Iteration – GCSE Maths exam questions with mark scheme

1Grade 6–7HigherCalculator4 marks

Using $x_{n+1} = \sqrt[3]{7 - 2x_n}$ with $x_0 = 1.5$

(a) find the values of $x_1$, $x_2$ and $x_3$. Give your answers correct to 4 decimal places. [3]

(b) $x_3$ is an estimate of a solution of an equation. Write down this equation in the form $x^3 + ax + b = 0$ [1]

(a) for substituting $x_0 = 1.5$ correctly: $x_1 = \sqrt[3]{7 - 2 \times 1.5} = 1.5874\ldots$M1
(a) for continuing the iteration with their $x_1$ to find $x_2$ ($1.5639\ldots$)M1
(a) $x_1 = 1.5874$, $x_2 = 1.5639$, $x_3 = 1.5703$A1
(b) $x^3 + 2x - 7 = 0$B1
Answer: (a) $1.5874$, $1.5639$, $1.5703$ (b) $x^3 + 2x - 7 = 0$
Examiner tip: Use your calculator's ANS key so you carry full accuracy into each step; only round the values you write down.

2Grade 6–7HigherCalculator3 marks

(a) Show that the equation $x^3 + 2x - 5 = 0$ has a solution between $x = 1$ and $x = 2$ [2]

(b) Show that the equation $x^3 + 2x - 5 = 0$ can be rearranged to give $x = \sqrt[3]{5 - 2x}$ [1]

(a) for substituting $x = 1$ and $x = 2$: $1^3 + 2 - 5 = -2$ and $2^3 + 4 - 5 = 7$M1
(a) for both values correct with a conclusion referring to the change of sign, e.g. 'one is negative and one is positive, so there is a solution between 1 and 2'A1
(b) for $x^3 = 5 - 2x$ then taking the cube root, $x = \sqrt[3]{5 - 2x}$, with no errorsC1
Answer: (a) $-2$ and $7$: sign change, so a root lies between 1 and 2 (b) shown
Examiner tip: Both values AND the words 'change of sign' (or equivalent) are needed for the second mark.

3Grade 8–9HigherCalculator6 marks

(a) Show that the equation $x^3 - 6x + 2 = 0$ can be written as $x = \dfrac{x^3 + 2}{6}$ [1]

(b) Starting with $x_0 = 0$, use the iteration formula $x_{n+1} = \dfrac{x_n^{\,3} + 2}{6}$ three times to find an estimate for a solution of $x^3 - 6x + 2 = 0$. Give your answer correct to 4 decimal places. [3]

(c) By substituting suitable values into $x^3 - 6x + 2$, show that this solution is $0.34$ correct to 2 decimal places. [2]

(a) for $6x = x^3 + 2$ then dividing by 6C1
(b) for $x_1 = \tfrac{2}{6} = 0.3333\ldots$M1
(b) for $x_2 = 0.3395\ldots$M1
(b) $x_3 = 0.3399$A1
(c) for substituting $x = 0.335$ and $x = 0.345$: $0.0275\ldots$ and $-0.0289\ldots$ (allow 1 sf)M1
(c) for a conclusion: there is a change of sign between $0.335$ and $0.345$, so the solution rounds to $0.34$A1
Answer: (a) shown (b) $0.3399$ (c) $f(0.335) > 0$, $f(0.345) < 0$, so solution $= 0.34$ (2 dp)
Examiner tip: To prove a value to 2 dp you test the BOUNDS ($0.335$ and $0.345$), not $0.34$ itself.

4Grade 6–7HigherCalculator4 marks

Using $x_{n+1} = \sqrt[3]{4x_n + 1}$ and $x_1 = 2$

(a) find the values of $x_2$ and $x_3$. Give your answers correct to 4 decimal places. [2]

(b) The values of $x_n$ approach a solution of an equation. Find this equation, giving your answer in the form $x^3 + ax + b = 0$, where $a$ and $b$ are integers. [2]

(a) for $x_2 = \sqrt[3]{4 \times 2 + 1} = 2.0800\ldots$M1
(a) $x_2 = 2.0801$ and $x_3 = 2.1045$A1
(b) for $x = \sqrt[3]{4x + 1}$ leading to $x^3 = 4x + 1$M1
(b) $x^3 - 4x - 1 = 0$A1
Answer: (a) $x_2 = 2.0801$, $x_3 = 2.1045$ (b) $x^3 - 4x - 1 = 0$
Examiner tip: Here the starting value is $x_1$, so the next value is $x_2$ – read the subscript you are given.

5Grade 6–7HigherCalculator4 marks

The number of fish in a lake at the start of year $n$ is $P_n$.

A model for the number of fish is

$$P_{n+1} = 1.2P_n - 50, \qquad P_0 = 500$$

(a) Work out the number of fish in the lake at the start of year 3. [2]

(b) For a different starting value $P_0$, the number of fish stays the same every year. Find this value of $P_0$. [2]

(a) for $P_1 = 1.2 \times 500 - 50 = 550$ and continuing, e.g. $P_2 = 610$M1
(a) $682$A1
(b) for $P = 1.2P - 50$M1
(b) $250$A1
Answer: (a) $682$ (b) $250$
Examiner tip: 'Start of year 3' is $P_3$ – that needs THREE applications of the formula from $P_0$.

6Grade 8–9HigherCalculator5 marks

(a) Show that the equation $x^3 - 2x^2 - 3 = 0$ can be rearranged to give $x = 2 + \dfrac{3}{x^2}$ [2]

(b) Using $x_{n+1} = 2 + \dfrac{3}{x_n^{\,2}}$ with $x_0 = 2$, find the values of $x_1$, $x_2$ and $x_3$. Give $x_3$ correct to 4 decimal places. [3]

(a) for $x^3 = 2x^2 + 3$ or dividing every term by $x^2$: $x - 2 - \dfrac{3}{x^2} = 0$M1
(a) for completing to $x = 2 + \dfrac{3}{x^2}$ with no errorsA1
(b) for $x_1 = 2 + \tfrac{3}{4} = 2.75$M1
(b) for $x_2 = 2.3966\ldots$ (ft their $x_1$)M1
(b) $x_3 = 2.5223$A1
Answer: (a) shown (b) $x_1 = 2.75$, $x_2 = 2.3967$, $x_3 = 2.5223$
Examiner tip: In (a) divide EVERY term by $x^2$ – a common slip is to leave the constant 3 undivided.

7Grade 4–5HigherNon-calculator3 marks

A sequence is defined by the iteration formula

$$x_{n+1} = 4 - \frac{1}{x_n}, \qquad x_0 = 1$$

Work out the value of $x_3$.

Give your answer as a fraction in its simplest form. [3]

for $x_1 = 4 - \tfrac{1}{1} = 3$M1
for $x_2 = 4 - \tfrac{1}{3} = \tfrac{11}{3}$ (ft their $x_1$; allow $3\tfrac{2}{3}$)M1
$x_3 = \dfrac{41}{11}$ (accept $3\tfrac{8}{11}$; do not accept a rounded decimal such as $3.73$)A1
Answer: $x_3 = \dfrac{41}{11}$
Examiner tip: The reciprocal of $\tfrac{11}{3}$ is $\tfrac{3}{11}$: students who write $x_3 = 4 - \tfrac{11}{3}$ instead of $4 - \tfrac{3}{11}$ lose the accuracy mark.

8Grade 6–7HigherCalculator5 marks

The equation $x^2 + x - 10 = 0$ has a positive solution.

This solution can be found using the iteration formula

$$x_{n+1} = \sqrt{10 - x_n}, \qquad x_0 = 3$$

(a) Use the iteration formula to find the positive solution correct to 2 decimal places. You must show enough values of $x_n$ to justify your answer. [3]

(b) Use the quadratic formula to find the exact positive solution of $x^2 + x - 10 = 0$, and use it to check your answer to part (a). [2]

(a) for $x_1 = \sqrt{10 - 3} = 2.6457\ldots$M1
(a) for continuing correctly, e.g. $x_2 = 2.7118\ldots$, $x_3 = 2.6996\ldots$, $x_4 = 2.7019\ldots$ (ft their previous value)M1
(a) $2.70$ with at least two consecutive values that both round to $2.70$ shown (e.g. $x_3$ and $x_4$)A1
(b) for correct substitution into the quadratic formula: $\dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-10)}}{2}$M1
(b) $\dfrac{-1 + \sqrt{41}}{2}$ $(= 2.7015\ldots)$ with a comment that this agrees with $2.70$ (ignore the negative root if also given)A1
Answer: (a) $2.70$ (b) $\dfrac{-1+\sqrt{41}}{2} = 2.70156\ldots$, which rounds to $2.70$
Examiner tip: Stop only when two successive values round to the same 2 dp answer; $x_2 = 2.71$ and $x_3 = 2.70$ disagree, so keep going.

9Grade 4–5HigherNon-calculator4 marks

The equation $x^3 - 2x = 30$ has exactly one solution.

(a) Find the two consecutive whole numbers between which the solution lies. You must show your working. [2]

(b) By substituting $x = 3.5$, decide whether the solution is closer to the lower or to the higher of your two whole numbers. Give a reason for your answer. [2]

(a) for substituting two consecutive integers into $x^3 - 2x$ and comparing with 30, e.g. $x = 3$: $21$ and $x = 4$: $56$ (or into $x^3 - 2x - 30$: $-9$ and $26$)M1
(a) 3 and 4, from $21 < 30 < 56$ (or a stated change of sign)A1
(b) for $3.5^3 - 2 \times 3.5 = 35.875$ (or $x^3 - 2x - 30 = 5.875$)M1
(b) closer to 3, because $35.875 > 30$ so the solution lies between 3 and 3.5 (ft their integers in (a))A1
Answer: (a) 3 and 4 (b) closer to 3, since the solution lies between 3 and 3.5
Examiner tip: Compare each value with 30 (or look for a sign change in $x^3 - 2x - 30$); just listing $21$ and $56$ without saying what they show can lose the accuracy mark.

10Grade 6–7HigherCalculator5 marks

The graph of $y = x^3 - 5x + 3$ is drawn for values of $x$ from $-3$ to $3$.

(a) Use the graph to write down the number of solutions of the equation $x^3 - 5x + 3 = 0$ [1]

(b) Use the graph to estimate the negative solution of $x^3 - 5x + 3 = 0$ [1]

(d) Starting with $x_0 = 2$, use the iteration formula $x_{n+1} = \sqrt[3]{5x_n - 3}$ to find $x_3$. Give your answer correct to 4 decimal places. [2]

(a) 3B1
(b) answer in the range $-2.6$ to $-2.4$B1
(c) for $x^3 = 5x - 3$ then taking the cube root of both sidesC1
(d) for $x_1 = \sqrt[3]{7} = 1.9129\ldots$M1
(d) $x_3 = 1.8530$ (accept $1.853$; values seen $x_2 = 1.8724\ldots$)A1
Answer: (a) 3 (b) about $-2.5$ (c) shown (d) $1.8530$
Examiner tip: The number of solutions is the number of times the curve crosses the $x$-axis, not the number of turning points.

11Grade 8–9HigherCalculator5 marks

The equation $x^2 - 5x + 2 = 0$ has one solution between 0 and 1 and one solution between 4 and 5.

(a) Show that $x^2 - 5x + 2 = 0$ can be rearranged to give $x = 5 - \dfrac{2}{x}$ [1]

(b) Using $x_{n+1} = 5 - \dfrac{2}{x_n}$ with $x_0 = 4$, find the solution between 4 and 5 correct to 2 decimal places. Show enough values of $x_n$ to justify your answer. [3]

His first three values are $3.6$, $2.992$ and $2.1904$

Sam says, “If I keep going, my values will reach the solution between 4 and 5.”

Is Sam correct? Give a reason for your answer. [1]

(a) for dividing every term by $x$: $x - 5 + \dfrac{2}{x} = 0$, then rearranging to $x = 5 - \dfrac{2}{x}$ (allow from $x^2 = 5x - 2$ then $\div x$)C1
(b) for $x_1 = 4.5$M1
(b) for $x_2 = 4.5555\ldots$ and $x_3 = 4.5609\ldots$ (ft their previous value)M1
(b) $4.56$ with at least two consecutive values rounding to $4.56$ shownA1
(c) No, with a reason, e.g. the values are decreasing / moving away from the solution between 4 and 5 (they are heading towards the solution between 0 and 1)C1
Answer: (a) shown (b) $4.56$ (c) No: the values move away from $4$–$5$ (they converge to the solution near $0.44$)
Examiner tip: Not every rearrangement finds every solution: always check whether the values are getting closer to the root you want.

12Grade 6–7HigherCalculator5 marks

(a) Show that the equation $x^3 + 4x - 3 = 0$ can be rearranged to give $x = \dfrac{3}{x^2 + 4}$ [2]

(b) Starting with $x_0 = 1$, use the iteration formula $x_{n+1} = \dfrac{3}{x_n^{\,2} + 4}$ to find the values of $x_1$, $x_2$ and $x_3$. Give $x_2$ and $x_3$ correct to 4 decimal places. [2]

(a) for $x^3 + 4x = 3$ and factorising to $x(x^2 + 4) = 3$M1
(a) for dividing both sides by $x^2 + 4$ to give $x = \dfrac{3}{x^2 + 4}$ with no errorsA1
(b) for $x_1 = \dfrac{3}{1 + 4} = 0.6$M1
(b) $x_2 = 0.6881$ and $x_3 = 0.6706$A1
(c) they are alternately above and below the solution (oscillate) and get closer each time (accept “zig-zag towards it”; do not accept just “they get closer”)B1
Answer: (a) shown (b) $0.6$, $0.6881$, $0.6706$ (c) alternately above and below, getting closer to the solution ($\approx 0.674$)
Examiner tip: In (a) you must factorise out $x$ first; dividing $x^3 + 4x = 3$ by $x^2$ does not lead to the given form.

13Grade 6–7HigherCalculator6 marks

A patient takes a 40 mg dose of a medicine every morning. Between one morning dose and the next, 30% of the medicine in the body is removed.

$D_n$ is the amount of medicine, in mg, in the body just after the dose on day $n$.

$$D_{n+1} = 0.7D_n + 40, \qquad D_0 = 40$$

(a) Explain why the formula uses $0.7$ [1]

(b) Find the first day on which $D_n$ is greater than 120 mg. [3]

Show that, using this model, it never will. [2]

(a) 30% is removed so 70% (= 0.7) of the medicine remains each dayB1
(b) for $D_1 = 0.7 \times 40 + 40 = 68$M1
(b) for continuing correctly, e.g. $D_2 = 87.6$, $D_3 = 101.32$, $D_4 = 110.92\ldots$M1
(b) day 6, with $D_5 = 117.6\ldots$ and $D_6 = 122.3\ldots$ seenA1
(c) for setting $D = 0.7D + 40$ (or $0.3D = 40$)M1
(c) $D = 133.3\ldots$ with a conclusion, e.g. the amount levels off at $133.3$ mg, which is less than 140 mg (allow an argument showing the values increase towards 133.3)A1
Answer: (a) 70% of the medicine remains each day (b) Day 6 ($D_6 = 122.35$) (c) it levels off at $133\tfrac{1}{3}$ mg $< 140$ mg
Examiner tip: Show $D_5$ as well as $D_6$: the mark depends on proving day 6 is the FIRST day above 120 mg.

14Grade 6–7HigherCalculator3 marks

Ben uses the iteration formula $x_{n+1} = \sqrt[3]{2x_n + 9}$ with $x_0 = 2$

Here is his working.

$x_1 = \sqrt[3]{2 \times 2 + 9} = 2.3513$
$x_2 = \sqrt[3]{2 \times 2.3513} + 9 = 10.6754$

(a) Describe the mistake Ben has made. [1]

(b) Work out the correct values of $x_2$ and $x_3$. Give your answers correct to 4 decimal places. [2]

(a) he added 9 after taking the cube root; the 9 should be added to $2x_1$ inside the cube root (accept “cube root should be of $2 \times 2.3513 + 9$”)B1
(b) for $x_2 = \sqrt[3]{2 \times 2.3513\ldots + 9} = 2.3929\ldots$M1
(b) $x_2 = 2.3930$ and $x_3 = 2.3978$A1
Answer: (a) he added 9 outside the cube root instead of inside it (b) $x_2 = 2.3930$, $x_3 = 2.3978$
Examiner tip: Use brackets on your calculator — type $\sqrt[3]{(2 \times \text{Ans} + 9)}$ — so the whole expression is inside the root.

15Grade 8–9HigherCalculator7 marks

A cuboid has a square base of side $x$ cm and a height of $(x + 3)$ cm.

The volume of the cuboid is 50 cm³.

(a) Show that $x = \sqrt{\dfrac{50}{x + 3}}$ [2]

(b) Starting with $x_1 = 3$, use the iteration formula $x_{n+1} = \sqrt{\dfrac{50}{x_n + 3}}$ to find the value of $x$ correct to 2 decimal places. Show enough values to justify your answer. [3]

(c) By substituting suitable values into $x^3 + 3x^2 - 50$, show that your answer to (b) is correct to 2 decimal places. [2]

(a) for forming $x^2(x + 3) = 50$M1
(a) for $x^2 = \dfrac{50}{x + 3}$ then square rooting, taking the positive root as $x$ is a lengthA1
(b) for $x_2 = \sqrt{\tfrac{50}{6}} = 2.8867\ldots$M1
(b) for continuing correctly, e.g. $x_3 = 2.9143\ldots$, $x_4 = 2.9075\ldots$, $x_5 = 2.9092\ldots$ (ft their previous value)M1
(b) $2.91$ with at least two consecutive values rounding to $2.91$ shownA1
(c) for substituting $x = 2.905$ and $x = 2.915$: $-0.16\ldots$ and $0.26\ldots$ (allow 1 sf)M1
(c) for a conclusion: change of sign between $2.905$ and $2.915$, so the solution is $2.91$ to 2 dpA1
Answer: (a) shown (b) $2.91$ (c) $f(2.905) = -0.168 < 0$, $f(2.915) = 0.261 > 0$: sign change, so $x = 2.91$ (2 dp)
Examiner tip: Testing $x = 2.91$ alone proves nothing: you must test the bounds $2.905$ and $2.915$ and state the sign change.

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