OCR MEI C3 — Question 7 7 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind stationary points coordinates
DifficultyStandard +0.8 This question requires finding stationary points using the product rule, then proving a geometric relationship. Students must differentiate e^(-x)sin(x), solve the transcendental equation for the maximum, identify where curves touch (sin x = 1), and verify the distance relationship. The multi-step nature, combination of calculus and geometric reasoning, and need to work with transcendental equations elevates this above routine stationary point questions.
Spec1.05g Exact trigonometric values: for standard angles1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives

7 Fig. 7 shows the graphs of the curves \(y = \mathrm { e } ^ { - x }\) and \(y = \mathrm { e } ^ { - x } \sin x\) for \(0 \leq x \leq \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-3_407_793_1085_740} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} The maximum point on \(y = \mathrm { e } ^ { - x } \sin x\) is at A , and the curves touch at B . \(\mathrm { A } ^ { \prime }\) and \(\mathrm { B } ^ { \prime }\) are the points on the \(x\)-axis such that \(\mathrm { A } ^ { \prime } \mathrm { A }\) and \(\mathrm { B } ^ { \prime } \mathrm { B }\) are parallel to the \(y\)-axis.
Show that \(\mathrm { OA } ^ { \prime } = \mathrm { A } ^ { \prime } \mathrm { B } ^ { \prime }\).

Question 7:
AnswerMarks Guidance
AnswerMark Guidance
\(y = e^{-x}\sin x \Rightarrow \frac{dy}{dx} = -e^{-x}\sin x + e^{-x}\cos x\)M1, A1 Product rule
\(= 0\) when \(\sin x = \cos x \Rightarrow x = \frac{\pi}{4}\) (in this range)M1 Set \(= 0\)
\(\Rightarrow OA' = \frac{\pi}{4}\)A1
For intersection: \(e^{-x}\sin x = e^{-x} \Rightarrow \sin x = 1\)M1 Solving
\(\Rightarrow x = \frac{\pi}{2}\) (in this range)A1
\(\Rightarrow OB' = \frac{\pi}{2} \Rightarrow A'B' = \frac{\pi}{4}\)A1
Total: 7
## Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = e^{-x}\sin x \Rightarrow \frac{dy}{dx} = -e^{-x}\sin x + e^{-x}\cos x$ | M1, A1 | Product rule |
| $= 0$ when $\sin x = \cos x \Rightarrow x = \frac{\pi}{4}$ (in this range) | M1 | Set $= 0$ |
| $\Rightarrow OA' = \frac{\pi}{4}$ | A1 | |
| For intersection: $e^{-x}\sin x = e^{-x} \Rightarrow \sin x = 1$ | M1 | Solving |
| $\Rightarrow x = \frac{\pi}{2}$ (in this range) | A1 | |
| $\Rightarrow OB' = \frac{\pi}{2} \Rightarrow A'B' = \frac{\pi}{4}$ | A1 | |
| | **Total: 7** | |

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7 Fig. 7 shows the graphs of the curves $y = \mathrm { e } ^ { - x }$ and $y = \mathrm { e } ^ { - x } \sin x$ for $0 \leq x \leq \pi$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-3_407_793_1085_740}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}

The maximum point on $y = \mathrm { e } ^ { - x } \sin x$ is at A , and the curves touch at B .\\
$\mathrm { A } ^ { \prime }$ and $\mathrm { B } ^ { \prime }$ are the points on the $x$-axis such that $\mathrm { A } ^ { \prime } \mathrm { A }$ and $\mathrm { B } ^ { \prime } \mathrm { B }$ are parallel to the $y$-axis.\\
Show that $\mathrm { OA } ^ { \prime } = \mathrm { A } ^ { \prime } \mathrm { B } ^ { \prime }$.

\hfill \mbox{\textit{OCR MEI C3  Q7 [7]}}