OCR Further Pure Core 2 2021 June — Question 5 33 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2021
SessionJune
Marks33
TopicComplex Numbers Argand & Loci
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
QuestionAnswerMarksAOGuidance
1(a)\(\mathbf { A } = \left( \begin{array} { l l } 30
01 \end{array} \right)\)
B1
[1]
1.1
(b)Stretch scale factor 1/3 parallel to \(x\)-axis \(\mathbf { A } ^ { - 1 } = \left( \begin{array} { l l } \frac { 1 } { 3 }0
01 \end{array} \right)\)
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)\(\begin{aligned}\mathbf { B A } = \left( \begin{array} { c c } 0- 1
- 10 \end{array} \right) \left( \begin{array} { l l } 30
01 \end{array} \right) = \ldots
\ldots = \left( \begin{array} { c c } 0- 1
- 30 \end{array} \right) \end{aligned}\)
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)\(( \mathbf { B A } ) ^ { - 1 } = - \frac { 1 } { 3 } \left( \begin{array} { l l } 01
30 \end{array} \right)\) \(\left. \begin{array} { r l } \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }= \left( \frac { 1 } { 3 } \right.
00
01 \end{array} \right) \left( \begin{array} { c c } 0- 1
- 10 \end{array} \right) = \left( \begin{array} { c c } 0- \frac { 1 } { 3 }
- 10 \end{array} \right) ~ \left( \begin{array} { l l } 01
30 \end{array} \right) = ( \mathbf { B } \mathbf { A } ) ^ { - 1 }\)A11.1aFor 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
QuestionAnswerMarksAOGuidance
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
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
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
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 \(0 . .\).E1ft3.1aIf \(D\) found to be non-zero in (b) then allow "Because there are not two points..."
QuestionAnswerMarksAOGuidance
...and so the lines must intersect.E1ft2.4A convincing demonstration that the two direction vectors are not parallel and "...and so the lines must be skew"
[2]
3(a)\(\begin{aligned}\mathbf { A B } = \left( \begin{array} { c c } 12
a- 1 \end{array} \right) \left( \begin{array} { c c } 2- 1
41 \end{array} \right) = \left( \begin{array} { c c } 101
2 a - 4- a - 1 \end{array} \right)
( \mathbf { A B } ) \mathbf { C } = \left( \begin{array} { c c } 101
2 a - 4- a - 1 \end{array} \right) \left( \begin{array} { c c } 50
- 22 \end{array} \right)
= \left( \begin{array} { c c } 482
12 a - 18- 2 a - 2 \end{array} \right)
\mathbf { B C } = \left( \begin{array} { c c } 2- 1
41 \end{array} \right) \left( \begin{array} { c c } 50
- 22 \end{array} \right) = \left( \begin{array} { c c } 12- 2
182 \end{array} \right)
\mathbf { A } ( \mathbf { B C } ) = \left( \begin{array} { c c } 12
a- 1 \end{array} \right) \left( \begin{array} { c c } 12- 2
182 \end{array} \right)
= \left( \begin{array} { c c } 482
12 a - 18- 2 a - 2 \end{array} \right) = ( \mathbf { A B } ) \mathbf { C } \text { (which }
\text { demonstrates associativity of matrix }
\text { multiplication) } \end{aligned}\)
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)\(\begin{aligned}\mathbf { A } \mathbf { C } = \left( \begin{array} { c c } 12
a- 1 \end{array} \right) \left( \begin{array} { c c } 50
- 22 \end{array} \right) = \left( \begin{array} { c c } 14
5 a + 2- 2 \end{array} \right)
\mathbf { C A } = \left( \begin{array} { c c } 50
- 22 \end{array} \right) \left( \begin{array} { c c } 12
a- 1 \end{array} \right) = \left( \begin{array} { c c } 510
2 a - 2- 6 \end{array} \right) \neq \mathbf { A C } \text { (so }
\text { matrix multiplication is not commutative) } \end{aligned}\)
M1
A1
[2]
1.1
2.1
Finding AC (or CA)
Finding the other and statement of non-equality
(c)\(\begin{aligned}\left( \begin{array} { c c } 12
a- 1 \end{array} \right) \binom { x } { y } = \binom { x + 2 y } { a x - y }
x + 2 y = 3 x = > y = x
a x - y = 3 y \text { and } y = x = > a = 4 \end{aligned}\)M1 >
A1
A1
\([ 3 ]\)
3.1a
2.2a
2.2a
Multiplying the vector into the matrix using the correct procedure
\begin{displayquote}
A1
A1
\([ 3 ]\)
\end{displayquote}
QuestionAnswerMarksAOGuidance
\multirow[t]{5}{*}{4}\(\begin{aligned}V = \pi \int _ { 1 } ^ { 3 } ( ( x - 3 ) \sqrt { \ln x } ) ^ { 2 } \mathrm {~d} x = \pi \int _ { 1 } ^ { 3 } ( x - 3 ) ^ { 2 } \ln x \mathrm {~d} x
V = \pi \left( \left[ \frac { 1 } { 3 } ( x - 3 ) ^ { 3 } \ln x \right] _ { 1 } ^ { 3 } - \int _ { 1 } ^ { 3 } \frac { 1 } { 3 } ( x - 3 ) ^ { 3 } \frac { 1 } { x } \mathrm {~d} x \right)
\frac { 1 } { x } ( x - 3 ) ^ { 3 } = x ^ { 2 } - 9 x + 27 - \frac { 27 } { x } \text { soi } \end{aligned}\)B13.1a\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}
A11.1
A11.1Completing 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 *M13.2aCorrectly dealing with limits
[7]
QuestionAnswerMarksAOGuidance
5\(\frac { z + 7 \mathrm { i } } { z - 24 } = \frac { x + \mathrm { i } y + 7 \mathrm { i } } { x - 24 + \mathrm { i } y } \times \frac { x - 24 - \mathrm { i } y } { x - 24 - \mathrm { i } y }\)M13.1aSubstituting \(z = x + \mathrm { i } y\) into \(\frac { z + 7 \mathrm { i } } { z - 24 }\)
\multirow{8}{*}{}\(\operatorname { Im } \frac { z + 7 \mathrm { i } } { z - 24 } = \frac { - x y + ( y + 7 ) ( x - 24 ) } { ( x - 24 ) ^ { 2 } + y ^ { 2 } } = \frac { 1 } { 4 }\)M12.1conjugate of bottom
\(28 x - 96 y - 672 = x ^ { 2 } - 48 x + 576 + y ^ { 2 }\)M11.1Multiplying out to get horizontal
\(0 = ( x - 38 ) ^ { 2 } - 1444 + ( y + 48 ) ^ { 2 } - 2304 + 1248\)M11.1Completing both squares with half signed coefficients of \(x\) and \(y\)
\(( x - 38 ) ^ { 2 } + ( y + 48 ) ^ { 2 } = 2500\)A12.2a
So the shape of \(C\) is a circle...E13.2a
...centre 38 - 48i, radius 50E13.2aOr \(( 38 , - 48 )\)
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]
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