Before You Begin
Read the concept card for each question before attempting it.
This quiz covers all five core topic areas of IB Math AA HL:
01 · Number & Algebra
02 · Functions
03 · Geometry & Trig
04 · Statistics & Prob
05 · Calculus
HL · Proof & Vectors
HL · Complex Numbers
HL · Matrices
Timer: 45 minutes · Select one answer per question · Immediate feedback after each selection.
Number & Algebra — Arithmetic Sequences
Arithmetic Sequences & Series
General term: \(u_n = u_1 + (n-1)d\) | Sum: \(S_n = \dfrac{n}{2}(2u_1 + (n-1)d)\)
Find \(S_{10}\) if \(u_1=2,\, d=3\): \(S_{10}=\frac{10}{2}(4+27)=155\)
Find the sum of the arithmetic series with first term \(u_1 = 3\) and common difference \(d = 4\), for \(n = 20\) terms.
\(S_{20} = 760\)
\(S_{20} = 820\)
\(S_{20} = 880\)
\(S_{20} = 740\)
B
Using \(S_n = \frac{n}{2}(2u_1+(n-1)d)\): \(S_{20}=\frac{20}{2}(2\cdot3+19\cdot4)=10(6+76)=10\times82=820\). Alternatively, the 20th term is \(u_{20}=3+19\times4=79\), so \(S_{20}=\frac{20}{2}(3+79)=10\times82=820\).
Number & Algebra — Geometric Series
Infinite Geometric Series
Sum to infinity: \(S_\infty = \dfrac{u_1}{1-r}\), valid when \(|r| < 1\)
If \(u_1=6,\,r=\frac{1}{3}\): \(S_\infty = \frac{6}{1-\frac{1}{3}} = 9\)
An infinite geometric series has first term \(u_1 = 8\) and common ratio \(r = \dfrac{1}{2}\). Find \(S_\infty\).
\(12\)
\(24\)
\(16\)
\(8\)
C
Since \(|r|=\frac{1}{2}<1\), the series converges. \(S_\infty = \dfrac{8}{1-\frac{1}{2}} = \dfrac{8}{\frac{1}{2}} = 16\).
Number & Algebra — Binomial Theorem
Binomial Expansion
General term: \(\binom{n}{r}a^{n-r}b^r\) | \((a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\)
Coefficient of \(x^2\) in \((1+x)^5\): \(\binom{5}{2}=10\)
Find the coefficient of \(x^3\) in the expansion of \((2+x)^5\).
\(40\)
\(80\)
\(10\)
\(20\)
A
The general term is \(\binom{5}{r}2^{5-r}x^r\). For the \(x^3\) term, set \(r=3\): \(\binom{5}{3}\cdot 2^{2}\cdot x^3 = 10\times4\times x^3 = 40x^3\). Coefficient \(= 40\).
Number & Algebra — Logarithms
Logarithm Laws
\(\log_a(xy)=\log_a x+\log_a y\) | \(\log_a x^n = n\log_a x\) | Change of base: \(\log_b x = \dfrac{\ln x}{\ln b}\)
Solve \(\log_2 x = 5\): \(x=2^5=32\)
Solve the equation \(\log_2(x+3) + \log_2(x-1) = 5\).
\(x = 3\)
\(x = 7\)
\(x = 5\)
\(x = 4\)
C
Combine logarithms: \(\log_2[(x+3)(x-1)]=5\), so \((x+3)(x-1)=2^5=32\). Expanding: \(x^2+2x-3=32 \Rightarrow x^2+2x-35=0 \Rightarrow (x+7)(x-5)=0\). Thus \(x=5\) or \(x=-7\). Since \(\log_2(x-1)\) requires \(x>1\), we reject \(x=-7\). Answer: \(x=5\).
Calculus — Differentiation
Product Rule & Derivatives
Product rule: \(\dfrac{d}{dx}[uv] = u'v + uv'\) | \(\dfrac{d}{dx}[e^x]=e^x\)
\(\frac{d}{dx}[xe^x]=e^x+xe^x=(1+x)e^x\)
Let \(f(x) = x^2 e^x\). Find \(f'(1)\).
\(e\)
\(3e\)
\(2e\)
\(4e\)
B
Using the product rule with \(u=x^2\), \(v=e^x\): \(f'(x)=2xe^x+x^2e^x=e^x(2x+x^2)\). At \(x=1\): \(f'(1)=e^1(2+1)=3e\).
Calculus — Integration
Definite Integration
\(\int_a^b f(x)\,dx = [F(x)]_a^b = F(b)-F(a)\) | \(\int \sin x\,dx = -\cos x + C\)
\(\int_0^{\pi/2}\cos x\,dx = [\sin x]_0^{\pi/2}=1\)
Evaluate \(\displaystyle\int_0^{\pi} \sin x \, dx\).
\(0\)
\(1\)
\(-2\)
\(2\)
D
\(\displaystyle\int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} = -\cos\pi-(-\cos 0)=-(-1)+1=1+1=2\).
HL — Complex Numbers
Modulus of a Complex Number
For \(z = a+bi\): \(|z| = \sqrt{a^2+b^2}\) | Argument: \(\arg z = \arctan\!\left(\dfrac{b}{a}\right)\)
\(|1+i|=\sqrt{1+1}=\sqrt{2}\)
Find the modulus \(|z|\) of the complex number \(z = 3 + 4i\).
\(5\)
\(7\)
\(\sqrt{7}\)
\(\sqrt{7}\)
A
\(|z| = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\).
Geometry & Trig — Vectors
Angle Between Two Vectors
\(\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\) | Dot product: \(\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3\)
If \(\mathbf{a}\cdot\mathbf{b}=0\), vectors are perpendicular.
Find the angle between vectors \(\mathbf{a}=(1,\,2,\,2)\) and \(\mathbf{b}=(2,\,1,\,-2)\).
\(60°\)
\(45°\)
\(90°\)
\(120°\)
C
\(\mathbf{a}\cdot\mathbf{b}=(1)(2)+(2)(1)+(2)(-2)=2+2-4=0\). Since the dot product is zero, the vectors are perpendicular. Angle \(= 90°\).
Statistics & Probability — Normal Distribution
Empirical Rule (68-95-99.7)
\(P(\mu-\sigma < X < \mu+\sigma) \approx 0.6827\) | \(P(\mu-2\sigma < X < \mu+2\sigma) \approx 0.9545\)
For \(X\sim N(50,4)\), \(P(48<X<52)\approx0.68\)
For a normal distribution \(X \sim N(\mu,\,\sigma^2)\), which value best approximates \(P(\mu - \sigma < X < \mu + \sigma)\)?
\(0.95\)
\(0.68\)
\(0.50\)
\(0.997\)
B
The empirical (68-95-99.7) rule states: approximately 68.27% of values fall within one standard deviation of the mean. Thus \(P(\mu-\sigma < X < \mu+\sigma) \approx 0.68\).
Functions — Inverse Functions
Finding the Inverse Function
To find \(f^{-1}(x)\): write \(y=f(x)\), swap \(x\) and \(y\), solve for \(y\).
If \(f(x)=2x+3\), then \(y=2x+3 \Rightarrow x=\frac{y-3}{2}\), so \(f^{-1}(x)=\frac{x-3}{2}\)
Find the inverse function of \(f(x) = \dfrac{2x+1}{x-3}\).
\(f^{-1}(x) = \dfrac{3x+1}{x-2}\)
\(f^{-1}(x) = \dfrac{x-3}{2x+1}\)
\(f^{-1}(x) = \dfrac{x+3}{2}\)
\(f^{-1}(x) = \dfrac{2x-1}{x+3}\)
A
Set \(y=\frac{2x+1}{x-3}\). Solve for \(x\): \(y(x-3)=2x+1 \Rightarrow xy-3y=2x+1 \Rightarrow xy-2x=3y+1 \Rightarrow x(y-2)=3y+1 \Rightarrow x=\frac{3y+1}{y-2}\). Therefore \(f^{-1}(x)=\dfrac{3x+1}{x-2}\).
Geometry & Trig — Exact Values
Exact Trigonometric Values
\(\sin\tfrac{\pi}{6}=\tfrac{1}{2},\quad\cos\tfrac{\pi}{3}=\tfrac{1}{2},\quad\tan\tfrac{\pi}{3}=\sqrt{3},\quad\tan\tfrac{\pi}{4}=1\)
From a 30-60-90 triangle: opp=\(\sqrt{3}\), adj=1, hyp=2
State the exact value of \(\tan\!\left(\dfrac{\pi}{3}\right)\).
\(\dfrac{1}{\sqrt{3}}\)
\(\dfrac{\sqrt{3}}{2}\)
\(\sqrt{3}\)
\(1\)
C
In a 30-60-90 right triangle with hypotenuse 2: the side opposite 60° is \(\sqrt{3}\), adjacent to 60° is 1. So \(\tan 60° = \tan\frac{\pi}{3} = \frac{\sqrt{3}}{1} = \sqrt{3}\).
Number & Algebra — Quadratic Equations
Discriminant & Real Roots
Discriminant: \(\Delta = b^2-4ac\) | Real roots \(\Leftrightarrow \Delta \geq 0\)
\(x^2+2x+k=0\): real roots when \(4-4k\geq0\), i.e. \(k\leq1\)
Find the values of \(k\) for which \(x^2 + kx + (k+3) = 0\) has real roots.
\(-2 \leq k \leq 6\)
\(k \geq 6\) only
\(k \leq -2\) only
\(k \leq -2\) or \(k \geq 6\)
D
For real roots: \(\Delta = k^2 - 4(k+3) \geq 0 \Rightarrow k^2-4k-12\geq0 \Rightarrow (k-6)(k+2)\geq0\). This holds when \(k\leq -2\) or \(k\geq 6\).
HL — Matrices
Determinant of a 2×2 Matrix
\(\det\!\begin{pmatrix}a&b\\c&d\end{pmatrix} = ad - bc\)
\(\det\!\begin{pmatrix}3&1\\2&4\end{pmatrix}=12-2=10\)
Find \(\det(M)\) where \(M = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}\).
\(2\)
\(-2\)
\(10\)
\(-10\)
B
\(\det(M)=2\times5-3\times4=10-12=-2\).
Calculus — Chain Rule
Chain Rule for Differentiation
\(\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\)
\(\dfrac{d}{dx}[\cos(x^2)]=−\sin(x^2)\cdot2x\)
Let \(f(x) = \sin(3x^2)\). Find \(f'(0)\).
\(0\)
\(6\)
\(3\)
\(1\)
A
By the chain rule: \(f'(x)=\cos(3x^2)\cdot6x\). At \(x=0\): \(f'(0)=\cos(0)\cdot0=1\cdot0=0\).
Calculus — Integration by Substitution
u-Substitution
Let \(u=g(x)\), then \(\displaystyle\int f(g(x))g'(x)\,dx = \int f(u)\,du\)
\(\int 2xe^{x^2}dx\): let \(u=x^2\), \(du=2x\,dx\) → \(\int e^u du = e^{x^2}+C\)
Find \(\displaystyle\int x\,e^{x^2}\,dx\).
\(x^2 e^{x^2} + C\)
\(2e^{x^2} + C\)
\(\dfrac{1}{2}e^{x^2} + C\)
\(e^{x^2} + C\)
C
Let \(u=x^2\), so \(du=2x\,dx\), i.e. \(x\,dx=\frac{du}{2}\). Then \(\displaystyle\int xe^{x^2}dx=\int e^u\cdot\frac{du}{2}=\frac{1}{2}e^u+C=\frac{1}{2}e^{x^2}+C\).
HL — Proof by Mathematical Induction
Proof by Mathematical Induction
Step 1: Prove base case. Step 2: Assume \(P(k)\) true, prove \(P(k+1)\) follows.
Sum \(1+3+\cdots+(2n-1)=n^2\): check \(n=1\): \(1=1^2\) ✓
Which formula gives the sum \(1 + 2 + 3 + \cdots + n\)?
\(\dfrac{n(n-1)}{2}\)
\(\dfrac{n(n+1)}{2}\)
\(\dfrac{(n+1)(n+2)}{2}\)
\(n^2\)
B
The sum of the first \(n\) positive integers is \(\dfrac{n(n+1)}{2}\). Verify: \(n=5\): \(\frac{5\cdot6}{2}=15=1+2+3+4+5\) ✓. Proved by induction: base case \(n=1\): \(\frac{1\cdot2}{2}=1\) ✓. Inductive step: assuming \(\sum_{i=1}^k i=\frac{k(k+1)}{2}\), adding \((k+1)\) gives \(\frac{(k+1)(k+2)}{2}\) ✓.
Statistics & Probability — Correlation
Pearson's Correlation Coefficient
\(r \in [-1,\, 1]\) | \(r\approx\pm1\): strong; \(r\approx0\): weak
As height increases, weight increases → positive correlation, \(r>0\)
A data set shows a strong negative linear correlation between two variables. Which value of the Pearson correlation coefficient \(r\) is most consistent with this?
\(r = 0.92\)
\(r = 0.12\)
\(r = -0.15\)
\(r = -0.92\)
D
"Strong" means \(|r|\) close to 1. "Negative" means \(r<0\). The value \(r=-0.92\) has \(|r|=0.92\approx1\) (strong) and is negative. Options A has strong but positive, C has negative but weak (\(|-0.15|\) is small).
Functions — Exponential Models
Exponential Growth & Natural Logarithm
\(P = P_0 e^{kt}\) | To solve for \(t\): \(t = \dfrac{\ln(P/P_0)}{k}\)
Double: \(P=2P_0 \Rightarrow 2P_0=P_0 e^{kt}\Rightarrow t=\frac{\ln2}{k}\)
A population grows according to \(P = 1000\,e^{0.05t}\). Find the time \(t\) at which the population first reaches 2000.
\(t = 20\ln 2\)
\(t = \ln 2\)
\(t = \dfrac{\ln 2}{0.05^2}\)
\(t = 0.05\ln 2\)
A
\(1000e^{0.05t}=2000 \Rightarrow e^{0.05t}=2 \Rightarrow 0.05t=\ln2 \Rightarrow t=\dfrac{\ln2}{0.05}=20\ln2 \approx 13.86\).
Calculus — Local Extrema
Second Derivative Test for Extrema
If \(f'(a)=0\): local min if \(f''(a)>0\); local max if \(f''(a)<0\)
\(f(x)=x^2\): \(f'(0)=0\), \(f''(0)=2>0\) → local min at \(x=0\)
For \(f(x) = x^3 - 3x\), which statement correctly identifies the local minimum?
\(x = -1,\; f(-1) = 2\)
\(x = 0,\; f(0) = 0\)
\(x = -1,\; f(-1) = -2\)
\(x = 1,\; f(1) = -2\)
D
\(f'(x)=3x^2-3=0 \Rightarrow x=\pm1\). Second derivative: \(f''(x)=6x\). At \(x=1\): \(f''(1)=6>0\) → local min. At \(x=-1\): \(f''(-1)=-6<0\) → local max. The local minimum is at \(x=1\), \(f(1)=1-3=-2\).
Calculus — Implicit Differentiation
Implicit Differentiation
Differentiate both sides w.r.t. \(x\), applying chain rule to \(y\) terms: \(\dfrac{d}{dx}[y^n]=ny^{n-1}\dfrac{dy}{dx}\)
\(xy=1\): differentiate → \(y+x\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}=-\frac{y}{x}\)
Given \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\) at the point \((3,\,4)\).
\(\dfrac{3}{4}\)
\(-\dfrac{3}{4}\)
\(-\dfrac{4}{3}\)
\(\dfrac{4}{3}\)
B
Differentiating \(x^2+y^2=25\) implicitly: \(2x+2y\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=-\dfrac{x}{y}\). At \((3,4)\): \(\dfrac{dy}{dx}=-\dfrac{3}{4}\).
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