OCR MEI C2 2012 June — Question 6 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs ln(x) linear graph
DifficultyModerate -0.3 This is a straightforward C2 question requiring students to recognize that a linear log-log graph implies a power law relationship y = ax^n, then read the gradient and intercept from the graph to find the constants. While it requires understanding of logarithmic relationships, it's a standard textbook exercise with clear visual information and a routine method, making it slightly easier than average.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

6 Fig. 6 shows the relationship between \(\log _ { 10 } x\) and \(\log _ { 10 } y\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-3_497_787_287_644} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} Find \(y\) in terms of \(x\).

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
gradient \(= 3\) seenB1 may be embedded
\(\log_{10} y - 5 = (\text{their } 3)(\log_{10} x - 1)\) or using \((5, 17)\)M1 or \(\log_{10} y = 3\log_{10} x + c\) and substitution of \((1, 5)\) or \((5, 17)\) for \(\log_{10} x\) and \(\log_{10} y\); condone omission of base throughout; NB may recover from e.g. \(Y = 3X + 2\)
\(\log_{10} y = 3\log_{10} x + 2\) oeA1
\(y = 10^{3\log_{10} x + 2}\) oeM1 or \(\log_{10} y = \log_{10} x^3 + \log_{10} 100\); or \(\log_{10}\frac{y}{x^3} = 2\) or \(\log_{10} y = \log_{10} 100x^3\)
\(y = 100x^3\)A1
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| gradient $= 3$ seen | B1 | may be embedded |
| $\log_{10} y - 5 = (\text{their } 3)(\log_{10} x - 1)$ or using $(5, 17)$ | M1 | or $\log_{10} y = 3\log_{10} x + c$ and substitution of $(1, 5)$ or $(5, 17)$ for $\log_{10} x$ and $\log_{10} y$; condone omission of base throughout; NB may recover from e.g. $Y = 3X + 2$ |
| $\log_{10} y = 3\log_{10} x + 2$ oe | A1 | |
| $y = 10^{3\log_{10} x + 2}$ oe | M1 | or $\log_{10} y = \log_{10} x^3 + \log_{10} 100$; or $\log_{10}\frac{y}{x^3} = 2$ or $\log_{10} y = \log_{10} 100x^3$ |
| $y = 100x^3$ | A1 | |

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6 Fig. 6 shows the relationship between $\log _ { 10 } x$ and $\log _ { 10 } y$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-3_497_787_287_644}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}

Find $y$ in terms of $x$.

\hfill \mbox{\textit{OCR MEI C2 2012 Q6 [5]}}