Analysis and Approaches Question Bank

15 exam-style questions with full mark schemes, aligned to the IB Mathematics: Analysis and Approaches syllabus (SL/HL). Each question is presented in table format for compact study, with worked solutions below.

An arithmetic sequence has first term $a_1 = 7$ and common difference $d = 4$. (a) Find the 20th term. [2 marks] (b) Find the sum of the first 20 terms. [3 marks]

Mark Scheme:

(a) $u_{20} = a_1 + 19d = 7 + 19(4) = 7 + 76 = \mathbf{83}$ ✓

(b) $S_{20} = \frac{20}{2}(a_1 + u_{20}) = 10(7 + 83) = 10(90) = \mathbf{900}$ ✓

medium — 5 marks total

The quadratic $f(x) = 2x^2 - 12x + 10$ can be written in the form $f(x) = a(x - h)^2 + k$. (a) Find the values of $a$, $h$, and $k$. [3 marks] (b) State the minimum value of $f(x)$ and the value of $x$ at which it occurs. [2 marks]

Mark Scheme:

(a) Completing the square: $f(x) = 2(x^2 - 6x + 5)$ $= 2[(x - 3)^2 - 9 + 5]$ $= 2[(x - 3)^2 - 4]$ $= \mathbf{2(x - 3)^2 - 8}$

So $a = 2$, $h = 3$, $k = -8$. ✓

(b) Since $a = 2 > 0$, the parabola opens upward. Minimum value is $\mathbf{-8}$ when $\mathbf{x = 3}$. ✓

medium — 5 marks total

Solve the equation $3^{2x - 1} = 5^{x + 1}$. Give your answer to three significant figures. [4 marks]

Mark Scheme:

Take logarithms (base 10 or natural): $(2x - 1)\ln 3 = (x + 1)\ln 5$ $2x \ln 3 - \ln 3 = x \ln 5 + \ln 5$ $x(2\ln 3 - \ln 5) = \ln 5 + \ln 3$ $x = \frac{\ln 5 + \ln 3}{2\ln 3 - \ln 5} = \frac{\ln 15}{\ln 9 - \ln 5} = \frac{\ln 15}{\ln(9/5)}$

$x \approx \frac{2.708}{0.5878} \approx \mathbf{4.61}$ (3 s.f.) ✓

hard — 4 marks total

Use the binomial theorem to find the coefficient of $x^3$ in the expansion of $(2x - 3)^5$. [3 marks]

Mark Scheme:

General term: $\binom{5}{r}(2x)^{5-r}(-3)^r$

For $x^3$: $5 - r = 3 \Rightarrow r = 2$.

Coefficient $= \binom{5}{2}(2)^3(-3)^2 = 10 \times 8 \times 9 = \mathbf{720}$ ✓

medium — 3 marks total

Solve $2\sin x - 1 = 0$ for $0° \leq x \leq 360°$. [4 marks]

Mark Scheme:

$2\sin x = 1 \Rightarrow \sin x = 0.5$

$x = 30°$ (principal value)

$\sin x = 0.5$ also in the second quadrant: $x = 180° - 30° = 150°$

Solutions: $\mathbf{x = 30°, 150°}$ ✓

easy — 4 marks total

In triangle ABC, AB = 8 cm, BC = 6 cm, and angle ABC = 110°. Find the length of AC. [3 marks]

Mark Scheme:

Using the cosine rule: $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\text{ABC})$ $AC^2 = 64 + 36 - 2(8)(6)\cos 110°$ $AC^2 = 100 - 96\cos 110°$

$\cos 110° \approx -0.3420$

$AC^2 = 100 - 96(-0.3420) = 100 + 32.83 = 132.83$

$AC = \sqrt{132.83} \approx \mathbf{11.5 \text{ cm}}$ (3 s.f.) ✓

medium — 3 marks total

Points A and B have position vectors $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 5\mathbf{i} - \mathbf{j} + 4\mathbf{k}$. (a) Find $\overrightarrow{AB}$. [1 mark] (b) Find $|\overrightarrow{AB}|$. [2 marks]

Mark Scheme:

(a) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5-2)\mathbf{i} + (-1-3)\mathbf{j} + (4-(-1))\mathbf{k} = \mathbf{3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}}$ ✓

(b) $|\overrightarrow{AB}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = \mathbf{5\sqrt{2}}$ ✓

medium — 3 marks total

The masses of apples are normally distributed with mean 150 g and standard deviation 12 g. (a) Find the probability that a randomly chosen apple has mass greater than 165 g. [2 marks] (b) Find the probability that a randomly chosen apple has mass between 135 g and 165 g. [3 marks]

Mark Scheme:

(a) $P(X > 165) = P\left(Z > \frac{165 - 150}{12}\right) = P(Z > 1.25)$

From tables: $P(Z < 1.25) \approx 0.8944$

$P(Z > 1.25) = 1 - 0.8944 = \mathbf{0.1056}$ ✓

(b) $P(135 < X < 165) = P(-1.25 < Z < 1.25) = 2(0.8944) - 1 = \mathbf{0.7888}$ ✓

medium — 5 marks total

In a school, 60% of students study Physics, 45% study Chemistry, and 25% study both. A student is chosen at random. Given that they study Chemistry, find the probability they also study Physics. [3 marks]

Mark Scheme:

$P(\text{Physics} \mid \text{Chemistry}) = \frac{P(\text{Physics} \cap \text{Chemistry})}{P(\text{Chemistry})} = \frac{0.25}{0.45} = \frac{25}{45} = \frac{5}{9} \approx \mathbf{0.556}$ ✓

| Part | Marks | Key Method | | ---- | ----- | ------------------------------------------ | ------------------------ | | — | 3 | Apply conditional probability formula $P(A | B) = P(A \cap B) / P(B)$ |

medium — 3 marks total

The following data shows the number of hours studied ($x$) and test score ($y$) for 5 students:

$x$	2	4	6	8	10
$y$	45	55	60	75	85

(a) Calculate Pearson”s product-moment correlation coefficient $r$. [3 marks] (b) Comment on the strength and direction of the correlation. [1 mark]

Mark Scheme:

$\bar{x} = 6$, $\bar{y} = 64$

$\sum x^2 = 220$, $\sum y^2 = 21600$, $\sum xy = 2100$, $n = 5$

$S_{xx} = 220 - 5(36) = 40$ $S_{yy} = 21600 - 5(4096) = 21600 - 20480 = 1120$ $S_{xy} = 2100 - 5(6)(64) = 2100 - 1920 = 180$

$r = \frac{180}{\sqrt{40 \times 1120}} = \frac{180}{\sqrt{44800}} = \frac{180}{211.7} \approx \mathbf{0.851}$

(b) Strong positive linear correlation — as study hours increase, test scores tend to increase. ✓

hard — 4 marks total

A committee of 4 people is to be selected from 7 men and 5 women. The committee must contain at least 2 women. In how many ways can this be done? [4 marks]

Mark Scheme:

Total people = 12. Need at least 2 women.

Case 1: 2 women, 2 men: $\binom{5}{2}\binom{7}{2} = 10 \times 21 = 210$

Case 2: 3 women, 1 man: $\binom{5}{3}\binom{7}{1} = 10 \times 7 = 70$

Case 3: 4 women, 0 men: $\binom{5}{4}\binom{7}{0} = 5 \times 1 = 5$

Total $= 210 + 70 + 5 = \mathbf{285}$ ✓

hard — 4 marks total

Use the definition of the derivative to show that the derivative of $f(x) = x^2$ is $f'(x) = 2x$. [4 marks]

Mark Scheme:

$f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h}$

$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$

$= \lim_{h \to 0} \frac{2xh + h^2}{h}$

$= \lim_{h \to 0}(2x + h) = \mathbf{2x}$ ✓

medium — 4 marks total

Find the area enclosed by the curve $y = x^2 - 4x + 3$, the x-axis, and the lines $x = 1$ and $x = 4$. [5 marks]

Mark Scheme:

First find where the curve crosses the x-axis between $x = 1$ and $x = 4$: $x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3) = 0 \Rightarrow x = 1, 3$

The curve is below the x-axis for $1 < x < 3$ and above for $3 < x < 4$.

Area $= \int_1^3 |x^2 - 4x + 3|\,dx + \int_3^4 (x^2 - 4x + 3)\,dx$

$\int (x^2 - 4x + 3)\,dx = \frac{x^3}{3} - 2x^2 + 3x$

First part (area = negative of integral): $-\left[\frac{x^3}{3} - 2x^2 + 3x\right]_1^3 = -\left[\left(9 - 18 + 9\right) - \left(\frac{1}{3} - 2 + 3\right)\right] = -(0 - \frac{4}{3}) = \frac{4}{3}$

Second part: $\left[\frac{x^3}{3} - 2x^2 + 3x\right]_3^4 = \left(\frac{64}{3} - 32 + 12\right) - 0 = \frac{64}{3} - 20 = \frac{4}{3}$

Total area $= \frac{4}{3} + \frac{4}{3} = \mathbf{\frac{8}{3}}$ ✓

hard — 5 marks total

A rectangular box with a square base has a volume of $500\text{ cm}^3$. The material for the base costs $0.10/\text{cm}^2$ and the material for the sides costs $0.05/\text{cm}^2$. Find the dimensions that minimise the total cost. [6 marks]

Mark Scheme:

Let base side length = $x$ cm, height = $h$ cm.

Volume: $x^2 h = 500 \Rightarrow h = \frac{500}{x^2}$

Base area = $x^2$. Side area = $4xh = 4x \cdot \frac{500}{x^2} = \frac{2000}{x}$.

Cost: $C = 0.10x^2 + 0.05 \times \frac{2000}{x} = 0.10x^2 + \frac{100}{x}$

$\frac{dC}{dx} = 0.20x - \frac{100}{x^2}$

Set to zero: $0.20x = \frac{100}{x^2} \Rightarrow 0.20x^3 = 100 \Rightarrow x^3 = 500 \Rightarrow x = \sqrt[3]{500} \approx \mathbf{7.94 \text{ cm}}$

$h = \frac{500}{500^{2/3}} = 500^{1/3} \approx \mathbf{7.94 \text{ cm}}$

$\frac{d^2C}{dx^2} = 0.20 + \frac{200}{x^3} > 0$ for all $x > 0$, confirming a minimum. ✓

hard — 6 marks total

A particle moves in a straight line with velocity $v(t) = 3t^2 - 8t + 4$ m/s. (a) Find when the particle is at rest. [2 marks] (b) Find the total distance travelled in the first 3 seconds. [4 marks]

Mark Scheme:

(a) At rest: $v = 0$: $3t^2 - 8t + 4 = 0 \Rightarrow (3t - 2)(t - 2) = 0 \Rightarrow t = \frac{2}{3}$ s and $t = 2$ s. ✓

(b) Displacement: $s(t) = \int v\,dt = t^3 - 4t^2 + 4t + C$

$s(0) = 0$ (take starting position as origin), so $C = 0$.

Check direction changes:

• $0 < t < \frac{2}{3}$: $v > 0$ (moving right)
• $\frac{2}{3} < t < 2$: $v < 0$ (moving left)
• $t > 2$: $v > 0$ (moving right)

$s\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - 4\left(\frac{2}{3}\right)^2 + 4\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{16}{9} + \frac{8}{3} = \frac{40}{27}$

$s(2) = 8 - 16 + 8 = 0$

$s(3) = 27 - 36 + 12 = 3$

Distance $= |s(\frac{2}{3}) - s(0)| + |s(2) - s(\frac{2}{3})| + |s(3) - s(2)|$ $= \frac{40}{27} + \frac{40}{27} + 3 = \frac{80}{27} + 3 = \frac{161}{27} \approx \mathbf{5.96 \text{ m}}$ ✓

hard — 6 marks total

flowchart TD
    A[Analysis And Approaches Question Bank] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

A question bank serves as a diagnostic tool. Each question tests a specific skill, and working through them reveals which areas need more practice. The mark schemes show not just the answer but the reasoning expected, teaching you to communicate mathematical arguments evidently. Exam questions are designed to distinguish between students who memorize procedures and those who understand concepts.

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.

• Calculus: Analysis covers calculus
• Functions: Functions are central to analysis
• Number and Algebra: Algebra underpins analysis