OCR MEI Further Pure Core 2020 November — Question 9 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2020
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind constant from invariant line or area condition
DifficultyStandard +0.8 This question requires understanding of eigenvalues (invariant lines exist when real eigenvalues exist), solving a quadratic inequality for part (a), then using determinant properties (det = -5 from area and orientation conditions) to find λ, followed by finding eigenvalues and eigenvectors. It combines multiple concepts but follows standard procedures once the connections are recognized.
Spec4.03g Invariant points and lines4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation
9 A linear transformation of the plane is represented by the matrix \(\mathbf { M } = \left( \begin{array} { r r } 1 & - 2 \\ \lambda & 3 \end{array} \right)\), where \(\lambda\) is a
constant. constant.
  1. Find the set of values of \(\lambda\) for which the linear transformation has no invariant lines through the origin.
  2. Given that the transformation multiplies areas by 5 and reverses orientation, find the invariant lines.
Question 9:
AnswerMarks Guidance
9(a) suppose y = mx is invariant
​​ ​ ​
λx + 3y = m(x – 2y)
​ ​​ ​ ​ ​​ ​​
⇒ λ + 3m = m(1− 2m)
​​ ​ ​ ​ ​ ​ ​
⇒ 2m2 ​+ 2m + λ = 0
​ ​ ​ ​ ​​
no solutions if discriminant < 0
⇒ 4 − 8λ < 0, λ > ½
AnswerMarks
​​ ​​B1
M1
A1
M1
A1
AnswerMarks
[5]2.1
2.1
1.1
3.1a
AnswerMarks
3.2aor y = mx + c
​​ ​ ​ ​
or λx + 3y = m(x – 2y)+c
​ ​ ​​ ​ ​ ​​ ​​ ​
AnswerMarks Guidance
9(b) det M = 3 + 2λ or det M = −5
​ ​ ​​ ​ ​
λ = −4
​
m2 ​+ m – 2 = 0, m = −2 or 1
​ ​ ​ ​ ​
so lines are y = x and y = −2x
AnswerMarks
​​ ​​ ​​ ​B1
B1
B1
AnswerMarks
[3]1.1
2.1
2.2a
PPMMTT
Y420/01 Mark SchemeNovember 2020
10 DR
M1 1.1
let u = 2πarcosh y, u’ = 2π/√(y2 ​– 1)
v’ = ​ ​ 1, v ​ = ​ y ​​ ​ ​ ​ ​ ​​ M1 3.1a integration by parts
​ ​​ ​
A1 2.1 condone missing 2π and
​ ​
incorrect limits
M1 1.1 subst u = y2 ​– 1 or inspection
​ ​ ​​
A1 1.1 A1 for √y2−1
= 2 ​ π ​ (2ln(2+√3) − √3) M1 1.1 use of arcosh x =
​​
A1cao 3.2a ln[x+√(x2−​ 1)]
​​ ​​
[7]
11 (a) DR
2, 2ei ​ π ​ /3,​ 2e2i ​ π ​ /3,​ −2, 2e4i ​ π ​ /3,​ 2e5i ​ π ​ /3 M1 2.5 modulus 2
​ ​ ​ ​
A1 2.5
[2]
11 (b) DR
modulus of G = √3 B1 3.1a
modulus of w = √3 B1 1.1
2
argument = π/6 B1 1.1
​ ​
B1 1.1
So
[4]
11 (c) DR
M1 1.1 taking the 6th ​power of one of
​
the midpoints
A1 1.1
[2]
so
11
Question 9:
9 | (a) | suppose y = mx is invariant
​​ ​ ​
λx + 3y = m(x – 2y)
​ ​​ ​ ​ ​​ ​​
⇒ λ + 3m = m(1− 2m)
​​ ​ ​ ​ ​ ​ ​
⇒ 2m2 ​+ 2m + λ = 0
​ ​ ​ ​ ​​
no solutions if discriminant < 0
⇒ 4 − 8λ < 0, λ > ½
​​ ​​ | B1
M1
A1
M1
A1
[5] | 2.1
2.1
1.1
3.1a
3.2a | or y = mx + c
​​ ​ ​ ​
or λx + 3y = m(x – 2y)+c
​ ​ ​​ ​ ​ ​​ ​​ ​
9 | (b) | det M = 3 + 2λ or det M = −5
​ ​ ​​ ​ ​
λ = −4
​
m2 ​+ m – 2 = 0, m = −2 or 1
​ ​ ​ ​ ​
so lines are y = x and y = −2x
​​ ​​ ​​ ​ | B1
B1
B1
[3] | 1.1
2.1
2.2a
PPMMTT
Y420/01 Mark SchemeNovember 2020
10 DR
M1 1.1
let u = 2πarcosh y, u’ = 2π/√(y2 ​– 1)
v’ = ​ ​ 1, v ​ = ​ y ​​ ​ ​ ​ ​ ​​ M1 3.1a integration by parts
​ ​​ ​
A1 2.1 condone missing 2π and
​ ​
incorrect limits
M1 1.1 subst u = y2 ​– 1 or inspection
​ ​ ​​
A1 1.1 A1 for √y2−1
= 2 ​ π ​ (2ln(2+√3) − √3) M1 1.1 use of arcosh x =
​​
A1cao 3.2a ln[x+√(x2−​ 1)]
​​ ​​
[7]
11 (a) DR
2, 2ei ​ π ​ /3,​ 2e2i ​ π ​ /3,​ −2, 2e4i ​ π ​ /3,​ 2e5i ​ π ​ /3 M1 2.5 modulus 2
​ ​ ​ ​
A1 2.5
[2]
11 (b) DR
modulus of G = √3 B1 3.1a
modulus of w = √3 B1 1.1
2
argument = π/6 B1 1.1
​ ​
B1 1.1
So
[4]
11 (c) DR
M1 1.1 taking the 6th ​power of one of
​
the midpoints
A1 1.1
[2]
so
11
9 A linear transformation of the plane is represented by the matrix $\mathbf { M } = \left( \begin{array} { r r } 1 & - 2 \\ \lambda & 3 \end{array} \right)$, where $\lambda$ is a\\
constant. constant.
\begin{enumerate}[label=(\alph*)]
\item Find the set of values of $\lambda$ for which the linear transformation has no invariant lines through the origin.
\item Given that the transformation multiplies areas by 5 and reverses orientation, find the invariant lines.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q9 [8]}}
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