| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find derivative of simple polynomial (integer powers) |
| Difficulty | Moderate -0.8 This is a straightforward P1 differentiation question requiring basic power rule application (part a), difference quotient calculation (part b), and recognition that the limit as h→0 gives the derivative (part c). All parts are standard textbook exercises with no problem-solving or novel insight required, making it easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{dy}{dx} = 4x\) at \(x = 2\) | M1 | Attempts to find value of \(\frac{dy}{dx} = ax\), \(a > 0\) at \(x = 2\) |
| At \(x = 2\), gradient of tangent \(= 8\) | A1 | For 8. No need to state this is the gradient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y_Q = 2(2+h)^2 + 5\) | B1 | |
| Gradient \(PQ = \frac{\text{their } y_Q - 13}{2 + h - 2}\) | M1 | Attempts \(\pm\frac{y_Q - y_P}{x_Q - x_P}\), condoning slips, but must be genuine attempt at \(y_Q\) |
| \(\left(= \frac{8h + 2h^2}{h}\right) = 8 + 2h\) | A1 | Gradient is \(8 + 2h\) with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States as \(h \to 0\), Gradient \(PQ \to 8 =\) Gradient of tangent | B1 | Reference to "limit" or "as \(h\) tends to 0" and linked to part (a). Be generous beyond these constraints |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{dy}{dx} = 4x$ at $x = 2$ | M1 | Attempts to find value of $\frac{dy}{dx} = ax$, $a > 0$ at $x = 2$ |
| At $x = 2$, gradient of tangent $= 8$ | A1 | For 8. No need to state this is the gradient |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y_Q = 2(2+h)^2 + 5$ | B1 | |
| Gradient $PQ = \frac{\text{their } y_Q - 13}{2 + h - 2}$ | M1 | Attempts $\pm\frac{y_Q - y_P}{x_Q - x_P}$, condoning slips, but must be genuine attempt at $y_Q$ |
| $\left(= \frac{8h + 2h^2}{h}\right) = 8 + 2h$ | A1 | Gradient is $8 + 2h$ with no errors seen |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States as $h \to 0$, Gradient $PQ \to 8 =$ Gradient of tangent | B1 | Reference to "limit" or "as $h$ tends to 0" and linked to part (a). Be generous beyond these constraints |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{50ec901b-b6b6-4b72-85bd-a084f313c99b-16_648_822_296_561}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows part of the curve with equation $y = 2 x ^ { 2 } + 5$
The point $P ( 2,13 )$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the tangent to the curve at $P$.
The point $Q$ with $x$ coordinate $2 + h$ also lies on the curve.
\item Find, in terms of $h$, the gradient of the line $P Q$. Give your answer in simplest form.
\item Explain briefly the relationship between the answer to (b) and the answer to (a).
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q7 [6]}}