OCR MEI C3 2012 June — Question 3 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2012
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyModerate -0.3 This is a straightforward implicit differentiation question requiring differentiation of exponentials, substitution of a given point, and solving for dy/dx. While it involves multiple steps, the techniques are standard C3 material with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.07s Parametric and implicit differentiation
3 Find the gradient at the point \(( 0 , \ln 2 )\) on the curve with equation \(\mathrm { e } ^ { 2 y } = 5 - \mathrm { e } ^ { - x }\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(e^{2y}=5-e^{-x}\)B1 \(2e^{2y}\frac{dy}{dx}=\ldots\) or \(y=\ln\sqrt{(5-e^{-x})}\) o.e.
\(\Rightarrow 2e^{2y}\frac{dy}{dx}=e^{-x}\)B1 \(=e^{-x}\); \(\Rightarrow dy/dx = e^{-x}/[2(5-e^{-x})]\) o.e. (but must be correct)
\(\Rightarrow \frac{dy}{dx}=\frac{e^{-x}}{2e^{2y}}\)
At \((0,\ln 2)\): \(\frac{dy}{dx}=\frac{e^0}{2e^{2\ln 2}}\)M1dep substituting \(x=0\), \(y=\ln 2\) into their \(dy/dx\); dep 1st B1 – allow one slip
\(=\frac{1}{8}\)A1cao [4] or substituting \(x=0\) into their correct \(dy/dx\) 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{2y}=5-e^{-x}$ | B1 | $2e^{2y}\frac{dy}{dx}=\ldots$ or $y=\ln\sqrt{(5-e^{-x})}$ o.e. |
| $\Rightarrow 2e^{2y}\frac{dy}{dx}=e^{-x}$ | B1 | $=e^{-x}$; $\Rightarrow dy/dx = e^{-x}/[2(5-e^{-x})]$ o.e. (but must be correct) |
| $\Rightarrow \frac{dy}{dx}=\frac{e^{-x}}{2e^{2y}}$ | | |
| At $(0,\ln 2)$: $\frac{dy}{dx}=\frac{e^0}{2e^{2\ln 2}}$ | M1dep | substituting $x=0$, $y=\ln 2$ into their $dy/dx$; dep 1st B1 – allow one slip |
| $=\frac{1}{8}$ | A1cao **[4]** | or substituting $x=0$ into their correct $dy/dx$ |

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3 Find the gradient at the point $( 0 , \ln 2 )$ on the curve with equation $\mathrm { e } ^ { 2 y } = 5 - \mathrm { e } ^ { - x }$.

\hfill \mbox{\textit{OCR MEI C3 2012 Q3 [4]}}
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