| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Verify point and find gradient |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question with standard techniques. Part (i) is simple substitution to verify a point. Part (ii) requires routine application of implicit differentiation rules (chain rule for sin 2x and cos y) followed by substitution of given coordinates. No novel insight or complex manipulation required, making it slightly easier than average. |
| Spec | 1.05g Exact trigonometric values: for standard angles1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin(\pi/3) + \cos(\pi/6) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}\) | B1 | Must be exact, must show working; not just \(\sin(\pi/3)+\cos(\pi/6)=\sqrt{3}\); if substituting for \(y\) and solving for \(x\) must evaluate \(\sin\pi/3\) e.g. not \(\arccos(\sqrt{3}-\sin\pi/3)\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\cos 2x - \sin y\frac{dy}{dx} = 0\) | M1 | Implicit differentiation; allow one error but must have \((\pm)\sin y\,dy/dx\); ignore \(dy/dx = \ldots\); \(2\cos 2x\,dx - \sin y\,dy = 0\) is M1A1 |
| \(\frac{dy}{dx} = \frac{2\cos 2x}{\sin y}\) | A1 | Correct expression; \(\frac{-2\cos 2x}{-\sin y}\) is A0 |
| When \(x = \pi/6,\ y = \pi/6\): | A1cao | |
| \(\frac{dy}{dx} = \frac{2\cos\pi/3}{\sin\pi/6} = 2\) | M1dep | Substituting dep 1st M1; www |
| A1 | ||
| [5] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin(\pi/3) + \cos(\pi/6) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$ | B1 | Must be exact, must show working; not just $\sin(\pi/3)+\cos(\pi/6)=\sqrt{3}$; if substituting for $y$ and solving for $x$ must evaluate $\sin\pi/3$ e.g. not $\arccos(\sqrt{3}-\sin\pi/3)$ |
| **[1]** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\cos 2x - \sin y\frac{dy}{dx} = 0$ | M1 | Implicit differentiation; allow one error but must have $(\pm)\sin y\,dy/dx$; ignore $dy/dx = \ldots$; $2\cos 2x\,dx - \sin y\,dy = 0$ is M1A1 |
| $\frac{dy}{dx} = \frac{2\cos 2x}{\sin y}$ | A1 | Correct expression; $\frac{-2\cos 2x}{-\sin y}$ is A0 |
| When $x = \pi/6,\ y = \pi/6$: | A1cao | |
| $\frac{dy}{dx} = \frac{2\cos\pi/3}{\sin\pi/6} = 2$ | M1dep | Substituting dep 1st M1; www |
| | A1 | |
| **[5]** | | |
---7 A curve is defined by the equation $\sin 2 x + \cos y = \sqrt { 3 }$.\\
(i) Verify that the point $\mathrm { P } \left( \frac { 1 } { 6 } \pi , \frac { 1 } { 6 } \pi \right)$ lies on the curve.\\
(ii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
Hence find the gradient of the curve at the point P .
\hfill \mbox{\textit{OCR MEI C3 Q7 [6]}}