| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (extended problem with normals, stationary points, or further geometry) |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring standard power rule integration (including negative powers) and using a point to find the constant of integration. Part (ii) involves basic differentiation and finding a minimum by setting the second derivative to zero. All techniques are routine with no problem-solving insight required, making it easier than average but not trivial due to the two-part structure. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{x^2}{2} - \frac{4}{x} + (c)\) | B1 B1 | co.co (ignore \(+c\) at this stage) |
| Uses \((4, 8) \rightarrow c = 1\) | M1 A1 [4] | Uses the point after integration for \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 1 - \frac{8}{x^3}\) | B1 | Co |
| \(= 0\) when \(x = 2\) | B1 | Sets to \(0\) + solution or verifies and states a conclusion (stationary or min) |
| \(\rightarrow\) gradient of \(3\) | B1 | Allow for \(x = 2\) into \(dy/dx\) |
| \(\frac{d}{dx}(1 - \frac{8}{x^3}) = \frac{24}{x^4} \rightarrow +\text{ve} \rightarrow\) Min. | B1 [4] | Any valid method - 3rd differential \(+\)ve; 2nd diff goes \(-0+\), or 1st goes \(>3, 3, >3\) |
# Question 10:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{x^2}{2} - \frac{4}{x} + (c)$ | B1 B1 | co.co (ignore $+c$ at this stage) |
| Uses $(4, 8) \rightarrow c = 1$ | M1 A1 [4] | Uses the point after integration for $c$ |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = 1 - \frac{8}{x^3}$ | B1 | Co |
| $= 0$ when $x = 2$ | B1 | Sets to $0$ + solution or verifies and states a conclusion (stationary or min) |
| $\rightarrow$ gradient of $3$ | B1 | Allow for $x = 2$ into $dy/dx$ |
| $\frac{d}{dx}(1 - \frac{8}{x^3}) = \frac{24}{x^4} \rightarrow +\text{ve} \rightarrow$ Min. | B1 [4] | Any valid method - 3rd differential $+$ve; 2nd diff goes $-0+$, or 1st goes $>3, 3, >3$ |
---10 A curve is defined for $x > 0$ and is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = x + \frac { 4 } { x ^ { 2 } }$. The point $P ( 4,8 )$ lies on the curve.\\
(i) Find the equation of the curve.\\
(ii) Show that the gradient of the curve has a minimum value when $x = 2$ and state this minimum value.
\hfill \mbox{\textit{CAIE P1 2012 Q10 [8]}}