| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | September |
| Marks | 11 |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.8 This is a multivariable calculus question requiring partial differentiation, solving a system of three simultaneous equations to find stationary points, and understanding local vs global extrema. While the differentiation is routine, solving the coupled nonlinear system (likely yielding xyz = constant relationships) requires algebraic manipulation beyond standard A-level. The global minimum part requires insight but is relatively straightforward once the stationary point is found. This is typical Further Maths content but more involved than standard optimization questions. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero |
| Answer | Marks | Guidance |
|---|---|---|
| \(f_x = 2xyz + 2y^2z + 3yz^2 - 24yz\) or \(yz(2x + 2y + 3z - 24)\) * | M1, A1 | Good attempt at (at least) one partial derivative |
| \(f_y = x^2z + 4xyz + 3xz^2 - 24xz\) or \(xz(x + 4y + 3z - 24)\) | A1 | |
| \(f_z = x^3y + 2xy^2 + 6xyz - 24xy\) or \(xy(x + 2y + 6z - 24)\) | A1 | |
| (i) (b) Setting all p.d.s to zero | M1 | |
| \(2x + 2y + 3z = 24\), \(x + 4y + 3z = 24\), \(x + 2y + 6z = 24\) | M1 | Factoring out the non-zero terms and setting up a solvable system of equations |
| \(x = 6, y = 3, z = 2\) | M1, A1 | Solving a 3×3 system of equations (BC, for instance); Correct |
| \(w = -216\) | A1 | cao (no need to mention these are \(a, b, c, d\)) |
| (i) (ii) Want (e.g.) \(x\) large and \(\to\infty\), \(y\) small and \(\to\infty\), \(z\) small and \(\to\infty\) e.g. \(x = -10, y = -1, z = 1\) so that \(w = -330 \times -216\) | M1, A1 | Correct idea; Demonstrated correctly |
**(i) (a)**
$f_x = 2xyz + 2y^2z + 3yz^2 - 24yz$ or $yz(2x + 2y + 3z - 24)$ * | M1, A1 | Good attempt at (at least) one partial derivative
$f_y = x^2z + 4xyz + 3xz^2 - 24xz$ or $xz(x + 4y + 3z - 24)$ | A1 |
$f_z = x^3y + 2xy^2 + 6xyz - 24xy$ or $xy(x + 2y + 6z - 24)$ | A1 |
**(i) (b)** Setting all p.d.s to zero | M1 |
$2x + 2y + 3z = 24$, $x + 4y + 3z = 24$, $x + 2y + 6z = 24$ | M1 | Factoring out the non-zero terms and setting up a solvable system of equations
$x = 6, y = 3, z = 2$ | M1, A1 | Solving a 3×3 system of equations (BC, for instance); Correct
$w = -216$ | A1 | cao (no need to mention these are $a, b, c, d$)
**(i) (ii)** Want (e.g.) $x$ large and $\to\infty$, $y$ small and $\to\infty$, $z$ small and $\to\infty$ e.g. $x = -10, y = -1, z = 1$ so that $w = -330 \times -216$ | M1, A1 | Correct idea; Demonstrated correctlyThe function $w = f(x, y, z)$ is given by $f(x, y, z) = x^2yz + 2xy^2z + 3xyz^2 - 24xyz$, for $x, y, z \neq 0$.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Find
\begin{itemize}
\item $f_x$,
\item $f_y$,
\item $f_z$. [4]
\end{itemize}
\item Hence find the values of $a$, $b$, $c$ and $d$ for which $w$ has a stationary value when $d = f(a, b, c)$. [5]
\end{enumerate}
\item You are given that this stationary value is a local minimum of $w$. Find values of $x$, $y$ and $z$ which show that it is not a global minimum of $w$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q3 [11]}}