The function $f(x) = (x-3)^2 - 1$ is defined for the restricted domain $x \ge 3$. Find an expression for its inverse, $f^{-1}(x)$.

1. Set the function equal to $y$: $y = (x-3)^2 - 1$. 2. Swap the $x$ and $y$ variables to find the inverse mapping: $x = (y-3)^2 - 1$. 3. Rearrange to isolate the squared term: $x + 1 = (y-3)^2$. 4. Square root both sides, remembering the $\pm$ symbol: $y - 3 = \pm\sqrt{x+1}$. 5. Select the correct root. Since the original domain was $x \ge 3$, the range of the inverse must be $y \ge 3$. Thus, we must take the positive root: $y - 3 = +\sqrt{x+1}$. 6. State the final inverse function: $\mathbf{f^{-1}(x) = \sqrt{x+1} + 3}$.

Solve the inequality $|2x - 5| \ge |x + 1|$ algebraically.

1. Square both sides of the inequality. Since both sides are absolute values and strictly non-negative, squaring preserves the inequality and removes the modulus: $(2x - 5)^2 \ge (x + 1)^2$. 2. Expand both binomials: $4x^2 - 20x + 25 \ge x^2 + 2x + 1$. 3. Rearrange all terms to one side to form a quadratic inequality: $3x^2 - 22x + 24 \ge 0$. 4. Factorise the quadratic to find the critical values: $(3x - 4)(x - 6) \ge 0$. 5. Identify the critical values where the expression equals zero: $x = \frac{4}{3}$ and $x = 6$. 6. Determine the valid regions. Since it is a "greater than or equal to" ($\ge$) inequality for a positive parabola, the solution lies outside the roots: $\mathbf{x \le \frac{4}{3}}$ or $\mathbf{x \ge 6}$.

Prove algebraically that the logarithmic function $g(x) = \ln\left(\frac{1+x}{1-x}\right)$ is an odd function for its entire domain $-1 < x < 1$ .

Substitute $-x$ : $g(-x) = \ln\left(\frac{1+(-x)}{1-(-x)}\right)$ M1 $g(-x) = \ln\left(\frac{1-x}{1+x}\right)$ A1 Use $\ln\left(\frac{a}{b}\right) = -\ln\left(\frac{b}{a}\right)$ : $g(-x) = -\ln\left(\frac{1+x}{1-x}\right)$ M1 So $g(-x) = -g(x)$ for all $-1 < x < 1$ (domain symmetric about 0) R1 Hence $g$ is odd AG

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IB Maths AA HL · Unit 2: Functions

IB Maths AA HL Odd, Even, Inverse, Inequalities and Modulus Questions

Updated October 2026

Exam-style IB Maths AA HL odd, even, inverse, inequalities and modulus questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step, showing the working that earns the marks — then try the practice questions and check your working against the mark scheme.

38 questions
Paper 1: 33
Paper 2: 5
11 easy
17 medium
9 hard
1 very hard
3 worked examples

Practise Odd, Even, Inverse, Inequalities and Modulus questions →

What you need to know

HL AA Paper 1 tests |x-2| < 3 or (x-1)/(x+2) ≥ 0. Sign diagrams are your friend — never multiply through by a variable expression. Inequalities — rational and modulus overview →

What's examined in AA HL odd, even, inverse, inequalities and modulus

The question bank covers these odd, even, inverse, inequalities and modulus question types (number of questions in brackets):

Function Types and Inverses (20)
Algebraic Equations & Inequalities (13)
Graphical Transformations (5)

Key formulas

Modulus & argument: $r = | z | = \sqrt{a^{2} + b^{2}}, θ = \arg z$

In the same notation as the IB formula booklet. All AA HL formulas →

Odd, Even, Inverse, Inequalities and Modulus worked examples

Worked example 1: Identifying odd/even functions algebraically · easy

Determine algebraically whether the function $f (x) = \frac{x^{3}}{x^{2} + 1}$ is odd, even, or neither.

Solution

6 steps

Examiner tip: Students often incorrectly classify functions as "neither" because they make sign errors when expanding negative variables raised to powers, especially

(- x)^{2}

. Remember that an even power always eliminates the negative.

Worked example 2: Inverses with restricted domains · medium

The function $f (x) = (x - 3)^{2} - 1$ is defined for the restricted domain $x \geq 3$ . Find an expression for its inverse, $f^{- 1} (x)$ .

Solution

6 steps

Examiner tip: Forgetting the

\pm

when taking the square root, and therefore failing to explicitly justify why the positive root is chosen based on the domain restriction, will cost you reasoning marks in Paper 1.

Worked example 3: Solving modulus inequalities · hard

Solve the inequality $| 2 x - 5 | \geq | x + 1 |$ algebraically.

Solution

6 steps · 1 to fill in

Examiner tip: Squaring both sides is the most efficient algebraic method when there is a modulus on both sides. Trying to split this into multiple piecewise cases often leads to messy intersecting domains and lost solutions.

Try these IB Maths AA HL odd, even, inverse, inequalities and modulus questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 2 marks · Paper 1

A function $f (x)$ has a domain of $1 \leq x \leq 8$ and a range of $- 5 \leq y \leq 10$ . Given that the inverse function $f^{- 1} (x)$ exists, state the domain and range of $f^{- 1} (x)$ .

Attempt it and see the mark scheme →

Question 2 · medium · 4 marks · Paper 1

Prove algebraically that the logarithmic function $g (x) = \ln (\frac{1 + x}{1 - x})$ is an odd function for its entire domain $- 1 < x < 1$ .

Full worked solution Free, no sign-in

Substitute $- x$ : $g (- x) = \ln (\frac{1 + (- x)}{1 - (- x)})$ M1
$g (- x) = \ln (\frac{1 - x}{1 + x})$ A1
Use $\ln (\frac{a}{b}) = - \ln (\frac{b}{a})$ : $g (- x) = - \ln (\frac{1 + x}{1 - x})$ M1
So $g (- x) = - g (x)$ for all $- 1 < x < 1$ (domain symmetric about 0) R1
Hence $g$ is odd AG

Every question here has a full worked solution. See them all →

Question 3 · hard · 6 marks · Paper 2

Solve the following rational modulus inequality: $| \frac{2 x - 1}{x + 3} | > 1$

Attempt it and see the mark scheme →

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All 38 odd, even, inverse, inequalities and modulus questions with mark schemes →

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FAQ

How many IB Maths AA HL odd, even, inverse, inequalities and modulus questions are there?

There are 38 exam-style odd, even, inverse, inequalities and modulus questions in the AA HL question bank (Paper 1: 33 · Paper 2: 5), graded 11 easy, 17 medium, 9 hard, 1 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is odd, even, inverse, inequalities and modulus on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 33 · Paper 2: 5. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA HL Unit 2 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

What else is free here

Save your answers and track your progressA free account remembers the questions you have done on every topic and brings you back to this page.

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Free ready-to-teach lesson (slides and teacher notes): Odd & even functions, self-inverse functions.

IB Maths AA HL Odd, Even, Inverse, Inequalities and Modulus Questions

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What you need to know

What's examined in AA HL odd, even, inverse, inequalities and modulus

Key formulas

Odd, Even, Inverse, Inequalities and Modulus worked examples

Worked example 1: Identifying odd/even functions algebraically · easy

Solution

Worked example 2: Inverses with restricted domains · medium

Solution

Worked example 3: Solving modulus inequalities · hard

Solution

Try these IB Maths AA HL odd, even, inverse, inequalities and modulus questions

Question 1 · easy · 2 marks · Paper 1

Question 2 · medium · 4 marks · Paper 1

Full worked solution Free, no sign-in

Question 3 · hard · 6 marks · Paper 2

Revise odd, even, inverse, inequalities and modulus for the exam

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