Overview
Analysis and Approaches HL (AA HL) is the most rigorous IB math course — equivalent to a first-year university pure mathematics course. It extends AA SL with complex numbers, proof by induction, vectors in 3D, advanced integration (by parts, substitution, partial fractions), differential equations, and series convergence. Applications and Interpretation HL (AI HL) extends AI SL with matrices, complex numbers, Poisson distribution, graph theory (networks), and deeper optimization and differential equations.
Exam: Paper 1 (no GDC, 120 min) + Paper 2 (GDC, 120 min) + Paper 3 (extended, 60 min). Internal Assessment 20%.Topic AA HL specifics AI HL specifics ~Marks 1 — Algebra Proof by induction, complex numbers Matrices, eigenvalues 20 2 — Functions Modulus, complex roots, partial fractions Additional regression 20 3 — Geometry & Trig 3D vectors, cross product, planes Graph theory, algorithms 25 4 — Statistics Unbiased estimators, CIs, hypothesis tests Poisson, Markov chains 30 5 — Calculus By parts, DEs, Maclaurin series Coupled DEs, numerics 40
Exam (both courses): Paper 1 (no GDC, 120 min, 110 marks) + Paper 2 (GDC, 120 min, 110 marks) + Paper 3 (60 min, 55 marks — investigative, GDC required).
Algebra, Proof & Complex Numbers
HL Topic 1 extends SL algebra with proof by induction and complex numbers — the foundation for much of higher mathematics.
Structure: (1) Base case: verify for \(n=1\) (or initial \(n\)). (2) Inductive hypothesis: assume true for \(n=k\). (3) Inductive step: prove true for \(n=k+1.\) (4) Conclude.
\[z=a+bi,\quad i^2=-1,\quad\text{Re}(z)=a,\quad\text{Im}(z)=b.\]\[|z|=\sqrt{a^2+b^2},\quad\arg(z)=\arctan(b/a)\text{ (adjusted for quadrant)}.\]\[\bar{z}=a-bi,\quad z\bar{z}=|z|^2,\quad\frac{1}{z}=\frac{\bar{z}}{|z|^2}.\]
\[z=r(\cos\theta+i\sin\theta)=re^{i\theta}\quad(r=|z|,\;\theta=\arg z).\]\[z^n=r^n(\cos n\theta+i\sin n\theta)\quad(\text{de Moivre's theorem}).\] \(n\)-th roots of unity: solutions to \(z^n=1\) are \(e^{2\pi ik/n}\) for \(k=0,1,\ldots,n-1.\)
\[\frac{p(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b},\quad\frac{p(x)}{(x-a)(x^2+b)}=\frac{A}{x-a}+\frac{Bx+C}{x^2+b}.\]
Problem. Prove by induction: \(\displaystyle\sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}{6}.\)
Base (\(n=1\)): LHS\(=1.\) RHS\(=\frac{1\cdot2\cdot3}{6}=1.\;\checkmark\)
Assume \(\displaystyle\sum_{r=1}^k r^2=\frac{k(k+1)(2k+1)}{6}.\)
For \(n=k+1\): \(\displaystyle\sum_{r=1}^{k+1}r^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2=(k+1)\!\left[\frac{k(2k+1)}{6}+(k+1)\right]\)
\(=(k+1)\cdot\frac{2k^2+7k+6}{6}=\frac{(k+1)(k+2)(2k+3)}{6}.\;\checkmark\)
Problem. Write \(z=1+i\sqrt{3}\) in polar form and find \(z^6.\)
\(|z|=\sqrt{1+3}=2.\quad\arg z=\arctan(\sqrt{3}/1)=\pi/3.\)
\(z=2e^{i\pi/3}.\quad z^6=2^6e^{i\cdot6\pi/3}=64\,e^{i2\pi}=64(\cos2\pi+i\sin2\pi)=64.\)
Problem. Decompose \(\dfrac{5x-3}{(x+1)(x-2)}\).
\(\dfrac{5x-3}{(x+1)(x-2)}=\dfrac{A}{x+1}+\dfrac{B}{x-2}.\)
Multiply through: \(5x-3=A(x-2)+B(x+1).\)
\(x=2:\;7=3B\Rightarrow B=7/3.\quad x=-1:\;-8=-3A\Rightarrow A=8/3.\)
Answer: \(\dfrac{8/3}{x+1}+\dfrac{7/3}{x-2}.\)
Express \(z=-1+i\sqrt{3}\) in polar form and find \(z^6.\) \(r=2,\quad\theta=\frac{2\pi}{3}.\quad z=2e^{i2\pi/3}.\) \(z^6=2^6e^{i4\pi}=64(\cos4\pi+i\sin4\pi)=64.\)Show solution
Worked Example · Complex number in polar form
Express \(z = -\sqrt{3}+i\) in modulus-argument form.
Solution: \(|z|=\sqrt{3+1}=2\). \(\arg(z)=\pi-\arctan\!\left(\dfrac{1}{\sqrt{3}}\right)=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\).
\(z = 2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right) = 2e^{i5\pi/6}\)
Worked Example · Proof by induction
Prove \(\sum_{r=1}^n r = \dfrac{n(n+1)}{2}\) by induction.
Solution: Base: \(n=1\): LHS\(=1\), RHS\(=\frac{2}{2}=1\). ✓
Step: Assume true for \(n=k\). Then \(\sum_{r=1}^{k+1}r = \frac{k(k+1)}{2}+(k+1)=\frac{(k+1)(k+2)}{2}\). ✓
Functions (HL Extensions)
HL functions extend SL content with the modulus function, rational functions with oblique asymptotes, and complex roots of polynomials.
\[|f(x)|:\text{ reflect negative part above }x\text{-axis}.\]\[f(|x|):\text{ reflect positive part across }y\text{-axis}.\] Equations \(|f(x)|=k\) give two cases: \(f(x)=k\) and \(f(x)=-k.\) Inequalities \(|f(x)|<k\) give \(-k<f(x)<k.\)
Complex conjugate root theorem: if \(p(x)\) has real coefficients and \(a+bi\) is a root, then \(a-bi\) is also a root. Relation between roots and coefficients for \(az^3+bz^2+cz+d=0\) with roots \(\alpha,\beta,\gamma\):
Problem. Sketch \(y=|2x-4|\) and solve \(|2x-4|=x+1.\)
V-shape with vertex at \((2,0).\)
Case 1: \(2x-4=x+1\Rightarrow x=5.\) Check: \(|6|=6=6.\;\checkmark\)
Case 2: \(2x-4=-(x+1)\Rightarrow3x=3\Rightarrow x=1.\) Check: \(|{-2}|=2=2.\;\checkmark\)
Problem. \(p(x)=x^3-3x^2+7x-5\) has root \(x=1.\) Find all roots.
Factor out \((x-1)\): \(p(x)=(x-1)(x^2-2x+5).\)
Quadratic: \(x=\dfrac{2\pm\sqrt{4-20}}{2}=1\pm2i.\)
Roots: \(x=1,\;1+2i,\;1-2i.\)
\(p(x)=x^4-2x^3+5x^2-8x+4\) has \(x=1\) as a double root. Find all roots. \((x-1)^2\) divides \(p(x)\). Divide: \(p(x)=(x-1)^2(x^2+4).\) \(x^2+4=0\Rightarrow x=\pm2i.\) Roots: \(1\) (double), \(\pm2i.\)Show solution
Worked Example · Rational function sketch
Sketch \(f(x)=\dfrac{2x^2-x-6}{x^2-4}\), stating all asymptotes and intercepts.
Solution: Factor: \(f(x)=\dfrac{(2x+3)(x-2)}{(x+2)(x-2)}=\dfrac{2x+3}{x+2}\) for \(x\neq2\).
Vertical asymptote: \(x=-2\). Horizontal: \(y\to2\). Hole at \(x=2\), \(y=\frac{7}{4}\). \(x\)-int: \(x=-\frac{3}{2}\). \(y\)-int: \(y=\frac{3}{2}\).
Worked Example · Solving modulus equation
Solve \(|2x-3|=x+1\).
Solution: Case 1: \(2x-3=x+1\implies x=4\). Check: \(|5|=5\). ✓
Case 2: \(2x-3=-(x+1)\implies 3x=2\implies x=\frac{2}{3}\). Check: \(|{-\frac{5}{3}}|=\frac{5}{3}=\frac{2}{3}+1\). ✓
Vectors & 3D Geometry
HL extends vectors to three dimensions, covering lines and planes in space, the cross product, and intersection problems — foundational for physics and engineering.
\[\mathbf{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix},\quad|\mathbf{v}|=\sqrt{v_1^2+v_2^2+v_3^2},\quad\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3.\]
\[\mathbf{u}\times\mathbf{v}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}=\begin{pmatrix}u_2v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{pmatrix}.\] \(\mathbf{u}\times\mathbf{v}\) is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}.\quad|\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta=\) area of parallelogram.
Line: \(\mathbf{r}=\mathbf{a}+t\mathbf{d}.\) Parametric: \(x=a_1+td_1,\;y=a_2+td_2,\;z=a_3+td_3.\) Plane: \(\mathbf{r}\cdot\hat{\mathbf{n}}=d,\) or \(ax+by+cz=d,\) where \(\mathbf{n}=\langle a,b,c\rangle\) is the normal vector. Angle between planes: \(\cos\theta=\dfrac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}.\) Distance from point \(P\) to plane \(ax+by+cz=d\): \(\dfrac{|aP_x+bP_y+cP_z-d|}{\sqrt{a^2+b^2+c^2}}.\)
Problem. Find \(\mathbf{u}\times\mathbf{v}\) for \(\mathbf{u}=\langle1,2,3\rangle,\;\mathbf{v}=\langle4,-1,2\rangle.\)
\(\mathbf{u}\times\mathbf{v}=\begin{pmatrix}2(2)-3(-1)\\3(4)-1(2)\\1(-1)-2(4)\end{pmatrix}=\begin{pmatrix}4+3\\12-2\\-1-8\end{pmatrix}=\begin{pmatrix}7\\10\\-9\end{pmatrix}.\)
Drag to rotate, scroll to zoom · loads a small 3D library on first use
| vector | ⟨x, y, z⟩ | magnitude |
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\(\mathbf u=\langle1,2,3\rangle\) fixed (blue), \(\mathbf v=\langle4,-1,v_3\rangle\) (amber). \(\mathbf u\times\mathbf v\) (green) always stays perpendicular to both.
Problem. Find the line of intersection of planes \(\pi_1:x+2y-z=3\) and \(\pi_2:2x-y+3z=1.\)
Direction of line: \(\mathbf{n}_1\times\mathbf{n}_2=\langle1,2,-1\rangle\times\langle2,-1,3\rangle=\langle6-1,-3-2,-1-4\rangle=\langle5,-5,-5\rangle\propto\langle1,-1,-1\rangle.\)
Find a point: set \(z=0\): \(x+2y=3,\;2x-y=1\). Solve: \(x=1,y=1.\) Point \((1,1,0).\)
Line: \(\mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+t\begin{pmatrix}1\\-1\\-1\end{pmatrix}.\)
Find the equation of the plane through \(A(1,2,3)\), \(B(4,0,1)\), \(C(2,3,0)\). \(\overrightarrow{AB}=\langle3,-2,-2\rangle,\;\overrightarrow{AC}=\langle1,1,-3\rangle.\) \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\langle8,7,5\rangle.\) Plane: \(8(x-1)+7(y-2)+5(z-3)=0\Rightarrow 8x+7y+5z=37.\)Show solution
Worked Example · Angle between two planes
Find the angle between planes \(\pi_1: 2x-y+2z=5\) and \(\pi_2: x+2y-2z=3\).
Solution: \(\mathbf{n}_1=(2,-1,2)\), \(\mathbf{n}_2=(1,2,-2)\).
\(\cos\theta=\dfrac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}=\dfrac{|2-2-4|}{3\cdot3}=\dfrac{4}{9}\implies\theta\approx63.6°\)
Worked Example · Line–plane intersection
Line: \(\mathbf{r}=(1,0,2)+t(1,1,-1)\). Plane: \(x+2y+z=4\). Find the intersection.
Solution: Substitute: \((1+t)+2t+(2-t)=4\implies 2t+3=4\implies t=\frac{1}{2}\).
Point: \(\left(\frac{3}{2},\frac{1}{2},\frac{3}{2}\right)\).
Statistics & Probability (HL)
HL extends SL statistics with unbiased estimators, confidence intervals, hypothesis testing, and (AI HL) the Poisson distribution.
\[\hat{\mu}=\bar{x}=\frac{\sum x_i}{n}\quad(\text{unbiased for }\mu).\]\[S^2=\frac{\sum(x_i-\bar{x})^2}{n-1}\quad(\text{unbiased for }\sigma^2;\;\text{use }s_{n-1}\text{ on GDC}).\]
\[\bar{x}\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\quad(\sigma\text{ known}),\qquad\bar{x}\pm t_{\alpha/2,n-1}\cdot\frac{s}{\sqrt{n}}\quad(\sigma\text{ unknown}).\] Common critical values: 90% CI → \(z=1.645\); 95% CI → \(z=1.96\); 99% CI → \(z=2.576.\)
\[X\sim\text{Po}(\lambda):\quad P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!},\quad\mu=\sigma^2=\lambda.\] Conditions: events occur independently at a constant average rate \(\lambda\) per unit interval. Poisson as approximation to Binomial: when \(n\) is large, \(p\) is small, and \(np=\lambda\) is moderate.
Problem. A sample of \(n=64\) gives \(\bar{x}=52.3\) and \(s=8.4.\) Construct a 95% CI for \(\mu.\)
Use \(t\)-interval (\(\sigma\) unknown): \(t_{0.025,\,63}\approx2.00\) (from tables or GDC).
\(CI=52.3\pm2.00\cdot\dfrac{8.4}{\sqrt{64}}=52.3\pm2.00(1.05)=52.3\pm2.1=(50.2,\;54.4).\)
Problem. Cars arrive at a toll booth at a rate of 3 per minute. Find: (a) \(P(X=5)\); (b) \(P(X\ge2).\)
(a) \(P(X=5)=\dfrac{e^{-3}\cdot3^5}{5!}=\dfrac{e^{-3}\cdot243}{120}\approx0.1008.\)
(b) \(P(X\ge2)=1-P(X=0)-P(X=1)=1-e^{-3}-3e^{-3}=1-4e^{-3}\approx1-0.1991=0.8009.\)
| k | P(X = k) |
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Default \(\lambda=3\) is the toll-booth rate from Worked Example 4.2 — the amber bar is exactly that example's answer.
A machine produces defective items at 2 per hour (Poisson). Find P(exactly 3 defective) in a 2-hour period. \(\lambda=4.\quad P(X=3)=\frac{e^{-4}\cdot4^3}{3!}=\frac{64e^{-4}}{6}\approx0.1954.\)Show solution
Worked Example · Hypothesis test for population mean
A sample of 25 gives \(\bar{x}=48.2\), \(s=6\). Test \(H_0:\mu=50\) vs \(H_1:\mu\neq50\) at 5% significance.
Solution: \(t=\dfrac{48.2-50}{6/\sqrt{25}}=\dfrac{-1.8}{1.2}=-1.5\).
Critical value \(t_{24,0.025}\approx2.064\). Since \(|-1.5|<2.064\), do not reject \(H_0\).
Worked Example · Confidence interval
Construct a 95% CI for the mean of a normal population with known \(\sigma=8\) given \(n=36\), \(\bar{x}=62\).
Solution: \(\text{CI} = \bar{x}\pm z_{\alpha/2}\cdot\dfrac{\sigma}{\sqrt{n}} = 62\pm1.96\cdot\dfrac{8}{6} = 62\pm2.61 = (59.4,\,64.6)\)
Calculus (HL)
HL calculus covers advanced integration techniques, differential equations, and Maclaurin series — the core of pure mathematics at undergraduate level.
Integration by substitution: \(\displaystyle\int f(g(x))g'(x)\,dx=\int f(u)\,du\) where \(u=g(x).\) Integration by parts: \(\displaystyle\int u\,dv=uv-\int v\,du.\) (Choose \(u\) by LIATE: Log, Inverse trig, Algebraic, Trig, Exponential.) Partial fractions: decompose rational integrand, then integrate term by term.
Separable: \(\dfrac{dy}{dx}=f(x)g(y)\Rightarrow\displaystyle\int\frac{dy}{g(y)}=\int f(x)\,dx.\) Homogeneous: \(\dfrac{dy}{dx}=f(y/x);\) substitute \(y=vx,\;\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.\) Linear first-order: \(\dfrac{dy}{dx}+P(x)y=Q(x).\) Integrating factor: \(\mu=e^{\int P(x)\,dx}.\)
\[f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots\] Standard series:
| x | cos(x) | approximation |
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\(n=3\) reproduces Worked Example 5.3 exactly (up to the \(x^4\) term). Watch the amber curve hug the blue one near \(x=0\) — and peel away further out.
Problem. Find \(\displaystyle\int x e^{2x}\,dx.\)
Let \(u=x,\;dv=e^{2x}\,dx\Rightarrow du=dx,\;v=\dfrac{e^{2x}}{2}.\)
\(\int xe^{2x}\,dx=\dfrac{xe^{2x}}{2}-\int\dfrac{e^{2x}}{2}\,dx=\dfrac{xe^{2x}}{2}-\dfrac{e^{2x}}{4}+C=\dfrac{e^{2x}(2x-1)}{4}+C.\)
Problem. Solve \(\dfrac{dy}{dx}=\dfrac{2x}{y+1}\), given \(y(0)=2.\)
Separate: \((y+1)\,dy=2x\,dx.\) Integrate: \(\dfrac{(y+1)^2}{2}=x^2+C.\)
Initial condition \(y(0)=2\): \(\dfrac{9}{2}=C.\) Solution: \((y+1)^2=2x^2+9.\)
Problem. Find the Maclaurin series for \(f(x)=\cos x\) up to the \(x^4\) term and use it to approximate \(\cos(0.1)\).
\(\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots\)
\(\cos(0.1)\approx1-\dfrac{0.01}{2}+\dfrac{0.0001}{24}=1-0.005+0.000004\approx0.995004.\)
Solve \(\dfrac{dy}{dx}+\dfrac{2}{x}y=x^2,\;x>0\) with \(y(1)=1.\) I.F. \(\mu=x^2.\quad\frac{d}{dx}(x^2y)=x^4.\) \(x^2y=\frac{x^5}{5}+C.\quad y(1)=1\Rightarrow C=\frac{4}{5}.\) \(y=\dfrac{x^3}{5}+\dfrac{4}{5x^2}.\)Show solution
Worked Example · Integration by parts
Evaluate \(\displaystyle\int x^2 e^x\,dx\).
Solution: Apply IBP twice. Let \(u=x^2,\,dv=e^x dx\):
\(= x^2 e^x - 2\displaystyle\int xe^x\,dx = x^2 e^x - 2(xe^x - e^x)+C = e^x(x^2-2x+2)+C\)
Worked Example · Differential equation
Solve \(\dfrac{dy}{dx}=\dfrac{x}{y}\) with \(y(0)=3\).
Solution: Separate: \(y\,dy = x\,dx\implies\dfrac{y^2}{2}=\dfrac{x^2}{2}+C\).
At \(x=0,y=3\): \(C=\frac{9}{2}\). So \(y^2 = x^2+9\implies y=\sqrt{x^2+9}\).
Worked Example · Maclaurin series
Write the Maclaurin series for \(e^x\sin x\) up to and including the \(x^4\) term.
Solution: \(e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\), \(\sin x=x-\frac{x^3}{6}+\cdots\)
Product: \(x+x^2+\frac{x^3}{2}-\frac{x^3}{6}+\cdots = x+x^2+\frac{x^3}{3}-\frac{x^4}{30}+\cdots\)
AA HL — Additional Topics
These topics are specific to Analysis and Approaches HL.
\[\text{Geometric series converges iff }|r|<1:\quad S_\infty=\frac{a}{1-r}.\]\[\text{Ratio test: if }\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\Rightarrow\text{series converges absolutely}.\]
\[\sin^{-1}:\;[-1,1]\to[-\pi/2,\pi/2],\quad\cos^{-1}:\;[-1,1]\to[0,\pi],\quad\tan^{-1}:\;\mathbb{R}\to(-\pi/2,\pi/2).\]\[\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}},\quad\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}.\]
Problem. Determine if \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}\) converges.
Ratio test: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n+1}{2^{n+1}}\cdot\dfrac{2^n}{n}=\dfrac{n+1}{2n}\to\dfrac{1}{2}<1.\) Series converges.
Find the Maclaurin series for \(\ln(1+2x)\) up to the \(x^3\) term. Use \(\ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots\) with \(u=2x\): \(\ln(1+2x)=2x-2x^2+\frac{8x^3}{3}-\cdots\quad(-\tfrac{1}{2}<x\le\tfrac{1}{2}).\)Show solution
Worked Example · Partial fractions
Decompose \(\dfrac{5x+1}{(x+1)(x-2)}\) into partial fractions.
Solution: \(\dfrac{A}{x+1}+\dfrac{B}{x-2}\). Multiply: \(5x+1=A(x-2)+B(x+1)\).
\(x=2\): \(11=3B\implies B=\frac{11}{3}\). \(x=-1\): \(-4=-3A\implies A=\frac{4}{3}\).
AI HL — Additional Topics
These topics are specific to Applications and Interpretation HL.
For \(2\times2\) matrix \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\): Solving \(A\mathbf{x}=\mathbf{b}\): \(\mathbf{x}=A^{-1}\mathbf{b}.\) Eigenvalues: \(\det(A-\lambda I)=0.\)
A graph consists of vertices and edges. Weighted graph: edges have assigned weights (distances, costs). Key algorithms: Kruskal's algorithm — minimum spanning tree (MST): add edges in increasing weight, skip cycles. Prim's algorithm — MST from a vertex; add cheapest edge to tree at each step. Dijkstra's algorithm — shortest path from a source vertex. Euler circuit: traverses every edge exactly once; requires all vertices to have even degree. Hamiltonian path: visits every vertex exactly once.
Problem. Solve using matrices: \(2x+3y=7,\;x-y=1.\)
\(A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\;\mathbf{b}=\begin{pmatrix}7\\1\end{pmatrix}.\quad\det A=2(-1)-3(1)=-5.\)
\(A^{-1}=\dfrac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}=\begin{pmatrix}0.2&0.6\\0.2&-0.4\end{pmatrix}.\)
\(\mathbf{x}=A^{-1}\mathbf{b}=\begin{pmatrix}0.2(7)+0.6(1)\\0.2(7)-0.4(1)\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}.\quad x=2,\;y=1.\)
Edges with weights: AB=3, BC=2, CD=4, DE=1, AC=7, BD=5, CE=6, AD=8. Use Kruskal\'s algorithm to find the MST and its total weight. Sorted edges: DE=1, BC=2, AB=3, CD=4, BD=5, CE=6, AC=7, AD=8. Add: DE(1)✓, BC(2)✓, AB(3)✓, CD(4)✓ — BD creates cycle, CE creates cycle, AC creates cycle. MST edges: DE, BC, AB, CD. Total weight = 1+2+3+4 = 10.Show solution
Worked Example · Coupled differential equations
Given \(\dot{x}=3x-y,\;\dot{y}=2x\). Find equilibrium and classify stability.
Solution: Equilibrium: \(3x-y=0,\;2x=0\implies (0,0)\).
Jacobian eigenvalues: \(\lambda^2-3\lambda+2=0\implies\lambda=1,2\). Both positive → unstable node.
Practice Set
Express \(z = -1 + i\sqrt{3}\) in polar form \(r(\cos\theta + i\sin\theta)\) and find \(z^6.\)
By De Moivre: \(z^6 = 2^6\!\left(\cos 4\pi + i\sin 4\pi\right) = 64(1+0i) = 64.\)
Prove by induction that \(5^n - 1\) is divisible by 4 for all positive integers \(n.\)
Inductive step: Assume \(4\mid(5^k-1)\), i.e. \(5^k=4m+1.\) Then \(5^{k+1}-1 = 5\cdot5^k-1 = 5(4m+1)-1 = 20m+4 = 4(5m+1).\) ✓
By induction, \(4\mid(5^n-1)\) for all \(n\in\mathbb{Z}^+.\)
Let \(\mathbf{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\4\\-2\end{pmatrix}.\) Find \(\mathbf{a}\times\mathbf{b}.\)
\(= \mathbf{i}(2-12) - \mathbf{j}(-4-3) + \mathbf{k}(8+1) = \begin{pmatrix}-10\\7\\9\end{pmatrix}.\)
Find \(\displaystyle\int_0^1 x\ln(x+1)\,dx.\)
\(=\Big[\tfrac{x^2}{2}\ln(x+1)\Big]_0^1 - \displaystyle\int_0^1\tfrac{x^2}{2(x+1)}\,dx.\)
First term: \(\tfrac{1}{2}\ln 2 - 0 = \tfrac{\ln 2}{2}.\) For the integral, write \(\tfrac{x^2}{x+1}=x-1+\tfrac{1}{x+1}\), so \(\displaystyle\int_0^1\tfrac{x^2}{2(x+1)}dx = \tfrac{1}{2}\Big[\tfrac{x^2}{2}-x+\ln(x+1)\Big]_0^1 = \tfrac{1}{2}(\tfrac{1}{2}-1+\ln2) = \tfrac{\ln2-\frac12}{2}.\)
Answer: \(\tfrac{\ln2}{2} - \tfrac{\ln2-\frac12}{2} = \tfrac{\ln2}{2} - \tfrac{\ln2}{2} + \tfrac{1}{4} = \tfrac{1}{4}.\)
A Poisson random variable \(X\sim\text{Po}(3).\) Find \(P(X\geq 2).\)
Solve the differential equation \(\dfrac{dy}{dx} = 2xy^2\) with initial condition \(y(0) = 1.\)
Apply IC: \(-1 = 0 + C\Rightarrow C=-1.\) So \(-\dfrac{1}{y} = x^2-1\Rightarrow y = \dfrac{1}{1-x^2},\;|x|<1.\)
Prove that \(\sqrt{2}\) is irrational.
Then \(2 = \dfrac{p^2}{q^2}\Rightarrow p^2=2q^2.\) So \(p^2\) is even, meaning \(p\) is even. Write \(p=2k\).
Then \(4k^2=2q^2\Rightarrow q^2=2k^2\), so \(q\) is also even.
This contradicts \(\gcd(p,q)=1.\) Therefore \(\sqrt{2}\) is irrational. \(\square\)
Find the equation of the plane passing through \(A(1,0,2),\;B(3,1,0),\;C(0,2,1).\)
Normal \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&-2\\-1&2&-1\end{vmatrix}=\begin{pmatrix}(-1+4)\\(2+2)\cdot(-1)\\(4+1)\end{pmatrix}\)
\(\mathbf{n}=\begin{pmatrix}3\\-4\\5\end{pmatrix}.\) Using point \(A(1,0,2)\): \(3(x-1)-4(y)+5(z-2)=0\Rightarrow 3x-4y+5z=13.\)
Use the binomial theorem to find the coefficient of \(x^4\) in the expansion of \((2x - \tfrac{1}{x})^8.\)
For \(x^4\): \(8-2r=4\Rightarrow r=2.\) Coefficient: \(\binom{8}{2}\cdot2^6\cdot(-1)^2 = 28\times64\times1 = 1792.\)
Find all complex roots of \(z^3 = -8\) and plot them on the Argand diagram (describe positions).
\(z_0 = 2e^{i\pi/3} = 2(\tfrac{1}{2}+i\tfrac{\sqrt3}{2}) = 1+i\sqrt3.\)
\(z_1 = 2e^{i\pi} = -2.\)
\(z_2 = 2e^{i5\pi/3} = 2(\tfrac{1}{2}-i\tfrac{\sqrt3}{2}) = 1-i\sqrt3.\)
On the Argand diagram, the three roots lie on a circle of radius 2, equally spaced at 120° apart, with \(z_1=-2\) on the negative real axis.
Solve the second-order ODE \(y'' - 5y' + 6y = 0\) with \(y(0)=1,\;y'(0)=3.\)
General solution: \(y=Ae^{2x}+Be^{3x}.\)
\(y(0)=1\Rightarrow A+B=1.\)
\(y'=2Ae^{2x}+3Be^{3x},\;y'(0)=3\Rightarrow 2A+3B=3.\)
Solving: \(A=0,\;B=1.\) So \(y = e^{3x}.\)
(IB AI HL) A transition matrix \(T = \begin{pmatrix}0.7&0.2\\0.3&0.8\end{pmatrix}\) represents weekly customer movement between two brands. Find the steady-state distribution.
\(\begin{pmatrix}-0.3&0.2\\0.3&-0.2\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\mathbf{0}\Rightarrow -0.3x_1+0.2x_2=0\Rightarrow x_2=\tfrac{3}{2}x_1.\)
Normalise: \(x_1+x_2=1\Rightarrow x_1+\tfrac{3}{2}x_1=1\Rightarrow x_1=\tfrac{2}{5},\;x_2=\tfrac{3}{5}.\)
Steady-state: Brand A holds 40%, Brand B holds 60% of customers.
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