OCR MEI C3 — Question 5 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring differentiation of exponentials, application of the chain rule, and substitution of a given point. It's slightly above average difficulty due to the exponential terms on both sides, but follows a standard procedure with no conceptual surprises.
Spec1.07s Parametric and implicit differentiation
5 Find the gradient at the point \(( 0 , \ln 2 )\) on the curve with equation \(\mathrm { e } ^ { 2 y } = 5 - \mathrm { e } ^ { - x }\).
Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(e^{2y} = 5 - e^{-x}\)B1 \(2e^{2y}\frac{dy}{dx} = \ldots\); or \(y = \ln\sqrt{5-e^{-x}}\) o.e. e.g. \(\frac{1}{2}\ln(5-e^{-x})\)
\(2e^{2y}\frac{dy}{dx} = e^{-x}\)B1 \(= e^{-x}\); \(\Rightarrow dy/dx = e^{-x}/[2(5-e^{-x})]\) o.e. B1 (but must be correct)
\(\frac{dy}{dx} = \frac{e^{-x}}{2e^{2y}}\)
At \((0, \ln 2)\): \(\frac{dy}{dx} = \frac{e^0}{2e^{2\ln 2}} = \frac{1}{8}\)M1dep Substituting \(x=0,\ y=\ln 2\) into their \(dy/dx\); dep 1st B1 — allow one slip; or substituting \(x=0\) into their correct \(dy/dx\)
\(= \frac{1}{8}\)A1cao
[4] 
## Question 5:

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

---
5 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  Q5 [4]}}
This paper (8 questions)
View full paper
Q1 6 Q2 6 Q3 4 Q4 8 Q5 4 Q6 4 Q7 6 Q8 7