| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find derivative of simple polynomial (integer powers) |
| Difficulty | Easy -1.3 This is a straightforward P1 differentiation question requiring basic polynomial differentiation (power rule) and understanding of the derivative as a limit. Part (a) is routine application of d/dx, part (b) uses the gradient formula with algebra simplification, and part (c) tests conceptual understanding that the chord gradient approaches the tangent gradient as h→0. All techniques are standard with no problem-solving insight required. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Attempts \(\frac{dy}{dx} = 2x+3\) at \(x=3\); gradient of tangent \(= 9\) | M1 A1 | M1: Attempts to find value of \(\left(\frac{dy}{dx}=\right) ax+3\), \(a>0\) at \(x=3\). Look for 3 substituted and proceeding to a value. A1: 9 (answer only scores both marks). |
| (b) \(y_Q = (3+h)^2 + 3(3+h) - 2\) | B1 | Seen or implied. |
| Gradient \(PQ = \frac{(3+h)^2+3(3+h)-2-16}{3+h-3} = \frac{9h+h^2}{h} = 9+h\) | M1 A1 | M1: Attempts \(\pm\frac{y_Q - 16}{x_Q - 3}\), condoning slips, genuine attempt at \(y_Q\). A1: \(9+h\) with no errors, not originating from calculus. |
| (c) States as \(h \to 0\), Gradient \(PQ \to 9\) = Gradient of tangent | B1 | Must have achieved \(9+h\) in (b) and 9 in (a). Reference to "limit" or "\(h\) tends to 0" linked to part (a). |
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** Attempts $\frac{dy}{dx} = 2x+3$ at $x=3$; gradient of tangent $= 9$ | M1 A1 | M1: Attempts to find value of $\left(\frac{dy}{dx}=\right) ax+3$, $a>0$ at $x=3$. Look for 3 substituted and proceeding to a value. A1: 9 (answer only scores both marks). |
| **(b)** $y_Q = (3+h)^2 + 3(3+h) - 2$ | B1 | Seen or implied. |
| Gradient $PQ = \frac{(3+h)^2+3(3+h)-2-16}{3+h-3} = \frac{9h+h^2}{h} = 9+h$ | M1 A1 | M1: Attempts $\pm\frac{y_Q - 16}{x_Q - 3}$, condoning slips, genuine attempt at $y_Q$. A1: $9+h$ with no errors, not originating from calculus. |
| **(c)** States as $h \to 0$, Gradient $PQ \to 9$ = Gradient of tangent | B1 | Must have achieved $9+h$ in (b) and 9 in (a). Reference to "limit" or "$h$ tends to 0" linked to part (a). |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{28839dd5-b9c1-4cbd-981e-8f79c43ba086-06_652_654_269_646}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $y = x ^ { 2 } + 3 x - 2$
The point $P ( 3,16 )$ 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 $3 + h$ also lies on the curve.
\item Find, in terms of $h$, the gradient of the line $P Q$. Write 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 2020 Q3 [6]}}