| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard techniques. Part (i) is routine application of the product and chain rules. Part (ii) requires setting dy/dx = 1 and solving simultaneous equations with the curve equation—methodical but not conceptually demanding. Slightly above average due to the algebraic manipulation needed in part (ii), but well within standard A-level scope. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State anywhere \((y^2)' = 2yy'\) | B1 | |
| Differentiate \(xy\) to obtain at least any one correct term – either \(xy'\) or \(y\) | M1 | |
| Obtain \(2x - xy' - y + 2yy' = 0\) | A1 | |
| Obtain \(y' = \frac{2x-y}{x-2y}\) | A1 | [4] |
| (ii) State or imply their \(\frac{2x-y}{x-2y} = 1\) | B1 | |
| Substitute \(x = -y\) or equiv into implicit eqn and attempt to solve for \(x\) or \(y\) | M1 | |
| Obtain \(x\) or \(y = \frac{1}{\sqrt{3}}\) or \(\pm 0.577\) | A1 | |
| Attempt to obtain \(y\) co-ordinates | M1 | |
| State \(\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)\) and \(\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) or decimal equiv | A1 | [5] |
(i) State anywhere $(y^2)' = 2yy'$ | B1 |
Differentiate $xy$ to obtain at least any one correct term – either $xy'$ or $y$ | M1 |
Obtain $2x - xy' - y + 2yy' = 0$ | A1 |
Obtain $y' = \frac{2x-y}{x-2y}$ | A1 | [4]
(ii) State or imply their $\frac{2x-y}{x-2y} = 1$ | B1 |
Substitute $x = -y$ or equiv into implicit eqn and attempt to solve for $x$ or $y$ | M1 |
Obtain $x$ or $y = \frac{1}{\sqrt{3}}$ or $\pm 0.577$ | A1 |
Attempt to obtain $y$ co-ordinates | M1 |
State $\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ and $\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ or decimal equiv | A1 | [5]A curve has equation $x^2 - xy + y^2 = 1$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ in terms of $x$ and $y$. [4]
\item Find the coordinates of the points on the curve in the second and fourth quadrants where the tangent is parallel to $y = x$. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q9 [9]}}