| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Sketch reciprocal function graphs |
| Difficulty | Moderate -0.3 Part (a) requires identifying asymptotes of cosec x where sin x = 0, which is straightforward recall. Part (b) involves standard tangent line calculation using given derivative, point coordinates, and algebraic manipulation—routine A-level calculus with no novel insight required, though the algebra is slightly involved. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Asymptotes \(x = 0\) (or \(y\)-axis), \(x = \pi\) and \(x = 2\pi\) | B1 [1] | Must have all three |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x = \dfrac{\pi}{3}\), \(y = \dfrac{2\sqrt{3}}{3}\) | B1 | soi; any exact form e.g. \(\dfrac{2}{\sqrt{3}}\) |
| When \(x = \dfrac{\pi}{3}\), \(\dfrac{dy}{dx} = -\text{cosec}\dfrac{\pi}{3}\cot\dfrac{\pi}{3} = -\dfrac{2}{3}\) | M1 | Uses the derivative when \(x = \dfrac{\pi}{3}\) |
| A1 | May be embedded in the tangent equation | |
| Equation of tangent: \(y - \dfrac{2\sqrt{3}}{3} = -\dfrac{2}{3}\!\left(x - \dfrac{\pi}{3}\right)\) | M1 | Uses both coordinates and gradient to find equation. If \(y = -\frac{2}{3}x + c\) used, there must be an attempt to find \(c\) using both coordinates and gradient |
| When \(y = 0\): \(-\dfrac{2\sqrt{3}}{3} = -\dfrac{2}{3}\!\left(x - \dfrac{\pi}{3}\right)\) | M1 | Substituting \(y = 0\) into their tangent |
| giving \(x = \dfrac{\pi}{3} + \sqrt{3}\) (AG) | A1 [6] | Working must be correct and exact throughout |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Asymptotes $x = 0$ (or $y$-axis), $x = \pi$ and $x = 2\pi$ | B1 [1] | Must have all three |
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## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x = \dfrac{\pi}{3}$, $y = \dfrac{2\sqrt{3}}{3}$ | B1 | soi; any exact form e.g. $\dfrac{2}{\sqrt{3}}$ |
| When $x = \dfrac{\pi}{3}$, $\dfrac{dy}{dx} = -\text{cosec}\dfrac{\pi}{3}\cot\dfrac{\pi}{3} = -\dfrac{2}{3}$ | M1 | Uses the derivative when $x = \dfrac{\pi}{3}$ |
| | A1 | May be embedded in the tangent equation |
| Equation of tangent: $y - \dfrac{2\sqrt{3}}{3} = -\dfrac{2}{3}\!\left(x - \dfrac{\pi}{3}\right)$ | M1 | Uses both coordinates and gradient to find equation. If $y = -\frac{2}{3}x + c$ used, there must be an attempt to find $c$ using both coordinates and gradient |
| When $y = 0$: $-\dfrac{2\sqrt{3}}{3} = -\dfrac{2}{3}\!\left(x - \dfrac{\pi}{3}\right)$ | M1 | Substituting $y = 0$ into their tangent |
| giving $x = \dfrac{\pi}{3} + \sqrt{3}$ **(AG)** | A1 [6] | Working must be correct and exact throughout |
---6
\begin{enumerate}[label=(\alph*)]
\item The diagram shows part of the graph of $\mathrm { y } = \operatorname { cosec } \mathrm { x }$, where $x$ is in radians.
State the equations of the three vertical asymptotes that can be seen.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-06_734_672_603_324}
The tangent to the graph at the point P with $x$-coordinate $\frac { \pi } { 3 }$ meets the $x$-axis at Q .
\item Show that the $x$-coordinate of Q is $\frac { \pi } { 3 } + \sqrt { 3 }$. (You may use without proof the result that the derivative of $\operatorname { cosec } x$ is $- \operatorname { cosec } x \cot x$.)
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q6 [7]}}