Unit 1
Number & Algebra
Key Concepts to Memorise
- Binomial Theorem: \((a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\)
- General term: \(T_{r+1} = \binom{n}{r}a^{n-r}b^r\)
- Complex number modulus: \(|a+bi| = \sqrt{a^2+b^2}\)
- De Moivre's Theorem: \((r\,\text{cis}\,\theta)^n = r^n\,\text{cis}(n\theta)\)
\[T_{r+1}=\tbinom{n}{r}a^{n-r}b^r\qquad z^n = r^n(\cos n\theta + i\sin n\theta)\]
Worked Example
Find the term independent of \(x\) in \((x + \tfrac{1}{x})^6\).
Solution: \(T_{r+1} = \binom{6}{r}x^{6-r}\cdot x^{-r} = \binom{6}{r}x^{6-2r}\). Set \(6-2r=0 \Rightarrow r=3\). Term \(= \binom{6}{3} = 20\).
01
Find the constant term in the expansion of \(\left(2x - \dfrac{1}{x}\right)^6\).
Binomial Theorem
Full Solution
\(T_{r+1} = \binom{6}{r}(2x)^{6-r}\!\left(-\tfrac{1}{x}\right)^r = \binom{6}{r}2^{6-r}(-1)^r x^{6-2r}\).
Constant term: \(6-2r=0 \Rightarrow r=2\).
\(T_3 = \binom{6}{2}\cdot 2^4\cdot(-1)^2 = 15\cdot 16\cdot 1 = \mathbf{240}\).
Constant term: \(6-2r=0 \Rightarrow r=2\).
\(T_3 = \binom{6}{2}\cdot 2^4\cdot(-1)^2 = 15\cdot 16\cdot 1 = \mathbf{240}\).
02
Let \(z = 1 + i\sqrt{3}\). Find \(z^6\).
Complex Numbers
Full Solution
\(|z|=\sqrt{1+3}=2,\; \arg(z)=\arctan(\sqrt{3})=\tfrac{\pi}{3}\).
By De Moivre: \(z^6 = 2^6\,\text{cis}\!\left(6\cdot\tfrac{\pi}{3}\right) = 64\,\text{cis}(2\pi) = 64\cdot 1 = \mathbf{64}\).
By De Moivre: \(z^6 = 2^6\,\text{cis}\!\left(6\cdot\tfrac{\pi}{3}\right) = 64\,\text{cis}(2\pi) = 64\cdot 1 = \mathbf{64}\).
Unit 2
Functions
Key Concepts to Memorise
- Inverse: swap \(x\) and \(y\), then solve for \(y\)
- Log laws: \(\log_a(mn)=\log_a m + \log_a n\); change of base: \(\log_a b = \dfrac{\ln b}{\ln a}\)
- Transformations: \(f(x-h)+k\) shifts right \(h\), up \(k\)
\[\log_a b = \frac{\ln b}{\ln a}\qquad (f\circ g)^{-1} = g^{-1}\circ f^{-1}\]
Worked Example
If \(f(x)=2x-3\), find \(f^{-1}(x)\).
Solution: Let \(y=2x-3\), swap: \(x=2y-3 \Rightarrow f^{-1}(x)=\dfrac{x+3}{2}\).
03
Let \(f(x) = \dfrac{2x+1}{x-3}\). Find \(f^{-1}(x)\).
Inverse Functions
Full Solution
Let \(y=\dfrac{2x+1}{x-3}\). Swap \(x\) and \(y\): \(x(y-3)=2y+1 \Rightarrow xy-3x=2y+1\).
\(y(x-2)=3x+1 \Rightarrow f^{-1}(x) = \mathbf{\dfrac{3x+1}{x-2}}\).
\(y(x-2)=3x+1 \Rightarrow f^{-1}(x) = \mathbf{\dfrac{3x+1}{x-2}}\).
04
Solve \(\log_2 x + \log_2(x-2) = 3\).
Logarithms
Full Solution
\(\log_2[x(x-2)]=3 \Rightarrow x(x-2)=8 \Rightarrow x^2-2x-8=0\).
\((x-4)(x+2)=0 \Rightarrow x=4\) or \(x=-2\).
Since \(\log_2 x\) requires \(x>0\) and \(x>2\), we reject \(x=-2\). Answer: \(\mathbf{x=4}\).
\((x-4)(x+2)=0 \Rightarrow x=4\) or \(x=-2\).
Since \(\log_2 x\) requires \(x>0\) and \(x>2\), we reject \(x=-2\). Answer: \(\mathbf{x=4}\).
Unit 3
Geometry & Trigonometry
Key Concepts to Memorise
- Dot product: \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\)
- Compound angle: \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\)
- Double angle: \(\cos 2A = 1-2\sin^2 A = 2\cos^2 A-1\)
\[\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\qquad \sin 75° = \sin(45°+30°)\]
Worked Example
Angle between \(\mathbf{a}=(1,0,0)\) and \(\mathbf{b}=(1,1,0)\):
\(\cos\theta = \dfrac{1}{\sqrt{2}} \Rightarrow \theta = 45°\).
05
Find the angle between vectors \(\mathbf{a}=(1,2,-1)\) and \(\mathbf{b}=(2,-1,2)\).
Vectors
Full Solution
\(\mathbf{a}\cdot\mathbf{b} = (1)(2)+(2)(-1)+(-1)(2) = 2-2-2 = -2\).
\(|\mathbf{a}|=\sqrt{1+4+1}=\sqrt{6},\quad|\mathbf{b}|=\sqrt{4+1+4}=3\).
\(\cos\theta = \dfrac{-2}{3\sqrt{6}} \Rightarrow \theta = \arccos\!\left(\dfrac{-2}{3\sqrt{6}}\right) \approx 105.8°\).
\(|\mathbf{a}|=\sqrt{1+4+1}=\sqrt{6},\quad|\mathbf{b}|=\sqrt{4+1+4}=3\).
\(\cos\theta = \dfrac{-2}{3\sqrt{6}} \Rightarrow \theta = \arccos\!\left(\dfrac{-2}{3\sqrt{6}}\right) \approx 105.8°\).
06
Find the exact value of \(\sin 75°\).
Trigonometry
Full Solution
\(\sin 75° = \sin(45°+30°) = \sin 45°\cos 30°+\cos 45°\sin 30°\)
\(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}}{4}+\dfrac{\sqrt{2}}{4} = \mathbf{\dfrac{\sqrt{6}+\sqrt{2}}{4}}\).
Note: Option C simplifies to the same value — but B is the standard IB form.
\(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}}{4}+\dfrac{\sqrt{2}}{4} = \mathbf{\dfrac{\sqrt{6}+\sqrt{2}}{4}}\).
Note: Option C simplifies to the same value — but B is the standard IB form.
Unit 4
Statistics & Probability
Key Concepts to Memorise
- Binomial: \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\), mean \(\mu=np\)
- Normal: \(X\sim N(\mu,\sigma^2)\), standardise \(Z=\dfrac{X-\mu}{\sigma}\)
- Bayes' theorem: \(P(A|B)=\dfrac{P(B|A)P(A)}{P(B)}\)
- Conditional probability: \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\)
\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\qquad Z=\frac{X-\mu}{\sigma}\]
Worked Example
\(X\sim B(5,0.3)\). Find \(P(X=2)\).
\(= \binom{5}{2}(0.3)^2(0.7)^3 = 10\times0.09\times0.343 \approx 0.309\).
07
If \(X \sim N(\mu, \sigma^2)\), find \(P(X < \mu + 1.5\sigma)\) to 4 decimal places.
Normal Distribution
Full Solution
Standardise: \(P\!\left(Z < \dfrac{\mu+1.5\sigma-\mu}{\sigma}\right) = P(Z < 1.5)\).
From standard normal tables: \(P(Z < 1.5) = \mathbf{0.9332}\).
From standard normal tables: \(P(Z < 1.5) = \mathbf{0.9332}\).
08
Let \(X \sim B(8, 0.4)\). Find \(P(X = 3)\) to 4 significant figures.
Binomial Distribution
Full Solution
\(P(X=3) = \binom{8}{3}(0.4)^3(0.6)^5 = 56 \times 0.064 \times 0.07776\)
\(= 56 \times 0.00497664 = \mathbf{0.2787}\) (4 s.f.).
\(= 56 \times 0.00497664 = \mathbf{0.2787}\) (4 s.f.).
Unit 5
Calculus
Key Concepts to Memorise
- Product rule: \((uv)' = u'v + uv'\)
- Chain rule: \(\dfrac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)\)
- Integration by parts: \(\displaystyle\int u\,dv = uv - \int v\,du\)
- Fundamental Theorem: \(\displaystyle\int_a^b f'(x)\,dx = f(b)-f(a)\)
\[\int u\,dv = uv - \int v\,du\qquad \frac{d}{dx}[\ln x]=\frac{1}{x},\quad \frac{d}{dx}[e^x]=e^x\]
Worked Example
Differentiate \(f(x)=x^3 e^x\).
\(f'(x) = 3x^2 e^x + x^3 e^x = x^2 e^x(3+x)\).
09
Find \(\dfrac{d}{dx}\left[x^2 \ln x\right]\).
Differentiation
Full Solution
Product rule with \(u=x^2,\; v=\ln x\):
\(\dfrac{d}{dx}[x^2\ln x] = 2x\cdot\ln x + x^2\cdot\dfrac{1}{x} = 2x\ln x + x = \mathbf{x(2\ln x+1)}\).
\(\dfrac{d}{dx}[x^2\ln x] = 2x\cdot\ln x + x^2\cdot\dfrac{1}{x} = 2x\ln x + x = \mathbf{x(2\ln x+1)}\).
10
Evaluate \(\displaystyle\int x e^x \, dx\).
Integration by Parts
Full Solution
Let \(u=x,\; dv=e^x dx \Rightarrow du=dx,\; v=e^x\).
\(\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = \mathbf{e^x(x-1)+C}\).
\(\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = \mathbf{e^x(x-1)+C}\).
11
Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin 3x}{x}\).
Limits
Full Solution
Use the standard result \(\displaystyle\lim_{x\to 0}\dfrac{\sin kx}{x} = k\).
Alternatively: \(\dfrac{\sin 3x}{x} = 3\cdot\dfrac{\sin 3x}{3x} \to 3\times 1 = \mathbf{3}\).
Alternatively: \(\dfrac{\sin 3x}{x} = 3\cdot\dfrac{\sin 3x}{3x} \to 3\times 1 = \mathbf{3}\).
12
The volume of a sphere is increasing at \(32\pi\) cm³/s. Find the rate of change of the radius when \(r = 2\) cm.
Related Rates
Full Solution
\(V = \tfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}\).
At \(r=2\): \(32\pi = 4\pi(4)\dfrac{dr}{dt} = 16\pi\dfrac{dr}{dt}\).
\(\dfrac{dr}{dt} = \dfrac{32\pi}{16\pi} = \mathbf{2 \text{ cm/s}}\).
At \(r=2\): \(32\pi = 4\pi(4)\dfrac{dr}{dt} = 16\pi\dfrac{dr}{dt}\).
\(\dfrac{dr}{dt} = \dfrac{32\pi}{16\pi} = \mathbf{2 \text{ cm/s}}\).
13
Write the Maclaurin series for \(e^x\) up to and including the term in \(x^3\).
Series Expansions
Full Solution
\(f(x)=e^x,\; f^{(n)}(0)=1\) for all \(n\).
Maclaurin: \(e^x = \displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!} = \mathbf{1 + x + \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots}\).
Note: \(2!=2\) and \(3!=6\), so the denominators are 2 and 6, not 2 and 3.
Maclaurin: \(e^x = \displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!} = \mathbf{1 + x + \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots}\).
Note: \(2!=2\) and \(3!=6\), so the denominators are 2 and 6, not 2 and 3.
14
Solve the differential equation \(\dfrac{dy}{dx} = 2xy\), given \(y(0) = 3\).
Differential Equations
Full Solution
Separate variables: \(\dfrac{dy}{y} = 2x\,dx\).
Integrate: \(\ln|y| = x^2 + C_1 \Rightarrow y = Ce^{x^2}\).
Apply \(y(0)=3\): \(3=Ce^0=C\). Therefore \(y = \mathbf{3e^{x^2}}\).
Integrate: \(\ln|y| = x^2 + C_1 \Rightarrow y = Ce^{x^2}\).
Apply \(y(0)=3\): \(3=Ce^0=C\). Therefore \(y = \mathbf{3e^{x^2}}\).
15
Find the inverse of \(A = \begin{pmatrix}3&1\\2&4\end{pmatrix}\).
Matrices (HL)
Full Solution
\(\det(A) = 3\cdot4 - 1\cdot2 = 12-2 = 10\).
For \(2\times2\) matrix: swap diagonal, negate off-diagonal.
\(A^{-1} = \mathbf{\dfrac{1}{10}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}}\).
For \(2\times2\) matrix: swap diagonal, negate off-diagonal.
\(A^{-1} = \mathbf{\dfrac{1}{10}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}}\).
16
Using proof by induction, which is the correct inductive step to prove \(\displaystyle\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}\)?
Proof by Induction
Full Solution
Inductive step: Assume \(\sum_{k=1}^{m}k = \dfrac{m(m+1)}{2}\) (inductive hypothesis).
Add \((m+1)\) to both sides: \(\dfrac{m(m+1)}{2}+(m+1) = \dfrac{m(m+1)+2(m+1)}{2} = \dfrac{(m+1)(m+2)}{2}\).
This matches the formula with \(n=m+1\). ✓ Option A is correct.
Add \((m+1)\) to both sides: \(\dfrac{m(m+1)}{2}+(m+1) = \dfrac{m(m+1)+2(m+1)}{2} = \dfrac{(m+1)(m+2)}{2}\).
This matches the formula with \(n=m+1\). ✓ Option A is correct.
17
A geometric series has first term \(3\) and common ratio \(\tfrac{1}{2}\). Find the sum to infinity.
Sequences & Series
Full Solution
\(S_\infty = \dfrac{a}{1-r}\) (valid for \(|r|<1\)).
\(S_\infty = \dfrac{3}{1-\tfrac{1}{2}} = \dfrac{3}{\tfrac{1}{2}} = \mathbf{6}\).
\(S_\infty = \dfrac{3}{1-\tfrac{1}{2}} = \dfrac{3}{\tfrac{1}{2}} = \mathbf{6}\).
18
Find the area enclosed between \(y = 4x - x^2\) and \(y = x^2\).
Area Between Curves
Full Solution
Intersection: \(4x-x^2=x^2 \Rightarrow 4x-2x^2=0 \Rightarrow 2x(2-x)=0 \Rightarrow x=0,\,2\).
Area \(= \displaystyle\int_0^2\!(4x-x^2-x^2)\,dx = \int_0^2(4x-2x^2)\,dx\)
\(= \Big[2x^2 - \tfrac{2x^3}{3}\Big]_0^2 = \left(8-\tfrac{16}{3}\right)-0 = \dfrac{24-16}{3} = \mathbf{\dfrac{8}{3}}\).
Area \(= \displaystyle\int_0^2\!(4x-x^2-x^2)\,dx = \int_0^2(4x-2x^2)\,dx\)
\(= \Big[2x^2 - \tfrac{2x^3}{3}\Big]_0^2 = \left(8-\tfrac{16}{3}\right)-0 = \dfrac{24-16}{3} = \mathbf{\dfrac{8}{3}}\).
19
A curve is defined parametrically by \(x = t^2,\; y = 2t\). Find \(\dfrac{dy}{dx}\).
Parametric Calculus
Full Solution
\(\dfrac{dx}{dt} = 2t,\quad \dfrac{dy}{dt} = 2\).
\(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2}{2t} = \mathbf{\dfrac{1}{t}}\).
\(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2}{2t} = \mathbf{\dfrac{1}{t}}\).
Unit 4 — Advanced
Conditional Probability
Key Formula
\(P(A|B) = \dfrac{P(A\cap B)}{P(B)}\) — the probability of \(A\) given that \(B\) has occurred.
20
Events \(A\) and \(B\) are such that \(P(A\cap B)=0.12\) and \(P(B)=0.3\). Find \(P(A|B)\).
Conditional Probability
Full Solution
\(P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.12}{0.3} = \mathbf{0.4}\).