Edexcel International A Level Maths · P3 · Pure Mathematics 3
Numerical methods
28 free, high-quality past-paper-style questions on Numerical methods 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 · 242 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-8-B-01P3 · A2Easy5 marks
f(x) = x4 − 3x − 5. The equation f(x) = 0 has a positive root α.
(a)
Show that α lies in the interval [1.7, 1.8].
(2)
(b)
Show that α = 1.795 correct to 3 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(1.7) = −1.75 and f(1.8) = 0.0976 (awrt 0.1), both attempted
M1
change of sign and f continuous (polynomial), so 1.7 < α < 1.8
A1
Part (b)
chooses the interval [1.7945, 1.7955]
M1
f(1.7945) = −0.0136 and f(1.7955) = 0.00652 (1 s.f. or better, correct signs)
A1
change of sign and f continuous, so 1.7945 < α < 1.7955 and α = 1.795 to 3 d.p.
A1
Exam tip: To prove a root to 3 d.p., test half a unit of the third decimal place either side: 1.7945 and 1.7955, not 1.794 and 1.796.
f(x) = 3 tan(x/2) + 4, where x is in radians and 0 < x < 2π, x ≠ π.
(a)
Find the values of f(3) and f(3.5), giving each to 3 significant figures.
(2)
(b)
A student says that, because of these values, the equation f(x) = 0 has a root between 3 and 3.5. Explain why the student is wrong.
(2)
(c)
Find the root of f(x) = 0 in the given range, giving your answer to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(3) = 3 tan 1.5 + 4 = 46.3
B1
f(3.5) = 3 tan 1.75 + 4 = −12.6
B1
Part (b)
tan(x/2) is undefined at x/2 = π/2, i.e. at x = π ≈ 3.14, which lies between 3 and 3.5
B1
so f is not continuous on [3, 3.5]: the graph has a vertical asymptote there and the sign changes without f being zero
B1
Part (c)
tan(x/2) = −4/3 with x/2 in (π/2, π): x/2 = π − arctan(4/3) = 2.214…
M1
x = 4.43
A1
Model sketch
Figure: the sign change between 3 and 3.5 is caused by the asymptote x = π, not a root
Common error: Quoting “change of sign, so there is a root” for a function with an asymptote in the interval. The change-of-sign test needs f to be continuous on the whole interval.
A cubic curve has equation y = f(x), where f(x) = x3 − 3x2 + 4. A student tests the interval [1.5, 2.5] for a root of f(x) = 0 by looking for a change of sign.
(a)
Find f(1.5) and f(2.5).
(2)
(b)
Show that f(x) = (x − 2)2(x + 1).
(2)
(c)
Hence explain why f(1.5) and f(2.5) have the same sign even though f(x) = 0 has a root between 1.5 and 2.5.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(1.5) = 0.625
B1
f(2.5) = 0.875
B1
Part (b)
(x − 2)2 = x2 − 4x + 4, multiplied by (x + 1)
M1
x3 − 4x2 + 4x + x2 − 4x + 4 = x3 − 3x2 + 4
A1
Part (c)
x = 2 is a repeated (double) root, so the curve touches the x-axis at x = 2 without crossing it
B1
(x − 2)2 ≥ 0 and x + 1 > 0 on [1.5, 2.5], so f ≥ 0 throughout: no change of sign occurs
B1
Model sketch
Figure: the curve touches the x-axis at x = 2, so there is no change of sign
Common error: Concluding “no change of sign, so no root”. A repeated root, or two roots close together, gives the same sign at both ends of the interval.
change of sign and f continuous for x ≥ 0, so 1.9 < α < 2.0
A1
Part (c)
chooses the interval [1.895, 1.905]
M1
f(1.895) = −0.0324 and f(1.905) = 0.00924 (correct signs, 1 s.f. or better)
A1
change of sign and f continuous, so 1.895 < α < 1.905 and α = 1.90 to 2 d.p.
A1
Exam tip: A graph tells you where to look, but it is never the proof. Read a starting interval from the grid, then confirm it with a sign change of a function you have defined.
f(x) = (x2 − 5)/(2x − 3), x ≥ 0, for all values of x where f is defined. Figure 1 shows the graph of y = f(x). The points A and B on the graph have x-coordinates 1 and 2, the graph crosses the x-axis at R, and the dashed line is a vertical asymptote.
(a)
Write down the equation of the asymptote.
(1)
(b)
Find f(1) and f(2).
(2)
(c)
Explain, with reference to Figure 1, why the change of sign between A and B does not show that f(x) = 0 has a root between x = 1 and x = 2.
(2)
(d)
Show that the x-coordinate of R lies in the interval [2.2, 2.3].
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
2x − 3 = 0, so x = 1.5
B1
Part (b)
f(1.0) = (1 − 5)/(2 − 3) = 4
B1
f(2.0) = (4 − 5)/(4 − 3) = −1
B1
Part (c)
f is not continuous on [1, 2]: it is undefined at x = 1.5, where the graph has its asymptote
B1
the curve goes off to +∞ before x = 1.5 and returns from −∞ after it, so it never meets the x-axis between A and B (the numerator x2 − 5 < 0 throughout [1, 2])
B1
Part (d)
f(2.2) = −0.114 and f(2.3) = 0.181
M1
change of sign and f continuous on [2.2, 2.3] (the denominator is not zero there), so R lies between 2.2 and 2.3
A1
Exam tip: Before trusting a sign change, look for any value of x that makes a denominator zero. If one lies inside the interval, the test tells you nothing about roots there.
Figure 1 shows part of the curve y = (2x + 7)1/3 and the line y = x, together with the first steps of the iteration xn+1 = (2xn + 7)1/3, starting with x0 = 1. The iteration converges to α.
(a)
Show that α is a root of the equation x3 − 2x − 7 = 0.
(2)
(b)
Find the values of x1, x2 and x3, giving each to 4 decimal places.
(3)
(c)
State the name of the type of diagram shown in Figure 1 and describe how the values of xn approach α.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
at the limit α = (2α + 7)1/3
M1
cubes: α3 = 2α + 7, so α3 − 2α − 7 = 0
A1
Part (b)
attempts x1 = 91/3
M1
x1 = 2.0801
A1
x2 = 2.2347, x3 = 2.2552
A1
Part (c)
a staircase: the values increase steadily, all below α, approaching it from one side in smaller and smaller steps
B1
Exam tip: In a staircase every iterate lies on the same side of the root, so two consecutive iterates do not trap the root between them. A sign change is still needed to fix its accuracy.
f(x) = 2.2 − x − 3e−x. Figure 1 shows part of the graph of y = f(x). The points C and D on the graph have x-coordinates 0.5 and 2, and M is the maximum point.
(a)
Find f(0.5) and f(2), giving each to 3 significant figures.
(2)
(b)
A student says that, since f(0.5) and f(2) have the same sign, f(x) = 0 has no root between 0.5 and 2. Use Figure 1 to explain why the student is wrong.
(1)
(c)
Show that the x-coordinate of M is ln 3.
(2)
(d)
Show that f(ln 3) = 1.2 − ln 3. Hence explain why f(x) = 0 has a root in each of the intervals [0.5, ln 3] and [ln 3, 2].
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(0.5) = −0.120
B1
f(2.0) = −0.206
B1
Part (b)
the graph rises above the x-axis between C and D and crosses it twice, so there are two roots: an even number of crossings leaves the end values with the same sign
f(0.5) < 0 < f(ln 3) and f(ln 3) > 0 > f(2), with f continuous, so there is a change of sign, and a root, in each interval
B1
Exam tip: When the ends of an interval give the same sign, test a point in between, such as a turning point. Splitting the interval there can produce a sign change on each side.
The equation x2 ln x − 5 = 0, x > 1, has a single root α.
(a)
Show that the equation can be rearranged to give x = √(5/ln x).
(2)
(b)
Starting with x0 = 2.4, use the iteration xn+1 = √(5/ln xn) to find x1, x2 and x3 to 4 decimal places.
(3)
(c)
Show that α = 2.39 correct to 2 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
x2 ln x = 5 ⇒ x2 = 5/ln x (ln x ≠ 0 as x > 1)
M1
x = √(5/ln x), positive root as x > 1
A1
Part (b)
attempts √(5/ln 2.4)
M1
x1 = 2.3898
A1
x2 = 2.3956, x3 = 2.3923
A1
Part (c)
uses f(x) = x2 ln x − 5 on [2.385, 2.395]
M1
f(2.385) = −0.0558, f(2.395) = 0.00975
A1
change of sign and f continuous on [2.385, 2.395], so α = 2.39 to 2 d.p.
A1
Exam tip: These iterates settle slowly and still disagree in the third decimal place, which is why the accuracy of the root is proved by a sign change and not by the iteration alone.
The curve with equation y = xex meets the line y = 4 at exactly one point, where x = α.
(a)
Show that α lies in the interval [1.2, 1.3].
(2)
(b)
Show that α satisfies the equation x = ln(4/x).
(2)
(c)
Using xn+1 = ln(4/xn) with x0 = 1, find x1, x2 and x3 to 4 decimal places.
(3)
(d)
State, with a reason, whether a staircase or a cobweb diagram would illustrate these iterates.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
with f(x) = xex − 4: f(1.2) = −0.0159 and f(1.3) = 0.770
M1
change of sign and f continuous, so 1.2 < α < 1.3
A1
Part (b)
xex = 4 ⇒ ex = 4/x
M1
takes natural logarithms: x = ln(4/x)
A1
Part (c)
x1 = ln 4 attempted
M1
x1 = 1.3863
A1
x2 = 1.0597, x3 = 1.3283
A1
Part (d)
cobweb, because the iterates lie alternately above and below α (1.3863 > α, 1.0597 < α, 1.3283 > α)
B1
Model sketch
Figure: the iterates of xₙ₊₁ = ln(4/xₙ) form a cobweb
Exam tip: When the values jump from one side of the root to the other, the diagram is a cobweb; when they move steadily in one direction, it is a staircase.
The concentration, C mg per litre, of a medicine in a patient's blood t hours after a dose is modelled by C = 8te−0.5t, t ≥ 0. The concentration first reaches 2 mg per litre at time t = T.
(a)
Show that T is a root of the equation t = ¼e0.5t.
(2)
(b)
Use the iteration formula tn+1 = ¼e0.5tn with t0 = 0.5 to find t1, t2 and t3, giving each to 4 decimal places.
change of sign and g continuous, so 0.2885 < T < 0.2895 and T = 0.289 to 3 d.p.
A1
Model sketch
Figure: the concentration first reaches 2 mg per litre at t = T
Exam tip: In a context question the variable may be t, not x, but the method is identical. Define the function you are testing before you quote its values.
f(x) = x3 − 3x2 − 2. The equation f(x) = 0 has a single real root, α.
(a)
Show that α lies between 3 and 4.
(2)
(b)
Show that f(x) = 0 can be written as x = 3 + 2/x2.
(2)
(c)
Use xn+1 = 3 + 2/xn2 with x0 = 3 to find x1, x2 and x3, giving each to 4 decimal places.
(3)
(d)
Another student uses xn+1 = √((xn3 − 2)/3) with x0 = 3.2, obtaining x1 = 3.2025 and x2 = 3.2065. Given that α = 3.1958 to 4 decimal places, comment on the student's iteration.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(3) = −2 and f(4) = 14
M1
change of sign and f continuous, so 3 < α < 4
A1
Part (b)
x3 = 3x2 + 2
M1
divides by x2 (x ≠ 0): x = 3 + 2/x2
A1
Part (c)
attempts 3 + 2/9
M1
x1 = 3.2222
A1
x2 = 3.1926, x3 = 3.1962
A1
Part (d)
x0, x1, x2 are all above α and increasing, so each is further from α than the last
B1
the iteration diverges (a staircase moving away from the root), so it cannot be used to find α
B1
Model sketch
Figure: cobweb diagram for xₙ₊₁ = 3 + 2/xₙ² from x₀ = 3
Exam tip: “Comment on” an iteration: describe what the numbers do relative to the root, then give the conclusion. Small steps away from the root still mean divergence.
The equation ex − x − 3 = 0 has two roots: a positive root α and a negative root β.
(a)
Show that α lies in the interval [1.5, 1.6].
(2)
(b)
Use the iteration xn+1 = ln(xn + 3) with x0 = 1 to find x1, x2 and x3 to 4 decimal places.
(3)
(c)
A different rearrangement gives xn+1 = exn − 3. Starting with x0 = 1, find x1, x2 and x3 to 4 decimal places, and describe what happens to the sequence.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(1.5) = −0.0183, f(1.6) = 0.353, where f(x) = ex − x − 3
M1
change of sign and f continuous, so 1.5 < α < 1.6
A1
Part (b)
x1 = ln 4 attempted
M1
x1 = 1.3863
A1
x2 = 1.4785, x3 = 1.4993
A1
Part (c)
x1 = −0.2817, x2 = −2.2455
B1
x3 = −2.8941
B1
the sequence moves away from α and converges to the negative root β (≈ −2.95) instead
B1
Model sketch
Figure: staircase diagram for xₙ₊₁ = ln(xₙ + 3) from x₀ = 1
Exam tip: An iteration that will not settle on one root may converge to a different root of the same equation. Describe exactly where the values are heading.
The equation (x − 1)ex = 1 has a single root, α. Figure 1 shows the curve y = 1 + e−x, the line y = x, and the iteration xn+1 = 1 + e−xn starting with x0 = 2.
(a)
Show that α lies in the interval [1.2, 1.3].
(2)
(b)
Show that the equation (x − 1)ex = 1 can be written as x = 1 + e−x.
(2)
(c)
Find x1, x2 and x3, giving each to 4 decimal places.
(3)
(d)
State the name of the type of diagram in Figure 1 and explain how it shows that the iterates converge to α.
Figure 1 shows a circle with centre O and radius r cm. The points A and B lie on the circle and angle AOB = θ radians, where 0 < θ < π. The shaded region S is bounded by the chord AB and the minor arc AB. The perimeter of S is 4r cm.
(a)
Show that the length of the chord AB is 2r sin(θ/2).
(2)
(b)
Show that θ satisfies the equation θ = 4 − 2 sin(θ/2).
(2)
(c)
Show that the root of this equation lies in the interval [2.2, 2.3].
(2)
(d)
Use the iteration θn+1 = 4 − 2 sin(θn/2) with θ0 = 2.2 to find θ1, θ2 and θ3, giving each to 4 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
the perpendicular from O to AB bisects AB and angle AOB, giving a right-angled triangle with angle θ/2 at O and hypotenuse r
A new sports club predicts its membership, N hundred members, t months after it opens using two models. Model A: N = 50 + 20t. Model B: N = 30e0.25t. Figure 1 shows both models for 0 ≤ t ≤ 9. Model B first predicts more members than model A at time t = T.
(a)
Use Figure 1 to write down the two consecutive whole numbers of months between which T lies.
(1)
(b)
Show that T is a root of the equation t = 4 ln((5 + 2t)/3).
(2)
(c)
Use the iteration tn+1 = 4 ln((5 + 2tn)/3) with t0 = 7 to find t1, t2 and t3, giving each to 4 decimal places.
(3)
(d)
Show that T = 7.65 correct to 2 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
7 < T < 8
B1
Part (b)
30e0.25t = 50 + 20t ⇒ e0.25t = (5 + 2t)/3
M1
0.25t = ln((5 + 2t)/3) ⇒ t = 4 ln((5 + 2t)/3)
A1
Part (c)
attempts 4 ln(19/3)
M1
t1 = 7.3833
A1
t2 = 7.5415, t3 = 7.6051
A1
Part (d)
uses h(t) = 30e0.25t − 50 − 20t on [7.645, 7.655]
M1
h(7.645) = −0.0539 and h(7.655) = 0.254
A1
change of sign and h continuous, so 7.645 < T < 7.655 and T = 7.65 to 2 d.p.
A1
Exam tip: These iterates creep up slowly from one side, so after three steps they are still not correct to 2 d.p. Use the sign change to settle the accuracy rather than iterating on and on.
Figure 1 shows the curve y = (x3 + 1)/4 and the line y = x, which meet at the points A, B and C. It also shows the diagram produced by the iteration xn+1 = (xn3 + 1)/4 with x0 = −2.
(a)
Show that the x-coordinates of A, B and C are the roots of the equation x3 − 4x + 1 = 0.
(1)
(b)
Find x1, x2 and x3, giving each to 4 decimal places.
(3)
(c)
State which of the points A, B and C the sequence converges to, and use Figure 1 to describe how the values approach it.
(2)
(d)
Without further iteration, explain why starting with x0 = 2 would not give the x-coordinate of C.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
at each point x = (x3 + 1)/4 ⇒ 4x = x3 + 1 ⇒ x3 − 4x + 1 = 0
B1
Part (b)
attempts ((−2)3 + 1)/4
M1
x1 = −1.7500
A1
x2 = −1.0898, x3 = −0.0736
A1
Part (c)
converges to the x-coordinate of B (≈ 0.254)
B1
a staircase: the values increase steadily, moving away from A and approaching B from below in smaller and smaller steps
B1
Part (d)
near C the curve is steeper than y = x (its gradient is 3x2/4 ≈ 2.6 > 1)
B1
2 lies to the right of C, where the curve is above the line, so each iterate is larger than the last (2 → 2.25 → …): the values move away from C and diverge
B1
Exam tip: A root can only be found by this iteration if the curve is less steep than y = x there. Here that is true only at B, which is why A and C both push the iterates away.
The equation x3 + 2x − 5 = 0 has a single real root, α. Two iteration formulae are proposed: (I) xn+1 = (5 − 2xn)1/3 (II) xn+1 = (5 − xn3)/2. Figure 1 shows two diagrams, P and Q. Each shows the line y = x, the curve y = g(x) for one of the formulae, and the diagram produced by that formula, starting from x0 = 1.3 in P and from x0 = 1.5 in Q.
(a)
Show that α lies in the interval [1.3, 1.4].
(2)
(b)
Show that the equation can be rearranged into the form used in each of the formulae (I) and (II).
(2)
(c)
State which diagram, P or Q, shows formula (I). Give a reason for your answer.
(2)
(d)
Use the formula whose diagram shows convergence, with x0 = 1.5, to find x1, x2 and x3, giving each to 4 decimal places.
reason, e.g. formula (II) with x0 = 1.3 gives x1 = 1.4015, the first step in P; or near α the curve for (I) is shallow (gradient about −0.4), as in Q, while the curve for (II) is steep (gradient −3x2/2 ≈ −2.6), as in P
B1
Part (d)
uses (I): attempts 21/3
M1
x1 = 1.2599
A1
x2 = 1.3536, x3 = 1.3186
A1
Exam tip: Both diagrams are cobwebs because both curves slope downwards. What decides convergence is how steep the curve is near the root compared with y = x, not its direction.
Figure 1 shows the curve y = 3x and the line y = 4x. They meet at the points P and Q, whose x-coordinates are p and q respectively, where p < q.
(a)
Show that p lies in the interval [0.3, 0.4] and that q lies in the interval [1.7, 1.8].
(3)
(b)
Use the iteration xn+1 = ln(4xn)/ln 3 with x0 = 1.8 to find x1, x2 and x3, giving each to 4 decimal places.
(3)
(c)
A student uses instead the iteration xn+1 = 3xn/4, also with x0 = 1.8. Explain, with reference to Figure 1, why this iteration will not find q.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(x) = 3x − 4x; at least two of f(0.3) = 0.190, f(0.4) = −0.0482, f(1.7) = −0.327, f(1.8) = 0.0247
M1
all four correct
A1
a change of sign in each interval and f continuous, so a root in each
A1
Part (b)
attempts ln 7.2/ln 3
M1
x1 = 1.7969
A1
x2 = 1.7953, x3 = 1.7945
A1
Part (c)
near Q the curve y = 3x is steeper than the line y = 4x, so y = 3x/4 is steeper than y = x there (gradient 3x ln 3/4 ≈ 2 > 1)
B1
for x > q the curve is above the line, so 3x/4 > x: x1 = 1.8062 > 1.8 and the iterates keep increasing, moving away from q
B1
Exam tip: The two formulae are inverses of each other. Where one is steeper than y = x, the other is shallower, so between them they can reach both roots.
The curve C has equation y = x2 − 4 sin x, where x is in radians. C has a stationary point P with x-coordinate α, where 0 < α < π/2.
(a)
Show that α is a root of the equation x = 2 cos x.
(2)
(b)
Show that α lies in the interval [1, 1.1].
(2)
(c)
Use xn+1 = 2 cos xn with x0 = 1 to find x1, x2 and x3 to 4 decimal places, and describe the behaviour of this iteration.
(3)
(d)
Use instead xn+1 = arccos(xn/2) with x0 = 1 to find x1, x2 and x3 to 4 decimal places.
(3)
(e)
Show that α = 1.030 correct to 3 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
dy/dx = 2x − 4 cos x
M1
2x − 4 cos x = 0 at P ⇒ x = 2 cos x
A1
Part (b)
with h(x) = x − 2 cos x: h(1.0) = −0.0806, h(1.1) = 0.193
M1
change of sign and h continuous, so 1 < α < 1.1
A1
Part (c)
x1 = 1.0806, x2 = 0.9416
B1
x3 = 1.1770
B1
the values alternate either side of α and get further from it: the iteration diverges (an unwinding cobweb)
B1
Part (d)
rearranges or uses arccos(1/2) in radians
M1
x1 = 1.0472
A1
x2 = 1.0197, x3 = 1.0358
A1
Part (e)
h on [1.0295, 1.0305]
M1
h(1.0295) = −0.000995, h(1.0305) = 0.00172
A1
change of sign and h continuous, so α = 1.030 to 3 d.p.
A1
Model sketch
Figure: cobweb for xₙ₊₁ = 2 cos xₙ from x₀ = 1, spiralling away from α
Exam tip: Two rearrangements of the same equation can behave in opposite ways. Near the root, 2 cos x is steeper than y = x (gradient about −1.7) but arccos(x/2) is shallower (about −0.6).
A circle has centre O and radius r cm. The chord AB subtends an angle of θ radians at O, where 0 < θ < π. The area of the minor segment cut off by AB is one third of the area of the circle.
Figure: the minor segment cut off by the chord AB; not to scale
(a)
Show that θ satisfies the equation θ = sin θ + 2π/3.
(3)
(b)
Show that the root of this equation lies in the interval [2.6, 2.7].
(2)
(c)
Use the iteration θn+1 = sin θn + 2π/3 with θ0 = 2.6 to find θ1, θ2 and θ3 to 4 decimal places.
(3)
(d)
Show that θ = 2.605 correct to 3 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
area of segment = ½r2θ − ½r2 sin θ (sector minus triangle)
M1
½r2(θ − sin θ) = ⅓πr2
M1
⇒ θ − sin θ = 2π/3 ⇒ θ = sin θ + 2π/3
A1
Part (b)
with f(θ) = θ − sin θ − 2π/3: f(2.6) = −0.00990, f(2.7) = 0.178
M1
change of sign and f continuous, so the root lies in [2.6, 2.7]
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Part (c)
attempts sin 2.6 + 2π/3 in radians
M1
θ1 = 2.6099
A1
θ2 = 2.6014, θ3 = 2.6087
A1
Part (d)
f on [2.6045, 2.6055]
M1
f(2.6045) = −0.00154, f(2.6055) = 0.000324
A1
change of sign and f continuous, so θ = 2.605 to 3 d.p.
A1
Exam tip: The iterates here jump either side of the root and settle slowly (a cobweb), so the third iterate is still not correct to 3 d.p. The sign change in the last part is what proves the answer.
change of sign and f continuous for x > −1, so 4 < α < 4.2
A1
Part (c)
attempts 9 − 3 ln 5
M1
x1 = 4.1717
A1
x2 = 4.0704, x3 = 4.1297
A1
Part (d)
f on [4.1075, 4.1085]
M1
f(4.1075) = −0.000123, f(4.1085) = 0.000406
A1
change of sign and f continuous, so α = 4.108 to 3 d.p.
A1
Part (e)
y = 3 − α/3 with 4.1075 < α < 4.1085 (or ln(α + 1))
M1
y = 1.63
A1
Exam tip: To find where two curves meet, set the expressions equal and rearrange into the given form. Use the line, not the curve, for the y-coordinate when the line is simpler: both give the same value at P.
A quintic function is given by f(x) = x5 − 5x + 1. Two different rearrangements of f(x) = 0 are to be used to approximate its roots by iteration.
(a)
Show that the equation f(x) = 0 has a root in each of the intervals [−2, −1], [0, 1] and [1, 2].
(3)
(b)
Use xn+1 = (xn5 + 1)/5 with x0 = 0 to find x1 and x2 to 4 decimal places.
(2)
(c)
The same iteration is used with x0 = 1.5 in an attempt to find the root in [1, 2]. Find x1 and x2 to 4 decimal places and explain why the attempt fails.
(3)
(d)
Use instead xn+1 = (5xn − 1)1/5 with x0 = 1.5 to find x1, x2 and x3 to 4 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(−2) = −21, f(−1) = 5, f(0) = 1, f(1.0) = −3, f(2) = 23 (at least four correct)
M1
all five values correct
A1
a change of sign in each interval and f continuous (polynomial), so a root in each
A1
Part (b)
x1 = 0.2000
B1
x2 = 0.2001
B1
Part (c)
x1 = 1.7188, x2 = 3.1998
B1
the values increase rapidly, moving away from the root in [1, 2]: the iteration diverges
B1
near that root the curve y = (x5 + 1)/5 is much steeper than y = x (gradient x4 ≈ 4), whereas near 0.2 it is almost flat
B1
Part (d)
attempts (6.5)1/5
M1
x1 = 1.4541
A1
x2 = 1.4436, x3 = 1.4412
A1
Model sketch
Figure: y = x⁵ − 5x + 1 has one root in each of [−2, −1], [0, 1] and [1, 2]
Exam tip: One rearrangement rarely finds every root. A formula that is flat near one root can be steep near another, so check the behaviour of the iterates for each root separately.
Figure 1 shows part of the curve with equation y = ln x, x > 0. The point P on the curve is the point of the curve closest to the origin O, and the x-coordinate of P is α.
(a)
The point Q(x, ln x) lies on the curve. Write down an expression for D, where D = OQ2.
(1)
(b)
Show that dD/dx = 0 when x = e−x2.
(3)
(c)
Use the iteration xn+1 = e−xn2 with x0 = 0.65 to find x1, x2 and x3, giving each to 4 decimal places.
(3)
(d)
Show that α = 0.653 correct to 3 decimal places.
(3)
(e)
Hence find the least distance from O to the curve, giving your answer to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
D = x2 + (ln x)2
B1
Part (b)
differentiates, using the chain rule on (ln x)2
M1
dD/dx = 2x + (2 ln x)/x
A1
= 0 ⇒ x2 + ln x = 0 ⇒ ln x = −x2 ⇒ x = e−x2
A1
Part (c)
attempts e−0.4225
M1
x1 = 0.6554
A1
x2 = 0.6508, x3 = 0.6547
A1
Part (d)
uses h(x) = x − e−x2 (or x2 + ln x) on [0.6525, 0.6535]
M1
h(0.6525) = −0.000776 and h(0.6535) = 0.00108
A1
change of sign and h continuous, so α = 0.653 to 3 d.p.
A1
Part (e)
OP = √(α2 + (ln α)2), which equals √(α2 + α4) since ln α = −α2, with α = 0.653 or better
M1
least distance = 0.780
A1
Exam tip: Minimise the squared distance rather than the distance itself: it has the same stationary point and avoids differentiating a square root.
Figure 1 shows part of the curve C with equation y = ex − 3x2. C has a maximum point A and a minimum point B. The x-coordinate of B is β.
(a)
Show that the x-coordinates of A and B both satisfy the equation x = ln(6x).
(2)
(b)
Show that β lies in the interval [2.8, 2.9].
(2)
(c)
Use the iteration xn+1 = ln(6xn) with x0 = 3 to find x1, x2 and x3, giving each to 4 decimal places.
(3)
(d)
Show that β = 2.833 correct to 3 decimal places.
(3)
(e)
Using eβ = 6β, show that the y-coordinate of B is 3β(2 − β), and hence find it to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
dy/dx = ex − 6x
M1
= 0 at a stationary point ⇒ ex = 6x ⇒ x = ln(6x)
A1
Part (b)
with h(x) = ex − 6x: h(2.8) = −0.355 and h(2.9) = 0.774
M1
change of sign and h continuous, so 2.8 < β < 2.9
A1
Part (c)
attempts ln 18
M1
x1 = 2.8904
A1
x2 = 2.8531, x3 = 2.8402
A1
Part (d)
h on [2.8325, 2.8335]
M1
h(2.8325) = −0.00712 and h(2.8335) = 0.00387
A1
change of sign and h continuous, so β = 2.833 to 3 d.p.
A1
Part (e)
y = eβ − 3β2 = 6β − 3β2 = 3β(2 − β)
M1
y = −7.08
A1
Exam tip: The iteration x = ln(6x) settles on B but not on A: near A the curve y = ln(6x) has gradient 1/x ≈ 5, far steeper than y = x. Always ask which root an iteration can actually reach.
Figure 1 shows the curves with equations y = 4 cos x and y = x2, where x is in radians and −2 ≤ x ≤ 2. The curves meet at the points where x = −α and x = α, where α > 0. The shaded region R is bounded by the two curves.
(a)
Show that α lies in the interval [1.15, 1.25].
(2)
(b)
Show that α satisfies x = arccos(x2/4), and use the iteration xn+1 = arccos(xn2/4) with x0 = 1 to find x1, x2 and x3, giving each to 4 decimal places.
(4)
(c)
Show that α = 1.202 correct to 3 decimal places.
(3)
(d)
Show that the area of R is 8 sin α − ⅔α3, and hence find the area of R, giving your answer to 3 significant figures.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
h(x) = x2 − 4 cos x: h(1.15) = −0.311 and h(1.25) = 0.301
M1
change of sign and h continuous, so 1.15 < α < 1.25
A1
Part (b)
4 cos x = x2 ⇒ cos x = x2/4 ⇒ x = arccos(x2/4), valid as 0 < α < π
B1
attempts arccos(0.25) in radians
M1
x1 = 1.3181
A1
x2 = 1.1215, x3 = 1.2509
A1
Part (c)
h on [1.2015, 1.2025]
M1
h(1.2015) = −0.000235 and h(1.2025) = 0.00590
A1
change of sign and h continuous, so α = 1.202 to 3 d.p.
A1
Part (d)
area = ∫ (4 cos x − x2) dx from −α to α
M1
= [4 sin x − x3/3] from −α to α
A1
= 8 sin α − ⅔α3
A1
area of R = 6.30
A1
Exam tip: Use symmetry to check: both curves are even, so the area is twice the area from 0 to α. Keep the calculator in radians for the integral as well as the iteration.
Figure 1 shows the curve y = 1 + 3 ln x, x > 0, and the line y = x. They meet at the point (1, 1) and at the point where x = β. The figure also shows the diagram for the iteration xn+1 = 1 + 3 ln xn, starting with x0 = 2.
(a)
Verify that x = 1 is a root of the equation x = 1 + 3 ln x, and show that β lies in the interval [6.7, 6.8].
(3)
(b)
Find x1, x2 and x3, giving each to 4 decimal places.
(3)
(c)
Using Figure 1, explain why this iteration cannot be used to find the root x = 1, however close to 1 the value of x0 is taken.
(2)
(d)
Show that β = 6.711 correct to 3 decimal places.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
1 + 3 ln 1 = 1 + 0 = 1
B1
f(x) = 1 + 3 ln x − x: f(6.7) = 0.00632 and f(6.8) = −0.0492
M1
change of sign and f continuous for x > 0, so 6.7 < β < 6.8
A1
Part (b)
attempts 1 + 3 ln 2
M1
x1 = 3.0794
A1
x2 = 4.3742, x3 = 5.4272
A1
Part (c)
at (1, 1) the curve is steeper than y = x (gradient 3/x = 3 > 1)
B1
for x0 just above 1 the curve is above the line, so the iterates increase away from 1 (towards β); for x0 just below 1 it is below the line, so they decrease away from 1
B1
Part (d)
f on [6.7105, 6.7115]
M1
f(6.7105) = 0.000520 and f(6.7115) = −0.0000326
A1
change of sign and f continuous, so β = 6.711 to 3 d.p.
A1
Common error: Assuming an iteration finds the root nearest to the starting value. Here x₀ = 2 is much closer to 1, yet the staircase climbs away to β because the curve is steep at x = 1.
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