| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show dy/dx equals given expression |
| Difficulty | Moderate -0.3 This is a standard C4 implicit differentiation question requiring the chain rule and product rule. Finding dy/dx from cos(2y) + ye^(3x) = 2π involves routine application of differentiation rules with no novel problem-solving required. The presence of point A suggests verification, making this slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\): \(\lambda = -1\) gives \(P = (5, -8, 2)\) | M1 | Finding coordinates of \(P\) |
| \(Q\): \(\mu = 2\) gives \(Q = (11, -14, 8)\) | M1 | Finding coordinates of \(Q\) |
| \(\overrightarrow{PQ} = \begin{pmatrix}6\\-6\\6\end{pmatrix} = 6\begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 | Parallel to given vector (scalar multiple) |
| Answer | Marks | Guidance |
|---|---|---|
| \(l_1\) and \(l_2\) intersect: \(4-\lambda = 7+2\mu\), \(-5+3\lambda = -8-3\mu\), \(3+\lambda = 6+\mu\) | M1 | Setting equations equal |
| From equations: solving gives \(\lambda = 1\), \(\mu = -2\) | M1 | |
| \(x=3\): \(4-1=3\) ✓; \(b = -5+3(1) = -2\) so \(b=-2\) | A1 | Showing \(b=-2\) and finding \(c\) |
| \(c = 3+1 = 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Line through \(P\) parallel to \(l_2\): \(\mathbf{r} = \begin{pmatrix}5\\-8\\2\end{pmatrix} + t\begin{pmatrix}2\\-3\\1\end{pmatrix}\) | M1 | Correct line through \(P\) |
| \(\overrightarrow{RS} = S - R\); \(S = (5+2t, -8-3t, 2+t)\) | ||
| \(\overrightarrow{RS} \cdot \begin{pmatrix}1\\-1\\1\end{pmatrix} = 0\) (perpendicular to \(PQ\)) | M1 | Perpendicularity condition |
| \((5+2t-3) + (8+3t-2) \cdot(-1)\cdot(-1) + (2+t-4) = 0\) ... solving \(t = -1\) | M1 | Solving for \(t\) |
| \(S = (3, -5, 1)\) | A1 |
# Question 6:
## Part (a):
| $P$: $\lambda = -1$ gives $P = (5, -8, 2)$ | M1 | Finding coordinates of $P$ |
|---|---|---|
| $Q$: $\mu = 2$ gives $Q = (11, -14, 8)$ | M1 | Finding coordinates of $Q$ |
| $\overrightarrow{PQ} = \begin{pmatrix}6\\-6\\6\end{pmatrix} = 6\begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | Parallel to given vector (scalar multiple) |
## Part (b)(i):
| $l_1$ and $l_2$ intersect: $4-\lambda = 7+2\mu$, $-5+3\lambda = -8-3\mu$, $3+\lambda = 6+\mu$ | M1 | Setting equations equal |
|---|---|---|
| From equations: solving gives $\lambda = 1$, $\mu = -2$ | M1 | |
| $x=3$: $4-1=3$ ✓; $b = -5+3(1) = -2$ so $b=-2$ | A1 | Showing $b=-2$ and finding $c$ |
| $c = 3+1 = 4$ | | |
## Part (b)(ii):
| Line through $P$ parallel to $l_2$: $\mathbf{r} = \begin{pmatrix}5\\-8\\2\end{pmatrix} + t\begin{pmatrix}2\\-3\\1\end{pmatrix}$ | M1 | Correct line through $P$ |
|---|---|---|
| $\overrightarrow{RS} = S - R$; $S = (5+2t, -8-3t, 2+t)$ | | |
| $\overrightarrow{RS} \cdot \begin{pmatrix}1\\-1\\1\end{pmatrix} = 0$ (perpendicular to $PQ$) | M1 | Perpendicularity condition |
| $(5+2t-3) + (8+3t-2) \cdot(-1)\cdot(-1) + (2+t-4) = 0$ ... solving $t = -1$ | M1 | Solving for $t$ |
| $S = (3, -5, 1)$ | A1 | |6 The line $l _ { 1 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 4 \\ - 5 \\ 3 \end{array} \right] + \lambda \left[ \begin{array} { r } - 1 \\ 3 \\ 1 \end{array} \right]$.\\
The line $l _ { 2 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 7 \\ - 8 \\ 6 \end{array} \right] + \mu \left[ \begin{array} { r } 2 \\ - 3 \\ 1 \end{array} \right]$.\\
The point $P$ lies on $l _ { 1 }$ where $\lambda = - 1$. The point $Q$ lies on $l _ { 2 }$ where $\mu = 2$.
\begin{enumerate}[label=(\alph*)]
\item Show that the vector $\overrightarrow { P Q }$ is parallel to $\left[ \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right]$.
\item The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $R ( 3 , b , c )$.
\begin{enumerate}[label=(\roman*)]
\item Show that $b = - 2$ and find the value of $c$.
\item The point $S$ lies on a line through $P$ that is parallel to $l _ { 2 }$. The line $R S$ is perpendicular to the line $P Q$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-16_887_1159_1320_443}
Find the coordinates of $S$.\\
$7 \quad$ A curve has equation $\cos 2 y + y \mathrm { e } ^ { 3 x } = 2 \pi$.\\
The point $A \left( \ln 2 , \frac { \pi } { 4 } \right)$ lies on this curve.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2014 Q6 [10]}}