OCR Further Pure Core 2 2021 June

Question 1

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 \(( \mathbf { B A } ) ^ { - 1 } = \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }\).

Question 2

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2

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 \(l _ { 2 }\). 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
constant.
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 \(a\).
  \includegraphics[max width=\textwidth, alt={}, center]{570dba92-2c81-43e8-a0a8-741c40718626-3_586_1024_187_404}
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 \(S\).

Question 5 33 marks

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5
\(C\) is the locus of numbers, \(z\), for which \(\operatorname { Im } \left( \frac { z + 7 i } { z - 24 } \right) = \frac { 1 } { 4 }\).
By writing \(z = x + \mathrm { i } y\) give a complete description of the shape of \(C\) on an Argand diagram. \section*{Total Marks for Question Set 2: 38} Mark scheme

Question

Answer

Marks

AO

Guidance

1

(a)

\(\mathbf { A } = \left( \begin{array} { l l } 3

0

1 \end{array} \right)\)

B1

[1]

1.1

(b)

Stretch scale factor 1/3 parallel to \(x\)-axis \(\mathbf { A } ^ { - 1 } = \left( \begin{array} { l l } \frac { 1 } { 3 }

0

1 \end{array} \right)\)

M1

A1

[2]

1.1

2.2a

Must be complete description (except no need to specify 2-D)

(c)

Reflection in the line \(y = - x\)

B1

[1]

1.2

(d)

\(\begin{aligned}

\mathbf { B A } = \left( \begin{array} { c c } 0

- 1

0 \end{array} \right) \left( \begin{array} { l l } 3

0

1 \end{array} \right) = \ldots

\ldots = \left( \begin{array} { c c } 0

- 1

- 3

0 \end{array} \right) \end{aligned}\)

M1

A1

[2]

1.1a

1.1

For understanding that the matrix representing successive transformations is the product in the correct order. ie \(\mathbf { B A }\), not \(\mathbf { A B }\)

(e)

\(( \mathbf { B A } ) ^ { - 1 } = - \frac { 1 } { 3 } \left( \begin{array} { l l } 0

1

3

0 \end{array} \right)\) \(\left. \begin{array} { r l } \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }

= \left( \frac { 1 } { 3 } \right.

0

1 \end{array} \right) \left( \begin{array} { c c } 0

- 1

0 \end{array} \right) = \left( \begin{array} { c c } 0

- \frac { 1 } { 3 }

- 1

0 \end{array} \right) ~ \left( \begin{array} { l l } 0

1

3

0 \end{array} \right) = ( \mathbf { B } \mathbf { A } ) ^ { - 1 }\)

A1

1.1a

For carrying out the procedure for inverting the matrix found in (d) (or BA worked out from scratch)

OR \(\mathbf { M 1 }\) find \(\mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }\) from scratch A1 demonstrate that \(( \mathbf { B A } ) \left( \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 } \right)\) is equal to I

Question

Answer

Marks

AO

Guidance

2

(a)

\(\begin{aligned}

3 x + 4 y = 28 \text { so } a = 3 , b = 4 , c = 28 \text { (or any non- }

\text { zero multiples) so } D = \frac { | 3 \times - 6 + 4 \times 4 - 28 | } { \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } }

D = 6 \end{aligned}\)

M1

A1

1.1

Identifying \(a , b\) and \(c\) and substituting \(a , b , c\) and \(\left( x _ { 1 } , y _ { 1 } \right)\) correctly into distance formula

Alternative solution \(y - 4 = \frac { 4 } { 3 } ( x + 6 ) \text { oe so }\) \(- 0.75 x + 7 - 4 = \frac { 4 } { 3 } ( x + 6 ) \Rightarrow x = - 2.4 , y = 8.8\)

\(D = \sqrt { ( - 2.4 - - 6 ) ^ { 2 } + ( 8.8 - 4 ) ^ { 2 } } = 6\)

М1

A1

Finding equation of perpendicular line through \(( - 6,4 )\) and solving simultaneously to find foot of perpendicular

[2]

(b)

\(\left( \begin{array} { c } 2
1
- 4 \end{array} \right) \times \left( \begin{array} { c } 3
- 1
1 \end{array} \right) = \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right)\)
\(D = \frac { \left. \left\lvert \, \left( \begin{array} { c } 11
- 1
5 \end{array} \right) - \left( \begin{array} { c } 4
3
- 2 \end{array} \right) \right. \right) \left. \cdot \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\rvert \, } { \left\| \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\| }\) or \(\frac { \left\| \left( \begin{array} { c } 7
- 4
7 \end{array} \right) \cdot \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\| } { \left\| \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\| }\)
\(D = 0\)

B1

М1

A1

1.1a

1.1

Correctly finding a mutual perpendicular BC

Correct substitution into distance formula

Alternative solution \(\begin{aligned}

4 + 2 \lambda = 11 + 3 \mu , 3 + \lambda = - 1 - \mu \text { and } - 2 - 4 \lambda = 5

+ \mu

\lambda = - 1 , \mu = - 3

\text { eg } - 2 - 4 ( - 1 ) = 2 = 5 + - 3 \text { so lines intersect so } D

= 0 \end{aligned}\)

М1

A1

Looking for a PoI so all 3 ( \(3 ^ { \text {rd } }\) might be seen later)

Correctly solving any 2 equations Must be checked in the unsolved equation.

Value of each side must be found, not just equality asserted.

[3]

(c)

There are two points, one on each line, such that the distance between the points is \(0 . .\).

E1ft

3.1a

If \(D\) found to be non-zero in (b) then allow "Because there are not two points..."

Question

Answer

Marks

AO

Guidance

...and so the lines must intersect.

E1ft

2.4

A convincing demonstration that the two direction vectors are not parallel and "...and so the lines must be skew"

[2]

3

(a)

\(\begin{aligned}

\mathbf { A B } = \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \left( \begin{array} { c c } 2

- 1

4

1 \end{array} \right) = \left( \begin{array} { c c } 10

1

2 a - 4

- a - 1 \end{array} \right)

( \mathbf { A B } ) \mathbf { C } = \left( \begin{array} { c c } 10

1

2 a - 4

- a - 1 \end{array} \right) \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right)

= \left( \begin{array} { c c } 48

2

12 a - 18

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

\mathbf { B C } = \left( \begin{array} { c c } 2

- 1

4

1 \end{array} \right) \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right) = \left( \begin{array} { c c } 12

- 2

18

2 \end{array} \right)

\mathbf { A } ( \mathbf { B C } ) = \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \left( \begin{array} { c c } 12

- 2

18

2 \end{array} \right)

= \left( \begin{array} { c c } 48

2

12 a - 18

- 2 a - 2 \end{array} \right) = ( \mathbf { A B } ) \mathbf { C } \text { (which }

\text { demonstrates associativity of matrix }

\text { multiplication) } \end{aligned}\)

M1

A1

M1

A1

[4]

3.1a

2.1

1.1

2.1

Finding \(\mathbf { A B }\) (or \(\mathbf { B C }\) )

Finding (AB)C (or \(\mathbf { A }\) (BC))

Finding \(\mathbf { B C }\) (or \(\mathbf { A B }\) )

Correct final matrix and statement of equality

(b)

\(\begin{aligned}

\mathbf { A } \mathbf { C } = \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right) = \left( \begin{array} { c c } 1

4

5 a + 2

- 2 \end{array} \right)

\mathbf { C A } = \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right) \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) = \left( \begin{array} { c c } 5

10

2 a - 2

- 6 \end{array} \right) \neq \mathbf { A C } \text { (so }

\text { matrix multiplication is not commutative) } \end{aligned}\)

M1

A1

[2]

1.1

2.1

Finding AC (or CA)

Finding the other and statement of non-equality

(c)

\(\begin{aligned}

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

2

a

- 1 \end{array} \right) \binom { x } { y } = \binom { x + 2 y } { a x - y }

x + 2 y = 3 x = > y = x

a x - y = 3 y \text { and } y = x = > a = 4 \end{aligned}\)

M1 >

A1

\([ 3 ]\)

3.1a

2.2a

Multiplying the vector into the matrix using the correct procedure

\begin{displayquote}

A1

\([ 3 ]\)

\end{displayquote}

Question

Answer

Marks

AO

Guidance

\multirow[t]{5}{*}{4}

\(\begin{aligned}

V = \pi \int _ { 1 } ^ { 3 } ( ( x - 3 ) \sqrt { \ln x } ) ^ { 2 } \mathrm {~d} x = \pi \int _ { 1 } ^ { 3 } ( x - 3 ) ^ { 2 } \ln x \mathrm {~d} x

V = \pi \left( \left[ \frac { 1 } { 3 } ( x - 3 ) ^ { 3 } \ln x \right] _ { 1 } ^ { 3 } - \int _ { 1 } ^ { 3 } \frac { 1 } { 3 } ( x - 3 ) ^ { 3 } \frac { 1 } { x } \mathrm {~d} x \right)

\frac { 1 } { x } ( x - 3 ) ^ { 3 } = x ^ { 2 } - 9 x + 27 - \frac { 27 } { x } \text { soi } \end{aligned}\)

B1

3.1a

\multirow{2}{*}{

Correct substitution into formula (ignore limits) and simplification to integrable (by parts) form Integration by parts with \(( x - 3 ) ^ { 2 }\) (may be expanded) being integrated.

May come implicitly from previously expanded form

}

\multirow{3}{*}{ie from \(\int _ { 1 } ^ { 3 } x ^ { 2 } \ln x - 6 x \ln x + 9 \ln x \mathrm {~d} x\) integrated by parts term by term}

A1

1.1

A1

1.1

Completing the integral. NB \(\left[ ( x - 3 ) ^ { 3 } \ln x \right] _ { 1 } ^ { 3 } = 0\) so may be omitted provided it is seen earlier

\(V = \frac { \pi } { 3 } \left( \left[ \begin{array} { l } ( x - 3 ) ^ { 3 } \ln x - \frac { x ^ { 3 } } { 3 } +

\frac { 9 x ^ { 2 } } { 2 } + 27 x - 27 \ln x \end{array} \right] _ { 1 } ^ { 3 } \right)\)

dep *M1

3.2a

Correctly dealing with limits

[7]

Question	Answer	Marks	AO	Guidance
5	\(\frac { z + 7 \mathrm { i } } { z - 24 } = \frac { x + \mathrm { i } y + 7 \mathrm { i } } { x - 24 + \mathrm { i } y } \times \frac { x - 24 - \mathrm { i } y } { x - 24 - \mathrm { i } y }\)	M1	3.1a	Substituting \(z = x + \mathrm { i } y\) into \(\frac { z + 7 \mathrm { i } } { z - 24 }\)
\multirow{8}{*}{}	\(\operatorname { Im } \frac { z + 7 \mathrm { i } } { z - 24 } = \frac { - x y + ( y + 7 ) ( x - 24 ) } { ( x - 24 ) ^ { 2 } + y ^ { 2 } } = \frac { 1 } { 4 }\)	M1	2.1	conjugate of bottom
	\(28 x - 96 y - 672 = x ^ { 2 } - 48 x + 576 + y ^ { 2 }\)	M1	1.1	Multiplying out to get horizontal
	\(0 = ( x - 38 ) ^ { 2 } - 1444 + ( y + 48 ) ^ { 2 } - 2304 + 1248\)	M1	1.1	Completing both squares with half signed coefficients of \(x\) and \(y\)
	\(( x - 38 ) ^ { 2 } + ( y + 48 ) ^ { 2 } = 2500\)	A1	2.2a
	So the shape of \(C\) is a circle...	E1	3.2a
	...centre 38 - 48i, radius 50	E1	3.2a	Or \(( 38 , - 48 )\)
	function \(\operatorname { Im } \left( \frac { z + 7 i } { z - 24 } \right)\) is undefined at this point on the circle)			Do not penalise either lack of
		[7]

OCR Further Pure Core 2 (Further Pure Core 2) 2021 June

Question 1

Question 2

Question 5 33 marks