| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Identify faulty reasoning |
| Difficulty | Moderate -0.5 This question tests understanding of differentiation from first principles with scaffolding. Part (a) requires identifying the error of substituting h=0 directly (standard misconception), part (b) is routine application of the limit process for a simple quadratic, and part (c) is standard normal line calculation. The conceptual insight needed is minimal as the faulty reasoning is clearly highlighted, making this easier than average. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{0}{0}\) is not equal to 1 | B1 | Statement identifying a fault — should be equivalent to one of these statements. Do not allow \(\frac{0}{0} = 0\) or \(\frac{0}{0} = 6\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lim_{h \to 0}\left(\frac{f(3+h)-f(3)}{h}\right)\) | M1 | Replaces \(h=0\) with the idea of limit for an expression which is not just \(\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)\); Allow for value 3 used or this function i.e. \(\lim_{h\to0}\left(\frac{(x+h)^2 - x^2}{h}\right)\) |
| Gradient of chord is \(6 + h\) | M1 | Simplifies the fraction |
| Gradient of the curve is 6 | A1 | Must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of the normal is \(-\frac{1}{6}\) | M1 | Allow for \(y = -\frac{1}{6}x + c\) seen even if their \(c\) is incorrect; FT their (b); Allow for a value found by an attempt to differentiate \(y = x^2\) not from first principles |
| Equation of the normal is \(y - 9 = -\frac{1}{6}(x-3)\) | A1 | Any form FT their (b); ISW; \(x + 6y = 57\) |
# Question 12:
## Part (a):
Letting $h = 0$ does not make sense as this has the two points coincident
Dividing by zero is not allowed
$\frac{0}{0}$ is not well defined
$\frac{0}{0}$ is not equal to 1 | **B1** | Statement identifying a fault — should be equivalent to one of these statements. Do not allow $\frac{0}{0} = 0$ or $\frac{0}{0} = 6$ etc
## Part (b):
$\lim_{h \to 0}\left(\frac{f(3+h)-f(3)}{h}\right)$ | **M1** | Replaces $h=0$ with the idea of limit for an expression which is not just $\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$; Allow for value 3 used or this function i.e. $\lim_{h\to0}\left(\frac{(x+h)^2 - x^2}{h}\right)$
Gradient of chord is $6 + h$ | **M1** | Simplifies the fraction
Gradient of the curve is 6 | **A1** | Must be seen
## Part (c):
Gradient of the normal is $-\frac{1}{6}$ | **M1** | Allow for $y = -\frac{1}{6}x + c$ seen even if their $c$ is incorrect; FT their (b); Allow for a value found by an attempt to differentiate $y = x^2$ not from first principles
Equation of the normal is $y - 9 = -\frac{1}{6}(x-3)$ | **A1** | Any form FT their (b); ISW; $x + 6y = 57$
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Please share the actual mark scheme pages if you'd like me to extract and format the marking content.12 Below is a faulty argument that appears to show that the gradient of the curve $y = x ^ { 2 }$ at the point $( 3,9 )$ is 1 .
Consider the chord joining $( 3,9 )$ to the point $\left( 3 + h , ( 3 + h ) ^ { 2 } \right)$\\
The gradient is $\frac { ( 3 + h ) ^ { 2 } - 9 } { h } = \frac { 6 h + h ^ { 2 } } { h }$\\
When $h = 0$ the gradient is $\frac { 0 } { 0 }$ so the gradient of the curve is 1
\begin{enumerate}[label=(\alph*)]
\item Identify a fault in the argument.
\item Write a valid first principles argument leading to the correct value for the gradient at (3, 9).
\item Find the equation of the normal to the curve at the point ( 3,9 ).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2022 Q12 [6]}}