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IB Math AA HL

HL

IB Mathematics: Analysis and Approaches Higher Level develops analytical thinking and problem-solving skills through algebra, functions, geometry, statistics and calculus.

Exam BoardIB
LevelHigher Level
AssessmentExternal Exams
yxf(x) = ax² + bx + c∫ x dx

Analysis & Approaches · HL Formula Booklet

Every formula provided in the IB exam, by topic

Papers

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Paper 1 – Sample Questions

5 sample questions on this topic.

EasyMediumHard
Topics
EasyQ1

Consider the function f(x) = x3x^3x3 ln⁡\lnln(x), for x > 0.

(a) Find f'(x). [3]

(b) Find the exact x-coordinate of the stationary point of f(x). [3]

[6 marks]
MediumQ2

(a) Evaluate ∫\int∫ u sin⁡\sinsin u du. [4 marks]

(b) By using an appropriate substitution and the result from part (a), evaluate ∫01\int_0^1∫01​ xarcsin⁡x1−x2\frac{x \arcsin x}{\sqrt{1-x^2}}1−x2​xarcsinx​ dx. Express your answer in the form A + Bπ\piπ, where A, B ∈\in∈ Q\mathbb{Q}Q. [6 marks]

[10 marks]
MediumQ3

Let f(x) = ln⁡\lnln(cos⁡\coscos x).

(a) Write down the Maclaurin series for cos⁡\coscos x up to and including the term in x4x^4x4. (1 mark)

(b) Using your answer from part (a) and the Maclaurin series for ln⁡\lnln(1+u) = u - u22+u33−…\frac{u^2}{2} + \frac{u^3}{3} - \dots2u2​+3u3​−…, find the first three non-zero terms of the Maclaurin series for f(x) = ln⁡\lnln(cos⁡\coscos x). (5 marks)

(c) Let g(x) = x f'(x). Find the first two non-zero terms of the Maclaurin series for g(x). (3 marks)

[9 marks]
HardQ4

A particle moves in a straight line such that its velocity, v m/s, at time t seconds, for t ≥\geq≥ 0, is given by v(t) = (t-1)(t-3)(t-5).

(a) Find the initial velocity of the particle.

(b) Determine the exact times when the particle is momentarily at rest.

(c) Find the acceleration of the particle when t=2 seconds.

(d) Calculate the total distance traveled by the particle in the interval 0 ≤\leq≤ t ≤\leq≤ 3 seconds.

[7 marks]
HardQ5

Consider the differential equation dydx=y2ln⁡xx(y−1)\frac{dy}{dx} = \frac{y^2 \ln x}{x(y-1)}dxdy​=x(y−1)y2lnx​ for x>0 and y ≠\ne= 0, y ≠\ne= 1.

(a) Show that the general solution of the differential equation is 1y+ln⁡\frac{1}{y} + \lny1​+ln y = 12(ln⁡x)2\frac{1}{2}(\ln x)^221​(lnx)2 + C, where C is an arbitrary constant.

(b) Find the particular solution given that y(e) = -1.

(c) Explain why y must be negative for the particular solution found in part (b).

[8 marks]
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