v = e^x$.
$\displaystyle\int x^2 e^x\,dx = x^2 e^x - \int 2xe^x\,dx$
Step 2 Apply parts again to $\displaystyle\int 2xe^x\,dx$: let $u = 2x$, $v = e^x$.
$\displaystyle\int 2xe^x\,dx = 2xe^x - 2e^x$
Step 3 Combine:
$\displaystyle\int x^2 e^x\,dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C$
Integration by Substitution
Worked Example 8.2 — Integration by Substitution
Find $\displaystyle\int x\sqrt{x^2 + 1}\,dx$.
Step 1 Let $u = x^2 + 1$. Then $\dfrac{du}{dx} = 2x$, so $x\,dx = \dfrac{du}{2}$.
$\displaystyle\int x\sqrt{x^2+1}\,dx = \int \sqrt{u} \cdot \frac{du}{2} = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{3}(x^2+1)^{3/2} + C$
Integration Using Partial Fractions
Worked Example 8.3 — Integration with Partial Fractions
Find $\displaystyle\int \frac{5x - 2}{(x+1)(x-3)}\,dx$.
Step 1 Write $\dfrac{5x-2}{(x+1)(x-3)} = \dfrac{A}{x+1} + \dfrac{B}{x-3}$.
$5x - 2 = A(x-3) + B(x+1)$. Let $x = 3$: $13 = 4B$, $B = \dfrac{13}{4}$.
Let $x = -1$: $-7 = -4A$, $A = \dfrac{7}{4}$.
Step 2 Integrate:
$\displaystyle\int \left(\frac{7/4}{x+1} + \frac{13/4}{x-3}\right) dx = \frac{7}{4}\ln|x+1| + \frac{13}{4}\ln|x-3| + C$
9. Differential Equations
A first-order separable differential equation has the form $\dfrac{dy}{dx} = f(x)g(y)$. Separate variables and integrate both sides.
Worked Example 9.1 — Separable Differential Equation
Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$, given that $y = 3$ when $x = 1$.
Step 1 Separate: $\dfrac{dy}{y} = \dfrac{dx}{x}$.
Step 2 Integrate: $\ln|y| = \ln|x| + C$. So $y = Ax$ (where $A = e^C$).
Step 3 Apply initial condition: $3 = A \cdot 1$, so $A = 3$. Solution: $y = 3x$.
Worked Example 9.2 — Modelling with Differential Equations
A population $P$ grows at a rate proportional to its current size: $\dfrac{dP}{dt} = kP$. Solve, given $P(0) = P_0$.
Step 1 Separate: $\dfrac{dP}{P} = k\,dt$. Integrate: $\ln P = kt + C$.
Step 2 $P = Ae^{kt}$. At $t = 0$: $P_0 = A$. So $P = P_0 e^{kt}$.
10. Numerical Methods
Locating Roots by Sign Change
If $f(a)$ and $f(b)$ have opposite signs and $f$ is continuous on $[a,b]$, then by the IVT there is a root in $(a,b)$.
Newton-Raphson Method
Newton-Raphson Iteration
Starting from an estimate $x_0$, successive approximations are given by:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
This converges quickly for good initial estimates, but fails if $f'(x_n) = 0$ or the initial estimate is far from the root.
Worked Example 10.1 — Newton-Raphson
Use Newton-Raphson to find the root of $f(x) = x^3 - x - 1$ near $x_0 = 1.5$, giving your answer to 4 d.p.
$f'(x) = 3x^2 - 1$.
$x_1 = 1.5 - \dfrac{f(1.5)}{f'(1.5)} = 1.5 - \dfrac{3.375 - 1.5 - 1}{6.75 - 1} = 1.5 - \dfrac{0.875}{5.75} = 1.5 - 0.1522 = 1.3478$
$x_2 = 1.3478 - \dfrac{f(1.3478)}{f'(1.3478)} = 1.3478 - \dfrac{0.1003}{4.448} \approx 1.3252$
$x_3 \approx 1.3247$, $x_4 \approx 1.3247$ (converged). Root $\approx 1.3247$.
PURE2·2 — Newton-Raphson on $f(x) = x^3 - x - 1$. Starting from $x_0 = 1.5$ (red dot), the tangent to the curve (red dashed) crosses the $x$-axis at $x_1 \approx 1.348$ (amber), and subsequent iterations converge to the root $x \approx 1.3247$ (green).
11. Practice Problems
Problem 1
Prove by contradiction that there are infinitely many prime numbers.
Show Solution
Assume there are finitely many primes: $p_1, p_2, \ldots, p_k$. Consider $N = p_1 p_2 \cdots p_k + 1$. For any prime $p_i$, $N \div p_i$ leaves remainder $1$, so no $p_i$ divides $N$. But every integer $> 1$ has a prime factor. So $N$ must have a prime factor not in our list — contradiction. Therefore there are infinitely many primes.
Problem 2
Express $\dfrac{3x^2 + x - 2}{(x-1)(x^2 + 1)}$ in partial fractions.
Show Solution
Form: $\dfrac{A}{x-1} + \dfrac{Bx + C}{x^2+1}$.
$3x^2 + x - 2 = A(x^2+1) + (Bx+C)(x-1)$.
$x=1$: $2 = 2A$, so $A = 1$.
Compare $x^2$: $3 = A + B$, so $B = 2$.
Compare constant: $-2 = A - C$, so $C = 3$.
$$\frac{1}{x-1} + \frac{2x+3}{x^2+1}$$
Problem 3
Find the first three non-zero terms of $\dfrac{1}{\sqrt{4 - x}}$, and state its range of validity.
Show Solution
$\dfrac{1}{\sqrt{4-x}} = (4-x)^{-1/2} = 4^{-1/2}\!\left(1 - \dfrac{x}{4}\right)^{-1/2} = \dfrac{1}{2}\!\left(1 - \dfrac{x}{4}\right)^{-1/2}$
$\approx \dfrac{1}{2}\!\left[1 + \dfrac{1}{2}\cdot\dfrac{x}{4} + \dfrac{(-1/2)(-3/2)}{2}\cdot\dfrac{x^2}{16} + \cdots\right]$
$= \dfrac{1}{2}\!\left[1 + \dfrac{x}{8} + \dfrac{3x^2}{128} + \cdots\right] = \dfrac{1}{2} + \dfrac{x}{16} + \dfrac{3x^2}{256} + \cdots$
Valid for $\left|\dfrac{x}{4}\right| < 1$, i.e. $|x| < 4$.
Problem 4
Differentiate implicitly: $e^{xy} + y^2 = x + 3$ to find $\dfrac{dy}{dx}$.
Show Solution
Differentiate each term w.r.t. $x$:
$e^{xy}\!\left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 1$
$ye^{xy} + xe^{xy}\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 1$
$\dfrac{dy}{dx}(xe^{xy} + 2y) = 1 - ye^{xy}$
$\dfrac{dy}{dx} = \dfrac{1 - ye^{xy}}{xe^{xy} + 2y}$
Problem 5
Evaluate $\displaystyle\int_0^{\pi/2} x\sin x\,dx$.
Show Solution
Integration by parts: $u = x$, $dv = \sin x\,dx$. So $du = dx$, $v = -\cos x$.
$\left[-x\cos x\right]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx = \left[-x\cos x + \sin x\right]_0^{\pi/2}$
$= \left(-\dfrac{\pi}{2}\cos\dfrac{\pi}{2} + \sin\dfrac{\pi}{2}\right) - (0 + 0) = 0 + 1 = 1$
Problem 6
Solve $\dfrac{dy}{dx} = \dfrac{x+1}{y}$, given $y = 2$ when $x = 0$. Express $y$ in terms of $x$.
Show Solution
Separate: $y\,dy = (x+1)\,dx$.
Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + x + C$.
Apply IC: $\dfrac{4}{2} = 0 + 0 + C$, so $C = 2$.
$\dfrac{y^2}{2} = \dfrac{x^2}{2} + x + 2 \Rightarrow y^2 = x^2 + 2x + 4$
$y = \sqrt{x^2 + 2x + 4}$ (taking positive root since $y=2>0$ initially).
Problem 7
Show that $f(x) = x^5 - 3x + 1$ has a root in $(1, 2)$, then perform two iterations of the Newton-Raphson method starting from $x_0 = 1.2$.
Show Solution
$f(1) = 1 - 3 + 1 = -1 < 0$. $f(2) = 32 - 6 + 1 = 27 > 0$. Sign change: root in $(1,2)$.
$f'(x) = 5x^4 - 3$.
$x_1 = 1.2 - \dfrac{f(1.2)}{f'(1.2)} = 1.2 - \dfrac{2.488 - 3.6 + 1}{5(2.074) - 3} = 1.2 - \dfrac{-0.112}{7.372} \approx 1.2152$
$x_2 = 1.2152 - \dfrac{f(1.2152)}{f'(1.2152)} \approx 1.2131$