AQA D1 2010 January — Question 6 8 marks

Exam BoardAQA
ModuleD1 (Decision Mathematics 1)
Year2010
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeApply iteration to find root (pure fixed point)
DifficultyEasy -1.8 This is an algorithm tracing exercise requiring only careful step-by-step execution of given instructions with simple arithmetic. No problem-solving, conceptual understanding, or mathematical insight is needed—just methodical bookkeeping of variable values through loops, making it significantly easier than average A-level questions.
Spec7.03c Working with algorithms: trace, interpret, adapt

6 A student is finding a numerical approximation for the area under a curve.
The algorithm that the student is using is as follows:
Line 10 Input \(A , B , N\) Line 20 Let \(T = 0\) Line 30 Let \(D = A\) Line \(40 \quad\) Let \(H = ( B - A ) / N\) Line \(50 \quad\) Let \(E = H / 2\) Line 60 Let \(T = T + A ^ { 3 } + B ^ { 3 }\) Line \(70 \quad\) Let \(D = D + H\) Line 80 If \(D = B\) then go to line 110
Line 90 Let \(T = T + 2 D ^ { 3 }\) Line 100 Go to line 70
Line \(110 \quad\) Print 'Area \(=\), \(T \times E\) Line 120 End
Trace the algorithm in the case where the input values are:
  1. \(A = 1 , B = 5 , N = 2\);
  2. \(A = 1 , B = 5 , N = 4\).

Question 6:
Part (a)
AnswerMarks Guidance
SCA trace with columns \((A),(B),(N),T,D,H,E\) values \((1),(5),(2)\); \(T=126\); \(T=180\); \(D\) values \(0,1,3,5\)M1, A1, m1 SCA trace as far as a second value for \(T\) with at least 1 value for all other variables; \(T=(180)\) trace as far as a third value for \(T\) and 2 values for \(D\)
("Area \(=\)") \(180\)A1 4
Part (b)
AnswerMarks Guidance
SCA as above with \((N)=(4)\); \(T=142\); \(T=196,324\); \(D\) values \(0,1,2,3,4,5\); \(H=0.5,1\)M1, A1, m1 SCA as above
("Area \(=\)") \(162\)A1 4
## Question 6:

### Part (a)
| SCA trace with columns $(A),(B),(N),T,D,H,E$ values $(1),(5),(2)$; $T=126$; $T=180$; $D$ values $0,1,3,5$ | M1, A1, m1 | SCA trace as far as a second value for $T$ with at least 1 value for all other variables; $T=(180)$ trace as far as a third value for $T$ and 2 values for $D$ |
| ("Area $=$") $180$ | A1 | 4 | All correct values including final value of 180 and no extra values, but including $A$, $B$, $N$ and their values $1, 5, 2$ |

### Part (b)
| SCA as above with $(N)=(4)$; $T=142$; $T=196,324$; $D$ values $0,1,2,3,4,5$; $H=0.5,1$ | M1, A1, m1 | SCA as above |
| ("Area $=$") $162$ | A1 | 4 | All correct values including final value of 162 and no extra values, but including $A$, $B$, $N$ and their values $1, 5, 4$ |

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6 A student is finding a numerical approximation for the area under a curve.\\
The algorithm that the student is using is as follows:\\
Line 10 Input $A , B , N$\\
Line 20 Let $T = 0$\\
Line 30 Let $D = A$\\
Line $40 \quad$ Let $H = ( B - A ) / N$\\
Line $50 \quad$ Let $E = H / 2$\\
Line 60 Let $T = T + A ^ { 3 } + B ^ { 3 }$\\
Line $70 \quad$ Let $D = D + H$\\
Line 80 If $D = B$ then go to line 110\\
Line 90 Let $T = T + 2 D ^ { 3 }$\\
Line 100 Go to line 70\\
Line $110 \quad$ Print 'Area $=$, $T \times E$\\
Line 120 End\\
Trace the algorithm in the case where the input values are:
\begin{enumerate}[label=(\alph*)]
\item $A = 1 , B = 5 , N = 2$;
\item $A = 1 , B = 5 , N = 4$.
\end{enumerate}

\hfill \mbox{\textit{AQA D1 2010 Q6 [8]}}