| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a standard C4 implicit differentiation question with routine application of the product rule and chain rule. Part (i) is straightforward differentiation, part (ii) requires substituting the line equation and interpreting dy/dx, and part (iii) is a direct application. While it involves multiple steps, each technique is standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Treat \((x+3)(y+4)\) or \(xy\) as a product | M1 | Attempting \(u\,dv + v\,du\) |
| \(\frac{d}{dx}(x+3)(y+4) = (x+3)\frac{dy}{dx} + (y+4)\) or \(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\) | A1 | |
| \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\) | B1 | |
| \(\frac{dy}{dx} = \frac{2x - y - 4}{x - 2y + 3}\) | B1 | AEF including \(-\frac{a}{b}, \frac{-a}{b}, \frac{a}{-b}\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply that denominator is zero | B1 | Provided denom is \(x - 2y + 3\) or \(-x + 2y - 3\) |
| Tangents are parallel to \(y\)-axis | B1 | Accept vertical or of the form \(x = k\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \((6,0)\) into their \(\frac{dy}{dx}\) \(\left(= \frac{8}{9}\right)\) | M1 | |
| \(8x - 9y = 48\); FT \(fx - gy = 6f\) | A1 FT | FT their numerical \(\frac{dy}{dx} = \frac{f}{g}\) www in this part |
| [2] |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Treat $(x+3)(y+4)$ or $xy$ as a product | M1 | Attempting $u\,dv + v\,du$ |
| $\frac{d}{dx}(x+3)(y+4) = (x+3)\frac{dy}{dx} + (y+4)$ or $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ | A1 | |
| $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ | B1 | |
| $\frac{dy}{dx} = \frac{2x - y - 4}{x - 2y + 3}$ | B1 | AEF including $-\frac{a}{b}, \frac{-a}{b}, \frac{a}{-b}$ |
| **[4]** | | |
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## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply that denominator is zero | B1 | Provided denom is $x - 2y + 3$ or $-x + 2y - 3$ |
| Tangents are parallel to $y$-axis | B1 | Accept vertical or of the form $x = k$ |
| **[2]** | | |
---
## Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $(6,0)$ into their $\frac{dy}{dx}$ $\left(= \frac{8}{9}\right)$ | M1 | |
| $8x - 9y = 48$; FT $fx - gy = 6f$ | A1 FT | FT their numerical $\frac{dy}{dx} = \frac{f}{g}$ www in this part |
| **[2]** | | |
---3 The equation of a curve $C$ is $( x + 3 ) ( y + 4 ) = x ^ { 2 } + y ^ { 2 }$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(ii) The line $2 y = x + 3$ meets $C$ at two points. What can be said about the tangents to $C$ at these points? Justify your answer.\\
(iii) Find the equation of the tangent at the point ( 6,0 ), giving your answer in the form $a x + b y = c$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C4 2012 Q3 [8]}}