9 Numerical methods — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 9. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

9.1 · Locate roots of f(x) = 0 by considering changes of sign of f(x) in an interval on which f(x) is sufficiently well behaved; understand how change of sign methods can fail.

Explanation

If a continuous function has values of opposite sign at the endpoints of an interval, the intermediate value theorem guarantees at least one root inside that interval.
Evaluate $f(a)$ and $f(b)$ accurately, record their signs, and conclude that a root lies in $(a,b)$ only after checking that $f$ is continuous there.
For example, $f(1)<0$ and $f(2)>0$ for a continuous $f$ , so the graph must cross the $x$ -axis at least once between $1$ and $2$ .
A sign change across a discontinuity need not contain a root, while a repeated root can touch the axis without changing sign; a common error is to treat the sign test as an equivalence.

Worked example

For $f(x)=x^3+x-5$ , show that there is exactly one root in $(1,2)$ .

1.Calculate $f(1)=1+1-5=-3$ and $f(2)=8+2-5=5$ .
2.Continuity gives at least one root in $(1,2)$ .
3.Also $f'(x)=3x^2+1$ , which is positive for every real $x$ , so $f$ is strictly increasing and cannot cross the axis more than once.

Answer: $f(1)=-3$ and $f(2)=5$ , so a root lies in $(1,2)$ .; Since $f'(x)=3x^2+1>0$ , $f$ is strictly increasing and the root is unique.

Common mistakes

Don't conclude that an even-multiplicity root cannot exist because endpoint values have the same sign.
Don't claim a root from a sign change without checking continuity, or claim uniqueness without a monotonicity argument.

Exam tip

To establish exactly one root, combine a sign change on a continuous interval with a separate uniqueness argument.

Tier 1 · Easy

1.
Let $f(x)=x^2-x-1$ . Use endpoint values to establish that $(1,2)$ brackets a zero of $f$ .
(2)
(Total for Question 1 is 2 marks)
2.
Let $p(x)=x^3-4x+1$ . Show that $p(x)=0$ has a root in the interval $(0,1)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The function $f(x)=(x-1)^2$ has a root in the interval $[0,2]$ . Explain why a search based only on a change of sign between the endpoints does not detect this root.
(3)
(Total for Question 1 is 3 marks)
2.
For $-1\leq x\leq0$ , let $f(x)=x+\cos x$ . Show that $f(x)=0$ has exactly one root in $(-1,0)$ .
(4)
(Total for Question 2 is 4 marks)
3.
For $f(x)=x^3+x-a$ , where $a$ is a constant, find the range of values of $a$ for which the sign-change test on the interval $(1,2)$ guarantees a root of $f(x)=0$ in that interval.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Two sign tests are proposed. For $h(x)=\frac{1}{x-2}$ , the values $h(1)$ and $h(3)$ have opposite signs. For $k(x)=(x-2)^2$ , the values $k(1.9)$ and $k(2.1)$ have the same sign. Explain why neither test gives the correct conclusion about roots in the stated intervals.
(4)
(Total for Question 1 is 4 marks)
2.
Let $g(x)=x^3-3x+1$ . By considering changes of sign on three disjoint intervals, prove that $g(x)=0$ has exactly three real roots.
(5)
(Total for Question 2 is 5 marks)
3.
Let $h(x)=e^x-3x$ . By considering endpoint values and the sign of $h'(x)$ , show that $h(x)=0$ has exactly two roots in the interval $(0,2)$ .
(5)
(Total for Question 3 is 5 marks)
4.
Let $f(x)=x^3+x-1$ . Establish the existence and uniqueness of a root in $(0.6,0.7)$ . By evaluating $f(0.68)$ and $f(0.685)$ , give this root to $2$ decimal places.
(6)
(Total for Question 4 is 6 marks)
5.
A function $g$ is continuous on $[-3,2]$ . Some of its values are shown: $g(-3)=2$ , $g(-2)=-1$ , $g(-1)=4$ , $g(0)=0$ , $g(1)=3$ and $g(2)=-2$ . Determine the least number of distinct roots that $g(x)=0$ must have in $[-3,2]$ . Identify where they occur and explain why the information does not establish the exact number of roots.
(5)
(Total for Question 5 is 5 marks)

9.2 · Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams.

Explanation

A fixed-point iteration rewrites an equation as $x=g(x)$ and generates approximations using $x_{n+1}=g(x_n)$ from a chosen starting value.
Calculate successive values without premature rounding; on a graph of $y=g(x)$ and $y=x$ , move vertically to $y=g(x)$ and horizontally to $y=x$ to display each iteration.
Near a fixed point $\alpha$ , iterations usually converge when $|g'(\alpha)|<1$ : a positive gradient gives a staircase pattern and a negative gradient gives a cobweb pattern.
Different rearrangements of the same equation can converge at different rates or diverge; a common error is to assume that obtaining $x=g(x)$ automatically produces a useful recurrence.

A staircase diagram alternates between

y=g(x)

and

y=x

to display successive fixed-point iterates.

Worked example

Use $x_{n+1}=\sqrt{7-x_n}$ with $x_0=2$ to calculate $x_1$ , $x_2$ and $x_3$ to $4$ decimal places. State the equation satisfied by any limiting value.

1.Repeated substitution gives $x_1=\sqrt5=2.236067\ldots$ , $x_2=\sqrt{7-2.236067\ldots}=2.182643\ldots$ and $x_3=\sqrt{7-2.182643\ldots}=2.194847\ldots$ .
2.If the sequence tends to $L$ , then $L=\sqrt{7-L}$ , so $L^2+L-7=0$ .

Answer: $x_1=2.2361$ , $x_2=2.1826$ and $x_3=2.1948$ .; A limiting value satisfies $x^2+x-7=0$ .

Common mistakes

Don't draw horizontal moves to $y=g(x)$ and vertical moves to $y=x$ , reversing the cobweb construction.
Don't calculate iterates and fail to state the fixed-point equation or assess convergence from the iteration graph.

Exam tip

Show each substituted iterate to the requested accuracy, then set the limiting value equal to the iteration function.

Tier 1 · Easy

1.
The recurrence $x_{n+1}=\frac{1}{2}\left(x_n+\frac{6}{x_n}\right)$ is used with $x_0=2$ . Find $x_1$ and $x_2$ .
(2)
(Total for Question 1 is 2 marks)
2.
The recurrence $x_{n+1}=\sqrt{11-x_n}$ is used with $x_0=3$ . Find $x_1$ and $x_2$ , giving each answer to $4$ decimal places.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
On a pair of axes, draw $y=x$ and $y=2+\dfrac14x$ . Starting at $x_0=0$ on the $x$ -axis, draw the first three steps of the fixed-point construction. State whether the construction is a staircase or a cobweb, and find its limiting value.
(4)
(Total for Question 1 is 4 marks)
2.
Starting with $x_0=0$ , use $x_{n+1}=1+\dfrac{6}{x_n+4}$ to find $x_1$ , $x_2$ and $x_3$ . Find the positive limiting value and state whether the associated construction is a staircase or a cobweb.
(4)
(Total for Question 2 is 4 marks)
3.
The recurrence $x_{n+1}=\ln(x_n+6)$ is used with $x_0=1$ and is known to converge to a fixed point $\alpha$ . Write down an equation of the form $f(x)=0$ whose root is $\alpha$ . Calculate $x_1$ , $x_2$ and $x_3$ , giving each value to $4$ decimal places. The values satisfy $x_0<x_1<x_2<x_3<\alpha$ . Without using derivatives, explain why the associated fixed-point path is a staircase rather than a cobweb.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For $x_{n+1}=3-0.4x_n$ with $x_0=1$ , find $x_1$ to $x_4$ . Determine the limiting value and state whether the graphical construction is a cobweb or a staircase, giving a reason.
(5)
(Total for Question 1 is 5 marks)
2.
The positive root $\alpha$ of $x^2+x-4=0$ lies in $(1.5,1.6)$ . Two proposed iterations are $x_{n+1}=4-x_n^2$ and $x_{n+1}=\sqrt{4-x_n}$ . Without carrying out a long sequence of iterations, determine which recurrence is locally convergent to $\alpha$ . State whether its graphical construction is a staircase or a cobweb.
(5)
(Total for Question 2 is 5 marks)
3.
The recurrence $x_{n+1}=\dfrac{3}{x_n+1}$ is used with $x_0=1$ . Use exact arithmetic to find $x_1$ , $x_2$ , $x_3$ and $x_4$ . Find the positive fixed point. Use the pattern in these four iterates to identify the type of fixed-point diagram and state whether convergence is indicated.
(5)
(Total for Question 3 is 5 marks)
4.
A fixed-point iteration has the linear form $x_{n+1}=a+bx_n$ . On its fixed-point diagram, the first vertical step from $x_0=1$ reaches $(1,1.8)$ , and the next vertical step reaches $(1.8,2.04)$ . Find $a$ and $b$ . Calculate $x_3$ , find the limiting value, and classify the diagram as a convergent staircase or a convergent cobweb.
(5)
(Total for Question 4 is 5 marks)
5.
The iteration $x_{n+1}=2-e^{-x_n}$ starts from $x_0=1$ . Use unrounded values in your working. Calculate $x_1$ to $x_6$ , giving each value to $6$ decimal places, and find the least value of $n$ for which $|x_{n+1}-x_n|<0.0005$ . Hence estimate the fixed point to $4$ decimal places and state the type of fixed-point diagram.
(6)
(Total for Question 5 is 6 marks)

9.3 · Solve equations using the Newton-Raphson method and other recurrence relations of the form xₙ₊₁ = g(xₙ); understand how such methods can fail.

Explanation

Newton-Raphson replaces the curve locally by its tangent, giving $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ for a root of $f(x)=0$ .
Differentiate first, substitute the current approximation into both $f$ and $f'$ , and retain extra calculator digits until the requested final accuracy.
For instance, applying Newton-Raphson to $f(x)=x^2-a$ gives $x_{n+1}=\frac12\left(x_n+\frac{a}{x_n}\right)$ .
The method can fail when $f'(x_n)=0$ , when a tangent sends the next value away from the desired root, or when the values enter a cycle; a common error is to continue without checking the iterates.
Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ .

Newton–Raphson uses the tangent at

x_n

; its intercept with the

x

-axis is the next iterate

x_{n+1}

Worked example

Derive a Newton-Raphson recurrence for $x^3+x-4=0$ . Starting with $x_0=1.3$ , find $x_1$ and $x_2$ to $5$ decimal places.

1.Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ .
2.Substitution in the Newton-Raphson formula gives $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .
3.Using $x_0=1.3$ gives $x_1=1.382866\ldots$ , and substituting this unrounded value gives $x_2=1.378806\ldots$ .

Answer: $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .; $x_1=1.38287$ and $x_2=1.37881$ .

Common mistakes

Don't choose a starting value where $f'(x_0)=0$ , making the first Newton step undefined.
Don't use the Newton formula with the derivative evaluated at the new iterate rather than the current one.

Exam tip

Write the Newton recurrence explicitly, retain guard digits between iterations and round only the reported values.

Tier 1 · Easy

1.
Use one Newton-Raphson step on $f(x)=x^2-7$ starting from $x_0=2.5$ . Find $x_1$ .
(2)
(Total for Question 1 is 2 marks)
2.
Use one Newton-Raphson step on $f(x)=e^x-3$ , starting with $x_0=1$ . Give $x_1$ to $4$ decimal places.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Newton-Raphson is applied to $f(x)=x^3-2x+2$ starting from $x_0=0$ . Calculate $x_1$ , $x_2$ and $x_3$ , and explain why the method fails to converge to a root from this starting value.
(4)
(Total for Question 1 is 4 marks)
2.
Derive a Newton-Raphson recurrence for $x^3+x^2-6=0$ . Starting with $x_0=1.5$ , find $x_1$ and $x_2$ , giving each answer to $4$ decimal places.
(4)
(Total for Question 2 is 4 marks)
3.
Newton-Raphson is used to solve $\ln x+x-3=0$ . Derive the recurrence and, starting with $x_0=2$ , find $x_1$ and $x_2$ , giving each value to $5$ decimal places. Use unrounded values in your working.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Newton-Raphson is applied to $f(x)=x^2-a$ , where $a>0$ , starting from $x_0=4$ . The first iterate is $x_1=\dfrac{13}{4}$ . Find $a$ . Then calculate $x_2$ exactly and $x_3$ to six decimal places, and hence give the positive root of $f(x)=0$ correct to four decimal places.
(5)
(Total for Question 1 is 5 marks)
2.
Newton-Raphson is to be used to solve $x^3-3x+3=0$ . Explain why $x_0=1$ cannot be used. Then start with $x_0=-2$ and calculate $x_1$ and $x_2$ , giving each value to $5$ decimal places.
(5)
(Total for Question 2 is 5 marks)
3.
Consider applying Newton-Raphson to $f(x)=\sqrt[3]{x}$ , whose only real root is $0$ . For $x_n\ne0$ , derive the recurrence in its simplest form. Starting with $x_0=\dfrac18$ , find $x_1$ , $x_2$ and $x_3$ , and explain why the method does not converge to the root.
(4)
(Total for Question 3 is 4 marks)
4.
Apply Newton-Raphson to $\cos x=x$ by first deriving a recurrence. From $x_0=0.7$ , obtain the first three iterates and report each to $4$ decimal places, retaining full calculator precision throughout. Verify by a change of sign between $0.73905$ and $0.73915$ that the root is $0.7391$ to $4$ decimal places.
(6)
(Total for Question 4 is 6 marks)
5.
Newton-Raphson is to be applied to $\dfrac1x-\dfrac15=0$ . Derive the recurrence in polynomial form. Starting with $x_0=1$ , find $x_1$ and $x_2$ exactly. Then show that the starting value $x_0=10$ makes the method fail after one step, and explain the failure.
(5)
(Total for Question 5 is 5 marks)

9.4 · Understand and use numerical integration of functions, including the use of the trapezium rule and estimating the approximate area under a curve and limits that it must lie between.

Explanation

The trapezium rule approximates an integral by replacing the curve over each strip with a straight chord and adding the resulting trapezium areas.
For ordinates $y_0,\ldots,y_n$ at equal spacing $h$ , use $\frac{h}{2}[y_0+y_n+2(y_1+\cdots+y_{n-1})]$ and include each interior ordinate twice.
A convex curve with $f''(x)>0$ lies below its chords, so the trapezium rule overestimates its integral; more, narrower strips usually improve the estimate.
Numerical area is not automatically the signed integral: a common error is to leave contributions below the $x$ -axis negative when the question asks for total area.

The trapezium rule joins consecutive ordinates with straight segments; convexity indicates whether this polygonal boundary lies above or below the curve.

Worked example

Use four equal strips and the trapezium rule to estimate $\int_0^2(x^2+1)\,dx$ . State whether the estimate is an overestimate or an underestimate, with a reason.

1.Here $h=0.5$ and the ordinates at $x=0,0.5,1,1.5,2$ are $1,1.25,2,3.25,5$ .
2.Thus the estimate is $\frac{0.5}{2}[1+5+2(1.25+2+3.25)]=0.25(19)=4.75$ .
3.Since $f''(x)=2>0$ , each chord lies above the curve, so the result is an overestimate.

Answer: $4.75$ ; It is an overestimate because $f''(x)=2>0$ , so the curve is convex.

Common mistakes

Don't apply weight two to the endpoints and weight one to the interior ordinates, reversing the trapezium-rule coefficients.
Don't use the number of ordinates as the number of strips, giving the wrong strip width and endpoint weights.

Exam tip

For the trapezium rule, calculate strip width from the interval and use half weight only on the two endpoints.

Tier 1 · Easy

1.
The values of a function are $y=1,3,9$ at $x=0,1,2$ respectively. Use two strips and the trapezium rule to estimate $\int_0^2 y\,dx$ .
(2)
(Total for Question 1 is 2 marks)
2.
At $x=0,1,2$ , the corresponding ordinates of a curve are $2,k,6$ . The trapezium rule with two strips gives an estimate of $10$ for the integral from $0$ to $2$ . Find $k$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The values of a function at $x=0,1,2,3,4$ are respectively $2,2.5,4,6.5,10$ . Use the trapezium rule to estimate $\displaystyle\int_0^4 f(x)\,\mathrm dx$ .
(3)
(Total for Question 1 is 3 marks)
2.
The table gives values of $y=\sqrt{x+1}$ : for $x=0,0.5,1,1.5,2$ , the corresponding values of $y$ are $1,1.225,1.414,1.581,1.732$ . Obtain a trapezium-rule estimate for $\int_0^2\sqrt{x+1}\,\mathrm dx$ . Report the result to $3$ decimal places and state, with a reason, whether it is an overestimate or an underestimate.
(4)
(Total for Question 2 is 4 marks)
3.
The corresponding values of $x$ and $y$ are $x=0,0.5,1.5,3$ and $y=2,4,3,7$ . Using a straight-line trapezium between each pair of adjacent data points, estimate $\int_0^3 y\,\mathrm dx$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For $f(x)=\frac1x$ on $1\leq x\leq2$ , use four equal strips to find a trapezium-rule estimate. By using left- and right-endpoint rectangles, also give lower and upper bounds for $\int_1^2\frac1x\,dx$ . Give decimals to $4$ decimal places.
(6)
(Total for Question 1 is 6 marks)
2.
A continuous curve passes through the points whose coordinates are given by $x=0,1,2,3$ and $y=-2,-1,1,4$ respectively. Use the trapezium rule to estimate the signed integral from $0$ to $3$ . Then, using the straight-line trapezium approximation between each pair of adjacent points, estimate the total area between the curve and the $x$ -axis.
(5)
(Total for Question 2 is 5 marks)
3.
At $x=0,1,2,3,4$ , the ordinates of $y=f(x)$ are $8,7,9,12,14$ and the ordinates of $y=g(x)$ are $2,3,4,6,9$ . Throughout the interval, $f(x)\geq g(x)$ . Use the trapezium rule with four strips to estimate the area between the two curves. Confirm the result by separately estimating $\int_0^4 f(x)\,\mathrm dx$ and $\int_0^4 g(x)\,\mathrm dx$ .
(5)
(Total for Question 3 is 5 marks)
4.
A solid of revolution has radius $r$ at distance $x$ along its axis. For successive unit distances $x=0$ through $x=4$ , the radii are $1,2,3,2,1$ . Using $V=\pi\int_0^4 r^2\,\mathrm dx$ , obtain a four-strip trapezium estimate of the volume. Give it exactly in terms of $\pi$ and show the transformed ordinates used.
(5)
(Total for Question 4 is 5 marks)
5.
Five function values are equally spaced from $x=0$ to $x=2$ in steps of $0.5$ . The ordinates at the first, third and fifth points are $2,5,8$ . Using only these three ordinates, verify that the two-strip trapezium estimate is $10$ . The four-strip estimate using all five ordinates is $9.5$ , and the fourth ordinate is twice the second. Find the two missing ordinates.
(5)
(Total for Question 5 is 5 marks)

9.5 · Use numerical methods to solve problems in context.

Explanation

A contextual numerical model translates a physical condition, such as a target value or accumulated quantity, into a root, recurrence or numerical integral.
Define the function and units before applying the method, choose an interval or starting value that is meaningful in the context, and state the requested accuracy.
For a threshold $M(t)=C$ , solve $M(t)-C=0$ numerically; for a varying rate $q(t)$ , approximate the accumulated change with $\int q(t)\,dt$ .
A calculator value is not a complete contextual answer: a common error is to omit units, ignore the model's domain or round in a way that makes a safety decision invalid.

Worked example

A culture is modelled by $P(t)=200e^{0.3t}$ , where $t$ is in hours. Use Newton-Raphson on $f(t)=200e^{0.3t}-1000$ with $t_0=5$ to find $t_1$ and $t_2$ . Hence estimate when the culture reaches $1000$ cells.

1.Differentiate to get $f'(t)=60e^{0.3t}$ .
2.Use $t_{n+1}=t_n-\frac{200e^{0.3t_n}-1000}{60e^{0.3t_n}}$ .
3.Substitution of $t_0=5$ gives $t_1=5.385502\ldots$ , and the next substitution gives $t_2=5.364857\ldots$ .
4.The positive time is therefore about $5.365$ hours.

Answer: $t_1=5.38550$ and $t_2=5.36486$ .; The culture reaches $1000$ cells after approximately $5.365$ hours.

Common mistakes

Don't select a root that satisfies the equation but lies outside the interval allowed by the context.
Don't carry out a numerical iteration but give no contextual time, units or check that the root is meaningful.

Exam tip

After numerical iteration, translate the root back into the context and report a sensible accuracy with units.

Tier 1 · Easy

1.
A greenhouse temperature is modelled by $T(t)=18+4t-t^2$ degrees Celsius, where $t$ is hours after noon. Show that the temperature returns to $20^\circ\mathrm{C}$ at a time between $3$ and $4$ hours after noon.
(2)
(Total for Question 1 is 2 marks)
2.
A sensor output is modelled by $S(t)=t^3+t$ volts for $t\geq0$ , where $t$ is measured in seconds. Show that the output reaches $20$ volts at a time between $2$ and $3$ seconds.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A particle's height is modelled by $h(t)=20t-4.9t^2$ metres. Show that the positive time when the particle returns to the ground lies in $(4.0,4.1)$ seconds and, by considering the sign of $h(4.05)$ , show that it lies in $(4.05,4.10)$ seconds.
(4)
(Total for Question 1 is 4 marks)
2.
A curing model predicts that the completion time $t$ hours satisfies $t=2.4+0.35\ln(t+1)$ . Starting with $t_0=3$ , use $t_{n+1}=2.4+0.35\ln(t_n+1)$ to find $t_1$ , $t_2$ and $t_3$ , giving each iterate to $6$ decimal places. Hence give the predicted completion time to $3$ decimal places.
(4)
(Total for Question 2 is 4 marks)
3.
A radio tag's reliable range $R$ metres is modelled by $R=40+4\cos(R/20)$ . Starting with $R_0=45$ , use $R_{n+1}=40+4\cos(R_n/20)$ to find $R_1$ to $R_5$ , giving each iterate to $4$ decimal places. Use unrounded values in your working. Hence estimate the reliable range to the nearest $0.1$ m.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Water enters a tank at rates $10,14,18,21,19,15,11$ litres per minute at times $0,5,10,15,20,25,30$ minutes. The tank initially contains $350$ litres and has capacity $800$ litres. Use the trapezium rule to estimate whether the tank overflows during the $30$ minutes, and estimate any excess at the end.
(5)
(Total for Question 1 is 5 marks)
2.
The volume of liquid in a vessel is modelled by $V(h)=h^3+4h^2$ litres when the depth is $h$ metres, for $h>0$ . Show that the depth at which $V=70$ litres lies between $3$ m and $4$ m. Starting with $h_0=3.2$ , apply Newton-Raphson twice, giving each iterate to $6$ decimal places, and hence give the required depth to $3$ significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
The distance $D$ metres travelled by an automated trolley after $t$ seconds is modelled by $D(t)=5t+8\sin(0.5t)$ for $0\leq t\leq8$ . Show that there is a unique time in $(5,6)$ at which $D=30$ . Derive a Newton-Raphson recurrence and, starting with $t_0=5.5$ , find $t_1$ and $t_2$ to $5$ decimal places. Use unrounded values in your working. Hence give the time to $3$ significant figures.
(6)
(Total for Question 3 is 6 marks)
4.
A drying-process output is modelled by $S(t)=t+2e^{-t}$ , where $t$ is the time in hours. The process is complete when $S=3.5$ . Show that the completion time lies in $(3,4)$ and is unique in this interval. Starting with this interval, repeatedly test the midpoint of the current bracket and retain the half containing the sign change; carry out five such steps. State the resulting interval, use its midpoint to estimate the completion time, and give the maximum possible error in this estimate.
(7)
(Total for Question 4 is 7 marks)
5.
An optical instrument has a calibration factor $c>1$ satisfying $c+\ln c=4$ . Show that there is exactly one possible factor between $2.9$ and $3$ . By testing suitable decimal values and using changes of sign, first locate $c$ in an interval of width $0.01$ and then refine this to an interval of width $0.001$ . Hence give the calibration factor to $3$ significant figures.
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

9.1 · Locate roots of f(x) = 0 by considering changes of sign of f(x) in an interval on which f(x) is sufficiently well behaved; understand how change of sign methods can fail.

Tier 1 · Easy

Mark scheme for 9.1 Tier 1 · Easy
Question	Scheme	Marks
1	$f(1)=-1$ and $f(2)=1$ , so continuity and the change of sign give a root in $(1,2)$ .	2
(2 marks)		2
Notes
Substitute the endpoints: $f(1)=1-1-1=-1$ and $f(2)=4-2-1=1$ . A polynomial is continuous, and the endpoint values have opposite signs, so the intermediate value theorem guarantees at least one root between $1$ and $2$ .
2	$p(0)=1$ and $p(1)=-2$ , so continuity and the change of sign give a root in $(0,1)$ .	2
(2 marks)		2
Notes
The polynomial $p$ is continuous. Also $p(0)=1$ and $p(1)=1-4+1=-2$ . Since these values have opposite signs, the intermediate value theorem guarantees at least one root in $(0,1)$ .

Tier 2 · Standard

Mark scheme for 9.1 Tier 2 · Standard
Question	Scheme	Marks
1	$f(0)=f(2)=1$ , so there is no endpoint sign change, although $f(1)=0$ . The graph touches the axis at the repeated (even-multiplicity) root instead of crossing it; any equivalent explanation that the sign stays positive on both sides is valid.	3
(3 marks)		3
Notes
At the endpoints, $f(0)=(-1)^2=1$ and $f(2)=1^2=1$ , so their product is positive and the usual sign-change test gives no evidence of a root. Nevertheless, $f(1)=0$ . Because the root is repeated, the non-negative graph touches the $x$ -axis at $x=1$ and turns back without changing sign.
2	$f(-1)=\cos1-1<0$ and $f(0)=1>0$ , so a root lies in $(-1,0)$ ; since $f'(x)=1-\sin x>0$ on this interval, the root is unique.	4
(4 marks)		4
Notes
$f$ is continuous on $[-1,0]$ . Its endpoint values are $f(-1)=\cos1-1<0$ and $f(0)=1>0$ , so there is at least one root in $(-1,0)$ . On this interval $\sin x\leq0$ , so $f'(x)=1-\sin x\geq1>0$ . Thus $f$ is strictly increasing and can have at most one root, proving uniqueness.
3	$2<a<10$	4
(4 marks)		4
Notes
The polynomial is continuous. At the endpoints, $f(1)=2-a$ and $f(2)=10-a$ . The sign-change test guarantees a root when $(2-a)(10-a)<0$ , which holds for $2<a<10$ . The endpoints are excluded because $a=2$ or $a=10$ places the root at an endpoint rather than inside $(1,2)$ .

Tier 3 · Hard

Mark scheme for 9.1 Tier 3 · Hard
Question	Scheme	Marks
1	$h$ changes sign across $x=2$ but has no root because it is discontinuous there. $k$ has the repeated root $x=2$ but does not change sign across it.	4
(4 marks)		4
Notes
For $h$ , $h(1)=-1$ and $h(3)=1$ , but $h$ is undefined at $x=2$ , so continuity fails and the sign change occurs across a vertical asymptote rather than a root. For $k$ , $k(2)=0$ , but squaring makes $k(x)\geq0$ on both sides of $2$ , so the graph touches the axis at a repeated root without crossing it.
2	There is one root in each of $(-2,-1)$ , $(0,1)$ and $(1,2)$ ; these are exactly the three real roots.	5
(5 marks)		5
Notes
$g$ is continuous. The values $g(-2)=-1$ and $g(-1)=3$ have opposite signs, as do $g(0)=1$ and $g(1)=-1$ , and $g(1)=-1$ and $g(2)=3$ . Hence there is a root in each of the three disjoint intervals $(-2,-1)$ , $(0,1)$ and $(1,2)$ . A non-zero cubic polynomial has at most three real roots, so these three roots are all the real roots of $g$ .
3	There is one root in $(0,1)$ and one root in $(1,2)$ , and these are the only roots in $(0,2)$ .	5
(5 marks)		5
Notes
$h$ is continuous, with $h(0)=1>0$ , $h(1)=e-3<0$ and $h(2)=e^2-6>0$ . Hence sign changes give a root in each of $(0,1)$ and $(1,2)$ . Also $h'(x)=e^x-3$ , which is negative for $x<\ln3$ and positive for $x>\ln3$ . Thus $h$ decreases and then increases, so it has at most one root on each side of its single turning point; the turning point itself is not a root, since $h(\ln3)=3-3\ln3<0$ . The two bracketed roots are therefore the only roots in $(0,2)$ .
4	$f(0.6)=-0.184$ and $f(0.7)=0.043$ , so a root lies in $(0.6,0.7)$ . $f'(x)=3x^2+1>0$ , so the root is unique. $f(0.68)=-0.005568$ and $f(0.685)=0.006419125$ , so the root is $0.68$ to $2$ decimal places.	6
(6 marks)		6
Notes
The polynomial $f$ is continuous. Since $f(0.6)=0.6^3+0.6-1=-0.184$ and $f(0.7)=0.7^3+0.7-1=0.043$ , a change of sign gives a root in $(0.6,0.7)$ . Also $f'(x)=3x^2+1>0$ for every real $x$ , so $f$ is strictly increasing and the root is unique. Now $f(0.68)=0.68^3+0.68-1=-0.005568$ , while $f(0.685)=0.685^3+0.685-1=0.006419125$ . Hence the root lies in $(0.68,0.685)$ and is $0.68$ to $2$ decimal places.
5	At least four distinct roots: one in each of $(-3,-2)$ , $(-2,-1)$ and $(1,2)$ , together with the root $x=0$ . The data give no uniqueness or monotonicity information, so additional roots may occur between the sampled points.	5
(5 marks)		5
Notes
Continuity and the sign changes between $-3$ and $-2$ , between $-2$ and $-1$ , and between $1$ and $2$ guarantee a root in each of those three disjoint open intervals. The stated value $g(0)=0$ supplies a fourth root, distinct from the three bracketed roots. Thus four is the least number forced by the data. A continuous graph could cross the axis additional times inside any sampled interval, so these values alone cannot prove that there are exactly four roots.

9.2 · Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams.

Tier 1 · Easy

Mark scheme for 9.2 Tier 1 · Easy
Question	Scheme	Marks
1	$x_1=2.5$ and $x_2=2.45$	2
(2 marks)		2
Notes
Substitute $x_0=2$ to obtain $x_1=\frac12(2+6/2)=2.5$ . Then $x_2=\frac12(2.5+6/2.5)=\frac12(4.9)=2.45$ .
2	$x_1=2.8284$ and $x_2=2.8586$	2
(2 marks)		2
Notes
$x_1=\sqrt{11-3}=\sqrt8=2.828427\ldots$ . Using this unrounded value, $x_2=\sqrt{11-2.828427\ldots}=2.858596\ldots$ . Therefore $x_1=2.8284$ and $x_2=2.8586$ to $4$ decimal places.

Tier 2 · Standard

Mark scheme for 9.2 Tier 2 · Standard
Question	Scheme	Marks
1	The construction moves vertically to $y=2+x/4$ and horizontally to $y=x$ , giving $x_1=2$ , $x_2=2.5$ and $x_3=2.625$ . It is a convergent staircase. The limiting value is $\dfrac83$ .	4
(4 marks)		4
Notes
The graphical steps follow $x_{n+1}=2+x_n/4$ : from $x_0=0$ , they give $x_1=2$ , $x_2=5/2$ and $x_3=21/8$ . The gradient $g'(x)=1/4$ is positive, so the path is a staircase; since $0<1/4<1$ , it converges. At the fixed point $L=2+L/4$ , so $L=8/3$ .
2	$x_1=\dfrac52$ , $x_2=\dfrac{25}{13}$ and $x_3=\dfrac{155}{77}$ . The positive limit is $2$ , and the construction is a convergent cobweb.	4
(4 marks)		4
Notes
Successive substitution gives $x_1=1+6/4=2.5$ , $x_2=1+6/6.5=25/13=1.923076\ldots$ and $x_3=1+6/(25/13+4)=155/77=2.012987\ldots$ . A limit $L$ satisfies $(L-1)(L+4)=6$ , so $L^2+3L-10=0$ and the positive solution is $L=2$ . Here $g'(x)=-6/(x+4)^2$ , so $g'(2)=-1/6$ : the negative gradient gives a cobweb and its magnitude below $1$ gives convergence.
3	$e^x-x-6=0$ $x_1=1.9459$ , $x_2=2.0727$ and $x_3=2.0885$ . The path is a staircase because the iterates remain on the same side of $\alpha$ and increase towards it instead of alternating across it.	4
(4 marks)		4
Notes
At a fixed point, $\alpha=\ln(\alpha+6)$ , hence $e^\alpha=\alpha+6$ and $e^x-x-6=0$ . Using unrounded values gives $x_1=\ln7=1.945910149\ldots$ , $x_2=\ln(6+x_1)=2.072657349\ldots$ and $x_3=\ln(6+x_2)=2.088482715\ldots$ , so the requested values are $1.9459$ , $2.0727$ and $2.0885$ . Since the iterates increase towards $\alpha$ while staying below it, the graphical steps approach the intersection from one side and form a staircase, not a cobweb.

Tier 3 · Hard

Mark scheme for 9.2 Tier 3 · Hard
Question	Scheme	Marks
1	$x_1=2.6$ , $x_2=1.96$ , $x_3=2.216$ and $x_4=2.1136$ . The limit is $\frac{15}{7}$ , and the construction is a convergent cobweb because $g'(x)=-0.4$ .	5
(5 marks)		5
Notes
Apply the recurrence successively: $x_1=3-0.4=2.6$ , $x_2=3-0.4(2.6)=1.96$ , $x_3=3-0.4(1.96)=2.216$ and $x_4=3-0.4(2.216)=2.1136$ . A limit $L$ obeys $L=3-0.4L$ , hence $L=3/1.4=15/7$ . Since the constant gradient is negative, the iterates alternate across the fixed point, producing a cobweb; since $\|-0.4\|<1$ , it converges.
2	$x_{n+1}=\sqrt{4-x_n}$ is locally convergent to $\alpha$ and gives a cobweb; $x_{n+1}=4-x_n^2$ is locally divergent.	5
(5 marks)		5
Notes
Both fixed-point equations rearrange to $x^2+x-4=0$ . For $g_1(x)=4-x^2$ , $\|g_1'(\alpha)\|=2\alpha>3>1$ , so this recurrence is locally divergent. For $g_2(x)=\sqrt{4-x}$ , $g_2'(x)=-1/(2\sqrt{4-x})$ . At the fixed point $\sqrt{4-\alpha}=\alpha$ , so $\|g_2'(\alpha)\|=1/(2\alpha)<1/3<1$ and the recurrence is locally convergent. Its derivative is negative, so successive graphical steps alternate across the fixed point and form a cobweb.
3	$x_1=\dfrac32$ , $x_2=\dfrac65$ , $x_3=\dfrac{15}{11}$ and $x_4=\dfrac{33}{26}$ . The positive fixed point is $\dfrac{-1+\sqrt{13}}2$ ; the construction is a cobweb and the iterates appear to converge.	5
(5 marks)		5
Notes
Exact substitution gives $x_1=3/2$ , $x_2=3/(5/2)=6/5$ , $x_3=3/(11/5)=15/11$ and $x_4=3/(26/11)=33/26$ . A fixed point $L$ satisfies $L=3/(L+1)$ , so $L^2+L-3=0$ and the positive solution is $(-1+\sqrt{13})/2=1.3027\ldots$ . The iterates lie alternately above and below this value, with decreasing deviations over the four steps, so the graphical path is a converging cobweb.
4	$a=1.5$ and $b=0.3$ , so $x_{n+1}=1.5+0.3x_n$ . $x_3=2.112$ , the limit is $\dfrac{15}{7}$ , and the diagram is a convergent staircase.	5
(5 marks)		5
Notes
The two vertical steps show that $g(1)=1.8$ and $g(1.8)=2.04$ . Hence $a+b=1.8$ and $a+1.8b=2.04$ . Subtracting gives $0.8b=0.24$ , so $b=0.3$ and $a=1.5$ . The second ordinate is $x_2=2.04$ , so $x_3=1.5+0.3(2.04)=2.112$ . A limit $L$ satisfies $L=1.5+0.3L$ , giving $L=15/7$ . Since $0<b<1$ , the steps approach the fixed point from one side as a convergent staircase.
5	$x_1=1.632121$ , $x_2=1.804485$ , $x_3=1.835441$ , $x_4=1.840457$ , $x_5=1.841255$ and $x_6=1.841382$ . The least value is $n=5$ , since $\|x_6-x_5\|=0.000126668\ldots<0.0005$ but $\|x_5-x_4\|=0.000798258\ldots>0.0005$ . The fixed point is approximately $1.8414$ , and the diagram is a convergent staircase.	6
(6 marks)		6
Notes
Repeated substitution, retaining the unrounded value each time, gives $x_1=1.632120558\ldots$ , $x_2=1.804485465\ldots$ , $x_3=1.835440893\ldots$ , $x_4=1.840456855\ldots$ , $x_5=1.841255113\ldots$ and $x_6=1.841381782\ldots$ . The first consecutive difference below $0.0005$ is therefore $\|x_6-x_5\|$ , so $n=5$ . Continuing the iteration gives the fixed point $1.841405660\ldots$ , hence $1.8414$ to $4$ decimal places. Here $g'(x)=e^{-x}$ is positive and is less than $1$ for these positive iterates, so the path is a convergent staircase.

9.3 · Solve equations using the Newton-Raphson method and other recurrence relations of the form xₙ₊₁ = g(xₙ); understand how such methods can fail.

Tier 1 · Easy

Mark scheme for 9.3 Tier 1 · Easy
Question	Scheme	Marks
1	$x_1=2.65$	2
(2 marks)		2
Notes
Here $f'(x)=2x$ . Therefore $x_1=2.5-\frac{2.5^2-7}{2(2.5)}=2.5-\frac{-0.75}{5}=2.65$ .
2	$x_1=1.1036$	2
(2 marks)		2
Notes
Since $f'(x)=e^x$ , Newton-Raphson gives $x_1=1-(e-3)/e=3/e=1.103638\ldots$ . Hence $x_1=1.1036$ to $4$ decimal places.

Tier 2 · Standard

Mark scheme for 9.3 Tier 2 · Standard
Question	Scheme	Marks
1	$x_1=1$ , $x_2=0$ , $x_3=1$ The iteration enters the cycle $0,1,0,1,\ldots$ .	4
(4 marks)		4
Notes
Here $f'(x)=3x^2-2$ . From $x_0=0$ , $x_1=0-f(0)/f'(0)=0-2/(-2)=1$ . Then $x_2=1-f(1)/f'(1)=1-1/1=0$ , so $x_3=1$ again. The values repeat in a two-cycle and neither $0$ nor $1$ is a root, so this starting value does not produce convergence.
2	$x_{n+1}=x_n-\dfrac{x_n^3+x_n^2-6}{3x_n^2+2x_n}$ ; $x_1=1.5385$ and $x_2=1.5377$ .	4
(4 marks)		4
Notes
Take $f(x)=x^3+x^2-6$ , so $f'(x)=3x^2+2x$ . Substitution in $x_{n+1}=x_n-f(x_n)/f'(x_n)$ gives the stated recurrence. With $x_0=1.5$ , $x_1=1.538461\ldots$ ; substituting this unrounded value gives $x_2=1.537656\ldots$ . Thus the requested values are $1.5385$ and $1.5377$ .
3	$x_{n+1}=x_n-\dfrac{\ln x_n+x_n-3}{1+1/x_n}$ ; $x_1=2.20457$ and $x_2=2.20794$ .	4
(4 marks)		4
Notes
For $f(x)=\ln x+x-3$ , $f'(x)=1/x+1$ . Substitution in the Newton-Raphson formula gives $x_{n+1}=x_n-(\ln x_n+x_n-3)/(1+1/x_n)$ . Starting from $x_0=2$ gives $x_1=2.204568546\ldots$ . Using this unrounded value gives $x_2=2.207939227\ldots$ , so the requested iterates are $2.20457$ and $2.20794$ .

Tier 3 · Hard

Mark scheme for 9.3 Tier 3 · Hard
Question	Scheme	Marks
1	$a=10$ . $x_2=\dfrac{329}{104}$ and $x_3=3.162278$ to six decimal places. The positive root is $3.1623$ correct to four decimal places.	5
(5 marks)		5
Notes
For $f(x)=x^2-a$ , Newton-Raphson gives $x_{n+1}=\frac12(x_n+a/x_n)$ . Since $x_0=4$ and $x_1=13/4$ , $13/4=\frac12(4+a/4)$ , so $13/2=4+a/4$ and $a=10$ . Then $x_2=\frac12(13/4+40/13)=329/104$ . One more iteration gives $x_3=\frac12(329/104+1040/329)=216401/68432=3.162277881\ldots$ . Hence the positive root is $\sqrt{10}=3.162277660\ldots=3.1623$ correct to four decimal places.
2	$x_0=1$ fails because $f'(1)=0$ , so the Newton step is undefined. Starting from $-2$ , $x_1=-2.11111$ and $x_2=-2.10384$ .	5
(5 marks)		5
Notes
Here $f'(x)=3x^2-3$ and $x_{n+1}=x_n-(x_n^3-3x_n+3)/(3x_n^2-3)$ . At $x_0=1$ the denominator is zero, so no Newton iterate exists. From $x_0=-2$ , $f(-2)=1$ and $f'(-2)=9$ , giving $x_1=-2-1/9=-2.111111\ldots$ . Substitution of this unrounded value gives $x_2=-2.103835\ldots$ , hence the stated rounded values.
3	$x_{n+1}=-2x_n$ ; $x_1=-\dfrac14$ , $x_2=\dfrac12$ and $x_3=-1$ . The iterates alternate in sign while their distance from $0$ doubles, so they diverge.	4
(4 marks)		4
Notes
For non-zero $x$ , $f'(x)=\frac13x^{-2/3}$ . Newton-Raphson therefore gives $x_{n+1}=x_n-x_n^{1/3}/(\frac13x_n^{-2/3})=x_n-3x_n=-2x_n$ . From $x_0=1/8$ , the next values are $-1/4$ , $1/2$ and $-1$ . Their signs alternate and their magnitudes double, so the sequence moves away from the root $0$ rather than converging to it.
4	$x_{n+1}=x_n+\dfrac{\cos x_n-x_n}{\sin x_n+1}$ . $x_1=0.7394$ , $x_2=0.7391$ and $x_3=0.7391$ . $\cos(0.73905)-0.73905>0$ and $\cos(0.73915)-0.73915<0$ , so the root is $0.7391$ to $4$ decimal places.	6
(6 marks)		6
Notes
Take $f(x)=\cos x-x$ , so $f'(x)=-\sin x-1$ . Newton-Raphson gives $x_{n+1}=x_n-f(x_n)/f'(x_n)=x_n+(\cos x_n-x_n)/(\sin x_n+1)$ . From $x_0=0.7$ , unrounded calculation gives $x_1=0.739436497\ldots$ , $x_2=0.739085160\ldots$ and $x_3=0.739085133\ldots$ . Also $f(0.73905)=0.0000587989\ldots>0$ and $f(0.73915)=-0.0001085633\ldots<0$ . Since $f'(x)<0$ on this interval, the root is unique there and the bracket pins it as $0.7391$ to $4$ decimal places.
5	$x_{n+1}=2x_n-\dfrac{x_n^2}{5}$ . For $x_0=1$ , $x_1=\dfrac95$ and $x_2=\dfrac{369}{125}$ . For $x_0=10$ , $x_1=0$ , where the original function and the next Newton-Raphson step are undefined.	5
(5 marks)		5
Notes
With $f(x)=1/x-1/5$ , $f'(x)=-1/x^2$ . Therefore $x_{n+1}=x_n-(1/x_n-1/5)/(-1/x_n^2)=2x_n-x_n^2/5$ . From $x_0=1$ , this gives $x_1=2-1/5=9/5$ and $x_2=18/5-81/125=369/125$ . If $x_0=10$ , then $x_1=20-100/5=0$ . Since $f(0)$ is undefined, the method cannot form another tangent step even though the equation itself has the root $x=5$ .

9.4 · Understand and use numerical integration of functions, including the use of the trapezium rule and estimating the approximate area under a curve and limits that it must lie between.

Tier 1 · Easy

Mark scheme for 9.4 Tier 1 · Easy
Question	Scheme	Marks
1	$8$	2
(2 marks)		2
Notes
The strip width is $h=1$ . The trapezium rule gives $\frac12[1+9+2(3)]=\frac12(16)=8$ .
2	$k=6$	3
(3 marks)		3
Notes
The strip width is $1$ , so the trapezium estimate is $\frac12[2+6+2k]=4+k$ . Equating this to $10$ gives $k=6$ .

Tier 2 · Standard

Mark scheme for 9.4 Tier 2 · Standard
Question	Scheme	Marks
1	$19$	3
(3 marks)		3
Notes
The strip width is $h=1$ . The trapezium rule gives $\dfrac12[2+10+2(2.5+4+6.5)]=\dfrac12[12+26]=19$ .
2	$2.793$ ; it is an underestimate because $y=\sqrt{x+1}$ is concave on the interval, so its chords lie below the curve.	4
(4 marks)		4
Notes
Here $h=0.5$ . The estimate is $\frac{0.5}{2}[1+1.732+2(1.225+1.414+1.581)]=2.793$ . Also $y''=-1/[4(x+1)^{3/2}]<0$ for $0\leq x\leq2$ , so the graph is concave and each trapezium chord lies below the curve. The estimate is therefore an underestimate.
3	$12.5$	4
(4 marks)		4
Notes
The unequal widths are $0.5$ , $1$ and $1.5$ , so the three trapezium areas are $\frac{0.5}{2}(2+4)=1.5$ , $\frac12(4+3)=3.5$ and $\frac{1.5}{2}(3+7)=7.5$ . Their sum is $1.5+3.5+7.5=12.5$ .

Tier 3 · Hard

Mark scheme for 9.4 Tier 3 · Hard
Question	Scheme	Marks
1	Trapezium estimate $=0.6970$ . $0.6345<\int_1^2\frac1x\,dx<0.7595$ .	6
(6 marks)		6
Notes
The width is $h=0.25$ and the ordinates are $1,\frac45,\frac23,\frac47,\frac12$ . The trapezium estimate is $\frac18[1+\frac12+2(\frac45+\frac23+\frac47)]=\frac{1171}{1680}=0.6970$ . Since $1/x$ is decreasing, right endpoints give the lower sum $\frac14(\frac45+\frac23+\frac47+\frac12)=\frac{533}{840}=0.6345$ , while left endpoints give the upper sum $\frac14(1+\frac45+\frac23+\frac47)=\frac{319}{420}=0.7595$ .
2	Estimated signed integral $=1$ . Estimated total area $=4.5$ square units.	5
(5 marks)		5
Notes
With $h=1$ , the signed estimate is $\frac12[-2+4+2(-1+1)]=1$ . The straight segment from $(1,-1)$ to $(2,1)$ crosses the axis halfway, at $x=1.5$ . The estimated area below the axis is $\frac12(2+1)(1)+\frac12(0.5)(1)=1.75$ . The estimated area above it is $\frac12(0.5)(1)+\frac12(1+4)(1)=2.75$ . Hence the total area is $1.75+2.75=4.5$ square units.
3	The estimated area between the curves is $20.5$ square units. The separate estimates are $39$ and $18.5$ , whose difference is $20.5$ .	5
(5 marks)		5
Notes
The ordinates of $f-g$ are $6,4,5,6,5$ . With width $1$ , the trapezium estimate is $\frac12[6+5+2(4+5+6)]=20.5$ . Separately, the estimate for $\int_0^4f(x)\,\mathrm dx$ is $\frac12[8+14+2(7+9+12)]=39$ , while that for $\int_0^4g(x)\,\mathrm dx$ is $\frac12[2+9+2(3+4+6)]=18.5$ . Their difference is $39-18.5=20.5$ square units.
4	The transformed ordinates are $1,4,9,4,1$ , giving $V\approx18\pi$ .	5
(5 marks)		5
Notes
The integrand is $r^2$ , not $r$ , so the ordinates for the trapezium rule are $1^2,2^2,3^2,2^2,1^2$ , namely $1,4,9,4,1$ . With strip width $h=1$ , $\int_0^4r^2\,\mathrm dx\approx(1/2)[1+2(4+9+4)+1]=18$ . Multiplying by $\pi$ gives the volume estimate $18\pi$ .
5	The two-strip estimate is $10$ . The missing ordinates are $3$ at $x=0.5$ and $6$ at $x=1.5$ .	5
(5 marks)		5
Notes
For two strips of width $1$ , the estimate is $(1/2)[2+2(5)+8]=10$ . Let the missing ordinates at $x=0.5$ and $x=1.5$ be $u$ and $2u$ . For four strips of width $0.5$ , the estimate is $(0.5/2)[2+2(u+5+2u)+8]=5+1.5u$ . Since this equals $9.5$ , $1.5u=4.5$ and $u=3$ . Thus the two missing ordinates are $3$ and $6$ .

9.5 · Use numerical methods to solve problems in context.

Tier 1 · Easy

Mark scheme for 9.5 Tier 1 · Easy
Question	Scheme	Marks
1	$T(3)-20=1$ and $T(4)-20=-2$ , so the return time lies between $3$ and $4$ hours after noon.	2
(2 marks)		2
Notes
Define $f(t)=T(t)-20=-2+4t-t^2$ . Then $f(3)=-2+12-9=1$ and $f(4)=-2+16-16=-2$ . The model is continuous and the sign changes, so a solution occurs in $(3,4)$ .
2	$S(2)-20=-10$ and $S(3)-20=10$ , so the model reaches $20$ volts for some $t\in(2,3)$ seconds.	2
(2 marks)		2
Notes
Let $f(t)=S(t)-20=t^3+t-20$ . Then $f(2)=8+2-20=-10$ and $f(3)=27+3-20=10$ . The model is continuous, so the change of sign guarantees a solution between $2$ and $3$ seconds.

Tier 2 · Standard

Mark scheme for 9.5 Tier 2 · Standard
Question	Scheme	Marks
1	$h(4.0)=1.6>0$ and $h(4.1)=-0.369<0$ $h(4.05)=0.62775>0$ , so the root lies in $(4.05,4.10)$ .	4
(4 marks)		4
Notes
Calculate $h(4.0)=20(4)-4.9(4^2)=1.6$ and $h(4.1)=20(4.1)-4.9(4.1^2)=-0.369$ . Continuity gives a root in $(4.0,4.1)$ . The midpoint is $4.05$ , and $h(4.05)=20(4.05)-4.9(4.05^2)=0.62775>0$ . The sign change is therefore between $4.05$ and $4.10$ seconds.
2	$t_1=2.885203$ , $t_2=2.875011$ and $t_3=2.874092$ . The predicted completion time is $2.874$ hours.	4
(4 marks)		4
Notes
Successive substitution gives $t_1=2.885203\ldots$ , $t_2=2.875011\ldots$ and $t_3=2.874092\ldots$ . Continuing gives $t_4=2.874009\ldots$ and the values settle to $2.8740007\ldots$ . Therefore the model predicts a completion time of $2.874$ hours to $3$ decimal places.
3	$R_1=37.4873$ , $R_2=38.8043$ , $R_3=38.5557$ , $R_4=38.6022$ and $R_5=38.5935$ . The reliable range is approximately $38.6$ m.	4
(4 marks)		4
Notes
Substituting unrounded values successively gives $R_1=37.4873055\ldots$ , $R_2=38.8042885\ldots$ , $R_3=38.5557086\ldots$ , $R_4=38.6021810\ldots$ and $R_5=38.5934763\ldots$ . The alternating values are settling near $38.6$ , so the model gives a reliable range of approximately $38.6$ m to the nearest $0.1$ m.

Tier 3 · Hard

Mark scheme for 9.5 Tier 3 · Hard
Question	Scheme	Marks
1	Estimated inflow $=487.5$ litres and estimated final volume $=837.5$ litres. The model predicts overflow, with $37.5$ litres in excess of capacity at $30$ minutes.	5
(5 marks)		5
Notes
With spacing $h=5$ , the trapezium estimate is $\frac52[10+11+2(14+18+21+19+15)]=\frac52(195)=487.5$ litres. Adding the initial $350$ litres gives $837.5$ litres, which is $837.5-800=37.5$ litres above capacity, so the tank must have overflowed by the end.
2	$V(3)-70=-7$ and $V(4)-70=58$ , so the depth lies in $(3,4)$ m. $h_1=3.133807$ , $h_2=3.132719$ , so the required depth is $3.13$ m to $3$ significant figures.	6
(6 marks)		6
Notes
Set $f(h)=h^3+4h^2-70$ . Since $f(3)=27+36-70=-7$ and $f(4)=64+64-70=58$ , continuity gives a root in $(3,4)$ . Also $f'(h)=3h^2+8h$ , so $h_{n+1}=h_n-(h_n^3+4h_n^2-70)/(3h_n^2+8h_n)$ . From $h_0=3.2$ , this gives $h_1=3.1338068\ldots=3.133807$ ( $6$ d.p.) and then $h_2=3.1327194\ldots=3.132719$ ( $6$ d.p.). Thus the required positive depth is $3.13$ m to $3$ significant figures.
3	$D(5)<30<D(6)$ and $D'(t)>0$ , so there is exactly one required time in $(5,6)$ . $t_{n+1}=t_n-\dfrac{5t_n+8\sin(0.5t_n)-30}{5+4\cos(0.5t_n)}$ ; $t_1=5.07531$ and $t_2=5.12236$ . The time is $5.12$ s to $3$ significant figures.	6
(6 marks)		6
Notes
Let $f(t)=5t+8\sin(0.5t)-30$ . Then $f(5)=8\sin2.5-5<0$ and $f(6)=8\sin3>0$ , so continuity gives a root in $(5,6)$ . Also $f'(t)=5+4\cos(0.5t)\geq1>0$ , so the root is unique. Newton-Raphson gives $t_{n+1}=t_n-[5t_n+8\sin(0.5t_n)-30]/[5+4\cos(0.5t_n)]$ . From $t_0=5.5$ , unrounded substitution gives $t_1=5.075305513\ldots$ and $t_2=5.122362632\ldots$ . Thus the estimated time is $5.12$ s to $3$ significant figures.
4	For $f(t)=t+2e^{-t}-3.5$ , $f(3)<0$ and $f(4)>0$ ; also $f'(t)=1-2e^{-t}>0$ on $(3,4)$ , so the root there is unique. The five tested midpoints are $3.5$ , $3.25$ , $3.375$ , $3.4375$ and $3.40625$ . The resulting interval is $(3.40625,3.4375)$ , so the midpoint estimate is $3.421875$ hours with maximum possible error $0.015625$ hours.	7
(7 marks)		7
Notes
Let $f(t)=t+2e^{-t}-3.5$ . Then $f(3)=-0.400425\ldots<0$ and $f(4)=0.536631\ldots>0$ . Since $f'(t)=1-2e^{-t}>0$ when $3<t<4$ , there is exactly one root in the interval. The midpoint signs are $f(3.5)>0$ , $f(3.25)<0$ , $f(3.375)<0$ , $f(3.4375)>0$ and $f(3.40625)<0$ . The successive brackets therefore end at $(3.40625,3.4375)$ . Its midpoint is $(3.40625+3.4375)/2=3.421875$ , and half the interval width is $(3.4375-3.40625)/2=0.015625$ , which is the maximum possible error.
5	For $f(c)=c+\ln c-4$ , $f(2.9)<0<f(3)$ and $f'(c)=1+1/c>0$ , so there is exactly one possible factor in $(2.9,3)$ . $f(2.92)<0<f(2.93)$ , and then $f(2.926)<0<f(2.927)$ , so $2.926<c<2.927$ . The calibration factor is $2.93$ to $3$ significant figures.	6
(6 marks)		6
Notes
Let $f(c)=c+\ln c-4$ . Then $f(2.9)=-0.035289\ldots$ and $f(3)=0.098612\ldots$ , so continuity gives a root in $(2.9,3)$ . Since $f'(c)=1+1/c>0$ for $c>1$ , this root is unique. Now $f(2.92)=-0.008416\ldots$ and $f(2.93)=0.005002\ldots$ , giving the width- $0.01$ bracket $(2.92,2.93)$ . Refining within it, $f(2.926)=-0.000363\ldots$ and $f(2.927)=0.000978\ldots$ , giving the width- $0.001$ bracket $(2.926,2.927)$ . Every number in this final bracket rounds to $2.93$ to $3$ significant figures.