| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard C4 implicit differentiation question with routine verification and algebraic manipulation. While it requires multiple techniques (implicit differentiation, substitution, solving equations), each step follows predictable patterns without requiring novel insight or particularly challenging algebra. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}\) | B1 | |
| Consider \(\frac{d}{dx}(xy)\) as a product | M1 | |
| \(= x\frac{dy}{dx} + y\) | A1 | Tolerate omission of '6' |
| \(\frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x}\) ISW AEF | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 = 2^4\) or \(16\) and \(y^3 = 2^5\) or \(32\) | \*B1 | |
| Satisfactory conclusion | dep\*B1 | AG |
| Substitute \(\left(2^{\frac{4}{3}}, 2^{\frac{5}{3}}\right)\) into their \(\frac{dy}{dx}\) | M1 | or the numerator of \(\frac{dy}{dx}\) |
| Show or use calc to demo that num \(= 0\), ignore denom AG | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \((a,a)\) into eqn of curve | M1 | & attempt to state \(a = \ldots\) |
| \(a=3\) only with clear ref to \(a\neq 0\) | A1 | |
| Substitute \((3,3)\) or (their \(a\), their \(a\)) into their \(\frac{dy}{dx}\) | M1 | |
| \(-1\) only WWW | A1 (4) from (their \(a\), their \(a\)) | |
| Total: 12 |
# Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$ | B1 | |
| Consider $\frac{d}{dx}(xy)$ as a product | M1 | |
| $= x\frac{dy}{dx} + y$ | A1 | Tolerate omission of '6' |
| $\frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x}$ ISW AEF | A1 **(4)** | |
---
# Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 = 2^4$ or $16$ and $y^3 = 2^5$ or $32$ | \*B1 | |
| Satisfactory conclusion | dep\*B1 | **AG** |
| Substitute $\left(2^{\frac{4}{3}}, 2^{\frac{5}{3}}\right)$ into their $\frac{dy}{dx}$ | M1 | or the numerator of $\frac{dy}{dx}$ |
| Show or use calc to demo that num $= 0$, ignore denom **AG** | A1 **(4)** | |
---
# Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $(a,a)$ into eqn of curve | M1 | & attempt to state $a = \ldots$ |
| $a=3$ only with clear ref to $a\neq 0$ | A1 | |
| Substitute $(3,3)$ or (their $a$, their $a$) into their $\frac{dy}{dx}$ | M1 | |
| $-1$ only WWW | A1 **(4)** from (their $a$, their $a$) |
| **Total: 12** | | |
---8 The equation of a curve is $x ^ { 3 } + y ^ { 3 } = 6 x y$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(ii) Show that the point $\left( 2 ^ { \frac { 4 } { 3 } } , 2 ^ { \frac { 5 } { 3 } } \right)$ lies on the curve and that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ at this point.\\
(iii) The point $( a , a )$, where $a > 0$, lies on the curve. Find the value of $a$ and the gradient of the curve at this point.
\hfill \mbox{\textit{OCR C4 2009 Q8 [12]}}