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IB Maths: Functions and Graphs – Practice Paper 1 (10 Questions)
IB Maths: Functions and Graphs – Practice Paper 1 (10 Questions)
Paper 1 starts with function notation: evaluating, domain and range, composite functions, inverses and the vertex of a quadratic, with answers and a one-line note for every item.
Quick answer Ten items, about twenty-five minutes. Items 01 to 04 cover substitution, domain and range; items 05 to 07 cover composites, inverses and completing the square; items 08 and 09 are restricted-domain inverses; item 10 redoes function notation.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: IB Maths: Functions and Graphs – Tutor Lesson Plan |
| Level | IB SL |
02 Questions (10 items)
- FoundationGiven f(x) = 2x + 5, find f(3).
- FoundationGiven g(x) = x squared minus 4, find g(-3).
- FoundationState the domain of f(x) = 1 / (x – 3).
- FoundationState the range of y = x squared plus 2 for real x.
- Intermediatef(x) = 2x + 5 and g(x) = x squared. Find f(g(2)).
- Intermediatef(x) = 3x – 4. Find the inverse and then f inverse of 8.
- IntermediateFind the coordinates of the vertex of y = x squared minus six x plus five.
- Challengef(x) = x squared plus 1 for x at least 0. Find the inverse and its domain.
- Challengeh(x) = (2x + 1) / (x – 3). Find the inverse of h.
- Error redoA student computes f(3) for f(x) = x squared as nine plus one equals ten. What is the correct value?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | f(3) = 11 | Substitute x = 3: two times three is six, plus five gives eleven. |
| 2 | g(-3) = 5 | Square first: minus three squared is nine, and nine minus four is five. |
| 3 | x is not equal to 3 | The denominator may not be zero, so x minus three is not zero and x cannot be three. |
| 4 | y is greater than or equal to 2 | A square is never negative, so the smallest value is two at x equals zero. |
| 5 | f(g(2)) = 13 | Work inside out: g(2) is four, then f(4) is eight plus five, that is thirteen. |
| 6 | inverse is (x + 4) / 3, and f inverse of 8 is 4 | Write y equals three x minus four and solve for x, then swap the letters; twelve over three is four. |
| 7 | (3, -4) | The axis of symmetry is x equals three, and substituting gives nine minus eighteen plus five, that is minus four. |
| 8 | inverse is the square root of x minus 1, with domain x at least 1 | Solving gives x equal to the positive root of y minus one; the old range becomes the new domain. |
| 9 | inverse is (3x + 1) / (x – 2), x not equal to 2 | Cross multiply, collect x on one side and divide; checking x equals one returns minus one point five in both directions. |
| 10 | f(3) = 9 | The notation asks for the value when x is three, so three squared is nine and nothing is added. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.
