OCR A Level Mathematics A: Modulus, functions, inverses and transformations (H240) – Revision Notes

Short recall notes for the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018, sections 1.02l, 1.02s, 1.02u, 1.02v, 1.02w and 1.02x. Full explanations and worked examples are in the study guide. When you are ready, move on to the practice questions. Related: OCR A Level Mathematics hub, the printable checklist and the free 10-minute diagnostics.

Stage 1 content is common with AS (H230). Stage 2 content is A Level only. Only 1.02w (single transformations) is Stage 1; everything else here is A Level only.

1. Modulus (1.02l and 1.02s, A Level only)

Fact	Detail
Definition
Two moduli
Less than
Greater than
Graph	y =

Method: equation |linear| = linear.

1. Write two cases: inside = right side, and −(inside) = right side.
2. Solve each.
3. Substitute back. Reject any answer that makes the right side negative.

Method: modulus on both sides.

1. Square both sides.
2. Move everything to one side and factorise.
3. For an inequality, sketch the quadratic and read off the region.

Method: sketch y = |ax + b|.

1. Sketch the line y = ax + b with both intercepts.
2. Reflect the part below the x-axis in the x-axis.
3. Label the vertex and the y-intercept.

Worked reminder. Solve |3x + 5| = x + 1.

3x + 5 = x + 1    →  x = −2   →  right side −1: reject
−3x − 5 = x + 1   →  x = −3/2 →  right side −1/2: reject

The equation has no solutions. A sketch would show the arm of the V never reaching the line.

Check by inequality: |3x + 5| = x + 1 needs x + 1 ≥ 0, so x ≥ −1, but both candidates are below −1.

Inequality reminder. Solve |x − 2| ≥ 3. Distance from 2 is at least 3, so x − 2 ≤ −3 or x − 2 ≥ 3. x ≤ −1 or x ≥ 5. Two separate regions, joined by “or”, not one chain.

2. Functions (1.02u, A Level only)

Term	Meaning
Function	a mapping where every x in the domain has exactly one image f(x)
Domain	the set of allowed inputs
Range	the set of all outputs f(x)
One-one	each output from exactly one input
Many-one	at least two inputs share an output (still a function)
One-many	one input has several outputs (not a function)

• State a range in terms of f(x) (or y), never x.
• Set notation example: {x ∈ ℝ : x > 3}.
• A quadratic on a restricted domain: complete the square, find the turning point, and check both ends of the domain.

Worked reminder. f(x) = x² + 2x − 2 for −3 ≤ x ≤ 2. Write f(x) = (x + 1)² − 3. The minimum −3 is at x = −1, inside the domain. f(−3) = 1 and f(2) = 6. Range: −3 ≤ f(x) ≤ 6.

3. Composite functions (1.02v, A Level only)

Notation	Meaning
fg(x)	f(g(x)): g first, then f
f²(x)	f(f(x))
gf	exists if the range of f lies inside the domain of g

Worked reminder. f(x) = 2x − 1 and g(x) = x² + 3, both on ℝ.

• fg(x) = 2(x² + 3) − 1 = 2x² + 5
• gf(x) = (2x − 1)² + 3 = 4x² − 4x + 4

They are different, so the order of composition matters.

• Solving fg(x) = a number means solve the composite as one equation; do not evaluate piece by piece unless you want to.
• The domain of fg is the part of the domain of g whose outputs are allowed by f.

4. Inverse functions (1.02v, A Level only)

Fact	Detail
Existence	f must be one-one on its domain
Swap	f⁻¹ takes the range of f as its domain, and the domain of f as its range
Identity	f⁻¹f(x) = x and ff⁻¹(x) = x
Graph	mirror image of y = f(x) in y = x; (p, q) on f gives (q, p) on f⁻¹
Meeting	for an increasing f, the graphs of f and f⁻¹ meet on y = x: solve f(x) = x

Method to find f⁻¹.

1. Write y = f(x).
2. Make x the subject (collect x terms and factorise if x appears twice).
3. Swap y for x. Write the domain from the range of f.

Worked reminder. f(x) = 4/(x − 3) + 1, x ≠ 3. Then y − 1 = 4/(x − 3), so x − 3 = 4/(y − 1) and x = 3 + 4/(y − 1) = (3y + 1)/(y − 1). So f⁻¹(x) = (3x + 1)/(x − 1), x ≠ 1. The value 1 is excluded because the range of f omits 1.

Quadratic reminder. To give x² − 12x + 7 an inverse, restrict to one side of the vertex: (x − 6)² − 29, so x ≥ 6 or x ≤ 6. Choose the root sign to match the side: x ≥ 6 gives 6 + √(x + 29), x ≤ 6 gives 6 − √(x + 29).

5. Graph transformations (1.02w Stage 1; 1.02x A Level only)

Equation	Effect on y = f(x)
y = f(x) + a	up by a: translation (0, a)
y = f(x + a)	a units left: translation (−a, 0)
y = af(x)	y-values times a: stretch parallel to the y-axis
y = f(ax)	x-values times 1/a: stretch parallel to the x-axis
negative a	adds a reflection (x-axis for af, y-axis for f(ax))

Point tracking for combinations. For y = p·f(qx + r) + s, a point (u, v) on y = f(x) goes to (X, Y) with qX + r = u and Y = pv + s. So X = (u − r)/q.

Worked reminder. (6, −2) lies on y = f(x). On y = 2f(x − 1) + 3 the image is X = 6 + 1 = 7 and Y = 2(−2) + 3 = −1, so (7, −1).

Describing a combination. y = f(2x + 6) can be written y = f(2(x + 3)). Stretch y = f(x) parallel to the x-axis with scale factor 1/2 to get f(2x), then translate by (−3, 0). The other order uses a translation of (−6, 0) first, then the stretch.

• Order matters when a stretch and a translation act on the same coordinate.
• Asymptotes, intercepts and turning points are all moved by the same rules.
• When describing, give the full name: “translation by the vector”, “stretch, scale factor, parallel to the axis”.
• You may be asked for an equation, a description, or a sketch with labelled points.

Must-know distinctions

Often confused	Difference
f(x + 3) and f(x) + 3	the first moves the graph 3 left; the second moves it 3 up
f(2x) and 2f(x)	the first halves x-values; the second doubles y-values
fg and gf	fg applies g first; the two are usually different
f⁻¹(x) and 1/f(x)	inverse function against reciprocal
Range and domain of f⁻¹	they are the range and domain of f, swapped
Many-one and one-many	a many-one mapping is a function; a one-many mapping is not

Quick self-test

1. Solve |2x − 9| = 5.
2. Solve |x + 3| < 4.
3. Solve |x − 1| = |3x + 5|.
4. State the vertex and the y-intercept of y = |4x − 12|.
5. Find the range of f(x) = x² + 2x − 2 for −3 ≤ x ≤ 2.
6. f(x) = 2x − 1 and g(x) = x² + 3. Find fg(x) and gf(x).
7. Find f⁻¹(x) for f(x) = 4/(x − 3) + 1, x ≠ 3, and state its domain.
8. Find the smallest k for which f(x) = x² − 12x + 7, x ≥ k, has an inverse.
9. A curve y = f(x) contains the point (6, −2). Give the new coordinates of this point after the curve is replaced by (a) y = f(x + 5), (b) y = f(3x).
10. Where does (6, −2) go when y = f(x) is replaced by y = 2f(x − 1) + 3?

Answers

1. 2x − 9 = 5 gives x = 7, and 2x − 9 = −5 gives x = 2. x = 2 or x = 7.
2. −4 < x + 3 < 4. −7 < x < 1.
3. Square: x² − 2x + 1 = 9x² + 30x + 25, so 8x² + 32x + 24 = 0 and x² + 4x + 3 = 0. x = −1 or x = −3.
4. The vertex is where 4x − 12 = 0. Vertex (3, 0); y-intercept 12.
5. (x + 1)² − 3. −3 ≤ f(x) ≤ 6.
6. fg(x) = 2x² + 5; gf(x) = 4x² − 4x + 4.
7. f⁻¹(x) = (3x + 1)/(x − 1), domain x ≠ 1.
8. f(x) = (x − 6)² − 29. k = 6.
9. (a) Translate 5 left: (1, −2). (b) Divide x by 3: (2, −2).
10. (7, −1).

Where marks are usually lost

• Not checking solutions of |linear| = linear for a negative right side.
• Squaring an inequality or equation in which only one side is a modulus without thinking about the sign.
• Shifting the graph right when the bracket says x + a.
• Giving the stretch factor for f(ax) as a instead of 1/a.
• Stating the range using x.
• Using only the end points of the domain when the turning point lies inside.
• Forgetting the domain and range statement for an inverse when the question asks for it.
• Choosing the wrong root sign when inverting a restricted quadratic.
• Describing a transformation without its vector or scale factor.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018, published by OCR.