Locating roots, iteration and the Newton-Raphson method - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Locating roots, iteration and the Newton-Raphson method for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

\[\mathrm{f}(x) = \dfrac{x^2 + 1}{x - 2}, \quad x ≠ 2\]
Find the values of \(\mathrm{f}(1)\) and \(\mathrm{f}(3)\).
Explain why, although \(\mathrm{f}(1)\) and \(\mathrm{f}(3)\) have different signs, the equation \(\mathrm{f}(x) = 0\) has no root in the interval \([1, 3]\).

Model answer

\(\mathrm{f}(1) = \dfrac{2}{-1} = -2\), \(\mathrm{f}(3) = \dfrac{10}{1} = 10\)
\(\mathrm{f}\) is not continuous on \([1, 3]\): it is undefined at \(x = 2\), where the graph has the vertical asymptote \(x = 2\). The sign changes because \(\mathrm{f}(x) \to -\infty\) as \(x \to 2\) from below and \(\mathrm{f}(x) \to +\infty\) as \(x \to 2\) from above, not because the graph crosses the \(x\)-axis.
In fact \(x^2 + 1 \geqslant 1 > 0\) for all \(x\), so the numerator is never 0 and \(\mathrm{f}(x) = 0\) has no roots at all.

Explanation

A change of sign only guarantees a root when the function is continuous on the whole interval. Here \(\mathrm{f}\) has a vertical asymptote at \(x = 2\), inside \([1, 3]\), and the sign changes across the asymptote, not at a root; since \(x^2 + 1\) is always positive, \(\mathrm{f}(x)\) is never 0. A common error is to answer only "there is an asymptote" without saying why that matters for the sign-change method.

Mark scheme

1 mark
B1 \(\mathrm{f}(1) = -2\) and \(\mathrm{f}(3) = 10\)
Both values needed. Correct values with no working score this mark.
1 mark
B1 \(\mathrm{f}\) is not continuous on \([1, 3]\): it is not defined at \(x = 2\), where the graph has a vertical asymptote, so the change of sign does not show that there is a root
Needs the discontinuity (or asymptote) at \(x = 2\) linked to the change of sign. Accept "the sign changes because the graph jumps from \(-\infty\) to \(+\infty\) at \(x = 2\)".
1 mark
B1 \(x^2 + 1 > 0\) for all \(x\) (it is at least 1), so the numerator is never zero and \(\mathrm{f}(x) ≠ 0\) for every \(x\): there is no root
Needs a reason why \(\mathrm{f}(x)\) is never 0. "\(x^2 + 1 = 0\) has no real solutions" is fine.

Question 2

4 marks

Show that the equation
\[\mathrm{e}^x + 2x = 5\]
has exactly one real root, and that this root lies in the interval \([1, 1.1]\).

Model answer

Let \(\mathrm{g}(x) = \mathrm{e}^x + 2x - 5\). Then \(\mathrm{g}'(x) = \mathrm{e}^x + 2 > 0\) for all \(x\), since \(\mathrm{e}^x > 0\).
So \(\mathrm{g}\) is increasing for all \(x\), and \(\mathrm{g}(x) = 0\) has at most one root.
\(\mathrm{g}(1) = \mathrm{e} + 2 - 5 = -0.2817... < 0\)
\(\mathrm{g}(1.1) = \mathrm{e}^{1.1} + 2.2 - 5 = 0.2041... > 0\)
There is a change of sign and \(\mathrm{g}\) is continuous, so there is a root in \([1, 1.1]\). As \(\mathrm{g}\) is increasing, this is the only real root: the equation has exactly one real root, and it lies in \([1, 1.1]\).

Explanation

"Exactly one" needs two facts: there is a root (a sign change of a continuous function) and there cannot be another (the function is increasing everywhere, because \(\mathrm{e}^x + 2 > 0\)). The sign change on \([1, 1.1]\) gives the location. A common error is to show only the sign change, which proves that there is a root but not that it is the only one.

Mark scheme

1 mark
M1 Considers \(\mathrm{g}(x) = \mathrm{e}^x + 2x - 5\) (or similar) and its derivative \(\mathrm{g}'(x) = \mathrm{e}^x + 2\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a show that. Alternative: a sketch argument that \(y = \mathrm{e}^x\) is increasing and \(y = 5 - 2x\) is decreasing, so they meet at most once, scores the first two marks if clearly explained.
1 mark
A1 \(\mathrm{g}'(x) = \mathrm{e}^x + 2 > 0\) for all \(x\), so \(\mathrm{g}\) is increasing and \(\mathrm{g}(x) = 0\) has at most one root
Needs the reason \(\mathrm{e}^x > 0\) (so \(\mathrm{g}'(x) > 0\)) and the deduction about the number of roots.
1 mark
M1 Evaluates \(\mathrm{g}(1)\) and \(\mathrm{g}(1.1)\)
For reference, \(\mathrm{g}(1) = -0.2817...\) and \(\mathrm{g}(1.1) = 0.2041...\)
1 mark
A1* \(\mathrm{g}(1) = -0.3\) and \(\mathrm{g}(1.1) = 0.2\) (1 significant figure), a change of sign, \(\mathrm{g}\) continuous, so there is a root in \([1, 1.1]\); with \(\mathrm{g}\) increasing, it is the only real root
The conclusion must bring together existence (sign change and continuity) and uniqueness (increasing). Do not accept "\(x\) is continuous".

Question 3

7 marks

The equation \(\mathrm{e}^{2x} = 4x + 1\) has exactly two real roots. One of the roots is \(x = 0\) and the other root is \(\alpha\), where \(\alpha > 0\).
By applying the Newton-Raphson method to \(\mathrm{f}(x) = \mathrm{e}^{2x} - 4x - 1\), show that successive approximations to a root are related by
\[x_{n+1} = \dfrac{(2x_n - 1)\mathrm{e}^{2x_n} + 1}{2\mathrm{e}^{2x_n} - 4}\]
Starting with \(x_1 = 0.7\), use the formula to find \(x_2\) and \(x_3\), giving each to 4 decimal places.
Using a suitable interval, show that \(\alpha = 0.628\) correct to 3 decimal places.

[The Newton-Raphson iteration for solving \(\mathrm{f}(x) = 0\): \(x_{n+1} = x_n - \dfrac{\mathrm{f}(x_n)}{\mathrm{f}'(x_n)}\)]

Model answer

\(\mathrm{f}'(x) = 2\mathrm{e}^{2x} - 4\)
\(x_{n+1} = x_n - \dfrac{\mathrm{e}^{2x_n} - 4x_n - 1}{2\mathrm{e}^{2x_n} - 4} = \dfrac{x_n\left(2\mathrm{e}^{2x_n} - 4\right) - \mathrm{e}^{2x_n} + 4x_n + 1}{2\mathrm{e}^{2x_n} - 4}\)
\(= \dfrac{2x_n\mathrm{e}^{2x_n} - \mathrm{e}^{2x_n} + 1}{2\mathrm{e}^{2x_n} - 4} = \dfrac{(2x_n - 1)\mathrm{e}^{2x_n} + 1}{2\mathrm{e}^{2x_n} - 4}\)
\(x_2 = \dfrac{0.4\mathrm{e}^{1.4} + 1}{2\mathrm{e}^{1.4} - 4} = 0.6379\) (4 d.p.)
\(x_3 = \dfrac{(2 \times 0.63791... - 1)\mathrm{e}^{1.27582...} + 1}{2\mathrm{e}^{1.27582...} - 4} = 0.6284\) (4 d.p.)
\(\mathrm{f}(0.6275) = \mathrm{e}^{1.255} - 2.51 - 1 = -0.00216\ldots < 0\)
\(\mathrm{f}(0.6285) = \mathrm{e}^{1.257} - 2.514 - 1 = 0.00086\ldots > 0\)
There is a change of sign and \(\mathrm{f}\) is continuous, so \(0.6275 < \alpha < 0.6285\), and \(\alpha = 0.628\) correct to 3 decimal places.

Explanation

Differentiate \(\mathrm{e}^{2x} - 4x - 1\) to get \(2\mathrm{e}^{2x} - 4\), substitute into the Newton-Raphson formula and put everything over the common denominator: the \(4x_n\) terms cancel, leaving the printed formula. From \(x_1 = 0.7\) the approximations fall quickly towards \(\alpha\), and a sign change of \(\mathrm{f}\) on [0.6275, 0.6285] proves that \(\alpha = 0.628\) to 3 decimal places. Common errors are differentiating \(\mathrm{e}^{2x}\) as \(\mathrm{e}^{2x}\) and using an interval such as [0.627, 0.629] that does not establish the 3 d.p. value.

Mark scheme

1 mark
B1 \(\mathrm{f}'(x) = 2\mathrm{e}^{2x} - 4\). An answer of \(x_2 = 0.6379\) and \(x_3 = 0.6284\) with no working scores 2 marks (the M1 and A1 for the iterates).
An answer of \(x_2 = 0.6379\) and \(x_3 = 0.6284\) with no working scores 2 marks (the M1 and A1 for the iterates); the show-that and the interval argument need working. Accept \(x_2\) in [0.63785, 0.63795) and \(x_3\) in [0.62835, 0.62845).
1 mark
M1 Substitutes into the Newton-Raphson formula: \(x_{n+1} = x_n - \dfrac{\mathrm{e}^{2x_n} - 4x_n - 1}{2\mathrm{e}^{2x_n} - 4}\)
Condone the omission of the subscripts for this mark only.
1 mark
A1* Combines over a common denominator: \(\dfrac{x_n\left(2\mathrm{e}^{2x_n} - 4\right) - \mathrm{e}^{2x_n} + 4x_n + 1}{2\mathrm{e}^{2x_n} - 4} = \dfrac{2x_n\mathrm{e}^{2x_n} - \mathrm{e}^{2x_n} + 1}{2\mathrm{e}^{2x_n} - 4}\), and factorises to reach the given formula
cso, with \(x_{n+1} =\) and the subscripts in the final line; the \(-4x_n\) and \(+4x_n\) terms must be seen to cancel. Jumping from the substitution to the given answer is A0*.
1 mark
M1 Uses the formula with \(x_1 = 0.7\) to find \(x_2\), and continues to find \(x_3\)
May use the given formula even if it was not shown. The values must round to 0.6379 and 0.6284, i.e. lie in [0.63785, 0.63795) and [0.62835, 0.62845), for the A mark.
1 mark
A1 \(x_2 = 0.6379\) and \(x_3 = 0.6284\)
Both needed, correct to 4 decimal places (from 0.637913... and 0.628427...). Accept awrt 0.628 for \(x_3\), i.e. in [0.6275, 0.6285), if \(x_2\) was rounded before use.
1 mark
M1 Chooses the interval [0.6275, 0.6285] (or a smaller interval that contains \(\alpha\) and lies inside it) and evaluates \(\mathrm{f}\) at both ends: \(\mathrm{f}(0.6275) = -0.00216\ldots\) and \(\mathrm{f}(0.6285) = 0.00086\ldots\)
Values may be rounded or truncated to 1 significant figure: \(-0.002\) and \(0.0009\) (or \(0.0008\)). Using an interval such as [0.627, 0.629], which does not show the 3 d.p. value, scores M0.
1 mark
A1 Both values correct with a change of sign, a reason (\(\mathrm{f}\) is continuous, and there is a sign change) and a conclusion that \(\alpha = 0.628\) correct to 3 decimal places, cso
All three elements are needed: correct values (to 1 s.f.), the sign change with continuity, and the conclusion. Do not accept 'so the root is 0.628' without reference to the sign change.

Question 4

8 marks

A capacitor in a circuit charges up and then slowly loses charge through a fault. In a model, the charge on the capacitor is \(Q\) microcoulombs at time \(t\) seconds after the circuit is switched on, where
\[Q = 30\left(1 - \mathrm{e}^{-0.5t}\right) - t, \quad 0 \leqslant t \leqslant 12\]
A warning light is on whenever the charge is at least 20 microcoulombs.

Show that the times at which \(Q = 20\) satisfy the equation
\[t = 10 - 30\mathrm{e}^{-0.5t}\]
Using a suitable interval, show that the warning light comes on between \(t = 2.87\) and \(t = 2.88\).
Using the iterative formula \(t_{n+1} = 10 - 30\mathrm{e}^{-0.5t_n}\) with \(t_1 = 9\), find \(t_2\) to 4 decimal places, and find, by repeated iteration, the time at which the warning light goes off, to 2 decimal places.
Hence find, to 1 decimal place, the length of time for which the warning light is on.

Model answer

\(30\left(1 - \mathrm{e}^{-0.5t}\right) - t = 20\)
\(30 - 30\mathrm{e}^{-0.5t} - t = 20\)
\(t = 10 - 30\mathrm{e}^{-0.5t}\)
Let \(\mathrm{f}(t) = 30\left(1 - \mathrm{e}^{-0.5t}\right) - t - 20\)
\(\mathrm{f}(2.87) = -0.0135... < 0\), \(\mathrm{f}(2.88) = 0.0122... > 0\)
There is a change of sign and \(\mathrm{f}\) is continuous, so \(Q = 20\) at a time between 2.87 and 2.88: the warning light comes on between these times.
\(t_2 = 10 - 30\mathrm{e}^{-4.5} = 9.6667\) (4 d.p.)
\(t_3 = 9.76121...\), \(t_4 = 9.77223...\), \(t_5 = 9.77348...\), \(t_6 = 9.77362...\)
The light goes off at \(t = 9.77\) s (2 d.p.)
Length of time \(= 9.77 - 2.88 = 6.89...\), so 6.9 seconds (1 d.p.)

Explanation

Setting \(Q = 20\) and rearranging gives the equation for both times: the charge rises past 20 microcoulombs, reaches a maximum and then falls back below 20. A sign change on \([2.87, 2.88]\) locates the first time, and the iteration from \(t_1 = 9\) converges quickly to the second, 9.77 s; the light is on for the 6.9 s between them. Common errors are leaving out the continuity statement and stopping the iteration before two successive values agree to 2 decimal places.

Mark scheme

1 mark
M1 Sets \(30\left(1 - \mathrm{e}^{-0.5t}\right) - t = 20\) and rearranges towards \(t = \ldots\). An answer with no working scores 0 marks for the show thats.
An answer with no working scores 0 marks for the two show thats; correct values \(t_2 = 9.6667\), 9.77 and 6.9 s with no working score the last 4 marks (the iteration is implied).
1 mark
A1* \(30 - 30\mathrm{e}^{-0.5t} - t = 20 \Rightarrow t = 10 - 30\mathrm{e}^{-0.5t}\) cso
1 mark
M1 Evaluates a suitable function, e.g. \(\mathrm{f}(t) = 30\left(1 - \mathrm{e}^{-0.5t}\right) - t - 20\) (or \(Q\) compared with 20), at \(t = 2.87\) and \(t = 2.88\)
For reference, \(\mathrm{f}(2.87) = -0.0135...\) and \(\mathrm{f}(2.88) = 0.0122...\) (\(Q(2.87) = 19.987...\), \(Q(2.88) = 20.012...\)). Accept any tighter interval containing the root.
1 mark
A1* \(\mathrm{f}(2.87) = -0.01\) and \(\mathrm{f}(2.88) = 0.01\) (1 significant figure), a change of sign, \(\mathrm{f}\) continuous, so \(Q = 20\) at a time between 2.87 and 2.88: the warning light comes on in this interval
Accept values of \(Q\) either side of 20 with continuity and a conclusion in context.
1 mark
M1 Uses the iterative formula with \(t_1 = 9\): \(t_2 = 10 - 30\mathrm{e}^{-4.5}\)
May use the given formula even if it was not shown.
1 mark
A1 \(t_2 = 9.6667\)
cao, 4 d.p. (from 9.666727...).
1 mark
A1 9.77 (seconds)
Requires the M mark for using the formula. cao, 2 d.p. (iterates 9.7612, 9.7722, 9.7735, 9.7736, ...).
1 mark
B1 6.9 seconds
From 9.77 − 2.87 or 9.77 − 2.88 or more accurate values (9.7736 − 2.8752 = 6.898...); all give 6.9 to 1 d.p. ft their time from the iteration minus a time between 2.87 and 2.88.

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