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AS A-level Pure Mathematics · Edexcel 8MA0 · free A-level maths worksheet

Functions

Domain and range, composite and inverse functions, the modulus function, combining transformations, solving modulus equations and inequalities.

12 original questions, 83 marks. Type your answers below for an instant mark and a score (algebra in any equivalent form, e.g. 2sqrt(3) or 2√3), or print the worksheet, mark it with the answers sheet, and use the solutions to see every step written out by hand.

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The questions

1

The functions f and g are defined by

f(x) = 3x − 5, x ∈ ℝ

g(x) = x2 + 1, x ∈ ℝ

  1. Find fg(2)
  2. Find gf(x), giving your answer in the form ax2 + bx + c
4 marks
Show answer
a) 10 b) 9x2 - 30x + 26
2

The function f is defined by

f(x) = 2x + 3x − 1, x > 1

  1. Find f−1(x)
  2. State the domain of f−1
4 marks
Show answer
a) f-1(x) = x + 3x - 2 b) x > 2
3
  1. Sketch the graph of y = |3x − 6|, showing the coordinates of any points where the graph meets the coordinate axes.
  2. Solve the equation |3x − 6| = 9
xy
5 marks
Show answer
a) V shape, vertex (2, 0), meets y-axis at (0, 6) b) x = -1 or x = 5
4

The function f is defined by

f(x) = x2 − 6x + 11, x ≥ 3

  1. Find the range of f.
  2. Find f−1(x), stating its domain.
  3. Solve the equation f(x) = f−1(x), giving your answer in the form a + √b2
7 marks
Show answer
a) f(x) ≥ 2 b) f-1(x) = 3 + √(x - 2), x ≥ 2 c) x = 7 + √52
5

The diagram shows a sketch of the graph with equation y = 2|x + 1| − 4

The vertex of the graph is the point P.

  1. Find the coordinates of P.
  2. Find the coordinates of the points where the graph meets the x-axis.
  3. Solve the equation 2|x + 1| − 4 = 12x + 1
  4. Find the set of values of the constant k for which the equation 2|x + 1| − 4 = k has exactly two solutions.
xyP
8 marks
Show answer
a) (-1, -4) b) (-3, 0) and (1, 0) c) x = -2.8 or x = 2 d) k > -4
6

The functions f and g are defined by

f(x) = e2x − 3, x ∈ ℝ

g(x) = ln(x + 4), x > −4

  1. State the range of f.
  2. Find f−1(x), stating its domain.
  3. Find gf(x), giving your answer in its simplest form.
  4. Solve gf(x) = 2, giving your answer in the form 12 ln a, where a is an exact value to be found.
7 marks
Show answer
a) f(x) > -3 b) f-1(x) = 12 ln(x + 3), x > -3 c) ln(e2x + 1) d) x = 12 ln(e2 - 1)
7
  1. On the same axes, sketch the graphs of y = |2x − 5| and y = |x + 1|, showing the coordinates of the points where each graph meets the coordinate axes.
  2. Solve the inequality |2x − 5| > |x + 1|
xy
7 marks
Show answer
a) y = |2x - 5|: vertex (2.5, 0), meets y-axis at (0, 5); y = |x + 1|: vertex (-1, 0), meets y-axis at (0, 1) b) x 6
8

The diagram shows a sketch of the curve with equation y = f(x). The curve meets the x-axis at (−1, 0) and (3, 0), meets the y-axis at (0, 3) and has a maximum point at (1, 4).

  1. Sketch the curve with equation y = |f(x)|, showing the coordinates of the points where it meets the axes and of the maximum point.
  2. Sketch the curve with equation y = f(|x|), showing the coordinates of the points where it meets the axes and of any maximum points.
  3. Find the set of values of k for which f(|x|) = k has exactly four solutions.
  4. State the coordinates of the maximum point of the curve with equation y = 2f(x + 1)
xy(-1, 0)(3, 0)(0, 3)(1, 4)y = f(x)xy(a) y = |f(x)|xy(b) y = f(|x|)
8 marks
Show answer
a) parts below the x-axis reflected in it; meets axes at (-1, 0), (3, 0), (0, 3); maximum (1, 4) b) right half reflected in the y-axis; maxima (-1, 4) and (1, 4); meets axes at (-3, 0), (3, 0), (0, 3) c) 3 < k < 4 d) (0, 8)
9

The function h is defined by

h(x) = 5 + 8x − 2x2, x ∈ ℝ

  1. Explain why h does not have an inverse.
  2. Write h(x) in the form p − q(x − s)2, where p, q and s are constants.

The function g is defined by g(x) = 5 + 8x − 2x2, x ≤ k

  1. State the largest value of k for which g has an inverse.
  2. Using this value of k, find g−1(x), stating its domain.
7 marks
Show answer
a) h is many-to-one, e.g. h(0) = h(4) = 5 b) 13 - 2(x - 2)2 c) k = 2 d) g-1(x) = 2 - √(13 - x2), x ≤ 13
10

The functions f and g are defined by

f(x) = 2x + 3, x ∈ ℝ

g(x) = 6x + 1, x > −1

  1. Show that gf(x) = 3x + 2
  2. Explain why gf(x) is only defined for x > −2, and state the range of gf.
  3. Show that the equation fg(x) = gf(x) has no solutions.
8 marks
Show answer
a) gf(x) = 3x + 2 b) f(x) must be in the domain of g: 2x + 3 > -1, so x > -2; range gf(x) > 0 c) the equation reduces to (x + 3)2 = 0, and x = -3 is not in the domain
11

The function f is defined by

f(x) = 2k − |x − k|, x ∈ ℝ

where k is a positive constant.

  1. Sketch the graph of y = f(x), stating, in terms of k, the coordinates of the maximum point and of the points where the graph meets the coordinate axes.
  2. Solve, in terms of k, the equation f(x) = 12x
  3. Hence find, in terms of k, the set of values of x for which f(x) > 12x
xy
9 marks
Show answer
a) inverted V, maximum (k, 2k), meets the axes at (-k, 0), (3k, 0) and (0, k) b) x = -2k or x = 2k c) -2k < x < 2k
12

Mia makes a cup of coffee. Its temperature, T °C, t minutes after it is made is modelled by

T = f(t) = 20 + 70e−0.04t, t ≥ 0

  1. State the range of f.
  2. Find f−1(T), stating its domain, and find the value of f−1(50), explaining what it means in this context.

A temperature of T °C is F °F where F = g(T) = 1.8T + 32

  1. Find gf(t) in its simplest form, and hence find the time at which the coffee is at 140 °F.
9 marks
Show answer
a) 20 < f(t) ≤ 90 b) f-1(T) = 25 ln(70T - 20), 20 < T ≤ 90; f-1(50) = 21.2: the coffee cools to 50 °C after 21.2 minutes c) gf(t) = 68 + 126e-0.04t; t = 14.0 minutes

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Every question here is original and written by Robinson Tuition. It follows the Edexcel AS specification and suits AQA and OCR AS maths too; this is not an exam-board paper.

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