FP1 · Further Pure Mathematics 1

Further Pure Mathematics 1 (FP1, paper WFM01/01, 1 hour 30 minutes, 75 marks) is compulsory for International AS and A Level Further Mathematics and for International AS and A Level Pure Mathematics. P1–P2 are assumed and may be tested, including sign-change location of roots, rotating shapes about the origin, and dividing a cubic or quartic by a quadratic. This reference covers complex numbers, roots of quadratic equations, numerical methods, the parabola and rectangular hyperbola, matrix algebra, transformations, finite series and proof by induction.

Must-know for FP1 (not in the booklet):

A complex number 复数 is $z = a + bi$ with $i^2 = -1$; $a$ is the real part 实部, $b$ the imaginary part 虚部. The conjugate 共轭复数 $\bar z = a - bi$ reverses the sign of the imaginary part, and $z\bar z = a^2 + b^2 = \lvert z\rvert^2$ — the trick that makes a quotient real: $\dfrac{z_1}{z_2} = \dfrac{z_1\bar z_2}{\lvert z_2\rvert^2}$. Two complex numbers are equal only when both real and imaginary parts match. The polar form $z = r(\cos\theta + i\sin\theta)$ pairs the modulus 模 $r$ with the argument 辐角 $\theta$; multiplying complex numbers multiplies their moduli ($\lvert z_1z_2\rvert = \lvert z_1\rvert\lvert z_2\rvert$).

Worked example — a real sum pins the imaginary part. $z_1 = r(\cos\tfrac{7\pi}6 + i\sin\tfrac{7\pi}6)$ with $r > 0$, and $z_2$ has modulus 5. Since $\lvert z_1z_2\rvert = \lvert z_1\rvert\lvert z_2\rvert = 5r = 15$, we get $r = 3$. So $z_1 = 3(-\tfrac{\sqrt3}2 - \tfrac12i) = -\tfrac{3\sqrt3}2 - \tfrac32 i$. If $z_1 + z_2$ is real, its imaginary part is zero, so $z_2 = a + \tfrac32 i$; then $a^2 + \tfrac94 = 25$ gives $a = \pm\tfrac{\sqrt{91}}2 \approx \pm4.77$. Both $z_2 = \tfrac{\sqrt{91}}2 + \tfrac32i$ and its reflection in the imaginary axis work.

On an Argand diagram 阿甘特图 (Argand diagram) plot $z = a + bi$ as the point $(a, b)$: addition is vector addition, the conjugate reflects in the real axis, and $\lvert z\rvert$ is the distance from the origin.

For a quadratic with real coefficients and negative discriminant, complete the square: $x^2 - 6x + 13 = (x-3)^2 + 4 = 0$ gives $x = 3 \pm 2i$. The roots are a conjugate pair — and this survives degree: if $f(z_1) = 0$ for real coefficients, then $f(\bar z_1) = 0$ too. So a real cubic has either three real roots or one real root plus one conjugate pair, and $(z - z_1)(z - \bar z_1) = z^2 - 2\operatorname{Re}(z_1)z + \lvert z_1\rvert^2$ is a real quadratic factor. Divide it out to reach the rest.

Worked example — a quartic with a given complex root (WFM01 June 2025 Q7). $f(z) = Pz^4 - 36z^3 + Qz^2 + 192z + 68$ has root $3 + 5i$, so $3 - 5i$ is also a root and

Writing $f(z) = (z^2 - 6z + 34)(Pz^2 + bz + c)$ and comparing: $34c = 68$ so $c = 2$; $-6c + 34b = 192$ so $34b = 204$, $b = 6$; then $b - 6P = -36$ gives $P = 7$ and $c - 6b + 34P = Q$ gives $Q = 204$. (Check against $f(3+5i) = 0$: the real equation $644P + 16Q + 7772$-type balance holds since $644(7) + 16(204) - 7772 = 0$.) The other roots solve $7z^2 + 6z + 2 = 0$:

Without a calculator: only $b^2 - 4ac < 0$ arithmetic and the conjugate fact are used.

For $ax^2 + bx + c = 0$ with roots $\alpha, \beta$: $\alpha + \beta = -\tfrac ba$ and $\alpha\beta = \tfrac ca$. Symmetric expressions in $\alpha, \beta$ then need no individual roots: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$, $\alpha^3 + \beta^3 \equiv (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$, and $\tfrac1\alpha + \tfrac1\beta = \tfrac{\alpha+\beta}{\alpha\beta}$. To build a new quadratic, express the new sum and product in terms of the old ones; the equation is $x^2 - (\text{sum})x + \text{product} = 0$.

Worked example — new roots without solving (WFM01 June 2025 Q6). $f(x) = 3x^2 + kx - 5$ has roots $\alpha, \beta$, so $\alpha\beta = -\tfrac53$. If $\alpha + \beta = 9\alpha\beta$, then $-\tfrac k3 = -15$, so $k = 45$ and $\alpha + \beta = -15$. For roots $\alpha^2 + \beta$ and $\alpha + \beta^2$:

The new equation is $9x^2 - 1920x - 31040 = 0$ (multiply $x^2 - \tfrac{640}3x - \tfrac{31040}9 = 0$ by 9).

Three iterative locators, all needing $f$ continuous on the interval: interval bisection 区间二分法 halves an interval carrying a sign change; linear interpolation 线性插值 chords from $(a, f(a))$ to $(b, f(b))$ and takes the chord's zero $\beta \approx a - \dfrac{(b-a)f(a)}{f(b)-f(a)}$; the Newton–Raphson method 牛顿-拉夫逊法 (formula given in the booklet) iterates $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$. Show every evaluation — "calculator-only" answers score nothing.

Worked example — bisection and interpolation (WFM01 June 2025 Q3). (i) $f(x) = x^2 + 5 - 8^{5x}$ has a single root $\alpha$ in $[0, 1]$: $f(0) = 4 > 0$, $f(0.5) = 5.25 - 8^{2.5} = -175.77 < 0$, so halve: $f(0.25) = 5.0625 - 8^{1.25} = -8.392 < 0$. The sign change sits in $[0, 0.25]$. (ii) $g(x) = 3^{\sin x} - 3\cos x$ (radians): $g(4) = 2.396 > 0$ and $g(5) = -0.502 < 0$, and $g$ is continuous, so a root $\beta$ lies in $[4, 5]$. Interpolating once: $\beta \approx 4 + \dfrac{2.396}{2.396 + 0.502} = 4.827$ (4 s.f.).

Newton–Raphson (original example). Solve $x^3 - 2x - 5 = 0$ from $x_0 = 2$: $f(2) = -1$, $f'(x) = 3x^2 - 2$ so $f'(2) = 10$ and $x_1 = 2 - \tfrac{-1}{10} = 2.1$; then $f(2.1) = 0.061$, $f'(2.1) = 11.23$, $x_2 = 2.0946$. The iterates home in on the true root $2.0946\ldots$ (check: $2.0946^3 - 2(2.0946) - 5 \approx +0.0005$).

The parabola 抛物线 $y^2 = 4ax$ has parametric form $x = at^2$, $y = 2at$; the rectangular hyperbola 直角双曲线 $xy = c^2$ has $x = ct$, $y = \tfrac ct$. Tangents and normals come from differentiating the Cartesian equation (parametric differentiation is not required): $y^2 = 4ax$ gives $\tfrac{dy}{dx} = \tfrac{2a}y = \tfrac1t$ at parameter $t$, and $xy = c^2$ gives $y = \tfrac{c^2}x$ so $\tfrac{dy}{dx} = -\tfrac{c^2}{x^2} = -\tfrac1{t^2}$ at $x = ct$.

Worked example — normal to a hyperbola (WFM01 June 2025 Q4). For $xy = 81$ at $P(9t, \tfrac9t)$: $\tfrac{dy}{dx} = -\tfrac{81}{x^2} = -\tfrac1{t^2}$, so the normal has gradient $t^2$:

At the $y$-axis, $A = (0, \tfrac{9(1-t^4)}t)$. With $t = \tfrac13$: $P = (3, 27)$ and $A = (0, \tfrac{80}3)$, so triangle $OPA$ has area $\tfrac12 \times 3 \times \tfrac{80}3 = 40$.

The parabola is the locus of points equidistant from the focus 焦点 $(a, 0)$ and the directrix 准线 $x = -a$. Distances, not algebra, unlock the geometry questions.

Worked example — focus-directrix distances (WFM01 June 2025 Q9). $P$ lies on the parabola 5 units from the focus, so $P$ is also 5 units from the directrix ($PF = PM = 5$, with $PM$ perpendicular to the directrix). Points $Q, R$ on the directrix satisfy $PQ = PR = 10$. Each right triangle $PQM$ gives $QM = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt3$, and $M$ lies between $Q$ and $R$, so $QR = 10\sqrt3$.

Matrices add entry-wise (same shape), scale entry-wise, and multiply row-by-column ($(\mathbf{AB})_{ik} = \sum_j A_{ij}B_{jk}$; $\mathbf{AB} \ne \mathbf{BA}$ in general). For $2\times 2$ matrices $\det\mathbf M = ad - bc$: zero means singular 奇异 (no inverse); otherwise $\mathbf M^{-1} = \dfrac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, and $\mathbf M\mathbf M^{-1} = \mathbf I$. Inverses reverse order: $(\mathbf{AB})^{-1} = \mathbf B^{-1}\mathbf A^{-1}$ — undo the last transformation first.

Worked example — an inverse that must stay consistent (WFM01 June 2025 Q1). $\mathbf M = \begin{pmatrix}1&a\\3&-5\end{pmatrix}$ with $a \ne -\tfrac53$:

Given $\mathbf M^{-1} = 2\mathbf M + 8\mathbf I = \begin{pmatrix}10&2a\\6&-2\end{pmatrix}$, compare entries: $\dfrac{-3}{-(5+3a)} = 6$ gives $5 + 3a = \tfrac12$, and every other entry confirms it, so $a = -\tfrac32$.

A $2\times 2$ matrix $\mathbf M$ transforms column vectors by $\mathbf y = \mathbf{Mx}$; its columns are the images of $\mathbf i$ and $\mathbf j$ — read any transformation straight off the matrix. The standard results:

For combinations, $\mathbf{AB}$ means "apply $\mathbf B$ first, then $\mathbf A$". The inverse transformation undoes it, and $\lvert\det\mathbf M\rvert$ is the area scale factor 面积比例因子: areas multiply by the determinant (negative determinants add a reflection).

Worked example — describing and combining (WFM01 June 2025 Q2). $\mathbf P = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$ maps $\mathbf i \to (0,-1)^T$ and $\mathbf j \to (1,0)^T$: a quarter-turn clockwise (270° anticlockwise) about $O$. $\mathbf Q = 5\mathbf I$ is an enlargement, scale factor 5, centre $O$. Reflection in $y = -x$ is $\mathbf R = \begin{pmatrix}0&-1\\-1&0\end{pmatrix}$, so $A = \mathbf R(4,3)^T = (-3,-4)$. If $\mathbf Q$ follows $\mathbf P$ and maps $B$ to $A$, then $B = \mathbf P^{-1}\mathbf Q^{-1}A = \mathbf P^{-1}\cdot\tfrac15(-3,-4)^T = \tfrac15(4,-3)^T = (0.8, -0.6)$ — $\mathbf P$'s own inverse is its transpose, so this is arithmetic only.

With $\sum_{r=1}^n r = \tfrac12 n(n+1)$ (and $\sum r^2 = \tfrac16 n(n+1)(2n+1)$ from the booklet), any polynomial summand splits into standard pieces — including sums to an even limit $2n$ (substitute $k = 2n$ into the standard formulae) and shifted sums like $\sum (r+1)^2$.

Worked example — a sum to 2n (WFM01 June 2025 Q8a). Show $\sum_{r=1}^{2n}(2r^2 - 1) = \tfrac43 n(n+1)(an + b)$:

and $4n^2 + 3n - 1 = (n+1)(4n - 1)$, so the sum is $\tfrac43n(n+1)(4n-1)$: $a = 4$, $b = -1$. (Check $n = 1$: $1 + 7 = 8$ and $\tfrac43\cdot1\cdot2\cdot3 = 8$.)

Induction 数学归纳法 proves a statement $P(n)$ for all positive integers in three moves: verify $P(1)$; assume $P(k)$; deduce $P(k+1)$ from it — then close with "true for $n = 1$, and truth for $n = k$ implies truth for $n = k+1$, so by induction true for all positive integers $n$". The closing sentence is part of the proof. Four assessed families: summation formulae, divisibility, sequences given by a recurrence, and matrix powers.

Worked example — a summation (WFM01 June 2025 Q8b). Prove $\sum_{r=1}^n r(3r-2)^2 = \tfrac14 n^2(n+1)(9n-7)$. For $n = 1$: LHS $= 1\cdot1^2 = 1$ and RHS $= \tfrac14\cdot1\cdot2\cdot2 = 1$ — true. Assume true for $n = k$; the $(k+1)$th term is $(k+1)(3k+1)^2$, so

Expand the bracket: $9k^3 - 7k^2 + 36k^2 + 24k + 4 = 9k^3 + 29k^2 + 24k + 4$. The target for $n = k+1$ is $\tfrac14(k+1)^2(k+2)(9k+2)$, and its bracket $(k+1)(k+2)(9k+2)$ expands as $(k^2+3k+2)(9k+2) = 9k^3 + 29k^2 + 24k + 4$ — identical. So the statement holds for $n = k+1$, and by induction for all positive integers $n$.

Divisibility (original example). Prove $n^3 + 2n$ is divisible by 3. For $n = 1$: $3$. Assume $k^3 + 2k = 3m$; then $(k+1)^3 + 2(k+1) = (k^3 + 2k) + 3k^2 + 3k + 3 = 3(m + k^2 + k + 1)$, still a multiple of 3. By induction, true for all $n$.

Matrix powers (original example). Prove $\begin{pmatrix}1&1\\0&1\end{pmatrix}^n = \begin{pmatrix}1&n\\0&1\end{pmatrix}$. True for $n = 1$. If true for $k$, then $\begin{pmatrix}1&1\\0&1\end{pmatrix}^{k+1} = \begin{pmatrix}1&k\\0&1\end{pmatrix} \begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&k+1\\0&1\end{pmatrix}$, completing the induction.

• Work with $a + bi$ and $r(\cos\theta + i\sin\theta)$: modulus, argument, conjugate; quotients made real.
• Use conjugate pairs to factor cubics and quartics with real coefficients; find all roots without a calculator.
• Convert symmetric expressions via $\alpha + \beta$ and $\alpha\beta$; form new quadratics for transformed roots.
• Locate roots by bisection and linear interpolation with every evaluation shown; iterate Newton–Raphson from a given start.
• Use $y^2 = 4ax$ and $xy = c^2$ parametrics; tangents/normals via Cartesian differentiation; focus-directrix distances.
• Invert $2\times2$ matrices (and $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$); solve matrix equations entry by entry.
• Identify reflections, rotations, stretches, enlargements from matrices; combine in the right order; use $|\det|$ as the area scale factor.
• Sum series to $n$ and $2n$ with the standard formulae; prove series, divisibility, recurrence and matrix results by induction with a closing statement.