| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show dy/dx equals given expression |
| Difficulty | Standard +0.3 Part (i) is a straightforward implicit differentiation exercise requiring the product rule and chain rule, then algebraic rearrangement—standard C4 technique. Part (ii) adds finding points on the curve (solving a quadratic), computing tangent equations, and finding their intersection, but follows a clear procedural path with no novel insight required. Slightly above average due to the multi-step nature of part (ii), but well within typical C4 scope. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\) | B1 | |
| \(\frac{d}{dx}(uv)=u\,dv+v\,du\) used on \((-7)xy\) | M1 | |
| \(\frac{d}{dx}(14x^2-7xy+y^2)=28x-7x\frac{dy}{dx}-7y+2y\frac{dy}{dx}\) | A1 | \((=0)\) |
| \(2y\frac{dy}{dx}-7x\frac{dy}{dx}=7y-28x\to\frac{dy}{dx}=\frac{28x-7y}{7x-2y}\) www AG | A1 4 | As AG, intermed step necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Subst \(x=1\) into eqn curve & solve quadratic eqn in \(y\) | M1 | (\(y=3\) or \(4\)) |
| Subst \(x=1\) and (one of) their \(y\)-value(s) into given \(\frac{dy}{dx}\) | M1 | \(\left(\frac{dy}{dx}=7\text{ or }0\right)\) |
| Find eqn of tgt, with their \(\frac{dy}{dx}\), going through \((1,\text{their }y)\) | *M1 | using (one of) \(y\) value(s) |
| Produce either \(y=7x-4\) or \(y=4\) | A1 | |
| Solve simultaneously their two equations | dep*M1 | provided they have two |
| Produce \(x=\frac{8}{7}\) | A1 6 |
# Question 8:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ | B1 | |
| $\frac{d}{dx}(uv)=u\,dv+v\,du$ used on $(-7)xy$ | M1 | |
| $\frac{d}{dx}(14x^2-7xy+y^2)=28x-7x\frac{dy}{dx}-7y+2y\frac{dy}{dx}$ | A1 | $(=0)$ |
| $2y\frac{dy}{dx}-7x\frac{dy}{dx}=7y-28x\to\frac{dy}{dx}=\frac{28x-7y}{7x-2y}$ www AG | A1 **4** | As AG, intermed step necessary |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Subst $x=1$ into eqn curve & solve quadratic eqn in $y$ | M1 | ($y=3$ or $4$) |
| Subst $x=1$ and (one of) their $y$-value(s) into given $\frac{dy}{dx}$ | M1 | $\left(\frac{dy}{dx}=7\text{ or }0\right)$ |
| Find eqn of tgt, with their $\frac{dy}{dx}$, going through $(1,\text{their }y)$ | *M1 | using (one of) $y$ value(s) |
| Produce either $y=7x-4$ or $y=4$ | A1 | |
| Solve simultaneously their two equations | dep*M1 | provided they have two |
| Produce $x=\frac{8}{7}$ | A1 **6** | |
---8 (i) Given that $14 x ^ { 2 } - 7 x y + y ^ { 2 } = 2$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 28 x - 7 y } { 7 x - 2 y }$.\\
(ii) The points $L$ and $M$ on the curve $14 x ^ { 2 } - 7 x y + y ^ { 2 } = 2$ each have $x$-coordinate 1 . The tangents to the curve at $L$ and $M$ meet at $N$. Find the coordinates of $N$.
\hfill \mbox{\textit{OCR C4 2009 Q8 [10]}}