Practise › Questions › Functions, inverses and the modulus

Functions, inverses and the modulus questions

A function maps each permitted input in its domain to exactly one output, and its range is the set of outputs it produces. This lesson adds the rules for composing two functions and for reversing one. The modulus makes negative values positive, and its V-shaped graphs give equations extra solutions.

7 original questions · 22 marks · the functions, inverses and the modulus notes · Algebra and functions

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

  With f(x) = 3x − 1 and g(x) = x², find fg(2) and gf(2).

  Worked answer

  fg means g first, so g(2) = 4 and then f(4) = 11. The other order gives f(2) = 5 and then g(5) = 25. B1 B1 for the two values. The two answers differ, as composites usually do, and reading fg from the left is the commonest way to lose both marks at once.

• 2 marks · warm-up

  With f(x) = 3x − 1 and g(x) = x², find fg(x) and gf(x) in general.

  Worked answer

  fg(x) = f(x²) = 3x² − 1, and gf(x) = (3x − 1)². B1 B1 for the two composites. The inner function's output is fed through the outer one whole, brackets and all. Writing gf(x) as 3x² − 1 as well is the slip that shows the bracket was dropped.

• 4 marks · find

  Find the inverse of f(x) = x² − 3 with domain x ≥ 0, stating the domain of the inverse.

  Worked answer

  Write y = x² − 3 and solve for x, giving x = √(y + 3). The positive root is the one to take, because the domain restricts x to be at least 0. Swapping letters, f⁻¹(x) = √(x + 3). The range of f is y ≥ −3, so the domain of f⁻¹ is x ≥ −3. Domain and range swap roles along with the letters, and the stated domain is worth a mark on its own. M1 for making x the subject, A1 for √(x + 3), A1 for taking the positive root, B1 for the domain x ≥ −3.

• 3 marks · find

  Find the inverse of f(x) = (2x + 1)/5.

  Worked answer

  y = (2x + 1)/5 unwinds a step at a time. Multiply by 5 to get 5y = 2x + 1, then x = (5y − 1)/2. Swapping letters, f⁻¹(x) = (5x − 1)/2. The inverse undoes the machine in reverse order, multiplying by 5 first where f divided by 5 last. M1 A1 for making x the subject, A1 for the inverse.

• 4 marks

  Solve |3x − 2| = x + 4.

  Worked answer

  Take the two arms separately. Original arm: 3x − 2 = x + 4 gives x = 3, and the check |7| = 7 holds. Reflected arm: −(3x − 2) = x + 4 gives 2 − 3x = x + 4, so x = −½, and |−3.5| = 3.5 = −½ + 4 holds too. Both solutions survive, so x = −½ and x = 3. M1 A1 for the first arm, M1 A1 for the reflected arm. Every arm's answer has to be checked back, because reflected algebra can invent crossings the graphs never make.

• 3 marks · explain · stretch

  Explain why f(x) = x² with domain all real numbers has no inverse, and state a restriction of the domain that repairs this.

  Worked answer

  An inverse has to send each output back to a single input, but x² sends both 3 and −3 to 9. The function is not one-to-one, so no inverse exists. Restricting the domain to x ≥ 0, or equally to x ≤ 0, keeps one branch of the parabola and the inverse appears. B1 for the two inputs giving one output, B1 for the function not being one-to-one, B1 for a suitable restriction. Saying “because it is a curve” names no property; one-to-one is the property being asked about.

• 4 marks · stretch

  Solve |2x + 1| = |x − 4|.

  Worked answer

  Two moduli means two cases, not four, because the overall sign of one side may be absorbed into the other. Same signs: 2x + 1 = x − 4 gives x = −5. Opposite signs: 2x + 1 = −(x − 4) = 4 − x gives 3x = 3, so x = 1. Checking, x = −5 gives |−9| = |−9| and x = 1 gives |3| = |−3|, so both hold and x = −5 or x = 1. M1 A1 for the same-signs case, M1 A1 for the opposite-signs case. Squaring both sides is the clean alternative, since (2x + 1)² = (x − 4)² leads to 3x² + 12x − 15 = 0 and the same pair; it also removes the need to check, because squaring cannot introduce a false root when both sides were already non-negative.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise functions, inverses and the modulus one question at a time

The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.