Questions - SPS SPS FM Pure

SPS SPS FM Pure 2025 February Q9

9. In this question, you must show detailed reasoning. Find the sum of all the integers from 1 to 999 inclusive that are not square or cube numbers.
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SPS SPS FM Pure 2025 February Q10

10. Three planes have equations $$\begin{array} { r } 4 x - 5 y + z = 8
3 x + 2 y - k z = 6
( k - 2 ) x + k y - 8 z = 6 \end{array}$$ where $k$ is a real constant.
The planes do not meet at a unique point.

Find the possible values of $k$.
For each value of $k$ found in part (a), identify the configuration of the given planes. Fully justify your answer, stating in each case whether or not the equations of the planes form a consistent system.
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SPS SPS FM Pure 2025 February Q11

11. The infinite series C and S are defined by $$\begin{aligned} & \mathrm { C } = \cos \theta + \frac { 1 } { 2 } \cos 5 \theta + \frac { 1 } { 4 } \cos 9 \theta + \frac { 1 } { 8 } \cos 13 \theta + \ldots
& \mathrm { S } = \sin \theta + \frac { 1 } { 2 } \sin 5 \theta + \frac { 1 } { 4 } \sin 9 \theta + \frac { 1 } { 8 } \sin 13 \theta + \ldots \end{aligned}$$ Given that the series C and S are both convergent,

show that $$C + i S = \frac { 2 e ^ { i \theta } } { 2 - e ^ { 4 i \theta } }$$
Hence show that $$\mathrm { S } = \frac { 4 \sin \theta + 2 \sin 3 \theta } { 5 - 4 \cos 4 \theta }$$ [BLANK PAGE]

SPS SPS FM Pure 2025 February Q12

12. The population density $P$, in suitable units, of a certain bacterium at time $t$ hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is 5 . The model uses the differential equation $$\frac { d P } { d t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \frac { t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }$$ Find $P$ as a function of $t$. [You may assume that $\lim _ { t \rightarrow \infty } t e ^ { - t } = 0$ ].
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SPS SPS FM Pure 2025 February Q13

13. (a) Write down the Maclaurin series of $\mathrm { e } ^ { x }$, in ascending power of $x$, up to and including the term in $x ^ { 3 }$
(b) Hence, without differentiating, determine the Maclaurin series of $$\mathrm { e } ^ { \left( \mathrm { e } ^ { x } - 1 \right) }$$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each coefficient in simplest form.
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SPS SPS FM Pure 2026 January Q1

1
6 \end{array} \right) + \lambda \left( \begin{array} { r }

SPS SPS FM Pure 2026 January Q2

2
- 3
1 \end{array} \right) \text { where } \lambda \text { is a scalar parameter. }$$ The point $P$ lies on $l _ { 1 }$. Given that $\overrightarrow { O P }$ is perpendicular to $l _ { 1 }$, calculate the coordinates of $P$.
(ii) Relative to a fixed origin $O$, the line $l _ { 2 }$ is given by the equation $$l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }

SPS SPS FM Pure 2026 January Q4

4
- 3
12 \end{array} \right) + \mu \left( \begin{array} { r }

SPS SPS FM Pure 2026 January Q6

6 \end{array} \right) + \lambda \left( \begin{array} { r } 2
- 3
1 \end{array} \right) \text { where } \lambda \text { is a scalar parameter. }$$ The point $P$ lies on $l _ { 1 }$. Given that $\overrightarrow { O P }$ is perpendicular to $l _ { 1 }$, calculate the coordinates of $P$.
(ii) Relative to a fixed origin $O$, the line $l _ { 2 }$ is given by the equation $$l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } 4
- 3

SPS SPS FM Pure 2026 January Q12

3 marks

12 \end{array} \right) + \mu \left( \begin{array} { r } 5
- 3
4 \end{array} \right) \text { where } \mu \text { is a scalar parameter. }$$ The point $A$ does not lie on $l _ { 2 }$. Given that the vector $\overrightarrow { O A }$ is parallel to the line $l _ { 2 }$ and $| \overrightarrow { O A } | = \sqrt { 2 }$ units, calculate the possible position vectors of the point $A$.
[0pt] [BLANK PAGE] Q7.
The simultaneous equations $$\begin{aligned} & 2 x - y = 1
& 3 x + k y = b \end{aligned}$$ are represented by the matrix equation $\mathbf { M } \binom { x } { y } = \binom { 1 } { b }$.

State the value of $k$ for which $\mathbf { M } ^ { - 1 }$ does not exist and find $\mathbf { M } ^ { - 1 }$ in terms of $k$ when $\mathbf { M } ^ { - 1 }$ exists.
Use $\mathbf { M } ^ { - 1 }$ to solve the simultaneous equations when $k = 5$ and $b = 21$.
The two equations can be interpreted as representing two lines in the $x - y$ plane. Describe the relationship between these two lines
(A) when $k = 5$ and $b = 21$,
(B) when $k = - \frac { 3 } { 2 }$ and $b = 1$,
(C) when $k = - \frac { 3 } { 2 }$ and $b = \frac { 3 } { 2 }$.
[0pt] [BLANK PAGE] Q8.
You are given the matrix $\mathbf { M } = \left( \begin{array} { r r } 0.8 & 0.6
0.6 & - 0.8 \end{array} \right)$.
Calculate $\mathbf { M } ^ { 2 }$. You are now given that the matrix $\mathbf { M }$ represents a reflection in a line through the origin.
Explain how your answer to part (i) relates to this information.
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By investigating the invariant points of the reflection, find the equation of the mirror line. [3]
Describe fully the transformation represented by the matrix $\mathbf { P } = \left( \begin{array} { r r } 0.8 & - 0.6
0.6 & 0.8 \end{array} \right)$.
A composite transformation is formed by the transformation represented by $\mathbf { P }$ followed by the transformation represented by $\mathbf { M }$. Find the single matrix that represents this composite transformation.
The composite transformation described in part (v) is equivalent to a single reflection. What is the equation of the mirror line of this reflection?
[0pt] [BLANK PAGE] Q9.
(a) Two points, $A$ and $B$, on an Argand diagram are represented by the complex numbers $2 + 3 \mathrm { i }$ and $- 4 - 5 \mathrm { i }$ respectively. Given that the points $A$ and $B$ are at the ends of a diameter of a circle $C _ { 1 }$, express the equation of $C _ { 1 }$ in the form $\left| z - z _ { 0 } \right| = k$.
(4 marks)
(b) A second circle, $C _ { 2 }$, is represented on the Argand diagram by the equation $| z - 5 + 4 i | = 4$. Sketch on one Argand diagram both $C _ { 1 }$ and $C _ { 2 }$.
(3 marks)
(c) The points representing the complex numbers $z _ { 1 }$ and $z _ { 2 }$ lie on $C _ { 1 }$ and $C _ { 2 }$ respectively and are such that $\left| z _ { 1 } - z _ { 2 } \right|$ has its maximum value. Find this maximum value, giving your answer in the form $a + b \sqrt { 5 }$.
(5 marks)
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SPS SPS FM Pure 2025 September Q2

2. Prove by induction that $11 \times 7 ^ { n } - 13 ^ { n } - 1$ is divisible by 3 , for all integers $n \geq 0$.
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SPS SPS FM Pure 2025 September Q4

6 marks

4. The curve $C$ has parametric equations $$x = 2 \cos t , \quad y = \sqrt { 3 } \cos 2 t , \quad 0 \leqslant t \leqslant \pi$$

Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. The point $P$ lies on $C$ where $t = \frac { 2 \pi } { 3 }$
The line $l$ is the normal to $C$ at $P$.
Show that an equation for $l$ is $$2 x - 2 \sqrt { 3 } y - 1 = 0$$ The line $l$ intersects the curve $C$ again at the point $Q$.
Find the exact coordinates of $Q$. You must show clearly how you obtained your answers.
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On the Argand diagram below, sketch the locus, $L$, of points satisfying the equation $$\arg ( z + \mathrm { i } ) = \frac { \pi } { 6 }$$ [2 marks]
\includegraphics[max width=\textwidth, alt={}, center]{75e73922-4324-441c-b942-c709a71d9025-12_1310_1354_513_463}
$\quad z _ { 1 }$ is a point on $L$ such that $| z |$ is a minimum. Find the exact value of $z _ { 1 }$ in the form $a + b \mathrm { i }$
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SPS SPS FM Pure 2025 September Q7

8 marks

7. (a) Prove the identity $\frac { \cos x } { \sec x + 1 } + \frac { \cos x } { \sec x - 1 } \equiv 2 \cot ^ { 2 } x$
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(b) Hence, solve the equation $$\frac { \cos \left( 2 \theta + \frac { \pi } { 3 } \right) } { \sec \left( 2 \theta + \frac { \pi } { 3 } \right) + 1 } = \cot \left( 2 \theta + \frac { \pi } { 3 } \right) - \frac { \cos \left( 2 \theta + \frac { \pi } { 3 } \right) } { \sec \left( 2 \theta + \frac { \pi } { 3 } \right) - 1 }$$ in the interval $0 \leq \theta \leq 2 \pi$, giving your values of $\theta$ to three significant figures where appropriate.
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SPS SPS FM Pure 2025 September Q8

8. A population of meerkats is being studied. The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 22 } P ( 11 - 2 P ) , \quad t \geqslant 0 , \quad 0 < P < 5.5$$ where $P$, in thousands, is the population of meerkats and $t$ is the time measured in years since the study began. Given that there were 1000 meerkats in the population when the study began, determine the time taken, in years, for this population of meerkats to double.
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SPS SPS FM Pure 2025 September Q9

6 marks

9. A curve $C$ has equation $y = \mathrm { f } ( x )$ where $$\mathrm { f } ( x ) = x + 2 \ln ( \mathrm { e } - x )$$

1. Show that the equation of the normal to $C$ at the point where $C$ crosses the $y$-axis is given by $$y = \left( \frac { \mathrm { e } } { 2 - \mathrm { e } } \right) x + 2$$
2. Find the exact area enclosed by the normal and the coordinate axes. Fully justify your answer.
The equation $\mathrm { f } ( x ) = 0$ has one positive root, $\alpha$.
1. Show that $\alpha$ lies between 2 and 3 Fully justify your answer.
2. Show that the roots of $\mathrm { f } ( x ) = 0$ satisfy the equation $$x = \mathrm { e } - \mathrm { e } ^ { - \frac { x } { 2 } }$$ [2 marks]
3. Use the recurrence relation $$x _ { n + 1 } = \mathrm { e } - \mathrm { e } ^ { - \frac { x _ { n } } { 2 } }$$ with $x _ { 1 } = 2$ to find the values of $x _ { 2 }$ and $x _ { 3 }$ giving your answers to three decimal places.
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4. Figure 1 below shows a sketch of the graphs of $y = e - e ^ { - \frac { x } { 2 } }$ and $y = x$, and the position of $x _ { 1 }$ On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x _ { 2 }$ and $x _ { 3 }$ on the $x$-axis.
  [0pt] [2 marks] \begin{figure}[h]
  \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{75e73922-4324-441c-b942-c709a71d9025-22_1236_1566_1519_360}
  \end{figure} [BLANK PAGE]
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SPS SPS FM Pure 2020 February Q1

1 In this question you must show detailed reasoning.

Determine the value of $$\int _ { 3 } ^ { 5 } \frac { 1 } { \sqrt { x ^ { 2 } - 9 } } d x$$ giving your answer as a single logarithm.
Determine the exact value of $$\int _ { 0 } ^ { \ln 3 } \sinh x \cosh x d x$$

SPS SPS FM Pure 2020 February Q2

2

Given that $$f ( x ) = \tan ^ { - 1 } ( x + 1 )$$ find $f ( 0 )$ and $f ^ { \prime } ( 0 )$, and show that $f ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 2 }$.
Hence find the first three terms in the Maclaurin series for $f ( x )$

SPS SPS FM Pure 2020 February Q3

3

Find the inverse of the matrix $\left( \begin{array} { l l l } 2 & 0 & 1
0 & 1 & a
1 & 3 & 0 \end{array} \right)$ in terms of $a$.
State the value of $a$ for which the matrix is singular.

SPS SPS FM Pure 2020 February Q4

4 The equation of a curve in polar coordinates is $r = \frac { 1 } { 2 } - \cos \theta$.

Sketch the polar graph of the curve.
Find the exact area enclosed by the curve between $\theta = \frac { 1 } { 3 } \pi$ and $\theta = \frac { 5 } { 3 } \pi$.

SPS SPS FM Pure 2020 February Q5

5 The roots of the equation $\quad 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x - 6 = 0$ are $\alpha , \beta$ and $\gamma$.

Find the value of $\alpha \beta \gamma$.
Hence, by making use of a suitable substitution, or otherwise, find a cubic equation whose roots are $\alpha \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) , \beta \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) , \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right)$.

SPS SPS FM Pure 2020 February Q6

6 Throughout this question, the complex number $z$ satisfies $\left| z - z _ { 0 } \right| \leq \sqrt { 2 }$, where $z _ { 0 } = 3 - \mathrm { i }$.

Draw an Argand diagram to illustrate the locus of $z$.
In this question you must show detailed reasoning. Show that the greatest possible argument of $z$ can be written as $\tan ^ { - 1 } \left( \frac { 1 } { n } \right)$, where $n$ is a positive integer to be determined and $\arg z \in ( - \pi , \pi ]$.

SPS SPS FM Pure 2020 February Q7

6 marks

7

Find the shortest distance from the point (9, 5, 2) to the plane $3 x + 2 y - 4 z = - 3$.
The vector equation of a second plane is given by $$\boldsymbol { r } = \lambda \left( \begin{array} { l } 3
0
1 \end{array} \right) + \mu \left( \begin{array} { c } 1
- 1
- 2 \end{array} \right)$$ By finding the Cartesian equation of the plane, calculate the obtuse angle between this plane and the plane $3 x + 2 y - 4 z = - 3$.
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SPS SPS FM Pure 2020 February Q8

8

Express $\frac { 8 x ^ { 2 } - x + 2 } { x \left( 4 x ^ { 2 } + 1 \right) }$ in partial fractions.
Hence find the general solution to the differential equation $$\frac { d y } { d x } + \frac { 2 y } { x } = \frac { 8 x ^ { 2 } - x + 2 } { x ^ { 3 } \left( 4 x ^ { 2 } + 1 \right) }$$

SPS SPS FM Pure 2020 February Q9

9

Show that $\cos 5 \theta = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta$.
Use the result of part (a) to prove that $\cos \frac { \pi } { 10 } = \sqrt { \frac { 5 + \sqrt { 5 } } { 8 } }$.

SPS SPS FM Pure 2020 February Q10

77 marks

10 The matrix $\left( \begin{array} { l l } 1 & 1
1 & 0 \end{array} \right)$ is denoted by $\mathbf { M }$.

Evaluate $\mathbf { M } ^ { 3 }$. The Fibonacci series $F _ { 0 } , F _ { 1 } , F _ { 2 } , F _ { 3 } , \ldots$ is defined by
$F _ { n + 1 } = F _ { n } + F _ { n - 1 }$ for $n \geq 1 , F _ { 0 } = 0 , F _ { 1 } = 1$.
Prove by mathematical induction that $\boldsymbol { M } ^ { n } = \left( \begin{array} { c c } F _ { n + 1 } & F _ { n }
F _ { n } & F _ { n - 1 } \end{array} \right)$ for $n \geq 1$.

Use the result of part (b) to find an expression for $F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }$ in terms of $n$, for $n \geq 1$.
\section*{ADDITIONAL ANSWER SPACE} If additional space is required, you should use the following lined page(s). The question number(s) must be clearly shown.

1(a)

$\left[ \ln \left\{ x + \sqrt { x ^ { 2 } - 9 } \right\} \right] _ { 3 } ^ { 5 }$ $= \ln ( 5 + 4 ) - \ln ( 3 ) = \ln ( 9 / 3 )$

$= \ln 3$

B1

M1

A1

[3]

Correct indefinite integral

Substitute and simplify to single logarithm ln 3 only

1(b)

\(\begin{gathered} { \left[ \frac { 1 } { 2 } \sinh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } \text { or } \left[ \frac { 1 } { 2 } \cosh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } }

\text { or } \left[ \frac { 1 } { 8 } \left( e ^ { 2 x } + e ^ { - 2 x } \right) \right] _ { 0 } ^ { \ln 3 }

= 1 / 2 \left[ 1 / 2 \left( \mathrm { e } ^ { \ln 3 } - \mathrm { e } ^ { - \ln 3 } \right) \right] ^ { 2 } - 0 \text { or equivalent }

= \frac { 1 } { 2 } \left[ \frac { 1 } { 2 } \left( 3 - \frac { 1 } { 3 } \right) \right] ^ { 2 } = \frac { 8 } { 9 } \end{gathered}\)

M1

A1

M1

A1

[4]

Correct method (e.g. factors can be wrong)

Correct indefinite integral

E.g. 25/9 - 1

Final answer 8/9, completely correct

2(a)

\(\begin{gathered} f ( 0 ) = \pi / 4

f ^ { \prime } ( x ) = \frac { 1 } { 1 + ( x + 1 ) ^ { 2 } }

f ^ { \prime } ( 0 ) = \frac { 1 } { 2 }

f ^ { \prime \prime } ( x ) = - \frac { 2 ( x + 1 ) } { \left( 1 + ( x + 1 ) ^ { 2 } \right) ^ { 2 } }

f ^ { \prime \prime } ( 0 ) = - \frac { 2 } { 4 } = - \frac { 1 } { 2 } \end{gathered}\)

B1

M1

A1ft

M1

A1

[5]

Correct f(0)

First derivative

Substitute $0 , \mathrm { ft }$ on their $f ^ { \prime } ( x )$

Second derivative (need not be expanded)

Substitute 0

2(b)

\(\begin{gathered} \frac { \pi } { 4 } + \frac { 1 } { 2 } x + \frac { 1 } { 2 } \left( - \frac { 1 } { 2 } \right) x ^ { 2 }

\frac { \pi } { 4 } + \frac { 1 } { 2 } x - \frac { 1 } { 4 } x ^ { 2 } \end{gathered}\)

M1

A1

Substitute into Maclaurin formula

CAO

3(a)

\(\frac { 1 } { 6 a + 1 } \left( \begin{array} { c c c } 3 a

- 3

1

- a

1

2 a

1

6

- 2 \end{array} \right)\)

M1

A1

M1

A1

[5]

Obtain cofactors

All cofactors correct and in right place

Divide by determinant

Determinant correct ( $- 6 a - 1$ )

Completely correct (expect to see all signs reversed)

3(b)

$- \frac { 1 } { 6 }$

A1ft

ft on their determinant

4(a)

\includegraphics[max width=\textwidth, alt={}]{6280d53b-3c1c-4dc9-a96b-2c58f2a7bf51-22_199_391_282_420}

B2

[2]

Give B1 if: just one obvious flaw in sketch, or if part of curve for negative $r$ given, e.g.

4(b)

$\frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 2 } - \cos \theta \right) ^ { 2 } d \theta$ \(\begin{gathered} = \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 4 } - \cos \theta + \cos ^ { 2 } \theta \right) d \theta

= \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 3 } { 4 } - \cos \theta + \frac { 1 } { 2 } \cos 2 \theta \right) d \theta

= \frac { 1 } { 2 } \left[ \frac { 3 } { 4 } \theta - \sin \theta + \frac { 1 } { 2 } \sin 2 \theta \right] _ { \pi / 3 } ^ { 5 \pi / 3 }

= \frac { 1 } { 2 } \pi + \frac { 3 } { 8 } \sqrt { 3 } \end{gathered}\)

M1

A1

[5]

Correct expression for area

Multiply out and use $\cos 2 \theta$

Correct integrand (ignore leading 1/2)

Correct indefinite integral (ignore leading 1/2)

Final answer, this or clear exact equivalent only

5(a)

2

A1

5(b)

Roots are $\alpha ( 1 + 1 / 2 ) , \beta ( 1 + 1 / 2 ) , \chi \left( 1 + \frac { 1 } { 2 } \right)$

Hence put $y = 3 x / 2$ $x = 2 y / 3$

$3 ( 2 y / 3 ) ^ { 3 } + 2 ( 2 y / 3 ) ^ { 2 } - 7 ( 2 y / 3 ) - 6 = 0$

$4 y ^ { 3 } + 4 y ^ { 2 } - 21 y - 27 = 0$

M1

A1ft

M1

A1

[5]

Use value of $\alpha \beta \gamma$ in $\alpha ( 1 + 1 / \alpha \beta \gamma )$ etc Correct new variable, ft on their $\alpha \beta \gamma$ Inverse function, ft Substitute inverse function

Complete equation including 0, any letter, any rational multiple

or

\(\begin{aligned}

\Sigma \alpha = - 2 / 3 , \Sigma \alpha \beta = - 7 / 3 , \alpha \beta \gamma = 2

\qquad \begin{array} { l } ( \alpha + \beta + \gamma ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ( = - 1 )

( \alpha \beta + \beta \gamma + \gamma \alpha ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 2 } \left( = - \frac { 21 } { 4 } \right) \end{array}

\qquad \alpha \beta \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 3 } \left( = \frac { 27 } { 4 } \right)

\text { Hence equation is } x ^ { 3 } + x ^ { 2 } - \frac { 21 } { 4 } x - \frac { 27 } { 4 } = 0 \end{aligned}\)

B1

M1

A1

All three correct

Correct formula for $\Sigma \alpha ^ { \prime }$ and substitute

Correct formula for $\Sigma \alpha ^ { \prime } \beta ^ { \prime }$ and substitute

Correct formula for $\alpha ^ { \prime } \beta ^ { \prime } \gamma ^ { \prime }$ and substitute

Correct final equation including 0, any letter, any rational multiple

\begin{table}[h]

\captionsetup{labelformat=empty} \caption{Further Mathematics Core (Pure) Mark Scheme (total 84)}

6(a)

Circle, centre $( 3 , - 1 )$

Radius $\sqrt { } 2$ indicated

Interior of circle indicated

M1

B1

A1

Allow +/- errors

Needs not to include origin

Any clear indication of interior

6(b)

Point of contact $( P )$ between circle and tangent through $O$ clearly identified

Form appropriate right-angled $\Delta$

$O C = \sqrt { 10 } , P C = \sqrt { 2 } \Rightarrow O P = 2 \sqrt { 2 }$ and $\angle P O C = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$

$\angle x O C = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$

$\angle x O P = \tan ^ { - 1 } \left( \frac { \frac { 1 } { 2 } - \frac { 1 } { 3 } } { 1 + \frac { 1 } { 2 } \cdot \frac { 1 } { 3 } } \right) = \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

B1

M1

A1

B1

M1

A1

[7]

Can be implied by correct working

Use exact method for a difference of angles Either $\tan ^ { - 1 } ( 1 / 7 )$ or $n = 7$.

If exactly 1/7 from calculator, give M0A0

or

Point of contact ( $P$ ) between circle and tangent through $O$ clearly identified

$y = m x$ meets $( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 2$

$( x - 3 ) ^ { 2 } + ( m x + 1 ) ^ { 2 } = 2$ has double roots $\left( m ^ { 2 } + 1 \right) x ^ { 2 } + 2 ( m - 3 ) x + 8 = 0$ has double roots

$4 ( m - 3 ) ^ { 2 } = 4.8 \left( m ^ { 2 } + 1 \right)$

$7 m ^ { 2 } + 6 m - 1 = 0$, so $m = \frac { 1 } { 7 }$ or - 1

But $m$ is not - 1 as this gives minimum, not maximum, argument

B1

M1

A1

M1

A1

Can be implied by correct working, provided final answer does not include - 1

Line and circle equations used

Discriminant set to zero

Obtain $\tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

Reject - 1 , with reason. Not enough to say " $m = \frac { 1 } { 7 }$ or - 1 so $n = 7$ "

or

Point of contact $( P )$ between circle and tangent through $O$ clearly identified
Appropriate right-angled $\Delta$ with centre $C$
$O C = \sqrt { 10 } , P C = \sqrt { 2 } \Rightarrow O P = 2 \sqrt { 2 }$
Circles, centre $O , r = 2 \sqrt { 2 }$, \	$C , r = \sqrt { 2 }$
$x ^ { 2 } + ( 3 x - 8 ) ^ { 2 } = 8,5 x ^ { 2 } - 24 x + 28 = 0$
$\Rightarrow P = \left( 2 \frac { 4 } { 5 } , \frac { 2 } { 5 } \right)$
$\theta = \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

B1

M1

B1

M1

A1

Can be implied by correct working

$x ^ { 2 } + y ^ { 2 } = 8$ and $( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 2$

Eliminate $y$, solve quadratic in $x$

Correct $P$, both cords. allow decimals

Either $\tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)$ or $n = 7$

\end{table}

7(a)

Distance is $\frac { | \mathbf { b } \mathbf { . n } - \mathbf { p } | } { | \mathbf { n } | }$ $\mathbf { b } = ( 9,5,2 ) , \mathbf { n } = ( 3,2 , - 4 )$

Scalar product is 29

Distance is $32 / \sqrt { } 29$

M2

A1

[4]

Use correct formula with correct $\mathbf { b }$ and $\mathbf { n }$

Correct scalar product

Answer (accept 5.94 (3s.f.))

7(b)

\(\begin{aligned}

( 3 \mathbf { i } + \mathbf { k } ) \times ( \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )

= \mathbf { i } + 7 \mathbf { j } - 3 \mathbf { k } \end{aligned}\) \(\begin{gathered} x + 7 y - 3 z = 0

\left( \begin{array} { c } 1

7

- 3 \end{array} \right) \cdot \left( \begin{array} { c } 3

2

- 4 \end{array} \right) \end{gathered}\) \(\begin{gathered} \cos \theta = \frac { 29 } { \sqrt { 59 } \sqrt { 29 } }

134.5 ^ { \circ } \text { or } 2.35 \text { radians } \end{gathered}\)

M1

A1

M1

A1ft

A1

[6]

Find perpendicular vector

Any multiple of this

Cartesian equation of second plane

Scalar product of normal vectors of 2 planes

Correct scalar product and moduli (ft their vectors)

Correct obtuse angle (awrt $135 ^ { \circ }$ or 2.35 rad )

8(a)

\(\begin{gathered} \frac { A } { x } + \frac { B x + C } { 4 x ^ { 2 } + 1 }

A \left( 4 x ^ { 2 } + 1 \right) + ( B x + C ) x \equiv 8 x ^ { 2 } - x + 2

A = 2 , B = 0 , C = - 1

\frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } \end{gathered}\)

M1

A1

[5]

Attempt at partial fractions

Set up identity

Compare coefficients or substitute x -values

All correct values

Final answer

8(b)

IF $\exp \left( \int 2 / x \mathrm {~d} x \right) = x ^ { 2 }$ \(\begin{gathered} y x ^ { 2 } = \int \frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } d x

y x ^ { 2 } = 2 \ln x - \frac { 1 } { 2 } \tan ^ { - 1 } ( 2 x ) + c

y = \frac { 4 \ln x - \tan ^ { - 1 } ( 2 x ) + c ^ { \prime } } { 2 x ^ { 2 } } \end{gathered}\)

M1

A1

M1

A1

[6]

Method for IF

$x ^ { 2 }$

Multiply through by IF and spot link to (a)

Use $\tan ^ { - 1 }$

$2 \ln x$ and $+ c$

Final answer, completely correct, any equivalent form, any correct way of writing constant (e.g. c or 2c)

\section*{Further Mathematics Core (Pure) Mark Scheme (total 84)}

9(a)

\(\begin{aligned}

( c + \mathrm { is } ) ^ { 5 } = c ^ { 5 } + 5 \mathrm { is } ^ { 4 } - 10 s ^ { 2 } c ^ { 2 } - 10 \mathrm { is } ^ { 3 } c ^ { 2 } + 5 s ^ { 4 } c + \mathrm { is } ^ { 5 }

\cos 5 \theta = \cos ^ { 5 } \theta - 10 \sin ^ { 2 } \theta \cos ^ { 3 } \theta + 5 \sin ^ { 4 } \theta \cos \theta

= c ^ { 5 } - 10 \left( 1 - c ^ { 2 } \right) c ^ { 3 } + 5 \left( 1 - c ^ { 2 } \right) ^ { 2 } c

= 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta \quad \text { AG } \end{aligned}\)

M1

A1

M1

A1

[4]

Use de Moivre (can ignore im parts if clear)

Correct equating of real parts

Use $\sin ^ { 2 } \theta = 1 - \cos ^ { 2 } \theta$

Correctly obtain given answer

9(b)

Using $\cos 5 \theta = 0$ gives $16 c ^ { 5 } - 20 c ^ { 4 } + 5 c = 0$ $c \neq 0 \text { so } 16 c ^ { 4 } - 20 c ^ { 2 } + 5 = 0$

$c ^ { 2 } = ( 20 \pm \sqrt { } 80 ) / 32$ $c = \pm \sqrt { \frac { 5 \pm \sqrt { 5 } } { 8 } }$

$\cos \pi / 10 > 0$ so first $\pm$ is +

$\cos \pi / 10 > \cos ( 3 \pi / 10 )$ so second $\pm$ is +

M1

B1

M1

A1

[6]

Use $\theta = \pi / 10$ and $\cos ( \pi / 2 ) = 0$

Justify cancellation of $c$

Solve quadratic in $c ^ { 2 }$ by exact method

Correct expression for $c$, any combination of $\pm$ and + signs

Justify outer +

Justify inner +

10(a)

\(\left( \begin{array} { l l } 3

2

1 \end{array} \right)\)

B2

[2]

If B0, give B1 if \(\mathbf { M } ^ { 2 } = \left( \begin{array} { l l } 2

1

1 \end{array} \right)\) seen

10(b)

Base case: \(n = 1 , \mathbf { M } = \left( \begin{array} { l l } 1	1
1	0 \end{array} \right) = \left( \begin{array} { l l } F _ { 2 }	F _ { 1 }
F _ { 1 }	F _ { 0 } \end{array} \right)\)
Inductive step: assume \(\mathbf { M } ^ { k } = \left( \begin{array} { c c } F _ { k + 1 }	F _ { k }
F _ { k }	F _ { k - 1 } \end{array} \right)\)
\(\Rightarrow \mathbf { M } ^ { k + 1 } = \left( \begin{array} { c c } F _ { k + 1 }	F _ { k }
F _ { k }	F _ { k - 1 } \end{array} \right) \left( \begin{array} { l l } 1	1
1	0 \end{array} \right)\) \(\begin{aligned}	= \left( \begin{array} { c c } F _ { k + 1 } + F _ { k }	F _ { k } + F _ { k - 1 }
F _ { k } + F _ { k - 1 }	F _ { k } \end{array} \right)
	\left( \begin{array} { c c } F _ { k + 2 }	F _ { k + 1 }
F _ { k + 1 }	F _ { k } \end{array} \right) \text { as required } \end{aligned}\)
Base case and inductive step both true, so statement true for all $n \in \mathbb { N }$ by mathematical induction

B1

M1

A1

[5]

Fully correct

Correct "assume" statement, allow $n$

Consider $\mathbf { M } \times$ hypothesised $\mathbf { M } ^ { k }$

Correctly show this result

Award only if all 4 other marks gained

10(c)

Det $\left( \mathbf { M } ^ { n } \right) = F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }$

$\operatorname { Det } \left( \mathbf { M } ^ { n } \right) = ( \operatorname { Det } \mathbf { M } ) ^ { n }$ $= ( - 1 ) ^ { n }$

M1

A1

[3]

Use determinant of $\mathbf { M } ^ { n }$

Relate $\operatorname { det } \mathbf { M } ^ { \boldsymbol { n } }$ to $\operatorname { det } \mathbf { M }$

Correct answer

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