AQA Further AS Paper 1 2018 June — Question 12 6 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeDeterminant calculation
DifficultyStandard +0.3 Part (a) is trivial verification (substitute k=1, compute determinant). Part (b) requires finding when det < 0, which involves solving a cubic inequality k(5-k) - 2(kΒ³+1) < 0, simplifying to -2kΒ³ - kΒ² + 5k - 2 < 0, then analyzing sign changes. While this requires factoring/testing values and justifying the solution set, it's a standard Further Maths technique with straightforward algebraic manipulation. The 5 marks reflect working rather than conceptual difficulty.
Spec4.03l Singular/non-singular matrices
  1. Show that the matrix \(\begin{pmatrix} 5 - k & 2 \\ k^3 + 1 & k \end{pmatrix}\) is singular when \(k = 1\). [1 mark]
  2. Find the values of \(k\) for which the matrix \(\begin{pmatrix} 5 - k & 2 \\ k^3 + 1 & k \end{pmatrix}\) has a negative determinant. Fully justify your answer. [5 marks]
Question 12:
AnswerMarks
12(a)Substitutes and correctly calculates the determinant,
and concludes that the matrix is singular.
AnswerMarks Guidance
π‘˜π‘˜ = 12.2a B1
the matrix is singular
= 4Γ—1βˆ’2Γ—2 = 0
AG
∴
AnswerMarks
12(b)Finds determinant in terms of .
Allow one error.
AnswerMarks Guidance
π‘˜π‘˜1.1a M1
3
= π‘˜π‘˜(5βˆ’π‘˜π‘˜)βˆ’2(π‘˜π‘˜ +1)
2 3
5π‘˜π‘˜βˆ’π‘˜π‘˜ βˆ’2π‘˜π‘˜ βˆ’2 < 0
3 2
2π‘˜π‘˜ +π‘˜π‘˜ βˆ’5π‘˜π‘˜+2 > 0
2
(π‘˜π‘˜βˆ’1)(2π‘˜π‘˜ +3π‘˜π‘˜βˆ’2) > 0
(π‘˜π‘˜βˆ’1)(2π‘˜π‘˜βˆ’1)(π‘˜π‘˜+2) > 0
βˆ’ + βˆ’ +
1
βˆ’2 1 π‘˜π‘˜
2
,
1
βˆ’2 < π‘˜π‘˜ < 2 π‘˜π‘˜ > 1
AnswerMarks Guidance
Obtains a correct inequality in .1.1b A1
π‘˜π‘˜
AnswerMarks Guidance
Obtains three correct critical values.1.1b A1
Deduces one correct region.
FT their three real distinct critical values if given as , (o.e.)
where
AnswerMarks Guidance
π‘Žπ‘Ž < π‘˜π‘˜ < 𝑏𝑏 π‘˜π‘˜ > 𝑐𝑐2.2a A1F
Deduceπ‘Žπ‘Žs< th𝑏𝑏e <oth𝑐𝑐er correct region.
FT their three real distinct critical values if given as , (o.e.)
where
π‘Žπ‘Ž < π‘˜π‘˜ < 𝑏𝑏 π‘˜π‘˜ > 𝑐𝑐
Condone the use of β€˜and’.
AnswerMarks Guidance
π‘Žπ‘Ž < 𝑏𝑏 < 𝑐𝑐2.2a A1F
Total6
(
AnswerMarks Guidance
1)(21)(π‘˜π‘˜ > 0
1
AnswerMarks Guidance
QMarking instructions AO 
Question 12:
--- 12(a) ---
12(a) | Substitutes and correctly calculates the determinant,
and concludes that the matrix is singular.
π‘˜π‘˜ = 1 | 2.2a | B1 | determinant
the matrix is singular
= 4Γ—1βˆ’2Γ—2 = 0
AG
∴
--- 12(b) ---
12(b) | Finds determinant in terms of .
Allow one error.
π‘˜π‘˜ | 1.1a | M1 | determinant
3
= π‘˜π‘˜(5βˆ’π‘˜π‘˜)βˆ’2(π‘˜π‘˜ +1)
2 3
5π‘˜π‘˜βˆ’π‘˜π‘˜ βˆ’2π‘˜π‘˜ βˆ’2 < 0
3 2
2π‘˜π‘˜ +π‘˜π‘˜ βˆ’5π‘˜π‘˜+2 > 0
2
(π‘˜π‘˜βˆ’1)(2π‘˜π‘˜ +3π‘˜π‘˜βˆ’2) > 0
(π‘˜π‘˜βˆ’1)(2π‘˜π‘˜βˆ’1)(π‘˜π‘˜+2) > 0
βˆ’ + βˆ’ +
1
βˆ’2 1 π‘˜π‘˜
2
,
1
βˆ’2 < π‘˜π‘˜ < 2 π‘˜π‘˜ > 1
Obtains a correct inequality in . | 1.1b | A1
π‘˜π‘˜
Obtains three correct critical values. | 1.1b | A1
Deduces one correct region.
FT their three real distinct critical values if given as , (o.e.)
where
π‘Žπ‘Ž < π‘˜π‘˜ < 𝑏𝑏 π‘˜π‘˜ > 𝑐𝑐 | 2.2a | A1F
Deduceπ‘Žπ‘Žs< th𝑏𝑏e <oth𝑐𝑐er correct region.
FT their three real distinct critical values if given as , (o.e.)
where
π‘Žπ‘Ž < π‘˜π‘˜ < 𝑏𝑏 π‘˜π‘˜ > 𝑐𝑐
Condone the use of β€˜and’.
π‘Žπ‘Ž < 𝑏𝑏 < 𝑐𝑐 | 2.2a | A1F
Total | 6
(
1)(2 | 1)(π‘˜π‘˜ | > 0
1
Q | Marking instructions | AO | Mark | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Show that the matrix $\begin{pmatrix} 5 - k & 2 \\ k^3 + 1 & k \end{pmatrix}$ is singular when $k = 1$.
[1 mark]

\item Find the values of $k$ for which the matrix $\begin{pmatrix} 5 - k & 2 \\ k^3 + 1 & k \end{pmatrix}$ has a negative determinant.

Fully justify your answer.
[5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q12 [6]}}
This paper (19 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 2 Q5 3 Q6 3 Q7 2 Q8 5 Q9 6 Q10 8 Q11 3 Q12 6 Q13 9 Q14 7 Q15 4 Q16 3 Q17 4 Q18 4 Q19 8