IB Math AI HL — Master Practice Quiz

§ Formula Reference

Core Concepts & Key Formulae

T1 Sequences & Series

Arithmetic: uₙ = u₁ + (n−1)d, Sₙ = n/2 · (2u₁ + (n−1)d) Geometric: uₙ = u₁ · rⁿ⁻¹, Sₙ = u₁(rⁿ−1)/(r−1) S∞ = u₁/(1−r), |r| < 1

Arithmetic: constant difference $d$ between terms
Geometric: constant ratio $r$ between terms
Infinite geometric sum exists only when $|r| < 1$

Example

Find $u_{15}$ of an AP where $u_1 = 3,\ d = 7$

$u_{15} = 3 + 14(7) = 101$

T2 Exponential & Logarithmic Functions

f(x) = a·bˣ (exponential growth/decay) log_b(x) = ln(x)/ln(b) (change of base) ln(eˣ) = x; log_b(bˣ) = x log(xy) = log x + log y; log(xⁿ) = n log x

Growth: $b > 1$; Decay: $0 < b < 1$
$\log_b(MN) = \log_b M + \log_b N$
$\log_b(M/N) = \log_b M - \log_b N$

Example

Solve $3 \cdot 2^x = 96$

$2^x = 32 \Rightarrow x = 5$

T3 Descriptive Statistics & Regression

r = Σ(xᵢ−x̄)(yᵢ−ȳ) / √[Σ(xᵢ−x̄)²·Σ(yᵢ−ȳ)²] LSRL: ŷ = a + bx, b = r·(sᵧ/sₓ), a = ȳ − b·x̄ Interpolation: within data range | Extrapolation: outside

$r$ ranges from $-1$ to $+1$; $r = \pm 1$ is perfect linear
$r^2$ = proportion of variance explained
LSRL always passes through $(\bar{x}, \bar{y})$

Example

If $\hat{y} = 1.5x + 1.6$ and $r = 0.98$, predict $y$ when $x = 7$

$\hat{y} = 1.5(7) + 1.6 = 12.1$

T4 Probability Distributions

Binomial X~B(n,p): P(X=k) = C(n,k)·pᵏ·(1−p)ⁿ⁻ᵏ E(X) = np, Var(X) = np(1−p) Poisson X~Po(λ): P(X=k) = e⁻λ·λᵏ/k! E(X) = Var(X) = λ Normal X~N(μ,σ²): Z = (X−μ)/σ

Binomial: fixed $n$, independent trials, constant $p$
Poisson: rare events in fixed time/space interval
Normal: symmetric, bell-shaped; 68-95-99.7 rule

Example

$X \sim B(10, 0.3)$. Find $P(X = 3)$

$P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7 \approx 0.2668$

T5 Differential Calculus

d/dx[xⁿ] = nxⁿ⁻¹ d/dx[eˣ] = eˣ d/dx[ln x] = 1/x d/dx[sin x] = cos x Product: (uv)' = u'v + uv' Chain: [f(g(x))]' = f'(g(x))·g'(x)

$f'(x) > 0$: increasing; $f'(x) < 0$: decreasing
$f'(a) = 0$: possible local max/min (test with $f''$)
Gradient of tangent at $x = a$ is $f'(a)$

Example

$f(x) = 3x^4 - 2x^3 + 5x - 1$. Find $f'(1)$

$f'(x) = 12x^3 - 6x^2 + 5 \Rightarrow f'(1) = 11$

T6 Integral Calculus

∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1) ∫eˣ dx = eˣ + C ∫(1/x)dx = ln|x| + C Area = ∫ₐᵇ f(x)dx or ∫ₐᵇ [f(x)−g(x)]dx

Definite integral gives signed area under curve
Area between curves: top minus bottom function
Fundamental Theorem: $\int_a^b f'(x)dx = f(b) - f(a)$

Example

$\int_0^2 (2x^3 + 3x^2 - x)\,dx$

$= \left[\frac{x^4}{2} + x^3 - \frac{x^2}{2}\right]_0^2 = 8 + 8 - 2 = 14$

T7 Geometry & Trigonometry

Sine Rule: a/sin A = b/sin B = c/sin C Cosine Rule: a² = b² + c² − 2bc·cos A Area = ½ab·sin C Vectors: a·b = |a||b|cos θ ← dot product

Sine rule: use when angle-side pairs are known
Cosine rule: use with 3 sides or 2 sides + included angle
$\sin^2\theta + \cos^2\theta = 1$ (Pythagorean identity)

Example

In $\triangle ABC$: $b=7$, $c=9$, $A=60°$. Find $a$

$a^2 = 49+81-63 = 67 \Rightarrow a \approx 8.19$

T8 Hypothesis Testing & Chi-Squared

Null hypothesis H₀ vs. Alternative H₁ t-test statistic: t = (x̄ − μ₀)/(s/√n) χ² statistic: Σ(O−E)²/E p-value < α ⟹ Reject H₀

$\chi^2$ test: goodness of fit or independence
Degrees of freedom for $r \times c$ table: $(r-1)(c-1)$
Expected frequency $E = \frac{\text{row total} \times \text{col total}}{\text{grand total}}$

Example

One-sample $t$-test: $n=25$, $\bar{x}=52$, $s=8$, $H_0\!:\mu=50$

$t = \frac{52-50}{8/\sqrt{25}} = 1.25;\ \ p \approx 0.223 > 0.05$ → Do not reject $H_0$

Q01 Sequences — Arithmetic [2 marks]

An arithmetic sequence has first term $u_1 = 3$ and common difference $d = 7$. What is the sum of the first 15 terms, $S_{15}$?

✓ Solution

Use $S_n = \dfrac{n}{2}(2u_1 + (n-1)d)$.
$S_{15} = \dfrac{15}{2}(2(3) + 14(7)) = \dfrac{15}{2}(6 + 98) = \dfrac{15}{2}(104) = 780$

Answer: B — 780

Q02 Sequences — Geometric [2 marks]

A geometric sequence has $u_1 = 2$ and common ratio $r = 3$. Find the sum of the first 6 terms, $S_6$.

✓ Solution

Use $S_n = \dfrac{u_1(r^n - 1)}{r - 1}$.
$S_6 = \dfrac{2(3^6 - 1)}{3 - 1} = \dfrac{2(729 - 1)}{2} = \dfrac{2 \times 728}{2} = 728$

Answer: C — 728

Q03 Exponential Functions [2 marks]

Solve for $x$: $\quad 3 \cdot 2^x = 96$

✓ Solution

Divide both sides by 3: $2^x = 32 = 2^5$.
Therefore $x = 5$.

Answer: B — $x = 5$

Q04 Logarithms [2 marks]

Evaluate: $\quad \log_3 81 + \log_3 \!\left(\dfrac{1}{9}\right)$

✓ Solution

$\log_3 81 = \log_3 3^4 = 4$.
$\log_3\!\left(\dfrac{1}{9}\right) = \log_3 3^{-2} = -2$.
Sum $= 4 + (-2) = 2$.

Answer: C — $2$

Q05 Linear Regression [3 marks]

Data points $(x, y)$: $(2,5),\,(4,8),\,(6,9),\,(8,14),\,(10,17)$ give the regression line $\hat{y} = 1.5x + 1.6$ with Pearson correlation $r = 0.983$. Predict $\hat{y}$ when $x = 7$.

✓ Solution

Substitute $x = 7$ into the regression line:
$\hat{y} = 1.5(7) + 1.6 = 10.5 + 1.6 = 12.1$.
Since $r = 0.983$ is very close to 1, the line fits well and this interpolation is reliable.

Answer: A — $12.1$

Q06 Normal Distribution [3 marks]

$X \sim N(100,\, 15^2)$. Find $P(X < 115)$.

✓ Solution

Standardise: $Z = \dfrac{115 - 100}{15} = \dfrac{15}{15} = 1$.
From the standard normal table: $P(Z < 1) = 0.8413$.

Answer: D — $0.8413$

Q07 Binomial Distribution [3 marks]

$X \sim B(10,\, 0.3)$. Find $P(X \leq 3)$.

✓ Solution

$P(X \le 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)$
$= 0.0282 + 0.1211 + 0.2335 + 0.2668 = 0.6496$
Note: $P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7 \approx 0.2668$ alone is option A — but the question asks for cumulative $P(X \le 3)$.

Answer: B — $0.6496$

Q08 Poisson Distribution [3 marks]

Cars pass a toll booth at a rate of 4 per minute. Assuming a Poisson model, $X \sim Po(4)$, find $P(X > 2)$.

✓ Solution

$P(X > 2) = 1 - P(X \le 2) = 1 - [P(0)+P(1)+P(2)]$
$P(0) = e^{-4} \approx 0.0183$
$P(1) = 4e^{-4} \approx 0.0733$
$P(2) = \dfrac{16e^{-4}}{2} \approx 0.1465$
$P(X > 2) = 1 - 0.2381 = 0.7619$

Answer: C — $0.7619$

Q09 Chi-Squared Test [3 marks]

A $2 \times 2$ contingency table has observed frequencies: $$\begin{array}{c|cc} & \text{Yes} & \text{No}\\\hline \text{Group A} & 30 & 20 \\ \text{Group B} & 10 & 40 \end{array}$$ The expected frequency for cell (Group A, Yes) is:

✓ Solution

Row total (A) $= 30 + 20 = 50$. Column total (Yes) $= 30 + 10 = 40$. Grand total $= 100$.
$E = \dfrac{50 \times 40}{100} = 20$

Answer: A — $20$

Q10 Differential Calculus [2 marks]

Let $f(x) = 3x^4 - 2x^3 + 5x - 1$. Find $f'(1)$.

✓ Solution

$f'(x) = 12x^3 - 6x^2 + 5$
$f'(1) = 12(1)^3 - 6(1)^2 + 5 = 12 - 6 + 5 = 11$

Answer: C — $11$

Q11 Integral Calculus — Definite [3 marks]

Evaluate $\displaystyle\int_0^2 \!(2x^3 + 3x^2 - x)\,dx$.

✓ Solution

$\displaystyle\int(2x^3+3x^2-x)\,dx = \frac{x^4}{2} + x^3 - \frac{x^2}{2}$
At $x=2$: $\dfrac{16}{2} + 8 - \dfrac{4}{2} = 8 + 8 - 2 = 14$
At $x=0$: $0$
Definite integral $= 14 - 0 = 14$

Answer: B — $14$

Q12 Trigonometry — Area of Triangle [2 marks]

In triangle $ABC$, $a = 8$, $b = 10$, and angle $C = 30°$. Find the area of the triangle.

✓ Solution

Area $= \dfrac{1}{2}ab\sin C = \dfrac{1}{2}(8)(10)\sin 30° = \dfrac{1}{2}(80)(0.5) = 20$

Answer: D — $20$

Q13 Cosine Rule [3 marks]

In triangle $ABC$, $b = 7$, $c = 9$, and $A = 60°$. Find the value of $a^2$.

✓ Solution

Cosine Rule: $a^2 = b^2 + c^2 - 2bc\cos A$
$a^2 = 49 + 81 - 2(7)(9)\cos 60°$
$= 130 - 126 \times \tfrac{1}{2}$
$= 130 - 63 = 67$

Answer: B — $67$

Q14 Matrices — Systems of Equations [3 marks]

Solve the system using matrices: $\begin{cases}2x + 3y = 12\\x + 4y = 11\end{cases}$

✓ Solution

$M = \begin{pmatrix}2&3\\1&4\end{pmatrix}$, $\det(M) = 8 - 3 = 5$
$M^{-1} = \dfrac{1}{5}\begin{pmatrix}4&-3\\-1&2\end{pmatrix}$
$\begin{pmatrix}x\\y\end{pmatrix} = M^{-1}\begin{pmatrix}12\\11\end{pmatrix} = \dfrac{1}{5}\begin{pmatrix}48-33\\-12+22\end{pmatrix} = \begin{pmatrix}3\\2\end{pmatrix}$
Check: $2(3)+3(2)=12$ ✓ and $3+4(2)=11$ ✓

Answer: A — $x=3,\ y=2$

Q15 Probability — Conditional [2 marks]

Events $A$ and $B$ satisfy $P(A) = 0.4$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Find $P(A \mid B)$.

✓ Solution

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4$
Note: $P(A\mid B) = P(A) = 0.4$, which confirms that $A$ and $B$ are independent events here.

Answer: C — $0.4$

Q16 Pearson Correlation [2 marks]

A dataset yields Pearson correlation coefficient $r = -0.92$. Which statement is the most accurate interpretation?

✓ Solution

$|r| = 0.92 > 0.85$ indicates a strong association. The negative sign means as $x$ increases, $y$ decreases — a strong negative linear relationship.

Answer: D — Strong negative linear relationship

Q17 Exponential Growth/Decay [3 marks]

A radioactive substance has a half-life of 10 years. A sample starts at 100 g. How many grams remain after 30 years?

✓ Solution

After each 10 years, the amount halves. 30 years = 3 half-lives.
$A = 100 \times \left(\dfrac{1}{2}\right)^3 = 100 \times \dfrac{1}{8} = 12.5$ g

Answer: B — $12.5$ g

Q18 Vectors — Dot Product & Angle [3 marks]

Vectors $\mathbf{a} = \begin{pmatrix}3\\4\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\-2\end{pmatrix}$. Find the angle $\theta$ between them, to the nearest degree.

✓ Solution

$\mathbf{a} \cdot \mathbf{b} = 3(1) + 4(-2) = 3 - 8 = -5$
$|\mathbf{a}| = \sqrt{9+16} = 5$, $\quad |\mathbf{b}| = \sqrt{1+4} = \sqrt{5}$
$\cos\theta = \dfrac{-5}{5\sqrt{5}} = \dfrac{-1}{\sqrt{5}} \approx -0.4472$
$\theta = \cos^{-1}(-0.4472) \approx 116.6° \approx 117°$

Answer: C — $117°$

Q19 Hypothesis Testing — t-test [3 marks]

A one-sample $t$-test is conducted with $n = 25$, $\bar{x} = 52$, $s = 8$, $H_0\!:\mu = 50$, $\alpha = 0.05$ (two-tailed). The $p$-value is approximately $0.223$. What is the correct conclusion?

✓ Solution

$t = \dfrac{52-50}{8/\sqrt{25}} = \dfrac{2}{1.6} = 1.25$
$p \approx 0.223 > \alpha = 0.05$, so we fail to reject $H_0$.
Important: we never "accept" $H_0$ or "prove" any value. We simply lack sufficient evidence against $H_0$.

Answer: B — Fail to reject $H_0$

Q20 Area Between Curves [4 marks]

Find the area of the region enclosed between $y = x + 2$ and $y = x^2$.

✓ Solution

Intersections: $x^2 = x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x = -1,\ 2$
Area $= \displaystyle\int_{-1}^{2}[(x+2)-x^2]\,dx$
$= \left[\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}\right]_{-1}^{2}$
At $x=2$: $2 + 4 - \dfrac{8}{3} = 6 - \dfrac{8}{3} = \dfrac{10}{3}$
At $x=-1$: $\dfrac{1}{2} - 2 + \dfrac{1}{3} = -\dfrac{7}{6}$
Area $= \dfrac{10}{3} - \left(-\dfrac{7}{6}\right) = \dfrac{20}{6} + \dfrac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2}$

Answer: D — $\dfrac{9}{2}$

IB Math AI HL
Master Quiz

T1 Sequences & Series

T2 Exponential & Logarithmic Functions

T3 Descriptive Statistics & Regression

T4 Probability Distributions

T5 Differential Calculus

T6 Integral Calculus

T7 Geometry & Trigonometry

T8 Hypothesis Testing & Chi-Squared

Answer Key & Detailed Explanations

Answer Key