Edexcel International A Level Maths · P3 · Pure Mathematics 3
Modulus and transformations of graphs
28 free, high-quality past-paper-style questions on Modulus and transformations of graphs 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 · 244 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-2-B-01P3 · A2Easy5 marks
Consider the modulus function f(x) = |5 − 2x|, defined for all real values of x.
(a)
Sketch the graph of y = f(x), showing the coordinates of the points where the graph meets the coordinate axes.
(2)
(b)
Solve f(x) = 7.
(2)
(c)
Hence solve f(x) < 7.
(1)
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Answer
Mark
Notes
Part (a)
V shape with its vertex on the x-axis
B1
vertex (5⁄2, 0) and y-intercept (0, 5)
B1
Part (b)
5 − 2x = 7 or 5 − 2x = −7
M1
x = −1 or x = 6
A1
Part (c)
−1 < x < 6, ft their critical values
B1
Model sketch
Figure: sketch of y = |5 − 2x|
Exam tip: A modulus less than a positive number gives one interval between the two solutions of the equation; a sketch with the line y = 7 shows this at a glance.
The curve C has equation y = f(x). C has a maximum point at P(2, 6), meets the y-axis at Q(0, 4) and has the line y = −1 as an asymptote. Write down the coordinates or equations asked for in each part.
Figure: one possible curve y = f(x); not to scale
(a)
For the curve y = 3f(x) − 2, the image of P and the equation of the asymptote.
(2)
(b)
For the curve y = f(½x), the images of P and of Q.
(2)
(c)
For the curve y = f(−x) + 5, the maximum point and the equation of the asymptote.
(2)
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Answer
Mark
Notes
Part (a)
(2, 16)
B1
y = −5
B1
Part (b)
(4, 6)
B1
(0, 4)
B1
Part (c)
(−2, 11)
B1
y = 4
B1
Common error: Applying the translation before the stretch in y = 3f(x) − 2. Multiply the y-coordinate by 3 first, then subtract 2: 6 → 18 → 16.
A quadratic function g is given by g(x) = x² − 6x + 5 for every real number x. The part of its graph for x ≥ 0 is kept and also reflected in the y-axis to form the curve y = g(|x|).
(a)
Sketch the graph of y = g(|x|), giving the coordinates of the points where it meets the coordinate axes and of its turning points.
(4)
(b)
Find the set of values of k for which the equation g(|x|) = k has exactly four solutions.
(2)
(c)
Solve g(|x|) = −3.
(1)
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Answer
Mark
Notes
Part (a)
for x ≥ 0 the graph is y = g(x), reflected in the y-axis: a W shape symmetrical about the y-axis
B1
meets the x-axis at (±1, 0) and (±5, 0)
B1
meets the y-axis at (0, 5)
B1
minimum points (±3, −4)
B1
Part (b)
lower limit −4 from the minimum points
B1
upper limit 5 from the y-intercept: −4 < k < 5
B1
Part (c)
x² − 6x + 8 = 0 for x ≥ 0 gives 2 and 4, so x = ±2, ±4
B1
Model sketch
Figure: sketch of y = g(|x|), symmetric about the y-axis
Exam tip: y = g(|x|) is always symmetrical about the y-axis: work with x ≥ 0 only, then add the mirror images.
Figure 1 shows the graph of y = |ax + b|, where a and b are positive constants, and the straight line L with equation y = 2x + 9. The two graphs meet at the points P and Q.
(a)
Using Figure 1, find the values of a and b.
(2)
(b)
Solve |ax + b| = 2x + 9 to find the x-coordinates of P and Q.
(3)
(c)
Hence solve |ax + b| > 2x + 9.
(1)
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Answer
Mark
Notes
Part (a)
the graph meets the y-axis at (0, 6), so b = 6
B1
the vertex is at (−2, 0), so −2a + 6 = 0 and a = 3
B1
Part (b)
3x + 6 = 2x + 9 gives x = 3
M1
−(3x + 6) = 2x + 9 gives 5x = −15, x = −3
M1
2x + 9 > 0 for both values, so both stand: P: x = −3, Q: x = 3
A1
Part (c)
x < −3 or x > 3, ft their critical values
B1
Exam tip: The figure shows two crossing points, so two valid roots are expected. Check each root in the right-hand side: a modulus can only equal a value that is not negative.
Figure 1 shows the curve y = f(x), which has a maximum point A and a minimum point B. It also shows, dashed, the curve y = g(x), where g(x) = pf(qx) for non-zero constants p and q. The curve y = g(x) has a maximum point A′ and a minimum point B′, the images of A and B. The coordinates of A, B, A′ and B′ are shown in Figure 1.
(a)
Find the values of p and q.
(2)
(b)
Hence describe fully a sequence of two transformations that maps the curve y = f(x) onto the curve y = g(x).
(2)
(c)
The curve y = f(x) is mapped onto the curve y = 3 − f(x). Write down the coordinates of the images of A and B, stating whether each is a maximum or a minimum point.
y-coordinates are doubled (4 → 8, −2 → −4), so p = 2
B1
Part (b)
reflection in the y-axis
B1
stretch parallel to the y-axis, scale factor 2 (either order)
B1
Part (c)
(−1, −1), a minimum point
B1
(3, 5), a maximum point
B1
Common error: Taking q = 1 and p = −2 because the curves look like reflections of each other. A reflection in the x-axis would send the maximum A to a minimum; here the maximum stays a maximum but moves to the other side of the y-axis.
Figure 1 shows three graphs, labelled A, B and C. Their equations, in a different order, are y = |2x − 4|, y = 2|x| − 4, y = |2|x| − 4|.
(a)
Match each equation to its graph, giving a reason for each choice.
(3)
(b)
Using the graph of y = |2|x| − 4|, find the set of values of k for which the equation |2|x| − 4| = k has exactly four solutions.
(2)
(c)
Solve |2|x| − 4| = 3.
(2)
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Answer
Mark
Notes
Part (a)
y = 2|x| − 4 is Graph A: symmetrical about the y-axis and takes negative values
B1
y = |2|x| − 4| is Graph B: symmetrical and never negative
B1
y = |2x − 4| is Graph C: a single V with vertex on the x-axis at (2, 0), not symmetrical about the y-axis
B1
Part (b)
the line y = k must lie above the x-axis and below the peak at (0, 4)
M1
0 < k < 4
A1
Part (c)
2|x| − 4 = 3 or 2|x| − 4 = −3, so |x| = 7⁄2 or |x| = ½
M1
x = ±7⁄2, x = ±½
A1
Exam tip: y = |f(x)| can never go below the x-axis, while y = f(|x|) is always symmetrical about the y-axis. Checking these two features sorts most modulus graphs quickly.
Figure 1 shows the graph of y = |4 sin x| for 0° ≤ x ≤ 360°, together with the line y = 2.
(a)
Write down the coordinates of the maximum points of the graph of y = |4 sin x| for 0° ≤ x ≤ 360°, and state the period of the function |4 sin x|.
(2)
(b)
Solve |4 sin x| = 2 for 0° ≤ x ≤ 360°.
(3)
(c)
Hence, using Figure 1, solve |4 sin x| > 2 for 0° ≤ x ≤ 360°.
(1)
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Answer
Mark
Notes
Part (a)
(90°, 4) and (270°, 4)
B1
period 180°
B1
Part (b)
sin x = ½ or sin x = −½
M1
x = 30°, 150°
A1
x = 210°, 330°
A1
Part (c)
30° < x < 150° or 210° < x < 330°
B1
Common error: Solving 4 sin x = 2 only and stopping at 30° and 150°. The part of the curve below the axis has been reflected, so 4 sin x = −2 gives two more solutions.
A hyperbola-shaped curve has equation y = f(x), where f(x) = 3 + 4⁄x − 2 and x ≠ 2. Two transformations of this curve are then considered.
(a)
State the equations of the asymptotes of y = f(x) and find the coordinates of the points where the curve meets the coordinate axes.
(3)
(b)
The curve y = f(x) is transformed to the curve y = ½f(−x). State the equations of its asymptotes and the coordinates of the points where it meets the coordinate axes.
(3)
(c)
The curve y = f(x) is transformed to the curve y = f(2x) + 1. State the equations of its asymptotes and the coordinates of the point where it meets the y-axis.
(2)
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Answer
Mark
Notes
Part (a)
x = 2 and y = 3
B1
f(0) = 3 − 2 = 1: (0, 1)
B1
4⁄x − 2 = −3 gives x − 2 = −4⁄3: (2⁄3, 0)
B1
Part (b)
x = −2 and y = 3⁄2
B1
(0, ½), ft
B1
(−2⁄3, 0), ft
B1
Part (c)
x = 1 and y = 4
B1
f(0) + 1: (0, 2)
B1
Model sketch
Figure: y = f(x) with its asymptotes
Exam tip: Treat each asymptote like a point: x = 2 goes through the x-rule and y = 3 through the y-rule of the transformation.
In this question x is in degrees and 0 ≤ x ≤ 360. The function f is defined by f(x) = sin x. The curve C has equation y = 3f(2x) − 1.
(a)
Describe fully a sequence of transformations that maps the curve y = f(x) onto C.
(3)
(b)
Find the coordinates of the maximum and minimum points of C.
(3)
(c)
Find the number of solutions of |3 sin 2x − 1| = 2 in the interval 0 ≤ x ≤ 360, giving a reason.
(2)
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Answer
Mark
Notes
Part (a)
stretch parallel to the x-axis, scale factor ½
B1
stretch parallel to the y-axis, scale factor 3
B1
then translation by the vector (0, −1), applied after the vertical stretch
B1
Part (b)
greatest value 3 − 1 = 2 and least value −3 − 1 = −4
B1
maxima (45, 2) and (225, 2)
B1
minima (135, −4) and (315, −4)
B1
Part (c)
3 sin 2x − 1 = 2 gives sin 2x = 1: 2 values (the line y = 2 touches C at its maxima); 3 sin 2x − 1 = −2 gives sin 2x = −1⁄3: 4 values (the line y = −2 cuts C twice in each period)
M1
6 solutions
A1
Model sketch
Figure: the curve C: y = 3 sin 2x − 1 for 0 ≤ x ≤ 360
Exam tip: A horizontal stretch with scale factor ½ halves the period: y = sin 2x completes two full cycles between 0 and 360.
On the same axes, sketch the graphs of y = |2x − k| and y = |x + k|, showing in terms of k the coordinates of the points where each graph meets the coordinate axes.
(3)
(b)
Solve |2x − k| = |x + k|, giving your answers in terms of k.
(3)
(c)
Hence solve |2x − k| < |x + k|.
(1)
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Answer
Mark
Notes
Part (a)
y = |2x − k|: V with vertex (k⁄2, 0) and y-intercept (0, k)
B1
y = |x + k|: V with vertex (−k, 0) and y-intercept (0, k)
B1
both pass through (0, k), with the arms of the first V steeper (gradients ±2 against ±1)
B1
Part (b)
4x² − 4kx + k² = x² + 2kx + k², or 2x − k = ±(x + k)
M1
3x² − 6kx = 0, or x − 2k = 0 and 3x = 0
A1
x = 0 or x = 2k
A1
Part (c)
0 < x < 2k
B1
Model sketch
Figure: drawn for one positive value of k
Common error: Dividing 3x² − 6kx = 0 by x and losing the solution x = 0. Factorise instead: 3x(x − 2k) = 0.
The function f is defined by f(x) = x2 + px + q, x ∈ ℝ, where p and q are constants. Figure 1 shows the graph of y = |f(x)| and the line y = 7.
(a)
Using Figure 1, find the values of p and q.
(2)
(b)
Solve |f(x)| = 7, giving your answers in exact form.
(4)
(c)
Hence solve |f(x)| < 7.
(2)
(d)
Write down the value of k for which the equation |f(x)| = k has exactly three solutions.
(1)
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Answer
Mark
Notes
Part (a)
f(x) = 0 at x = −2 and x = 4, and the coefficient of x2 is 1, so f(x) = (x + 2)(x − 4)
M1
p = −2, q = −8
A1
Part (b)
x2 − 2x − 8 = 7, so x2 − 2x − 15 = 0
M1
(x − 5)(x + 3) = 0: x = 5, x = −3
A1
−(x2 − 2x − 8) = 7, so x2 − 2x − 1 = 0
M1
x = 1 ± √2
A1
Part (c)
the graph is below the line between each outer solution and the neighbouring inner one
M1
−3 < x < 1 − √2 or 1 + √2 < x < 5
A1
Part (d)
the line must pass through the local maximum (1, 9): k = 9
B1
Exam tip: Use the figure to check the count: the line y = 7 lies below the local maximum (1, 9), so it must cut the graph four times. If your algebra gives only two answers, a case is missing.
Figure 1 shows the graph of y = |x − 3| + |x + 1|. The graph is made of three straight-line pieces, which meet at the two points marked.
(a)
Show that y = 4 for −1 ≤ x ≤ 3.
(2)
(b)
Write y without modulus signs for x > 3 and for x < −1.
(2)
(c)
Solve |x − 3| + |x + 1| = 7.
(2)
(d)
Hence solve |x − 3| + |x + 1| ≤ 7.
(1)
(e)
State the set of values of k for which the equation |x − 3| + |x + 1| = k has no solutions.
(1)
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Answer
Mark
Notes
Part (a)
for −1 ≤ x ≤ 3: |x − 3| = 3 − x and |x + 1| = x + 1
M1
y = (3 − x) + (x + 1) = 4
A1
Part (b)
for x > 3: y = (x − 3) + (x + 1) = 2x − 2
B1
for x < −1: y = (3 − x) − (x + 1) = 2 − 2x
B1
Part (c)
7 > 4, so the solutions lie on the sloping pieces: 2x − 2 = 7 or 2 − 2x = 7
M1
x = 4.5 or x = −2.5
A1
Part (d)
−2.5 ≤ x ≤ 4.5
B1
Part (e)
the least value of y is 4: k < 4
B1
Exam tip: With two moduli, split the number line at the values that make each modulus zero (here −1 and 3) and write the expression without modulus signs on each piece.
Figure 1 shows the graphs of y = |x − a| and y = |bx|, where a and b are positive constants. The graphs intersect at the points P(−2, 4) and Q.
(a)
Find the values of a and b.
(2)
(b)
Find the exact coordinates of Q.
(2)
(c)
Hence solve |x − a| > |bx|.
(2)
(d)
Find the value of the constant c for which the equation |x − a| = |bx| + c has exactly one solution.
(2)
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Answer
Mark
Notes
Part (a)
|−2 − a| = 4 with a > 0 gives a = 2
B1
|−2b| = 4 with b > 0 gives b = 2
B1
Part (b)
at Q, 0 < x < 2, so 2 − x = 2x
M1
Q(2⁄3, 4⁄3)
A1
Part (c)
the graph of y = |x − 2| is above y = 2|x| between P and Q
M1
−2 < x < 2⁄3
A1
Part (d)
|x − 2| − 2|x| has greatest value 2, at x = 0 (it equals 2 + x for x < 0 and 2 − 3x for 0 ≤ x ≤ 2)
M1
c = 2
A1
Common error: Using x − 2 = 2x for Q and getting x = −2 again. Read from the figure which arm each graph is on: at Q the first graph is on its left arm, where |x − 2| = 2 − x.
Figure 1 shows the curve y = f(x), where f(x) = ax + b⁄x + c, x ≠ −c, and a, b and c are constants. The asymptotes of the curve and the points where it meets the coordinate axes are marked.
(a)
Using Figure 1, find the values of a, b and c.
(3)
(b)
The curve y = rf(x) + s, where r and s are constants, has the line y = 1 as an asymptote and passes through the origin. Find the values of r and s.
(3)
(c)
The curve y = f(tx), where t is a positive constant, has the line x = −½ as an asymptote. Find the value of t, and write down the coordinates of the point where this curve meets the x-axis.
(2)
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Answer
Mark
Notes
Part (a)
vertical asymptote x = −2, so c = 2
B1
f(x) → a as x → ±∞ and the horizontal asymptote is y = 2, so a = 2
B1
f(0) = b/2 = −3, so b = −6 (consistent with (3, 0))
B1
Part (b)
asymptote: 2r + s = 1; through O: −3r + s = 0
M1
r = 1⁄5
A1
s = 3⁄5
A1
Part (c)
x-coordinates are divided by t: −2/t = −½, so t = 4
B1
(3⁄4, 0)
B1
Exam tip: A vertical stretch and a translation change only the y-values, so the horizontal asymptote y = 2 moves to y = 2r + s, while the vertical asymptote stays at x = −2.
Figure 1 shows the curve C with equation y = ae−x + b, where a and b are constants. The curve meets the y-axis at (0, 1) and has the line y = −2 as an asymptote.
(a)
Write down the value of b and find the value of a.
(2)
(b)
Describe fully a sequence of three transformations that maps the curve y = ex onto C.
(3)
(c)
Find the exact coordinates of the point where C meets the x-axis.
(2)
(d)
Find the set of values of k for which the equation |3e−x − 2| = k has exactly two solutions.
(2)
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Answer
Mark
Notes
Part (a)
e−x → 0 as x → ∞, so b = −2
B1
a + b = 1, so a = 3
B1
Part (b)
reflection in the y-axis
B1
stretch parallel to the y-axis, scale factor 3
B1
translation by the vector (0, −2), applied after the stretch
B1
Part (c)
3e−x = 2, so ex = 3⁄2
M1
(ln 3⁄2, 0)
A1
Part (d)
the graph of y = |3e−x − 2| falls from +∞ to 0 at x = ln 3⁄2, then rises towards the asymptote y = 2 without reaching it
M1
0 < k < 2
A1
Common error: Keeping the asymptote at y = −2 for the modulus graph. Reflecting the part below the x-axis moves the asymptote to y = 2, and the curve approaches it from below.
Two robots move at constant speeds along a straight track, each passing through a charging point C on the track. The distance of robot A from C at time t minutes is dA = |p − qt| metres, where p and q are positive constants. The distance of robot B from C is dB = |2t − 6| metres. Figure 1 shows the graphs of dA and dB against t for 0 ≤ t ≤ 8.
(a)
Using Figure 1, find the values of p and q.
(2)
(b)
Find the times at which the two robots are the same distance from C.
(3)
(c)
Find the set of times, during 0 ≤ t ≤ 8, for which robot A is closer to C than robot B is.
(2)
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Answer
Mark
Notes
Part (a)
dA = 18 when t = 0, so p = 18
B1
robot A reaches C at t = 4.5, so 18 − 4.5q = 0 and q = 4
B1
Part (b)
|18 − 4t| = |2t − 6|, so 18 − 4t = ±(2t − 6)
M1
18 − 4t = 2t − 6 gives t = 4
A1
18 − 4t = 6 − 2t gives t = 6
A1
Part (c)
|18 − 4t| < |2t − 6|: the solid graph is below the dashed graph between the crossing points
M1
4 < t < 6
A1
Exam tip: Both sides are moduli, so squaring both sides is a safe alternative: (18 − 4t)² = (2t − 6)² gives 12t² − 120t + 288 = 0, that is t² − 10t + 24 = 0.
Let f(x) = x² − 4x − 5 for real x. The straight line l has equation y = x + 1 and passes through one of the roots of f.
(a)
Sketch the graph of y = |f(x)|, giving the coordinates of the points where it meets the coordinate axes and of its local maximum point.
(3)
(b)
Solve |x² − 4x − 5| = x + 1.
(5)
(c)
Hence solve |x² − 4x − 5| < x + 1.
(2)
(d)
Find the set of values of k for which |f(x)| = k has exactly two solutions.
(2)
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Answer
Mark
Notes
Part (a)
meets the x-axis at (−1, 0) and (5, 0), with the part between them reflected upwards
B1
meets the y-axis at (0, 5)
B1
local maximum (2, 9)
B1
Part (b)
x² − 4x − 5 = x + 1, so x² − 5x − 6 = 0
M1
(x − 6)(x + 1) = 0, x = 6 or x = −1
A1
−(x² − 4x − 5) = x + 1, so x² − 3x − 4 = 0
M1
(x − 4)(x + 1) = 0, x = 4 or x = −1
A1
all roots give x + 1 ≥ 0: x = −1, 4, 6
A1
Part (c)
line above the graph only between consecutive critical values 4 and 6
M1
4 < x < 6
A1
Part (d)
k = 0
B1
k > 9
B1
Model sketch
Figure: sketch of y = |x² − 4x − 5|
Exam tip: Both graphs pass through (−1, 0), so x = −1 appears in both cases. List it once, and use the sketch to see that the line is above the curve only between 4 and 6.
The rational function f is given by f(x) = 3x − 1⁄x + 1, with domain all real numbers except −1.
(a)
Show that f(x) can be written as 3 − 4⁄x + 1.
(2)
(b)
State the equations of the asymptotes of y = f(x) and the coordinates of the points where the curve meets the coordinate axes.
(3)
(c)
The curve y = f(x) is transformed to the curve y = f(−x) − 3. State the equations of its asymptotes and the coordinates of the point where it meets the y-axis, and explain why it does not meet the x-axis.
(3)
(d)
Sketch the graph of y = f(|x|). Explain why this graph has no vertical asymptote, and state the equation of its asymptote and the coordinates of the points where it meets the coordinate axes.
(3)
(e)
Solve |f(x)| = 1.
(2)
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Answer
Mark
Notes
Part (a)
3x − 1 = 3(x + 1) − 4
M1
so f(x) = 3(x + 1) − 4⁄x + 1 = 3 − 4⁄x + 1
A1
Part (b)
x = −1 and y = 3
B1
(0, −1)
B1
(1⁄3, 0)
B1
Part (c)
x = 1 and y = 0
B1
(0, −4)
B1
the equation is y = −4⁄1 − x = 4⁄x − 1, which is never 0 (or: the x-axis is its asymptote)
B1
Part (d)
the vertical asymptote x = −1 lies in the part with x < 0, which is discarded; for x ≥ 0 the curve is continuous
B1
asymptote y = 3
B1
(0, −1) and (±1⁄3, 0)
B1
Part (e)
3x − 1 = x + 1 or 3x − 1 = −(x + 1)
M1
x = 1 or x = 0
A1
Model sketch
Figure: y = f(|x|) is symmetric about the y-axis and has no vertical asymptote
Exam tip: Writing a rational function as a constant plus a simple fraction shows its asymptotes immediately: 3 − 4/(x + 1) tends to 3, and blows up at x = −1.
An exponential curve has equation y = f(x), where f(x) = ex − 3 for all real x.
(a)
Sketch the graph of y = f(x), giving the coordinates of the points where it meets the coordinate axes and the equation of its asymptote.
(3)
(b)
Sketch the graph of y = |f(x)|, giving the coordinates of the points where it meets the coordinate axes and the equation of its asymptote.
(2)
(c)
Solve |f(x)| = 2, giving your answers in exact form.
(3)
(d)
Hence solve |f(x)| < 2.
(1)
(e)
The curve y = f(x) is transformed to the curve y = 2f(−x) + 4. State the equation of the asymptote of the new curve and show that it passes through the origin.
(2)
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Answer
Mark
Notes
Part (a)
(0, −2)
B1
(ln 3, 0)
B1
asymptote y = −3
B1
Part (b)
part for x < ln 3 reflected, with asymptote y = 3
B1
(0, 2) and a sharp corner at (ln 3, 0)
B1
Part (c)
ex − 3 = 2 or ex − 3 = −2
M1
x = ln 5
A1
x = 0
A1
Part (d)
0 < x < ln 5
B1
Part (e)
asymptote y = 2(−3) + 4 = −2
B1
at x = 0: y = 2(e0 − 3) + 4 = 0, so the curve passes through O
B1
Model sketch
Figure: sketch of y = |eˣ − 3|
Common error: Keeping the asymptote y = −3 for y = |f(x)|. The part of the curve near that asymptote is reflected, so the asymptote becomes y = 3.
The graph of y = f(x), −4 ≤ x ≤ 6, consists of three straight line segments joining the points A(−4, −2), B(−1, 4), C(3, −4) and D(6, 2), in that order.
Figure: the graph of y = f(x); not to scale
(a)
Find the coordinates of the points where the graph of y = f(x) meets the x-axis.
(3)
(b)
Sketch the graph of y = |f(x)|, labelling the coordinates of the end points, the turning points and the points on the x-axis.
(3)
(c)
The graph of y = f(x) is transformed to the graph of y = ½f(2x) + 1. State the set of x-values on which the new graph is defined, and find the images of B and C.
(3)
(d)
Find all the values of x for which |f(x)| = 2.
(3)
(e)
The graph of y = f(|x|) is defined for −6 ≤ x ≤ 6. Find the value of k for which the equation f(|x|) = k has exactly two solutions.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
AB: y = 2x + 6, so (−3, 0)
B1
BC: y = −2x + 2, so (1, 0)
B1
CD: y = 2x − 10, so (5, 0)
B1
Part (b)
parts below the x-axis reflected, with sharp corners on the axis
B1
end points (−4, 2) and (6, 2); peaks (−1, 4) and (3, 4)
B1
on the x-axis at (−3, 0), (1, 0), (5, 0)
B1
Part (c)
−2 ≤ x ≤ 3
B1
B → (−½, 3)
B1
C → (3⁄2, −1)
B1
Part (d)
f(x) = 2: x = −2, 0, 6
M1
f(x) = −2: x = −4, 2, 4
M1
x = −4, −2, 0, 2, 4, 6
A1
Part (e)
uses f on 0 ≤ x ≤ 6 (from (0, 2) down to C(3, −4) and up to (6, 2)) reflected in the y-axis
M1
k = −4 (the solutions are x = ±3)
A1
Model sketch
Figure: sketch of y = |f(x)|
Exam tip: With a graph made of line segments, find the equation of each segment once and use it for every part: intercepts, solutions and images.
Figure 1 shows the curve y = f(x), where f(x) is a cubic polynomial. The curve crosses the x-axis at the three points marked and meets the y-axis at (0, 6).
(a)
Using Figure 1, show that f(x) = ½(x3 − 2x2 − 11x + 12).
(3)
(b)
Sketch the graph of y = |f(x)|, giving the coordinates of the points where it meets the coordinate axes.
(2)
(c)
Sketch the graph of y = f(|x|), giving the coordinates of the points where it meets the coordinate axes.
(2)
(d)
Solve |f(x)| = 6, giving your answers in exact form.
(4)
(e)
Write down the coordinates of the points where the curve y = 2f(−x) meets the coordinate axes.
f(x) = −6: x3 − 2x2 − 11x + 24 = 0; x = 3 is a root, so (x − 3)(x2 + x − 8) = 0
M1
x = 3, x = −1 ± √33⁄2
A1
Part (e)
(3, 0), (−1, 0), (−4, 0) and (0, 12)
B1
Exam tip: The figure shows that the line y = 6 meets the curve three times and y = −6 meets it three times (the minimum is below −6), so |f(x)| = 6 must have six solutions. Use that count to check your algebra.
the dashed graph is above the solid graph between −4 and −1 and between 1 and 4
M1
−4 < x < −1 or 1 < x < 4
A1
Part (d)
the line must pass through the local maximum: k = 4
B1
x2 − 4 = 4 or x2 − 4 = −4
M1
x = 0, x = ±2√2
A1
Exam tip: When both sides are even functions, solve for x ≥ 0 only, where |x| = x, and then add the negatives. This halves the work and avoids sign slips.
In this question x is in radians. Figure 1 shows the graphs of y = |2 sin x − 1| and y = |sin x| for 0 ≤ x ≤ 2π.
(a)
Describe fully a sequence of two transformations that maps the curve y = sin x onto the curve y = 2 sin x − 1.
(2)
(b)
Find the exact coordinates of the points where the graph of y = |2 sin x − 1| meets the x-axis, and write down the coordinates of its two local maximum points for 0 < x < 2π.
(3)
(c)
State, with a reason, the number of solutions of the equation |2 sin x − 1| = k, for 0 ≤ x ≤ 2π, when 1 < k < 3.
(2)
(d)
Solve the equation |2 sin x − 1| = |sin x| for 0 ≤ x ≤ 2π. Give exact answers where possible and other answers to 3 significant figures.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
stretch parallel to the y-axis, scale factor 2
B1
followed by a translation by the vector (0, −1)
B1
Part (b)
sin x = ½: (π⁄6, 0) and (5π⁄6, 0)
B1
(π⁄2, 1)
B1
(3π⁄2, 3)
B1
Part (c)
for 1 < k < 3 the line y = k only meets the large arch between π and 2π, which it cuts twice (since 2 sin x − 1 = k would need sin x > 1)
M1
2 solutions
A1
Part (d)
2 sin x − 1 = sin x, so sin x = 1
M1
x = π⁄2
A1
2 sin x − 1 = −sin x, so sin x = ⅓
M1
x = 0.340, x = 2.80
A1
Exam tip: The figure shows that the two graphs touch at (π/2, 1) and cross twice near the first arch, so the equation has three solutions; none lie between π and 2π, where |2 sin x − 1| ≥ 1 ≥ |sin x| and the two are never equal.
The function f is defined by f(x) = 3 + 2x − x2, −2 ≤ x ≤ 4. Figure 1 shows the graph of y = f(x), with end points A(−2, −5) and D(4, −5) and maximum point M(1, 4).
(a)
The graph of y = f(x) is transformed to the graph of y = 2f(½x) − 1. State the set of values of x for which the new graph is defined, and find the images of M, A and D.
(3)
(b)
Sketch the graph of y = |f(x)|, giving the coordinates of its end points, its local maximum point and the points where it meets the x-axis.
(2)
(c)
Sketch the graph of y = f(|x|), stating the set of values of x for which it is defined and giving the coordinates of its maximum points, end points and the points where it meets the coordinate axes.
(3)
(d)
Solve |f(x)| = 3, giving your answers in exact form.
(3)
(e)
Find the set of values of k for which the equation f(|x|) = k has exactly two solutions.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
x-coordinates are doubled: −4 ≤ x ≤ 8
B1
M → (2, 7)
B1
A → (−4, −11), D → (8, −11)
B1
Part (b)
the two end sections below the x-axis reflected upwards, with sharp corners on the axis; end points (−2, 5) and (4, 5)
B1
local maximum (1, 4); on the x-axis at (−1, 0) and (3, 0)
B1
Part (c)
defined for −4 ≤ x ≤ 4 (|x| must not exceed 4)
B1
maxima (±1, 4), with a sharp local minimum at (0, 3)
B1
meets the x-axis at (±3, 0); end points (±4, −5)
B1
Part (d)
3 + 2x − x2 = 3 or 3 + 2x − x2 = −3
M1
x = 0, x = 2
A1
x2 − 2x − 6 = 0: x = 1 ± √7, both in −2 ≤ x ≤ 4
A1
Part (e)
k = 4 (the solutions are x = ±1)
B1
for k < 3 the line meets each outer branch once, down to the end points at y = −5
M1
−5 ≤ k < 3 (k = 3 gives three solutions)
A1
Common error: Forgetting the domain. f(|x|) needs |x| to lie in −2 ≤ |x| ≤ 4, which allows −4 ≤ x ≤ 4, and the end points (±4, −5) must be included when counting solutions.
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