OCR Further Pure Core 1 2024 June — Question 3 8 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeVolume/area scale factors
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring matrix squaring, determinant calculation with a parameter, and then applying volume scale factors with error bounds. The final part requires careful consideration of how rounding errors propagate through the determinant and volume calculation, which elevates it above routine matrix operations to require genuine problem-solving and understanding of bounds.
Spec4.03b Matrix operations: addition, multiplication, scalar4.03h Determinant 2x2: calculation4.03j Determinant 3x3: calculation
3 A transformation T is represented by the matrix \(\mathbf { N } = \left( \begin{array} { l l l } a & 4 & 2 \\ 5 & 1 & 0 \\ 3 & 6 & 3 \end{array} \right)\), where \(a\) is a constant.
  1. Find \(\mathbf { N } ^ { 2 }\) in terms of \(a\).
  2. Find det \(\mathbf { N }\) in terms of \(a\). The value of \(a\) is 13 to the nearest integer.
    A shape \(S _ { 1 }\) has volume 11.6 to 1 decimal place. Shape \(S _ { 1 }\) is mapped to shape \(S _ { 2 }\) by the transformation T . A student claims that the volume of \(S _ { 2 }\) is less than 400 .
  3. Comment on the student's claim.
Question 3:
AnswerMarks Guidance
3(a)  a 4 2 
For reference: N = 5 1 0
3 6 3
 2 a + 2 6 4 a + 1 6 2 a + 6 
N 2 = a 5 + 5 2 1 1 0
AnswerMarks
a 3 + 3 9 3 6 1 5M1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1A 3 by 3 matrix with at least three correct (simplified)
entries.
A 3 by 3 matrix with at least two correctly simplified rows
or at least two correctly simplified columns.
cao
AnswerMarks
(b)1 0 5 0 5 1
d e t N = a − 4 + 2
6 3 3 3 3 6
AnswerMarks
= 3a – 6M1
A1
AnswerMarks
[2]1.1
1.1Correct method. May see a(3– 0) – 4(15 – 0) + 2(30 – 3).
Ignore sign errors in 2  2 determinant calculations, but
cofactors must have correct signs. Do look out for
expanding by other rows and columns.
If using Sarrus’ method, must see correct 66 and 3a+60.
a 4 2 a k 
1
         
Or for 5 1  0 = 5 k (oe) with at least one
         2
         
3 6 3 3  k 
3
of k =3,k =0,k =−2 correct.
1 2 3
cao
AnswerMarks
(c)11.6 × det N
For example,
Upper bound is 11.65 × (3×13.5 – 6) = 401.925
Lower bound is 11.55 × (3×12.5 – 6) = 363.825
So, the student’s claim (that the volume is less than 400) is not
AnswerMarks
necessarily true.M1
A1
A1
AnswerMarks
[3]1.1
3.1a
AnswerMarks
2.2bOr k × det N where 11.55 k 11.65 and
1 2 .5 a 1 3 .5 in their det N from part (b). If not
calculating the correct UB and LB (from a correct
determinant) then we must see either a correct calculation
(for their determinant from part (b)) or they must state, the
values they’ve used in their calculation for both a and the
volume of S to award any marks.
1
For either correct upper or lower bound, or any correct V
2
for any value of a between 12.5 and 13.5, and 11.55 < V <
1
11.65. Allow answers rounded or truncated to the nearest
integer or greater degree of accuracy.
Demonstrates that there are values of 12.5 a13.5 and
1 1 .5 5 k  1 1 .6 5 such that both V > 400, and V < 400 and
concludes that the volume may be greater or less than 400
(depending on a more accurately determined value of a and
the volume of S ). Conclusion must indicate that the claim
1
could be correct but not necessarily so (oe).
If M0 then SC1 for LB = 144.375 and UB = 157.275 (using
13 for det N).
Question 3:
3 | (a) |  a 4 2 
For reference: N = 5 1 0
3 6 3
 2 a + 2 6 4 a + 1 6 2 a + 6 
N 2 = a 5 + 5 2 1 1 0
a 3 + 3 9 3 6 1 5 | M1
M1
A1
[3] | 1.1
1.1
1.1 | A 3 by 3 matrix with at least three correct (simplified)
entries.
A 3 by 3 matrix with at least two correctly simplified rows
or at least two correctly simplified columns.
cao
(b) | 1 0 5 0 5 1
d e t N = a − 4 + 2
6 3 3 3 3 6
= 3a – 6 | M1
A1
[2] | 1.1
1.1 | Correct method. May see a(3– 0) – 4(15 – 0) + 2(30 – 3).
Ignore sign errors in 2  2 determinant calculations, but
cofactors must have correct signs. Do look out for
expanding by other rows and columns.
If using Sarrus’ method, must see correct 66 and 3a+60.
a 4 2 a k 
1
         
Or for 5 1  0 = 5 k (oe) with at least one
         2
         
3 6 3 3  k 
3
of k =3,k =0,k =−2 correct.
1 2 3
cao
(c) | 11.6 × det N
For example,
Upper bound is 11.65 × (3×13.5 – 6) = 401.925
Lower bound is 11.55 × (3×12.5 – 6) = 363.825
So, the student’s claim (that the volume is less than 400) is not
necessarily true. | M1
A1
A1
[3] | 1.1
3.1a
2.2b | Or k × det N where 11.55 k 11.65 and
1 2 .5 a 1 3 .5 in their det N from part (b). If not
calculating the correct UB and LB (from a correct
determinant) then we must see either a correct calculation
(for their determinant from part (b)) or they must state, the
values they’ve used in their calculation for both a and the
volume of S to award any marks.
1
For either correct upper or lower bound, or any correct V
2
for any value of a between 12.5 and 13.5, and 11.55 < V <
1
11.65. Allow answers rounded or truncated to the nearest
integer or greater degree of accuracy.
Demonstrates that there are values of 12.5 a13.5 and
1 1 .5 5 k  1 1 .6 5 such that both V > 400, and V < 400 and
concludes that the volume may be greater or less than 400
(depending on a more accurately determined value of a and
the volume of S ). Conclusion must indicate that the claim
1
could be correct but not necessarily so (oe).
If M0 then SC1 for LB = 144.375 and UB = 157.275 (using
13 for det N).
3 A transformation T is represented by the matrix $\mathbf { N } = \left( \begin{array} { l l l } a & 4 & 2 \\ 5 & 1 & 0 \\ 3 & 6 & 3 \end{array} \right)$, where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { N } ^ { 2 }$ in terms of $a$.
\item Find det $\mathbf { N }$ in terms of $a$.

The value of $a$ is 13 to the nearest integer.\\
A shape $S _ { 1 }$ has volume 11.6 to 1 decimal place. Shape $S _ { 1 }$ is mapped to shape $S _ { 2 }$ by the transformation T .

A student claims that the volume of $S _ { 2 }$ is less than 400 .
\item Comment on the student's claim.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2024 Q3 [8]}}
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