| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Gradient function and tangent/normal |
| Difficulty | Moderate -0.8 This is a structured, multi-part question that guides students through the first principles definition of a derivative using a simple quadratic function. While it has 12 marks total, each part is routine: (i) is basic coordinate substitution, (ii) is algebraic manipulation following a standard template, (iii) applies a limit concept but the answer is given by the previous part, (iv) is standard tangent equation finding, and (v) requires solving two equations and finding a distance. The question requires no novel insight—it's a pedagogical exercise teaching differentiation from first principles with a straightforward quadratic. This is easier than average for A-level as it's highly scaffolded and uses only basic algebraic techniques. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| 4 | i |
| Answer | Marks |
|---|---|
| v | 6.1 |
| Answer | Marks |
|---|---|
| 0.021 cao | 2 |
| Answer | Marks |
|---|---|
| A1 | ( ) ( ) |
| Answer | Marks |
|---|---|
| 6 may be obtained from | 2 |
Question 4:
4 | i
ii
iii
iv
v | 6.1
( ( 3+h )2−7 ) − ( 32 −7 )
h
numerator = 6h + h2
6 + h
as h tends to 0,
grad. tends to 6 o.e. f.t.from “6”+h
y − 2 = “6” (x − 3) o.e.
y = 6x − 16
At P, x = 16/6 o.e. or ft
At Q, x = 7
0.021 cao | 2
M1
M1
A1
M1
A1
M1
A1
M1
M1
A1 | ( ) ( )
3.12 −7 − 32 −7
M1 for o.e.
3.1−3
s.o.i.
6 may be obtained from | 2
3
2
2
3\begin{enumerate}[label=(\roman*)]
\item Calculate the gradient of the chord joining the points on the curve $y = x^2 - 7$ for which $x = 3$ and $x = 3.1$. [2]
\item Given that $f(x) = x^2 - 7$, find and simplify $\frac{f(3 + h) - f(3)}{h}$. [3]
\item Use your result in part (ii) to find the gradient of $y = x^2 - 7$ at the point where $x = 3$, showing your reasoning. [2]
\item Find the equation of the tangent to the curve $y = x^2 - 7$ at the point where $x = 3$. [2]
\item This tangent crosses the $x$-axis at the point P. The curve crosses the positive $x$-axis at the point Q. Find the distance PQ, giving your answer correct to 3 decimal places. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q4 [12]}}