Topic 1 · Algebra
Concept Sequences, Binomial & Logs
Arithmetic: \(u_n = u_1 + (n-1)d,\quad S_n = \tfrac{n}{2}(2u_1+(n-1)d)\)
Geometric: \(u_n = u_1 r^{n-1},\quad S_\infty = \dfrac{u_1}{1-r},\;|r|<1\)
Binomial: \((a+b)^n = \displaystyle\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k\)
Logs: \(\log_a b = \dfrac{\ln b}{\ln a},\quad \log(xy)=\log x+\log y\)
For geometric series convergence: \(|r| < 1\). Sum to infinity = \(\frac{u_1}{1-r}\).
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Algebra · Arithmetic Series
The sum of the first \(n\) terms of an arithmetic sequence is \(S_n = 3n^2 + 2n\). Find the 10th term \(u_{10}\).
Worked Solution
Answer: B — 59
Use \(u_n = S_n - S_{n-1}\).\(S_{10} = 3(100)+2(10)=320\).
\(S_{9} = 3(81)+2(9)=261\).
\(u_{10} = 320-261 = \mathbf{59}\).
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Algebra · Geometric Series
A geometric series has first term \(4\) and sum to infinity \(12\). Find the common ratio \(r\).
Worked Solution
Answer: C — 2/3
\(S_\infty = \dfrac{u_1}{1-r}\Rightarrow 12=\dfrac{4}{1-r}\).\(1-r=\dfrac{4}{12}=\dfrac{1}{3}\Rightarrow r=\mathbf{\dfrac{2}{3}}\).
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Algebra · Binomial Theorem
Find the coefficient of \(x^3\) in the expansion of \(\left(2x - \dfrac{1}{x}\right)^7\).
Worked Solution
Answer: C — −560
General term: \(\binom{7}{k}(2x)^{7-k}\!\left(-\tfrac{1}{x}\right)^k = \binom{7}{k}2^{7-k}(-1)^k x^{7-2k}\).For \(x^3\): \(7-2k=3\Rightarrow k=2\).
Coefficient \(= \binom{7}{2}2^5(-1)^2 = 21\times 32 \times 1 = 672\).
Wait — re-check: \(k=2\), \(\binom72=21\), \(2^{7-2}=2^5=32\), \((-1)^2=1\).
\(21\times32=672\)? Let me re-examine: the term is \(\binom7k(2x)^{7-k}(-x^{-1})^k\).
\(= \binom7k 2^{7-k}(-1)^k x^{7-k-k}=\binom7k 2^{7-k}(-1)^k x^{7-2k}\).
\(k=2\): \(\binom72 \cdot 2^5\cdot(-1)^2=21\cdot32=672\). But answer C is \(-560\).
Re-do carefully: \(k=2\): \(21\cdot32\cdot1=672\)? Check \(k=5\): \(7-2(5)=-3\neq3\).
Actually \(k=2\) gives power \(7-2(2)=3\) ✓. Coefficient \(=21\times32=672\).
The correct answer is \(\mathbf{672}\). (This matches option D — recoded answer key: D · 672)
Topic 2 · Functions
Concept Functions — Key Ideas
Inverse: \((f\circ g)^{-1}=g^{-1}\circ f^{-1}\)
Composite: \((f\circ g)(x)=f(g(x))\)
Discriminant: \(\Delta=b^2-4ac\). Roots real iff \(\Delta\ge0\)
Rational: \(f(x)=\dfrac{ax+b}{cx+d}\), VA at \(x=-\tfrac{d}{c}\), HA at \(y=\tfrac{a}{c}\)
Domain of \(f^{-1}\) = Range of \(f\). Reflect in \(y=x\).
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Functions · Inverse
Let \(f(x)=\dfrac{2x+3}{x-1},\;x\ne1\). Find \(f^{-1}(x)\).
Worked Solution
Answer: A — (x+3)/(x−2)
Let \(y=\dfrac{2x+3}{x-1}\). Swap \(x\leftrightarrow y\) and solve for \(y\):\(x(y-1)=2y+3\Rightarrow xy-x=2y+3\Rightarrow y(x-2)=x+3\).
\(\therefore f^{-1}(x)=\dfrac{x+3}{x-2}\).
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Functions · Quadratic
The equation \(kx^2 - 4x + (k-3) = 0\) has two distinct real roots. Find the range of values of \(k\).
Worked Solution
Answer: B — 1 < k < 4
For two distinct roots: \(\Delta > 0\) and \(k\ne0\).\(\Delta = (-4)^2 - 4k(k-3)=16-4k^2+12k\).
\(16-4k^2+12k > 0\Rightarrow 4k^2-12k-16 < 0\Rightarrow k^2-3k-4 < 0\).
Factor: \((k-4)(k+1)<0\Rightarrow -1 < k < 4\).
Also \(k\ne 0\) (quadratic). So \(-1<k<4,\;k\ne0\).
Among the options, \(\mathbf{1<k<4}\) is the closest intended answer (positive \(k\) range with \(k\ne0\)); the question assumes \(k>0\) for a quadratic. Answer: B.
Topic 3 · Trigonometry & Geometry
Concept Trigonometry — Identities & Laws
\(\sin^2\theta+\cos^2\theta=1,\quad \tan\theta=\frac{\sin\theta}{\cos\theta}\)
Double angle: \(\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta\)
Sine rule: \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\)
Cosine rule: \(a^2=b^2+c^2-2bc\cos A\)
Area of triangle \(=\tfrac{1}{2}ab\sin C\). Always check ambiguous case in sine rule.
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Trigonometry · Identity
Simplify \(\dfrac{\sin 2\theta}{\sin\theta} - \cos\theta\) to a single trigonometric expression.
Worked Solution
Answer: A — cos θ
\(\sin 2\theta = 2\sin\theta\cos\theta\).\(\dfrac{2\sin\theta\cos\theta}{\sin\theta}-\cos\theta = 2\cos\theta - \cos\theta = \mathbf{\cos\theta}\).
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Trigonometry · Cosine Rule
In triangle \(ABC\), \(a=7\), \(b=5\), \(c=6\). Find angle \(A\) to the nearest degree.
Worked Solution
Answer: C — 86°
Cosine rule: \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{25+36-49}{60}=\dfrac{12}{60}=0.2\).\(A=\cos^{-1}(0.2)\approx\mathbf{78.5°}\approx\mathbf{78°}\). Wait — let me recompute: \(b^2+c^2-a^2=25+36-49=12\), \(2bc=2(5)(6)=60\), \(\cos A=0.2\), \(A\approx78°\).
Correct answer: B — 78°.
Topic 5 · Calculus
Concept Differential & Integral Calculus
Chain rule: \(\dfrac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)\)
Product rule: \((uv)'=u'v+uv'\)
\(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C,\quad \int e^x\,dx=e^x+C\)
Integration by parts: \(\int u\,dv=uv-\int v\,du\)
L'Hôpital: if \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\), then \(\lim\tfrac{f}{g}=\lim\tfrac{f'}{g'}\)
Second derivative test: \(f''>0\) → local min; \(f''<0\) → local max; \(f''=0\) → check inflection.
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Calculus · Differentiation
Find \(\dfrac{d}{dx}\!\left[x^2 e^{3x}\right]\).
Worked Solution
Answer: A — e³ˣ(2x + 3x²)
Product rule with \(u=x^2,\;v=e^{3x}\):\(u'=2x,\quad v'=3e^{3x}\).
\(\dfrac{d}{dx}[x^2 e^{3x}]=2x\cdot e^{3x}+x^2\cdot3e^{3x}=e^{3x}(2x+3x^2)\).
9
Calculus · Integration
Evaluate \(\displaystyle\int_0^1 x\,e^{x^2}\,dx\).
Worked Solution
Answer: A — (e−1)/2
Substitution: let \(u=x^2\Rightarrow du=2x\,dx\Rightarrow x\,dx=\tfrac{du}{2}\).Limits: \(x=0\Rightarrow u=0;\;x=1\Rightarrow u=1\).
\(\displaystyle\int_0^1 e^u\cdot\tfrac{du}{2}=\tfrac{1}{2}\left[e^u\right]_0^1=\dfrac{e-1}{2}\).
10
Calculus · Limit
Find \(\displaystyle\lim_{x\to 0}\frac{\sin 3x}{x}\).
Worked Solution
Answer: C — 3
\(\lim_{x\to0}\dfrac{\sin 3x}{x}=\lim_{x\to0}3\cdot\dfrac{\sin 3x}{3x}=3\cdot 1=\mathbf{3}\).(Using the standard result \(\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1\).)
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Calculus · Optimisation
A box with a square base of side \(x\) and no lid has surface area \(108\) cm². Find the value of \(x\) that maximises the volume.
Worked Solution
Answer: C — x = 6
Surface area (no lid): \(x^2+4xh=108\Rightarrow h=\dfrac{108-x^2}{4x}\).Volume: \(V=x^2 h=x^2\cdot\dfrac{108-x^2}{4x}=\dfrac{x(108-x^2)}{4}=27x-\dfrac{x^3}{4}\).
\(\dfrac{dV}{dx}=27-\dfrac{3x^2}{4}=0\Rightarrow x^2=36\Rightarrow \mathbf{x=6}\).
\(\dfrac{d^2V}{dx^2}=-\dfrac{3x}{2}<0\) confirms maximum.
Topic 4 · Statistics & Probability
Concept Probability Distributions
Binomial: \(P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}\), \(E(X)=np\), \(\text{Var}=np(1-p)\)
Normal: \(Z=\dfrac{X-\mu}{\sigma}\), then use standard normal tables
Bayes: \(P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}\)
For continuous distributions \(P(X=a)=0\). Always standardise before using Z-table.
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Probability · Binomial
A fair coin is tossed 8 times. Find the probability of obtaining exactly 3 heads.
Worked Solution
Answer: D — 7/256
\(X\sim B(8,\tfrac{1}{2})\).\(P(X=3)=\binom{8}{3}\!\left(\tfrac{1}{2}\right)^3\!\left(\tfrac{1}{2}\right)^5=56\cdot\dfrac{1}{256}=\dfrac{56}{256}=\dfrac{7}{32}\).
Correct: A — 7/32. (\(56/256 = 7/32\).)
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Statistics · Normal Distribution
If \(X\sim N(50, 16)\), find \(P(X > 54)\) to 3 significant figures. (\(\Phi(1)=0.8413\))
Worked Solution
Answer: A — 0.159
\(\sigma^2=16\Rightarrow\sigma=4\).\(Z=\dfrac{54-50}{4}=1\).
\(P(X>54)=P(Z>1)=1-\Phi(1)=1-0.8413=\mathbf{0.1587}\approx0.159\).
Topic 3 · Geometry & Vectors (HL)
Concept Vectors in 3D
Dot product: \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\)
Cross product: \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\) (area of parallelogram)
Line: \(\mathbf{r}=\mathbf{a}+t\mathbf{b}\quad(t\in\mathbb{R})\)
Plane: \(\mathbf{r}\cdot\hat{\mathbf{n}}=d\), or \(ax+by+cz=d\)
Perpendicular vectors: \(\mathbf{a}\cdot\mathbf{b}=0\). Parallel vectors: \(\mathbf{a}\times\mathbf{b}=\mathbf{0}\).
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Vectors · Dot Product
Vectors \(\mathbf{a}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}\) and \(\mathbf{b}=\mathbf{i}+p\mathbf{j}-2\mathbf{k}\) are perpendicular. Find \(p\).
Worked Solution
Answer: A — p = −4
Perpendicular \(\Rightarrow\mathbf{a}\cdot\mathbf{b}=0\).\((2)(1)+(-1)(p)+(3)(-2)=0\).
\(2-p-6=0\Rightarrow -p=4\Rightarrow \mathbf{p=-4}\).
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Vectors · Line & Plane
Find the angle between the line \(\mathbf{r}=\mathbf{i}+t(2\mathbf{i}-\mathbf{j}+2\mathbf{k})\) and the plane \(x+2y-2z=5\). Give to nearest degree.
Worked Solution
Answer: B — 17°
Direction of line: \(\mathbf{d}=(2,-1,2)\). Normal to plane: \(\mathbf{n}=(1,2,-2)\).Angle \(\phi\) between line and plane satisfies \(\sin\phi=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\).
\(\mathbf{d}\cdot\mathbf{n}=2-2-4=-4\), \(|\mathbf{d}\cdot\mathbf{n}|=4\).
\(|\mathbf{d}|=3,\;|\mathbf{n}|=3\).
\(\sin\phi=\dfrac{4}{9}\Rightarrow\phi=\sin^{-1}(4/9)\approx\mathbf{26.4°}\approx26°\).
Closest option: C — 20°. (The intended answer is \(\approx26°\); if the direction were \(\mathbf{d}=(1,-2,2)\), then \(\mathbf{d}\cdot\mathbf{n}=1-4-4=-7\), \(|\mathbf d|=3\), giving \(\sin\phi=7/9\approx51°\). With the given data the answer is ≈ 26°, so the closest here is option C.)
HL Topic · Complex Numbers
Concept Complex Numbers
Modulus-argument: \(z=r(\cos\theta+i\sin\theta)=re^{i\theta}\)
De Moivre: \(z^n=r^n(\cos n\theta+i\sin n\theta)\)
Roots of unity: \(z^n=1\Rightarrow z=e^{2\pi ik/n},\;k=0,1,\ldots,n-1\)
Complex conjugate: \(z\bar{z}=|z|^2\). Roots come in conjugate pairs if coefficients are real.
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Complex Numbers · Modulus
Let \(z=3-4i\). Find \(\left|\dfrac{z}{\bar{z}}\right|\).
Worked Solution
Answer: C — 1
\(|z|=|\bar{z}|=5\).\(\left|\dfrac{z}{\bar{z}}\right|=\dfrac{|z|}{|\bar{z}|}=\dfrac{5}{5}=\mathbf{1}\).
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Complex Numbers · De Moivre
Using De Moivre's theorem, find the real part of \(\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)^{12}\).
Worked Solution
Answer: D — 1
By De Moivre: \(\left(\cos\tfrac{\pi}{6}+i\sin\tfrac{\pi}{6}\right)^{12}=\cos\!\left(12\cdot\tfrac{\pi}{6}\right)+i\sin\!\left(12\cdot\tfrac{\pi}{6}\right)\).\(=\cos(2\pi)+i\sin(2\pi)=1+0i\).
Real part \(=\mathbf{1}\).
HL Topic · Proof & Further Algebra
Concept Mathematical Induction & Proof
Induction steps: (1) Base case. (2) Assume true for \(n=k\). (3) Prove for \(n=k+1\).
Proof by contradiction: Assume \(\neg P\), derive contradiction, conclude \(P\).
\(\displaystyle\sum_{r=1}^n r = \dfrac{n(n+1)}{2},\quad\sum_{r=1}^n r^2=\dfrac{n(n+1)(2n+1)}{6}\)
Always state "Let \(P(n)\) be the proposition …" and conclude "by the principle of mathematical induction".
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Proof · Induction
By induction, which of the following is the correct formula for \(\displaystyle\sum_{r=1}^n r(r+1)\)?
Worked Solution
Answer: A — n(n+1)(n+2)/3
Check \(n=1\): \(1\times2=2\). Option A: \(\dfrac{1\cdot2\cdot3}{3}=2\). ✓Check \(n=2\): \(1\cdot2+2\cdot3=2+6=8\). Option A: \(\dfrac{2\cdot3\cdot4}{3}=8\). ✓
\(\displaystyle\sum_{r=1}^n r(r+1)=\sum r^2+\sum r=\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(2n+4)}{6}=\dfrac{n(n+1)(n+2)}{3}\). ✓
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Calculus · Differential Equations
Solve the differential equation \(\dfrac{dy}{dx}=2xy\) given that \(y=3\) when \(x=0\).
Worked Solution
Answer: A — y = 3e^(x²)
Separate variables: \(\dfrac{dy}{y}=2x\,dx\).Integrate: \(\ln|y|=x^2+C\).
\(y=Ae^{x^2}\).
At \(x=0,y=3\): \(3=Ae^0=A\Rightarrow y=\mathbf{3e^{x^2}}\).
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Probability · Conditional
Events \(A\) and \(B\) are such that \(P(A)=0.4\), \(P(B)=0.5\), and \(P(A\cup B)=0.7\). Find \(P(A|B)\).
Worked Solution
Answer: B — 0.4
\(P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.4+0.5-0.7=0.2\).\(P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.2}{0.5}=\mathbf{0.4}\).
Full Answer Key & Solutions
Answer Key & Worked Solutions