Edexcel International A Level Maths · P3 · Pure Mathematics 3
Algebraic fractions and functions
28 free, high-quality past-paper-style questions on Algebraic fractions and functions for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 241 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-1-B-01P3 · A2Easy6 marks
Simplify each of the following fully. In each case, assume that x takes only values for which every denominator is non-zero.
(a)
(x2 − 16) / (x2 + x − 12)
(2)
(b)
(6x2 − x − 2) / (9x2 − 4)
(2)
(c)
4x/(x2 − 9) ÷ 2x2/(x + 3)
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x − 4)(x + 4) / ((x + 4)(x − 3))
M1
= (x − 4)/(x − 3)
A1
Part (b)
(2x + 1)(3x − 2) / ((3x − 2)(3x + 2))
M1
= (2x + 1)/(3x + 2)
A1
Part (c)
multiplies by the reciprocal and factorises: 4x/((x − 3)(x + 3)) × (x + 3)/(2x2)
M1
= 2/(x(x − 3)) (accept 2/(x2 − 3x))
A1
Common error: Cancelling terms such as the x² in (x² − 16)/(x² + x − 12). Only factors of the whole numerator and the whole denominator can be cancelled, so factorise first.
Exam tip: Factorise every denominator before choosing a common denominator: x² − 4 already contains (x − 2), so the lowest common denominator is just (x − 2)(x + 2).
Figure 1 shows mapping diagrams for two functions, f and g. The domain of f is {−1, 0, 1, 2, 3} and the domain of g is {−1, 0, 3}. Each arrow joins an input to its output.
(a)
State the range of f, and explain why f does not have an inverse.
(2)
(b)
Find gf(2).
(2)
(c)
Explain why fg(−1) is not defined.
(1)
(d)
The function g has an inverse. Write down the domain of g−1 and the value of g−1(4).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
range of f: {−1, 0, 3}
B1
f is many-one: for example 0 and 2 are both mapped to 0, so an inverse could not send 0 back to a single input
B1
Part (b)
f(2) = 0
M1
g(0) = 4, so gf(2) = 4
A1
Part (c)
g(−1) = 5, and 5 is not in the domain of f, so f cannot be applied to it
B1
Part (d)
domain of g−1: {1, 4, 5}
B1
g−1(4) = 0
B1
Common error: Giving every number in the output oval (including 1 and 8) as the range. The range contains only the outputs that an arrow actually reaches.
Figure 1 shows a pyramid of algebraic fractions. Each brick is equal to the product of the two bricks directly below it. Assume that x takes only values for which every denominator is non-zero (x ≠ 0, ±2, −3).
(a)
Find, in its simplest form, the left-hand brick of the middle row.
(2)
(b)
Find, in its simplest form, the right-hand brick of the middle row.
(2)
(c)
Show that the top brick is equal to the middle brick of the bottom row.
The function f is defined by f(x) = (3x − 1)/(x + 2), x > −2.
(a)
Show that f(x) = 3 − 7/(x + 2), and hence find the range of f.
(3)
(b)
Find f−1(x), stating its domain.
(3)
(c)
Show that the equation f(x) = x has no real solutions, and state what this tells you about the graphs of y = f(x) and y = f−1(x).
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
3x − 1 = 3(x + 2) − 7, giving the result
B1
for x > −2, x + 2 takes all positive values, so 7/(x + 2) takes all positive values
M1
f(x) < 3
A1
Part (b)
y(x + 2) = 3x − 1 ⇒ x(y − 3) = −1 − 2y
M1
f−1(x) = (2x + 1)/(3 − x)
A1
domain x < 3 (ft their range)
B1
Part (c)
3x − 1 = x(x + 2) ⇒ x2 − x + 1 = 0
M1
discriminant 1 − 4 = −3 < 0, so no real solutions
A1
f is increasing, so any intersection of the two graphs would lie on y = x; hence the graphs of y = f(x) and y = f−1(x) do not meet
B1
Model sketch
Figure: y = f(x) and y = f⁻¹(x) with the line y = x; neither curve meets y = x
Exam tip: Splitting an improper fraction into a constant plus a proper fraction shows the asymptote at once, and the asymptote is usually the boundary of the range.
A cyclist rides from A to B and then from B to C along the route shown in Figure 1. On each stage the speed is constant, with the value shown, where x > 0.
(a)
Show that the total time taken, T hours, is given by T = (50x + 72)/(x(x + 4)).
(3)
(b)
Given that the whole journey takes 3 hours 30 minutes, find the value of x.
(3)
(c)
Show that the average speed for the whole journey is 25x(x + 4)/(25x + 36) km h−1, and find its value, to 3 significant figures, for your value of x.
Figure 1 shows the graph of y = f(x), −2 ≤ x ≤ 4, which consists of two straight line segments, and the graph of y = g(x), −3 ≤ x ≤ 3, which is a single straight line segment. The labelled points are the end points of the segments.
(a)
State the range of f.
(1)
(b)
Find fg(−1).
(2)
(c)
Solve the equation f(x) = 3.
(2)
(d)
Explain why f does not have an inverse.
(1)
(e)
The function h is defined by h(x) = f(x), 1 ≤ x ≤ 4. Find h−1(x), stating its domain.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
−2 ≤ f(x) ≤ 4
B1
Part (b)
g(−1) = 2 (from the graph, or g(x) = 1 − x)
M1
f(2) = 3, so fg(−1) = 3
A1
Part (c)
finds the equation of a segment and solves: 2x + 2 = 3 gives x = ½
M1
5 − x = 3 gives x = 2: x = ½ and x = 2
A1
Part (d)
f is many-one: by part (c), f(½) = f(2) = 3
B1
Part (e)
h(x) = 5 − x, so h−1(x) = 5 − x
B1
domain 1 ≤ x ≤ 4
B1
Exam tip: For a graph made of straight segments, find the equation of each segment from its labelled end points before solving. A horizontal line can cross more than one segment.
Figure 1 shows the graph of y = f(x), x ∈ ℝ, where f(x) = p − q(x − r)2 and p, q and r are positive constants. The curve has a maximum point V(1, 7) and passes through the points A(3, −1) and B(−1, −1).
(a)
Find the values of p, q and r.
(2)
(b)
Use the points A and B to explain why f does not have an inverse.
(1)
(c)
The function g is defined by g(x) = f(x), x ≥ 1. Find g−1(x), stating its domain.
(4)
(d)
The curve y = g(x) meets the line y = x at the point Q. Find the coordinates of Q.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the maximum point V(1, 7) gives r = 1 and p = 7
B1
A: −1 = 7 − 4q, so q = 2
B1
Part (b)
f(−1) = f(3) = −1: two inputs give the same output, so f is many-one
B1
Part (c)
y = 7 − 2(x − 1)2 ⇒ (x − 1)2 = (7 − y)/2
M1
x − 1 = +√((7 − y)/2), the positive root because x ≥ 1
M1
g−1(x) = 1 + √((7 − x)/2)
A1
domain x ≤ 7
B1
Part (d)
7 − 2(x − 1)2 = x ⇒ 2x2 − 3x − 5 = 0
M1
(2x − 5)(x + 1) = 0 and x ≥ 1, so Q is (5/2, 5/2)
A1
Exam tip: Read the maximum point straight into p − q(x − r)²: the vertex (r, p) gives two constants at once, and any other labelled point then gives q.
Figure 1 shows the curve with equation y = (ax + b)/(x + c), x ≠ −c, where a, b and c are constants. The asymptotes of the curve are x = −1 and y = 2, and the curve passes through the points (0, −3) and (1.5, 0).
(a)
Write down the values of a and c, and find the value of b.
(2)
(b)
The function f is defined by f(x) = (ax + b)/(x + c), x < −1. By writing f(x) in the form 2 + k/(x + 1), find the range of f.
(3)
(c)
Write down the equations of the asymptotes of the curve y = f−1(x).
(2)
(d)
Without finding an expression for f−1(x), solve the equation f−1(x) = −6.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
c = 1 and a = 2, from the asymptotes x = −1 and y = 2
B1
(0, −3) gives b/c = −3, so b = −3
B1
Part (b)
2x − 3 = 2(x + 1) − 5, so k = −5
B1
for x < −1, x + 1 < 0, so −5/(x + 1) takes every positive value
M1
f(x) > 2
A1
Part (c)
x = 2
B1
y = −1
B1
Part (d)
f−1(x) = −6 means x = f(−6)
M1
x = (−12 − 3)/(−6 + 1) = 3
A1
Exam tip: Reflecting in y = x swaps x and y, so a vertical asymptote x = k of y = f(x) becomes a horizontal asymptote y = k of y = f⁻¹(x), and the other way round.
Figure 1 shows a water tank with two inlet pipes, A and B, and an outlet pipe, C. The times shown are for each pipe working on its own, where x > 0. Each pipe works at a constant rate.
(a)
Show that, with A and B both open and C closed, the fraction of the tank filled in one hour is (2x + 3)/(x(x + 3)).
(2)
(b)
Starting with the tank empty, A and B together, with C closed, fill the tank in exactly 2 hours. Find the value of x.
(3)
(c)
Show that, with all three pipes open, the fraction of the tank filled in one hour is (x2 + 2x + 3)/(x(x + 1)(x + 3)). Hence find how long it takes to fill the empty tank with all three pipes open.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
1/x + 1/(x + 3)
M1
= (x + 3 + x)/(x(x + 3)) = (2x + 3)/(x(x + 3))
A1
Part (b)
(2x + 3)/(x(x + 3)) = 1/2
M1
x2 + 3x = 4x + 6, so x2 − x − 6 = 0
A1
(x − 3)(x + 2) = 0, so x = 3
A1
Part (c)
1/x + 1/(x + 3) − 1/(x + 1) over the common denominator x(x + 1)(x + 3)
at x = 3 the fraction is 18/72 = 1/4, so the tank fills in 4 hours
A1
Common error: Adding the times instead of the rates. Each pipe contributes 1 ÷ (its time) of the tank per hour, and an outlet pipe contributes a negative rate.
Let f(x) = x2 − 4x for x ≥ 2, and let g(x) = 3/(x + 1) + 2 for x ≥ 0.
(a)
State the range of f.
(1)
(b)
Find f−1(x) and state its domain.
(4)
(c)
Find the range of g.
(2)
(d)
Explain why the composite function fg exists, and show that fg(x) = 9/(x + 1)2 − 4.
(3)
(e)
Solve the equation fg(x) = 5.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(x) ≥ −4
B1
Part (b)
y = (x − 2)2 − 4
M1
x − 2 = +√(y + 4), the positive root because x ≥ 2
M1
f−1(x) = 2 + √(x + 4)
A1
domain x ≥ −4
B1
Part (c)
x ≥ 0 ⇒ x + 1 ≥ 1 ⇒ 0 < 3/(x + 1) ≤ 3
M1
2 < g(x) ≤ 5
A1
Part (d)
the range of g (2 < g(x) ≤ 5) lies inside the domain of f (x ≥ 2)
B1
fg(x) = (g(x) − 2)2 − 4, or substitutes g(x) into x2 − 4x
M1
= (3/(x + 1))2 − 4 = 9/(x + 1)2 − 4
A1
Part (e)
9/(x + 1)2 = 9 ⇒ (x + 1)2 = 1 ⇒ x + 1 = ±1
M1
x = 0 only (x = −2 is rejected because x ≥ 0)
A1
Exam tip: Before forming fg, check that every output of g is an allowed input of f. Comparing the range of g with the domain of f is the whole explanation.
The function f is defined by f(x) = (x2 + 3)/4, x ≥ 0.
(a)
State the range of f.
(1)
(b)
Find f−1(x), stating its domain.
(3)
(c)
Describe the geometrical relationship between the graphs of y = f(x) and y = f−1(x).
(1)
(d)
Given that f is an increasing function, so that any point where the two graphs meet lies on y = x, find the coordinates of the points where the graphs of y = f(x) and y = f−1(x) meet.
(4)
(e)
Find the exact value of f−1f−1(7).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(x) ≥ 3/4
B1
Part (b)
x2 = 4y − 3
M1
f−1(x) = √(4x − 3), the positive root because x ≥ 0
A1
domain x ≥ 3/4
B1
Part (c)
each is the reflection of the other in the line y = x
B1
Part (d)
(x2 + 3)/4 = x
M1
x2 − 4x + 3 = 0
A1
(x − 1)(x − 3) = 0, x = 1 or 3, both in the domains
A1
(1, 1) and (3, 3)
A1
Part (e)
f−1(7) = √25 = 5
M1
f−1(5) = √17
A1
Model sketch
Figure: y = f(x) and y = f⁻¹(x) meet on the line y = x
Common error: Squaring f⁻¹ twice or working out f⁻¹(7) and then f(5). The composite f⁻¹f⁻¹ applies the inverse twice, starting from the inside.
t + 1/t − 2 = (t − 1)2/t ≥ 0, since (t − 1)2 ≥ 0 and t > 0
M1
so t + 1/t ≥ 2 and h(x) ≥ 3
A1
equality when t = 1, i.e. x = −1
B1
Part (c)
finds two inputs with the same output, e.g. h(0) = 7/2
M1
and h(−3/2) = 7/2, so h is many-one and has no inverse
A1
Part (d)
gh(x) = 2h(x) − 6
M1
= 2x + 2/(x + 2)
A1
= (2x2 + 4x + 2)/(x + 2) = 2(x + 1)2/(x + 2)
A1
Part (e)
h(x) ≥ 3 so gh(x) = 2h(x) − 6 ≥ 0: gh(x) ≥ 0
B1
Model sketch
Figure: y = h(x) for x > −2 has least value 3 at x = −1
Exam tip: A minimum value with an equality case proves the range and shows many-one at the same time: values just above the minimum are reached on both sides of x = −1.
The function f is defined by f(x) = 2x + k, x ∈ ℝ, where k is a constant. Figure 1 shows the curves y = f(x) and y = f−1(x) and the line y = x. The curve y = f(x) passes through (0, −1) and A(3, 6), and has the asymptote y = −2.
(a)
Show that k = −2, and state the range of f.
(2)
(b)
Find f−1(x), stating its domain.
(3)
(c)
Write down the coordinates of the point on y = f−1(x) that is the reflection of A in the line y = x.
(1)
(d)
Find the exact value of x for which ff(x) = 14.
(3)
(e)
The function g is defined by g(x) = x2 + 1, x ∈ ℝ. Solve the equation fg(x) = 30.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(0) = 1 + k = −1, so k = −2
B1
f(x) > −2
B1
Part (b)
y = 2x − 2 ⇒ 2x = y + 2
M1
f−1(x) = log2(x + 2)
A1
domain x > −2
B1
Part (c)
(6, 3)
B1
Part (d)
f(u) = 14 ⇒ 2u = 16 ⇒ u = 4
M1
so f(x) = 4
A1
2x = 6, x = log2 6
A1
Part (e)
fg(x) = 2x² + 1 − 2
M1
2x² + 1 = 32, so x2 + 1 = 5
A1
x = ±2
A1
Exam tip: To solve ff(x) = c, peel off one f at a time: first find the value u with f(u) = c, then solve f(x) = u.
Figure 1 shows part of the curve C with equation y = (2x3 − x2 + 3)/(x2 − 1), x ≠ ±1. The dashed lines are the asymptotes of C: the line x = 1 and the line L. The hollow circle at P marks a single point that is missing from C.
(a)
Express (2x3 − x2 + 3)/(x2 − 1) in the form Ax + B + (Dx + E)/(x2 − 1), where A, B, D and E are constants.
(3)
(b)
Hence show that, for x ≠ ±1, y = 2x − 1 + 2/(x − 1).
(2)
(c)
Find the coordinates of P, and state the equation of L.
(3)
(d)
Show that the line y = 5 meets C at exactly one point, and find the coordinates of this point.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
divides by x2 − 1 (or compares coefficients); first term 2x: 2x3 − x2 + 3 − 2x(x2 − 1) = −x2 + 2x + 3
M1
next term −1: −x2 + 2x + 3 + (x2 − 1) = 2x + 2
A1
2x − 1 + (2x + 2)/(x2 − 1), so A = 2, B = −1, D = 2, E = 2
A1
Part (b)
2x + 2 = 2(x + 1) and x2 − 1 = (x − 1)(x + 1)
M1
cancels (x + 1), allowed because x ≠ −1, to give 2/(x − 1)
A1
Part (c)
P is at x = −1, where the original fraction is undefined
a single repeated root x = 2 (an allowed value), so the line touches C only at (2, 5)
A1
Common error: Treating P as a point of the curve. Cancelling (x + 1) gives a simpler formula, but the original fraction is still undefined at x = −1, so C has a gap there.
Figure 1 shows the graph of y = f(x), where f(x) = x3 − 3x, −2 ≤ x ≤ 3. The graph has a maximum point at (−1, 2) and a minimum point at (1, −2), and its end points are A(−2, −2) and B(3, 18).
(a)
Find the range of f.
(2)
(b)
Explain why f does not have an inverse.
(1)
(c)
The function g is defined by g(x) = f(x), k ≤ x ≤ 3. State the least value of k for which g has an inverse, and the domain of g−1 for this value of k.
(2)
(d)
With this value of k, find g−1(2).
(3)
(e)
The function h is defined by h(x) = 6/(x + 4), x > −4. Explain why the composite function hg exists, and find the range of hg.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
least value −2 (at A and at the minimum point) and greatest value 18 (at B)
M1
−2 ≤ f(x) ≤ 18
A1
Part (b)
f(−2) = −2 and f is also −2 at x = 1, so f is many-one
B1
Part (c)
k = 1
B1
domain of g−1: −2 ≤ x ≤ 18
B1
Part (d)
g−1(2) is the solution of x3 − 3x = 2 with 1 ≤ x ≤ 3
M1
x3 − 3x − 2 = (x − 2)(x + 1)2 = 0
M1
x = −1 is outside the domain of g, so g−1(2) = 2
A1
Part (e)
the range of g, −2 ≤ g(x) ≤ 18, lies inside the domain of h (x > −4)
B1
h is decreasing, with h(−2) = 3 and h(18) = 6/22 = 3/11
M1
3/11 ≤ hg(x) ≤ 3
A1
Exam tip: Read the least k from the graph: the restricted piece must not contain a turning point inside it, so it starts at the minimum point at x = 1.
Figure 1 shows a square ABCD with sides of length 6 cm. The point P lies on AB with AP = x cm and the point Q lies on BC with BQ = 2x cm. P may coincide with A, and Q may coincide with C. The area of triangle DPQ is f(x) cm2.
(a)
Show that f(x) = x2 − 3x + 18.
(3)
(b)
State the domain of f, giving a reason.
(1)
(c)
Find the range of f.
(3)
(d)
Find the values of x for which the area of triangle DPQ is 16 cm2, and hence explain why f does not have an inverse.
(3)
(e)
The function g is defined by g(x) = f(x), 0 ≤ x ≤ 3/2. Find g−1(x), stating its domain.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
subtracts the three corner triangles from the square: ADP = 3x, PBQ = ½(6 − x)(2x) = x(6 − x)
M1
QCD = ½ × 6 × (6 − 2x) = 18 − 6x
A1
f(x) = 36 − 3x − 6x + x2 − 18 + 6x = x2 − 3x + 18
A1
Part (b)
0 ≤ x ≤ 3, because BQ = 2x cannot be more than BC = 6
B1
Part (c)
f(x) = (x − 3/2)2 + 63/4
M1
least value 63/4 at x = 3/2, and f(0) = f(3) = 18
A1
63/4 ≤ f(x) ≤ 18
A1
Part (d)
x2 − 3x + 18 = 16 ⇒ x2 − 3x + 2 = 0
M1
(x − 1)(x − 2) = 0, so x = 1 or x = 2, both in the domain
A1
two different values of x give the same area, so f is many-one and has no inverse
B1
Part (e)
y = (x − 3/2)2 + 63/4 ⇒ x − 3/2 = −√(y − 63/4), the negative root because x ≤ 3/2
M1
g−1(x) = 3/2 − √(x − 63/4)
A1
domain 63/4 ≤ x ≤ 18
B1
Common error: Forgetting that a domain set by a diagram has an upper limit as well as a lower one. Q must stay on BC, which caps x at 3 and makes f(0) and f(3) the end values of the range.
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