Questions - OCR Further Pure Core 2

OCR Further Pure Core 2 2018 September Q1

1 Line $l _ { 1 }$ has Cartesian equation $$l _ { 1 } : \quad \frac { - x } { 2 } = \frac { y - 5 } { 2 } = \frac { - z - 6 } { 7 } .$$

Find a vector equation for $l _ { 1 }$. Line $l _ { 2 }$ has vector equation $$l _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { c }

OCR Further Pure Core 2 2018 September Q2

2
7
- 1 \end{array} \right) + \mu \left( \begin{array} { c } 1
- 2
4 \end{array} \right) .$$ (ii) Find the point of intersection of $l _ { 1 }$ and $l _ { 2 }$.
(iii) Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$. 2 In this question you must show detailed reasoning.
(i) Find $\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } 2 \tan x \mathrm {~d} x$ giving your answer in the form $\ln p$.
(ii) Show that $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } 2 \tan x \mathrm {~d} x$ is undefined explaining your reasoning.

OCR Further Pure Core 2 2018 September Q4

4 \end{array} \right) .$$ (ii) Find the point of intersection of $l _ { 1 }$ and $l _ { 2 }$.
(iii) Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$. 2 In this question you must show detailed reasoning.
(i) Find $\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } 2 \tan x \mathrm {~d} x$ giving your answer in the form $\ln p$.
(ii) Show that $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } 2 \tan x \mathrm {~d} x$ is undefined explaining your reasoning. 3 The equation of a plane, $\Pi$, is $$\Pi : \quad \mathbf { r } = \left( \begin{array} { c } 2
- 3

OCR Further Pure Core 2 2018 September Q5

5 \end{array} \right) + \lambda \left( \begin{array} { l } 1
1
3 \end{array} \right) + \mu \left( \begin{array} { c } - 1
2
1 \end{array} \right) .$$

Find a vector which is perpendicular to $\Pi$.
Hence find an equation for $\Pi$ in the form r.n $= p$.
Find in the form $\sqrt { q }$ the shortest distance between $\Pi$ and the origin, where $q$ is a rational number. 4 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { c r c } a & 2 & 3
4 & 4 & 6
- 2 & 2 & 9 \end{array} \right)$ where $a$ is a constant. It is given that if $\mathbf { A }$ is not singular then $$\mathbf { A } ^ { - 1 } = \frac { 1 } { 24 a - 48 } \left( \begin{array} { c c c } 24 & - 12 & 0
- 48 & 9 a + 6 & 12 - 6 a
16 & - 2 a - 4 & 4 a - 8 \end{array} \right)$$
Use $\mathbf { A } ^ { - 1 }$ to solve the simultaneous equations below, giving your answer in terms of $k$. $$\begin{array} { r } x + 2 y + 3 z = 6
4 x + 4 y + 6 z = 8
- 2 x + 2 y + 9 z = k \end{array}$$
Consider the equations below where $a$ takes the value which makes $\mathbf { A }$ singular. $$\begin{aligned} a x + 2 y + 3 z & = b
4 x + 4 y + 6 z & = 10
- 2 x + 2 y + 9 z & = - 13 \end{aligned}$$ $b$ takes the value for which the equations have an infinite number of solutions.
- Determine the value of $b$.
- Find the solutions for $y$ and $z$ in terms of $x$.
- For the equations in part (ii) with the values of $a$ and $b$ found in part (ii) describe fully the geometrical arrangement of the planes represented by the equations.
5 The region $R$ between the $x$-axis, the curve $y = \frac { 1 } { \sqrt { p + x ^ { 2 } } }$ and the lines $x = \sqrt { p }$ and $x = \sqrt { 3 p }$, where $p$ is a positive parameter, is rotated by $2 \pi$ radians about the $x$-axis to form a solid of revolution $S$.
Find and simplify an algebraic expression, in terms of $p$, for the exact volume of $S$.
Given that $R$ must lie entirely between the lines $x = 1$ and $x = \sqrt { 48 }$ find in exact form
- the greatest possible value of the volume of $S$
- the least possible value of the volume of $S$.

OCR Further Pure Core 2 2018 September Q7

7
- 1 \end{array} \right) + \mu \left( \begin{array} { c } 1
- 2
4 \end{array} \right) .$$ (ii) Find the point of intersection of $l _ { 1 }$ and $l _ { 2 }$.
(iii) Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$. 2 In this question you must show detailed reasoning.
(i) Find $\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } 2 \tan x \mathrm {~d} x$ giving your answer in the form $\ln p$.
(ii) Show that $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } 2 \tan x \mathrm {~d} x$ is undefined explaining your reasoning. 3 The equation of a plane, $\Pi$, is $$\Pi : \quad \mathbf { r } = \left( \begin{array} { c } 2
- 3
5 \end{array} \right) + \lambda \left( \begin{array} { l } 1
1
3 \end{array} \right) + \mu \left( \begin{array} { c } - 1
2
1 \end{array} \right) .$$ (i) Find a vector which is perpendicular to $\Pi$.
(ii) Hence find an equation for $\Pi$ in the form r.n $= p$.
(iii) Find in the form $\sqrt { q }$ the shortest distance between $\Pi$ and the origin, where $q$ is a rational number. 4 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { c r c } a & 2 & 3
4 & 4 & 6
- 2 & 2 & 9 \end{array} \right)$ where $a$ is a constant. It is given that if $\mathbf { A }$ is not singular then $$\mathbf { A } ^ { - 1 } = \frac { 1 } { 24 a - 48 } \left( \begin{array} { c c c } 24 & - 12 & 0
- 48 & 9 a + 6 & 12 - 6 a
16 & - 2 a - 4 & 4 a - 8 \end{array} \right)$$ (i) Use $\mathbf { A } ^ { - 1 }$ to solve the simultaneous equations below, giving your answer in terms of $k$. $$\begin{array} { r } x + 2 y + 3 z = 6
4 x + 4 y + 6 z = 8
- 2 x + 2 y + 9 z = k \end{array}$$ (ii) Consider the equations below where $a$ takes the value which makes $\mathbf { A }$ singular. $$\begin{aligned} a x + 2 y + 3 z & = b
4 x + 4 y + 6 z & = 10
- 2 x + 2 y + 9 z & = - 13 \end{aligned}$$ $b$ takes the value for which the equations have an infinite number of solutions.

Determine the value of $b$.
Find the solutions for $y$ and $z$ in terms of $x$.
(iii) For the equations in part (ii) with the values of $a$ and $b$ found in part (ii) describe fully the geometrical arrangement of the planes represented by the equations.

5 The region $R$ between the $x$-axis, the curve $y = \frac { 1 } { \sqrt { p + x ^ { 2 } } }$ and the lines $x = \sqrt { p }$ and $x = \sqrt { 3 p }$, where $p$ is a positive parameter, is rotated by $2 \pi$ radians about the $x$-axis to form a solid of revolution $S$.
(i) Find and simplify an algebraic expression, in terms of $p$, for the exact volume of $S$.
(ii) Given that $R$ must lie entirely between the lines $x = 1$ and $x = \sqrt { 48 }$ find in exact form

the greatest possible value of the volume of $S$
the least possible value of the volume of $S$.

6 (i) By considering $\sum _ { r = 1 } ^ { n } \left( ( r + 1 ) ^ { 5 } - r ^ { 5 } \right)$ show that $\sum _ { r = 1 } ^ { n } r ^ { 4 } = \frac { 1 } { 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ { 2 } + 3 n - 1 \right)$.
(ii) Use the formula given in part (i) to find $50 ^ { 4 } + 51 ^ { 4 } + \ldots + 80 ^ { 4 }$. 7 The roots of the equation $a x ^ { 2 } + b x + c = 0$, where $a , b$ and $c$ are positive integers, are $\alpha$ and $\beta$.
(i) Find a quadratic equation with integer coefficients whose roots are $\alpha + \beta$ and $\alpha \beta$.
(ii) Show that it is not possible for the original equation and the equation found in part (i) both to have repeated roots.
(iii) Show that the discriminant of the equation found in part (i) is always positive.

OCR Further Pure Core 2 2018 September Q8

8 In this question you must show detailed reasoning.

Express $( 6 + 5 \mathrm { i } ) ( 7 + 5 \mathrm { i } )$ in the form $a + b \mathrm { i }$.
You are given that $17 ^ { 2 } + 65 ^ { 2 } = 4514$. Using the result in part (i) and by considering (6-5i)(7-5i) express 4514 as a product of its prime factors.

OCR Further Pure Core 2 2018 September Q9

9 The quantity of grass on an island at time $t$ years is $x$, in appropriate units. At time $t = 0$ some rabbits are introduced to the island. The population of rabbits on the island at time $t$ years is $y$, in units of 100s of rabbits. An ecologist who is studying the island suggests that the following pair of simultaneous first order differential equations can be used to model the population of rabbits and quantity of grass for $t \geqslant 0$. $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 3 x - 2 y ,
& \frac { \mathrm {~d} y } { \mathrm {~d} t } = y + 5 x \end{aligned}$$

(a) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = a \frac { \mathrm {~d} x } { \mathrm {~d} t } + b x$ where $a$ and $b$ are constants which should be found.
(b) Find the general solution for $x$ in real form.
Find the corresponding general solution for $y$. At time $t = 0$ the quantity of grass on the island was 4 units. The number of rabbits introduced at this time was 500 .
Find the particular solutions for $x$ and $y$.
The ecologist finds that the model predicts that there will be no grass at time $T$, when there are still rabbits on the island. Find the value of $T$.
State one way in which the model is not appropriate for modelling the quantity of grass and the population of rabbits for $0 \leqslant t \leqslant T$. \section*{OCR} \section*{Oxford Cambridge and RSA}

OCR Further Pure Core 2 2018 December Q1

1

The Argand diagram below shows the two points which represent two complex numbers, $z _ { 1 }$ and $z _ { 2 }$.
\includegraphics[max width=\textwidth, alt={}, center]{e792797e-6d20-4fc3-9733-43db10f764d7-2_371_689_429_360} On the copy of the diagram in the Printed Answer Booklet
- draw an appropriate shape to illustrate the geometrical effect of adding $z _ { 1 }$ and $z _ { 2 }$,
- indicate with a cross $( \times )$ the location of the point representing the complex number $z _ { 1 } + z _ { 2 }$.
- You are given that $\arg z _ { 3 } = \frac { 1 } { 4 } \pi$ and $\arg z _ { 4 } = \frac { 3 } { 8 } \pi$.
In each part, sketch and label the points representing the numbers $z _ { 3 } , z _ { 4 }$ and $z _ { 3 } z _ { 4 }$ on the diagram in the Printed Answer Booklet. You should join each point to the origin with a straight line.
(i) $\left| z _ { 3 } \right| = 1.5$ and $\left| z _ { 4 } \right| = 1.2$
(ii) $\left| z _ { 3 } \right| = 0.7$ and $\left| z _ { 4 } \right| = 0.5$

OCR Further Pure Core 2 2018 December Q2

2 In this question you must show detailed reasoning. S is the 2-D transformation which is a stretch of scale factor 3 parallel to the $x$-axis. $\mathbf { A }$ is the matrix which represents S .

Write down $\mathbf { A }$.
By considering the transformation represented by $\mathbf { A } ^ { - 1 }$, determine the matrix $\mathbf { A } ^ { - 1 }$. Matrix $\mathbf { B }$ is given by $\mathbf { B } = \left( \begin{array} { c c } 0 & - 1
- 1 & 0 \end{array} \right)$. T is the transformation represented by $\mathbf { B }$.
Describe T.
Determine the matrix which represents the transformation S followed by T .
Demonstrate, by direct calculation, that $( \mathbf { B A } ) ^ { - 1 } = \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }$.

OCR Further Pure Core 2 2018 December Q3

3 In this question you must show detailed reasoning. Solve the equation $2 \cosh ^ { 2 } x + 5 \sinh x - 5 = 0$ giving each answer in the form $\ln ( p + q \sqrt { r } )$ where $p$ and $q$ are rational numbers, and $r$ is an integer, whose values are to be determined.

OCR Further Pure Core 2 2018 December Q4

4 You are given that the matrix $\mathbf { A } = \left( \begin{array} { c c c } 1 & 0 & 0
0 & \frac { 2 a - a ^ { 2 } } { 3 } & 0
0 & 0 & 1 \end{array} \right)$, where $a$ is a positive constant, represents the transformation R which is a reflection in 3-D.

State the plane of reflection of R .
Determine the value of $a$.
With reference to R explain why $\mathbf { A } ^ { 2 } = \mathbf { I }$, the $3 \times 3$ identity matrix.

OCR Further Pure Core 2 2018 December Q5

5

Find the shortest distance between the point ( $- 6,4$ ) and the line $y = - 0.75 x + 7$. Two lines, $l _ { 1 }$ and $l _ { 2 }$, are given by
$l _ { 1 } : \mathbf { r } = \left( \begin{array} { c } 4
3
- 2 \end{array} \right) + \lambda \left( \begin{array} { c } 2
1
- 4 \end{array} \right)$ and $l _ { 2 } : \mathbf { r } = \left( \begin{array} { c } 11
- 1
5 \end{array} \right) + \mu \left( \begin{array} { c } 3
- 1
1 \end{array} \right)$.
Find the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.
Hence determine the geometrical arrangement of $l _ { 1 }$ and $l _ { 2 }$.

OCR Further Pure Core 2 2018 December Q6

6 Three matrices, $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$, are given by $\mathbf { A } = \left( \begin{array} { c c } 1 & 2
a & - 1 \end{array} \right) , \mathbf { B } = \left( \begin{array} { c c } 2 & - 1
4 & 1 \end{array} \right)$ and $\mathbf { C } = \left( \begin{array} { c c } 5 & 0
- 2 & 2 \end{array} \right)$ where $a$ is a

Using $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$ in that order demonstrate explicitly the associativity property of matrix multiplication.
Use $\mathbf { A }$ and $\mathbf { C }$ to disprove by counterexample the proposition 'Matrix multiplication is commutative'. For a certain value of $a , \mathbf { A } \binom { x } { y } = 3 \binom { x } { y }$.
Find
- $y$ in terms of $x$,
- the value of $a$.
  $7 C$ is the locus of numbers, $z$, for which $\operatorname { Im } \left( \frac { z + 7 \mathrm { i } } { z - 24 } \right) = \frac { 1 } { 4 }$.
  By writing $z = x + \mathrm { i } y$ give a complete description of the shape of $C$ on an Argand diagram.

OCR Further Pure Core 2 2018 December Q8

8
\includegraphics[max width=\textwidth, alt={}, center]{e792797e-6d20-4fc3-9733-43db10f764d7-4_677_1182_587_440} The figure shows part of the graph of $y = ( x - 3 ) \sqrt { \ln x }$. The portion of the graph below the $x$-axis is rotated by $2 \pi$ radians around the $x$-axis to form a solid of revolution, $S$. Determine the exact volume of $S$.

OCR Further Pure Core 2 2018 December Q9

9

By using Euler's formula show that $\cosh ( \mathrm { i } z ) = \cos z$.
Hence, find, in logarithmic form, a root of the equation $\cos z = 2$. [You may assume that $\cos z = 2$ has complex roots.]

OCR Further Pure Core 2 2018 December Q10

10 A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it. The extent to which the door is open at any time, $t$ seconds, is measured by the angle at the hinge, $\theta$, which the plane of the door makes with the plane of the equilibrium position. See the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{e792797e-6d20-4fc3-9733-43db10f764d7-5_367_1116_625_246} In an initial model of the motion of a certain swing door it is suggested that $\theta$ satisfies the following differential equation.
$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 25 \theta = 0$

1. Write down the general solution to (\textit{).
2. With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (}) is unlikely to be realistic. In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes
  $4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + \lambda \frac { \mathrm { d } \theta } { \mathrm { d } t } + 25 \theta = 0 \quad ( \dagger )$
  where $\lambda$ is a positive constant.
In the case where $\lambda = 16$ the door is held open at an angle of 0.9 radians and then released from rest at time $t = 0$.
1. Find, in a real form, the general solution of ( $\dagger$ ).
2. Find the particular solution of ( $\dagger$ ).
3. With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in ( $\dagger$ ) improves the model.
Find the value of $\lambda$ for which the door is critically damped.

OCR Further Pure Core 2 2017 Specimen Q2

4 marks

2 In this question you must show detailed reasoning. The finite region $R$ is enclosed by the curve with equation $y = \frac { 8 } { \sqrt { 16 + x ^ { 2 } } }$, the $x$-axis and the lines $x = 0$ and $x = 4$. Region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis. Find the exact value of the volume generated. [4]

OCR Further Pure Core 2 2017 Specimen Q3

3

Find $\sum _ { r = 1 } ^ { n } \left( \frac { 1 } { r } - \frac { 1 } { r + 2 } \right)$.
What does the sum in part (i) tend to as $n \rightarrow \infty$ ? Justify your answer.

OCR Further Pure Core 2 2017 Specimen Q4

4 Express $\frac { 5 x ^ { 2 } + x + 12 } { x ^ { 3 } + 4 x }$ in partial fractions.

OCR Further Pure Core 2 2017 Specimen Q5

5 In this question you must show detailed reasoning. Evaluate $\int _ { 0 } ^ { \infty } 2 x \mathrm { e } ^ { - x } \mathrm {~d} x$.
[0pt] [You may use the result $\lim _ { x \rightarrow \infty } x \mathrm { e } ^ { - x } = 0$.]

OCR Further Pure Core 2 2017 Specimen Q6

6 The equation of a plane $\Pi$ is $x - 2 y - z = 30$.

Find the acute angle between the line $\mathbf { r } = \left( \begin{array} { c } 3
2
- 5 \end{array} \right) + \lambda \left( \begin{array} { r } - 5
3
2 \end{array} \right)$ and $\Pi$.
Determine the geometrical relationship between the line $\mathbf { r } = \left( \begin{array} { l } 1
4
2 \end{array} \right) + \mu \left( \begin{array} { r } 3
- 1
5 \end{array} \right)$ and $\Pi$.

OCR Further Pure Core 2 2017 Specimen Q7

7

Use the Maclaurin series for $\sin x$ to work out the series expansion of $\sin x \sin 2 x \sin 4 x$ up to and including the term in $x ^ { 3 }$.
Hence find, in exact surd form, an approximation to the least positive root of the equation $2 \sin x \sin 2 x \sin 4 x = x$.

OCR Further Pure Core 2 2021 June Q1

1

The Argand diagram below shows the two points which represent two complex numbers, $z _ { 1 }$ and $z _ { 2 }$.
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-02_321_592_276_347} On the copy of the diagram in the Resource Materials.
- draw an appropriate shape to illustrate the geometrical effect of adding $z _ { 1 }$ and $z _ { 2 }$,
- indicate with a cross $( \times )$ the location of the point representing the complex number $z _ { 1 } + z _ { 2 }$.
- You are given that $\arg z _ { 3 } = \frac { 1 } { 4 } \pi$ and $\arg z _ { 4 } = \frac { 3 } { 8 } \pi$.
In each part, sketch and label the points representing the numbers $z _ { 3 } , z _ { 4 }$ and $z _ { 3 } z _ { 4 }$ on the diagram in the Resource Materials. You should join each point to the origin with a straight line.
(i) $\left| z _ { 3 } \right| = 1.5$ and $\left| z _ { 4 } \right| = 1.2$
(ii) $\left| z _ { 3 } \right| = 0.7$ and $\left| z _ { 4 } \right| = 0.5$

OCR Further Pure Core 2 2021 June Q2

35 marks

2 In this question you must show detailed reasoning.
Solve the equation $2 \cosh ^ { 2 } x + 5 \sinh x - 5 = 0$ giving each answer in the form $\ln ( p + q \sqrt { r } )$ where $p$ and $q$ are rational numbers, and $r$ is an integer, whose values are to be determined. You are given that the matrix $\mathbf { A } = \left( \begin{array} { c c c } 1 & 0 & 0
0 & \frac { 2 a - a ^ { 2 } } { 3 } & 0
0 & 0 & 1 \end{array} \right)$, where $a$ is a positive constant, represents the transformation R which is a reflection in 3-D.

State the plane of reflection of $R$.
Determine the value of $a$.
With reference to R explain why $\mathbf { A } ^ { 2 } = \mathbf { I }$, the $3 \times 3$ identity matrix.
By using Euler's formula show that $\cosh ( \mathrm { iz } ) = \cos z$.
Hence, find, in logarithmic form, a root of the equation $\cos z = 2$. [You may assume that $\cos z = 2$ has complex roots.] A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it. The extent to which the door is open at any time, $t$ seconds, is measured by the angle at the hinge, $\theta$, which the plane of the door makes with the plane of the equilibrium position. See the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-03_317_954_497_255} In an initial model of the motion of a certain swing door it is suggested that $\theta$ satisfies the following differential equation. $$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 25 \theta = 0$$
1. Write down the general solution to (\textit{).
2. With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (}) is unlikely to be realistic. In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes $$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + \lambda \frac { \mathrm { d } \theta } { \mathrm {~d} t } + 25 \theta = 0$$ where $\lambda$ is a positive constant.
In the case where $\lambda = 16$ the door is held open at an angle of 0.9 radians and then released from rest at time $t = 0$.
1. Find, in a real form, the general solution of ( $\dagger$ ).
2. Find the particular solution of ( $\dagger$ ).
3. With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in ( $\dagger$ ) improves the model.

Find the value of $\lambda$ for which the door is critically damped. \section*{Total Marks for Question Set 1: 37} \section*{Resource Materials} 1(a)
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-04_344_621_212_794} 1(b)(i)
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-04_355_556_648_685} 1(b)(ii)
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-04_355_556_1160_685} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Marks

AO

Guidance

1

(a)

\includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_395_674_200_575}

B1

[2]

2.2a

4 lines drawn as shown to complete a parallelogram

Or 2 lines drawn to form a triangle which is either the upper or lower half of the parallelogram (split by the leading diagonal). eg

(b)

(i)

\includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_374_630_712_568}

B1

[2]

1.1

2.2a

$z _ { 3 }$ and $z _ { 4 }$ approximately correctly positioned and labelled.

Approximate correct length (eg $z _ { 4 }$ length increased by 50\%) and angle (about a quarter of the way round the $2 ^ { \text {nd } }$ quadrant).

If no labels shown then B1B1 can only follow if there is no ambiguity between points (eg magnitudes shown).

$r = 1.8 , \theta = \frac { 5 } { 8 } \pi$

(b)

(ii)

\includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_385_634_1154_578}

B1

[2]

1.1

2.2a

$z _ { 3 }$ and $z _ { 4 }$ approximately correctly positioned and labelled.

Approximate correct length (eg $z _ { 3 }$ length halved) and either the same angle as part (b)(i) or about a quarter of the way round the $2 ^ { \text {nd } }$ quadrant.

If no labels shown then B1B1 can only follow if there is no ambiguity between points (eg magnitudes shown). $r = 0.35 , \theta = \frac { 5 } { 8 } \pi$

Question

Answer

Marks

AO

Guidance

2

$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$

M1

1.1a

Reduction to 3 term quadratic in $\sinh x$ or $\cosh ^ { 2 } x$

A1

1.1

M1

1.1

Use of $\ln$ formula for $\sinh ^ { - 1 }$ or $\cosh ^ { - }$ 1

\(\begin{aligned}

\sinh x = 1 / 2 \text { or } - 3

x = \ln \left( \frac { 1 } { 2 } + \sqrt { \frac { 5 } { 4 } } \right)

x = \ln \left( \frac { 1 } { 2 } + \frac { 1 } { 2 } \sqrt { 5 } \right) \end{aligned}\)

Must be in the correct form but

$x = \ln ( - 3 + \sqrt { 10 } )$

A1

1.1

[6]

3

(a)

The $x$ - $z$ plane

B1

[1]

2.2a

or $y = 0$

(b)

\(\begin{aligned}

\frac { 2 a - a ^ { 2 } } { 3 } = - 1

a ^ { 2 } - 2 a - 3 = 0 \Rightarrow a = - 1,3

a > 0 \Rightarrow a = 3 \end{aligned}\)

B1

1.1

\multirow{3}{*}{BC. Rearranging the quadratic equation and solving. discarding $a = - 1$}

\multirow{3}{*}{}

М1

3.1a

A1 [3]

2.3

\multirow[t]{3}{*}{(c)}

\multirow[t]{3}{*}{Any reflection is self-inverse... oe $\text { …so } \mathbf { A } ^ { 2 } = \mathbf { A } \mathbf { A } ^ { - 1 } = \mathbf { I }$}

B1

2.4

eg "If you do a reflection twice it gets back to where it started"

B1

2.4

[2]

Question

Answer

Marks

AO

Guidance

4

(a)

\(\begin{aligned}

\cosh ( \mathrm { i } z ) = \frac { \mathrm { e } ^ { \mathrm { i } z } + \mathrm { e } ^ { - \mathrm { i } z } } { 2 }

= \frac { \cos z + \mathrm { i } \sin z + \cos z - \mathrm { i } \sin z } { 2 }

= \frac { 2 \cos z } { 2 } = \cos z \quad \mathbf { A G } \end{aligned}\)

M1

М1

A1

[3]

2.1

1.1

2.1

Use of correct exponential form for cosh

Correct use of Euler's formula at least once

Both $\mathbf { M }$ marks must be awarded. Must have $\cosh ( \mathrm { i } z ) =$ or LHS =

Proof must be complete for A1

(b)

\(\begin{aligned}

\cos z = 2 = > \cosh ( \mathrm { i } z ) = 2 = > z = \left( \cosh ^ { - 1 } 2 \right) / \mathrm { i }

= - \mathrm { i } \ln ( 2 + \sqrt { 3 } ) \end{aligned}\)

M1

A1

[2]

3.1a 1.1

± inside or outside the $\ln$ (ie allow eg $i \ln ( 2 + \sqrt { 3 } )$ or $i \ln ( 2 - \sqrt { 3 } )$ and condone eg $\pm \mathrm { i } \ln ( 2 + \sqrt { } 3 )$ www)

or $2 \pi n \pm \mathrm { i } \ln ( 2 + \sqrt { } 3 )$ for any integer $n$

5

(a)

(i)

$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - \left( \frac { 5 } { 2 } \right) ^ { 2 } \theta$

$\theta = A \cos \omega t + B \sin \omega t$ or $R \cos ( \omega t + \phi )$ with any positive value for $\omega$ $\theta = A \cos \frac { 5 } { 2 } t + B \sin \frac { 5 } { 2 } t \text { or } R \cos \left( \frac { 5 } { 2 } t + \phi \right)$

М1

A1

[2]

1.1

If $\mathbf { M 0 }$ then $\mathbf { S C 1 }$ for $\theta = A \cos \frac { 5 } { 2 } t$ or $\theta = A \sin \frac { 5 } { 2 } t$

(a)

(ii)

The model predicts infinite oscillations of the same amplitude; in practice the amplitude must decrease over time.

E1

[1]

3.5b

Question

Answer

Marks

AO

Guidance

(b)

(i)

AE: $4 m ^ { 2 } + 16 m + 25 = 0$

M1

1.1a

Writing down the AE correctly or using $\theta = A \mathrm { e } ^ { m t }$ and substituting into (*) to derive a three term quadratic AE.

\(\begin{aligned}

- 2 \pm \frac { 3 } { 2 } \mathrm { i }

\theta = \mathrm { e } ^ { - 2 t } \left( A \cos \frac { 3 } { 2 } t + B \sin \frac { 3 } { 2 } t \right) \end{aligned}\)

A1ft

1.1

Their $\mathrm { e } ^ { p t } ( A \cos q t + B \sin q t )$ for solution of $\mathrm { AE } = p \pm q \mathrm { i }$

[3]

(b)

(ii)

\(\begin{aligned}

t = 0 , \theta = 0.9 \Rightarrow A = 0.9

\frac { \mathrm {~d} \theta } { \mathrm {~d} t } = - 2 \mathrm { e } ^ { - 2 t } \left( A \cos \frac { 3 } { 2 } t + B \sin \frac { 3 } { 2 } t \right)

+ \mathrm { e } ^ { - 2 t } \left( - \frac { 3 } { 2 } A \sin \frac { 3 } { 2 } t + \frac { 3 } { 2 } B \cos \frac { 3 } { 2 } t \right)

t = 0 , \frac { \mathrm {~d} \theta } { \mathrm {~d} t } = 0 \Rightarrow - 2 A + \frac { 3 } { 2 } B = 0

B = 1.2

\theta = \mathrm { e } ^ { - 2 t } \left( 0.9 \cos \frac { 3 } { 2 } t + 1.2 \sin \frac { 3 } { 2 } t \right) \end{aligned}\)

B1 M1

3.4 1.1 a

Attempt to differentiate using the product and chain rules ( $A$ may be replaced by a number).

Substituting $t = 0$ into $\frac { \mathrm { d } \theta } { \mathrm { d } t }$ to derive an equation in ( $A$ and) $B$

A1

1.1

[4]

(b)

(iii)

In the modified model $\theta \rightarrow 0$ as $t \rightarrow \infty$ oe This is the behaviour we would expect to observe with a real swing door and so the model is an improvement.

B1 B1

3.5a 3.5c

ie the amplitude decays etc

(c)

Need $4 m ^ { 2 } + \lambda m + 25 = 0$ to have repeated solutions so $\lambda ^ { 2 } - 4 \times 4 \times 25 = 0$

$\lambda > 0 \Rightarrow \lambda = 20$

M1

3.5c

Using " $b ^ { 2 } - 4 a c$ " $= 0$ directly

or $\left( 2 m + \frac { \lambda } { 4 } \right) ^ { 2 } + 25 - \frac { \lambda ^ { 2 } } { 16 } = 0$ and equating part outside brackets to 0

A1

3.3

Not -20 or $\pm 20$

[2]

OCR Further Pure Core 2 2021 June Q1

1 In this question you must show detailed reasoning.
S is the 2-D transformation which is a stretch of scale factor 3 parallel to the $x$-axis. A is the matrix which represents S .

Write down $\mathbf { A }$.
By considering the transformation represented by $\mathbf { A } ^ { - 1 }$, determine the matrix $\mathbf { A } ^ { - 1 }$. Matrix $\mathbf { B }$ is given by $\mathbf { B } = \left( \begin{array} { c c } 0 & - 1
- 1 & 0 \end{array} \right)$. T is the transformation represented by $\mathbf { B }$.
Describe T.
Determine the matrix which represents the transformation S followed by T .
Demonstrate, by direct calculation, that $( \mathbf { B A } ) ^ { - 1 } = \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }$.

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