| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (straightforward integration + point) |
| Difficulty | Easy -1.2 This is a straightforward integration question requiring only the power law for integration (converting √x to x^(1/2), integrating to get (2/3)x^(3/2), then using the given point to find the constant). It's a single-step technique with no problem-solving required, making it easier than average but not trivial since students must handle fractional powers correctly. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
5 The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = 12 \sqrt { x }$. The curve passes through the point (4,50). Find the equation of the curve.
\hfill \mbox{\textit{OCR C2 2008 Q5 [6]}}