| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Complete numerical table |
| Difficulty | Moderate -0.8 This is a straightforward introduction to differentiation from first principles with highly scaffolded steps: (a) requires simple arithmetic substitution into a partially completed table, (b) asks for conceptual understanding that's directly illustrated by the given data, and (c) is routine differentiation using the power rule. The question guides students through the limiting process without requiring them to manipulate the difference quotient algebraically. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation1.07h Differentiation from first principles: for sin(x) and cos(x)1.07i Differentiate x^n: for rational n and sums |
| \(x\) for P | \(y\) for P | \(h\) | \(x\) for Q | \(y\) for Q | change in \(y\) | gradient PQ |
| 1 | 1 | 1 | ||||
| 1 | 1 | 0.1 | 1.1 | 1.048809 | 0.048809 | 0.488088 |
| 1 | 1 | 0.01 | 1.01 | 1.004988 | 0.004988 | 0.498756 |
| 1 | 1 | 0.001 | 1.001 | 1.000500 | 0.000500 | 0.499875 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 2 \ | 1.414214 \ | 0.414214 \ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The limit of the sequence of gradients as \(h\) tends to zero is the gradient of (the tangent to) the curve. (The sequence of gradients tends to 0.5.) | B1 [1] (AO 2.4) | Must communicate the idea of a limit as \(h\) tends to zero but need not be expressed in that way. Do not allow "as \(h\) decreases, gradient increases" without "towards 0.5" or "towards a limit" or "towards the gradient of the curve/tangent" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\) | M1 (AO 1.1a) | Differentiating |
| When \(x=1\), \(\frac{dy}{dx} = \frac{1}{2}\times1^{-\frac{1}{2}} = \frac{1}{2}\) | A1 [2] (AO 1.1) |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 2 \| 1.414214 \| 0.414214 \| 0.414214 | B1 [1] (AO 1.1a) | cao Must be to 6 dp; condone truncated to 6 dp (1.414213 etc.) |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The limit of the sequence of gradients as $h$ tends to zero is the gradient of (the tangent to) the curve. (The sequence of gradients tends to 0.5.) | B1 [1] (AO 2.4) | Must communicate the idea of a limit as $h$ tends to zero but need not be expressed in that way. Do not allow "as $h$ decreases, gradient increases" without "towards 0.5" or "towards a limit" or "towards the gradient of the curve/tangent" |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ | M1 (AO 1.1a) | Differentiating |
| When $x=1$, $\frac{dy}{dx} = \frac{1}{2}\times1^{-\frac{1}{2}} = \frac{1}{2}$ | A1 [2] (AO 1.1) | |
---5 Fig. 5.1 shows part of the curve $y = x ^ { \frac { 1 } { 2 } }$. P is the point $( 1,1 )$ and $Q$ is the point on the curve with $x$-coordinate $1 + h$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a1b6c827-7d74-4527-9b60-58872e3d5ef7-4_451_611_991_242}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{figure}
Table 5.2 shows, for different values of $h$, the coordinates of P , the coordinates of Q , the change in $y$ from P to Q and the gradient of the chord PQ .
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
$x$ for P & $y$ for P & $h$ & $x$ for Q & $y$ for Q & change in $y$ & gradient PQ \\
\hline
1 & 1 & 1 & & & & \\
\hline
1 & 1 & 0.1 & 1.1 & 1.048809 & 0.048809 & 0.488088 \\
\hline
1 & 1 & 0.01 & 1.01 & 1.004988 & 0.004988 & 0.498756 \\
\hline
1 & 1 & 0.001 & 1.001 & 1.000500 & 0.000500 & 0.499875 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 5.2}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Fill in the missing values for the case $h = 1$ in the copy of Table 5.2 in the Printed Answer Booklet. Give your answers correct to 6 decimal places where necessary.
\item Explain how the sequence of values in the last column of Table 5.2 relates to the gradient of the curve $y = x ^ { \frac { 1 } { 2 } }$ at the point $P$.
\item Use calculus to find the gradient of the curve at the point P .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q5 [4]}}