Locating roots, iteration and the Newton-Raphson method - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Locating roots, iteration and the Newton-Raphson method for Edexcel A-Level Maths (Paper 1). 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.

18 questions
2–10 marks each
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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

2 marks

\[\mathrm{f}(x) = x^3 - 3x^2 + 4\]
Show that \(\mathrm{f}(2) = 0\).
Explain why a change of sign method cannot be used to locate the root of \(\mathrm{f}(x) = 0\) at \(x = 2\).

Model answer

\(\mathrm{f}(2) = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0\)
Since \(\mathrm{f}(2) = 0\), \((x - 2)\) is a factor: \(\mathrm{f}(x) = (x - 2)^2(x + 1)\).
So \(x = 2\) is a repeated root: the curve \(y = \mathrm{f}(x)\) touches the x-axis at \(x = 2\) (a minimum point) and \(\mathrm{f}(x) \geqslant 0\) on both sides of it. There is no change of sign, so a change of sign method cannot locate this root.

Explanation

A change of sign method only finds roots where the curve crosses the x-axis. At a repeated root such as \(x = 2\) here, the curve touches the axis at a turning point, so \(\mathrm{f}(x)\) has the same sign on both sides. A common wrong reason is that the function is not continuous; it is a polynomial, so it is continuous everywhere.

Mark scheme

1 mark
B1 \(\mathrm{f}(2) = 8 - 12 + 4 = 0\)
The substitution must be seen.
1 mark
B1 \(\mathrm{f}(x) = (x - 2)^2(x + 1)\), so \(x = 2\) is a repeated root: \(\mathrm{f}(x) \geqslant 0\) on both sides of \(x = 2\) (the curve touches the x-axis at a minimum point), so there is no change of sign
Accept a numerical demonstration with a reason, e.g. \(\mathrm{f}(1.9) = 0.029\) and \(\mathrm{f}(2.1) = 0.031\) are both positive because the curve touches the x-axis at \(x = 2\). Accept "the x-axis is a tangent to the curve at \(x = 2\)". Do not accept "\(\mathrm{f}\) is not continuous".

Question 2

4 marks

The equation \(x^2 - \sin x - 2 = 0\), where \(x\) is in radians, has a root \(\beta\) in the interval \([1.5, 2]\).
The iterative formula
\[x_{n+1} = \sqrt{2 + \sin x_n}\]
is used with \(x_1 = 1.5\) to find \(\beta\).
Find the values of \(x_2\) and \(x_3\), giving each answer to 4 decimal places.
By repeated iteration, find the value of \(\beta\) correct to 3 decimal places.

Model answer

\(x_2 = \sqrt{2 + \sin 1.5} = 1.7313\) (4 d.p.)
\(x_3 = \sqrt{2 + \sin 1.73133...} = 1.7283\) (4 d.p.)
Continuing: \(x_4 = 1.7285\), \(x_5 = 1.7285\), ...
The values settle at \(\beta = 1.728\) (3 d.p.)

Explanation

Substitute each value back into the formula, with the calculator in radian mode and keeping the unrounded value each time, until the value is stable to 3 decimal places. Here the iterates oscillate either side of the root and settle quickly at 1.728. Common errors are working in degrees and not giving the number of decimal places asked for.

Mark scheme

1 mark
M1 Attempts the iteration in radians: \(x_2 = \sqrt{2 + \sin 1.5}\); or correct answers \(x_2 = 1.7313\), \(x_3 = 1.7283\) and \(\beta = 1.728\) with no working (4 marks)
An answer of \(x_2 = 1.7313\), \(x_3 = 1.7283\) and \(\beta = 1.728\) with no working scores 4 marks, as the iteration is implied by the values. M1 may be implied by awrt 1.731, i.e. [1.7305, 1.7315). Working in degrees (\(x_2 = 1.4234\)) is M0.
1 mark
A1 \(x_2 = 1.7313\)
cao, to 4 decimal places: 1.731 or 1.73133 alone is A0.
1 mark
A1 \(x_3 = 1.7283\)
cao, to 4 decimal places. Requires the M mark for the iteration (it does not depend on the previous A mark).
1 mark
A1 \(\beta = 1.728\)
cao. Requires the M mark. The iterates oscillate either side of the root (1.7313, 1.7283, 1.7285, 1.7285, ...) and settle quickly.

Question 3

6 marks

The curves with equations \(y = \ln(x + 2)\) and \(y = \dfrac{1}{x}\), \(x > 0\), meet at the point \(P\).
Show that the x-coordinate of \(P\) satisfies the equation
\[x = \dfrac{1}{\ln(x + 2)}\]
Using the iterative formula \(x_{n+1} = \dfrac{1}{\ln(x_n + 2)}\) with \(x_1 = 1\), find \(x_2\) and \(x_3\), giving each answer to 4 decimal places.
By repeated iteration, find the coordinates of \(P\), giving each coordinate to 3 decimal places.

Model answer

At \(P\): \(\ln(x + 2) = \dfrac{1}{x}\), so \(x\ln(x + 2) = 1\) and \(x = \dfrac{1}{\ln(x + 2)}\)
\(x_2 = \dfrac{1}{\ln 3} = 0.9102\) (4 d.p.)
\(x_3 = \dfrac{1}{\ln 2.91023...} = 0.9361\) (4 d.p.)
Continuing: 0.9284, 0.9307, 0.9300, 0.9302, 0.9302, ...
x-coordinate of \(P\) = 0.930 (3 d.p.), and \(y = \dfrac{1}{0.93017...} = 1.075\) (3 d.p.)
\(P(0.930, 1.075)\)

Explanation

At an intersection the two expressions for \(y\) are equal, which rearranges to the iterative form. The values alternate above and below the root while closing in on it, so iterate until the third decimal place is stable, then find \(y\) from either equation using the unrounded \(x\). A common error is giving 0.93 instead of 0.930.

Mark scheme

1 mark
B1 Sets \(\ln(x + 2) = \dfrac{1}{x}\) and rearranges to \(x = \dfrac{1}{\ln(x + 2)}\) with no errors. An answer with no working scores 0 marks for this mark.
An answer with no working scores 0 marks for the show that. Correct values of \(x_2\), \(x_3\) and the coordinates of \(P\) with no working still score the remaining 5 marks, as the iteration is implied.
1 mark
M1 Attempts the iteration: \(x_2 = \dfrac{1}{\ln 3}\)
May be implied by awrt 0.910, i.e. [0.9095, 0.9105).
1 mark
A1 \(x_2 = 0.9102\)
cao, to 4 decimal places.
1 mark
A1 \(x_3 = 0.9361\)
cao, to 4 decimal places. Requires the M mark for the iteration (it does not depend on the previous A mark).
1 mark
A1 x-coordinate of \(P\): 0.930
cao. Requires the M mark. 0.93 is A0: 3 decimal places are required.
1 mark
B1 y-coordinate of \(P\): 1.075, so \(P(0.930, 1.075)\)
From \(\dfrac{1}{0.93017...}\) or \(\ln(2.93017...)\); accept awrt 1.075, i.e. [1.0745, 1.0755).

Question 4

7 marks

A garden pond is modelled as a hemisphere of radius 1.5 m, with its flat face horizontal at the top. When the water in the pond is \(h\) metres deep, the volume of water, \(V\) m\(^3\), is given by
\[V = \dfrac{\pi}{3}h^2(4.5 - h), \quad 0 \leqslant h \leqslant 1.5\]
The pond contains 4 m\(^3\) of water.
Show that \(h\) satisfies the equation
\[h = \sqrt{\dfrac{12}{\pi(4.5 - h)}}\]
Using the iterative formula \(h_{n+1} = \sqrt{\dfrac{12}{\pi(4.5 - h_n)}}\) with \(h_1 = 1\), find \(h_2\) and \(h_3\), giving each to 4 decimal places.
By repeated iteration, find the depth of water in the pond, in metres to 3 decimal places.
The equation \(\dfrac{\pi}{3}h^2(4.5 - h) = 4\) also has a root between 4 and 5. Explain why this root does not give the depth of water in the pond.

Model answer

\(\dfrac{\pi}{3}h^2(4.5 - h) = 4 \Rightarrow h^2 = \dfrac{12}{\pi(4.5 - h)}\)
Since \(h > 0\), \(h = \sqrt{\dfrac{12}{\pi(4.5 - h)}}\)
\(h_2 = \sqrt{\dfrac{12}{\pi(3.5)}} = 1.0447\) (4 d.p.)
\(h_3 = \sqrt{\dfrac{12}{\pi(4.5 - 1.04467...)}} = 1.0514\) (4 d.p.)
Continuing: 1.0524, 1.0526, 1.0526, ...
Depth of water = 1.053 m (3 d.p.)
The pond is a hemisphere of radius 1.5 m, so the depth of water cannot be more than 1.5 m; a root between 4 and 5 is outside \(0 \leqslant h \leqslant 1.5\), so it cannot be the depth.

Explanation

Set the volume formula equal to 4 and rearrange to \(h^2 = \dfrac{12}{\pi(4.5 - h)}\), taking the positive square root because \(h\) is a depth. The iteration converges quickly to 1.053 m. A cubic equation from a model often has roots that make no sense in context, and here the formula only applies for \(0 \leqslant h \leqslant 1.5\).

Mark scheme

1 mark
M1 Sets \(\dfrac{\pi}{3}h^2(4.5 - h) = 4\) and rearranges to make \(h^2\) the subject: \(h^2 = \dfrac{12}{\pi(4.5 - h)}\). An answer with no working scores 0 marks for the show that.
An answer with no working scores 0 marks for the show that; correct values \(h_2 = 1.0447\), \(h_3 = 1.0514\) and the depth 1.053 m with no working still score the 4 iteration marks, as the iteration is implied. The last mark is marked on its own.
1 mark
A1* \(h = \sqrt{\dfrac{12}{\pi(4.5 - h)}}\), taking the positive square root as \(h > 0\), cso
The printed equation must follow from exact working.
1 mark
M1 Attempts the iteration: \(h_2 = \sqrt{\dfrac{12}{3.5\pi}}\)
May be implied by awrt 1.045, i.e. [1.0445, 1.0455).
1 mark
A1 \(h_2 = 1.0447\)
cao, to 4 decimal places.
1 mark
A1 \(h_3 = 1.0514\)
cao, to 4 decimal places. Requires the M mark for the iteration (it does not depend on the previous A mark).
1 mark
A1 Depth = 1.053 m
cao. Requires the M mark. Units are not required. 1.05 is A0: 3 decimal places are required.
1 mark
B1 The depth cannot be more than 1.5 m, the radius of the hemisphere (the formula is only valid for \(0 \leqslant h \leqslant 1.5\)), so a root between 4 and 5 is impossible as a depth
Needs a reference to the maximum depth 1.5 m (the radius) or the stated range of \(h\).

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