| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Given one function find others |
| Difficulty | Standard +0.8 This question requires geometric insight to set up relationships between the nested right triangles, then uses reciprocal trig functions and fractional indices. While the steps are guided, it demands spatial reasoning to establish that AC = x sec θ, AD = x sec² θ, etc., followed by algebraic manipulation of sec³ θ = 2 to find the ratio 2^(2/3):1. This goes beyond routine application of identities. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AC = x\sec\theta\) | B1 | Accept any equivalent form (e.g. \(AC\cos\theta = x\)). If AC not seen then must be a diagram as evidence of correct sides; \(x\sec\theta\) with no AC is B0 |
| \(AD = x\sec^2\theta\) and \(AE = x\sec^3\theta\) | B1 | Accept \(2x = x\sec^3\theta\) (as \(AE = 2x\)) or any equivalent form. Accept \(\cos^3\theta = x/AC \times AC/AD \times AD/2x\) for first two marks |
| \(\Rightarrow x\sec^3\theta = 2x\) \(\Rightarrow \sec^3\theta = 2\)* | B1 | This line (oe) must be seen before \(x\)'s cancelled. NB AG – dependent on all previous marks |
| OR \(AD = 2x\cos\theta\), \(AC = 2x\cos^2\theta\), \(AB = 2x\cos^3\theta\), \(2x\cos^3\theta = x \Rightarrow \sec^3\theta = 2\)* | B1 B1 B1 | Same principles as above; or \(x = 2x\cos^3\theta\) (as \(AB=x\)). Must see \(2x\cos^3\theta = x\) (oe) before given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(ED = 2x\sin\theta\) | B1 | oe e.g. \(ED = \sqrt{4x^2 - AD^2}\) or \(ED = AD\tan\theta\) with AD correctly expressed in terms of \(x\) and \(\theta\) (or using \(\theta = 37.5\) or better). Allow \(ED = 1.22x\) (or better) but B0 if \(ED = \ldots\) missing |
| \(CB = x\tan\theta\) | B1 | oe e.g. \(CB = \sqrt{AC^2 - x^2}\) or \(CB = AC\sin\theta\) with AC correctly expressed. Allow \(CB = 0.77x\) (or better) but B0 if \(CB = \ldots\) missing |
| \(\frac{ED}{CB} = \frac{2x\sin\theta}{x\tan\theta} = 2\cos\theta\) | B1 | www must come from exact working (not using \(\theta = 37.46\ldots\) oe); accept \(\frac{ED}{CB} = \frac{2}{\sec\theta}\) or \(\frac{ED}{CB} = \sec^2\theta\) (oe) (as from (i): \(\sec^3\theta = 2\)) |
| \(= 2/2^{\frac{1}{3}} = 2^{\frac{2}{3}}\)* | B1 | NB AG – dependent on all previous marks in (ii); must be one step of intermediate working from \(2\cos\theta\) to given answer |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AC = x\sec\theta$ | B1 | Accept any equivalent form (e.g. $AC\cos\theta = x$). If AC not seen then must be a diagram as evidence of correct sides; $x\sec\theta$ with no AC is B0 |
| $AD = x\sec^2\theta$ and $AE = x\sec^3\theta$ | B1 | Accept $2x = x\sec^3\theta$ (as $AE = 2x$) or any equivalent form. Accept $\cos^3\theta = x/AC \times AC/AD \times AD/2x$ for first two marks |
| $\Rightarrow x\sec^3\theta = 2x$ $\Rightarrow \sec^3\theta = 2$* | B1 | This line (oe) must be seen before $x$'s cancelled. **NB AG** – dependent on all previous marks |
| **OR** $AD = 2x\cos\theta$, $AC = 2x\cos^2\theta$, $AB = 2x\cos^3\theta$, $2x\cos^3\theta = x \Rightarrow \sec^3\theta = 2$* | B1 B1 B1 | Same principles as above; or $x = 2x\cos^3\theta$ (as $AB=x$). Must see $2x\cos^3\theta = x$ (oe) before given answer |
---
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ED = 2x\sin\theta$ | B1 | oe e.g. $ED = \sqrt{4x^2 - AD^2}$ or $ED = AD\tan\theta$ with AD correctly expressed in terms of $x$ and $\theta$ (or using $\theta = 37.5$ or better). Allow $ED = 1.22x$ (or better) but B0 if $ED = \ldots$ missing |
| $CB = x\tan\theta$ | B1 | oe e.g. $CB = \sqrt{AC^2 - x^2}$ or $CB = AC\sin\theta$ with AC correctly expressed. Allow $CB = 0.77x$ (or better) but B0 if $CB = \ldots$ missing |
| $\frac{ED}{CB} = \frac{2x\sin\theta}{x\tan\theta} = 2\cos\theta$ | B1 | www must come from exact working (not using $\theta = 37.46\ldots$ oe); accept $\frac{ED}{CB} = \frac{2}{\sec\theta}$ or $\frac{ED}{CB} = \sec^2\theta$ (oe) (as from (i): $\sec^3\theta = 2$) |
| $= 2/2^{\frac{1}{3}} = 2^{\frac{2}{3}}$* | B1 | **NB AG** – dependent on all previous marks in (ii); must be one step of intermediate working from $2\cos\theta$ to given answer |5 In Fig. 5, triangles $\mathrm { ABC } , \mathrm { ACD }$ and ADE are all right-angled, and angles $\mathrm { BAC } , \mathrm { CAD }$ and DAE are all $\theta$. $\mathrm { AB } = x$ and $\mathrm { AE } = 2 x$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8b807b2e-777b-4c9a-b3dd-890d21d33174-2_567_465_1905_799}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(i) Show that $\sec ^ { 3 } \theta = 2$.\\
(ii) Hence show the ratio of lengths ED to CB is $2 ^ { \frac { 2 } { 3 } } : 1$.
\hfill \mbox{\textit{OCR MEI C4 2016 Q5 [7]}}