AQA FP1 2013 January — Question 7 7 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear regression
TypeLinearize non-linear relationships
DifficultyStandard +0.8 This is a Further Maths statistics question requiring students to identify appropriate transformations to linearize data, then use the linearized graph to determine unknown parameters in the original non-linear model. It combines algebraic manipulation with graphical interpretation and requires understanding of logarithmic transformations, placing it moderately above average difficulty.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines

  1. Show that there is a linear relationship between \(Y\) and \(X\).
  2. The graph of \(Y\) against \(X\) is shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{cf9337b9-b766-4ce5-967c-5d7522e2aa42-4_748_858_849_593} Find the value of \(n\) and the value of \(a\).

Question 7:
Part (a):
\(y = ax^n \Rightarrow \log_{10} y = \log_{10} ax^n\)
AnswerMarks Guidance
\(\log_{10} y = \log_{10} a + \log_{10} x^n\)M1 Take logs and apply one log law correctly. PI. Apply a further log law correctly.
\(\log_{10} y = \log_{10} a + n\log_{10} x\)m1
\(Y = \log_{10} a + nX\) (linear relationship between \(Y\) and \(X\))A1 Correct eqn. with base 10 (or lg or later evidence of use of base 10 if log without base here)
Part (b):
AnswerMarks Guidance
\(n =\) gradient of lineM1 Stated or used. Accept \(n = \pm\frac{2}{3}\) OE as evidence
\(n = -\dfrac{2}{3}\)A1 \(n = -\frac{2}{3}\) (OE 3sf)
\(\log_{10} a = 4\)M1 Equating c's constant term [must involve a log] in c's (a) eqn. to the \(Y\)-intercept value PI by correct value of \(a\)
\(a = 10^4\ (= 10\,000)\)A1
## Question 7:

### Part (a):
$y = ax^n \Rightarrow \log_{10} y = \log_{10} ax^n$

$\log_{10} y = \log_{10} a + \log_{10} x^n$ | M1 | Take logs and apply one log law correctly. PI. Apply a further log law correctly.

$\log_{10} y = \log_{10} a + n\log_{10} x$ | m1 |

$Y = \log_{10} a + nX$ (linear relationship between $Y$ and $X$) | A1 | Correct eqn. with base 10 (or lg or later evidence of use of base 10 if log without base here)

### Part (b):
$n =$ gradient of line | M1 | Stated or used. Accept $n = \pm\frac{2}{3}$ OE as evidence

$n = -\dfrac{2}{3}$ | A1 | $n = -\frac{2}{3}$ (OE 3sf)

$\log_{10} a = 4$ | M1 | Equating c's constant term [must involve a log] in c's (a) eqn. to the $Y$-intercept value PI by correct value of $a$

$a = 10^4\ (= 10\,000)$ | A1 |

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\begin{enumerate}[label=(\alph*)]
\item Show that there is a linear relationship between $Y$ and $X$.
\item The graph of $Y$ against $X$ is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{cf9337b9-b766-4ce5-967c-5d7522e2aa42-4_748_858_849_593}

Find the value of $n$ and the value of $a$.
\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2013 Q7 [7]}}