| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding curve equation from derivative |
| Difficulty | Moderate -0.8 This is a straightforward integration problem requiring a simple substitution (u = 1+2x) followed by using a point to find the constant, then solving a basic inequality. All steps are routine P1 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{3(1 + 2x)^{-1}}{-1} + (c)\) | B1 | |
| \(y = \frac{3(1 + 2x)^{-1}}{-2} + (c)\) | B1(indep) | Division by 2 y = necessary |
| Sub (1, (1/2)) \(\frac{1}{2} = \frac{3}{-6} + c \Rightarrow c = 1\) | M1 | Dependent on c present |
| A1 [4] | Use of \(y = mx + c\) etc. gets 0/4 | |
| (ii) \((1 + 2x)^{(>)}9\) or \(4x^2 + 4x - 8(>)0\) OE | M1 | |
| 1, −2 | A1 | |
| \(x > 1, x < -2\) ISW | A1 [3] |
**(i)** $\frac{3(1 + 2x)^{-1}}{-1} + (c)$ | B1 |
$y = \frac{3(1 + 2x)^{-1}}{-2} + (c)$ | B1(indep) | Division by 2 y = necessary
Sub (1, (1/2)) $\frac{1}{2} = \frac{3}{-6} + c \Rightarrow c = 1$ | M1 | Dependent on c present
| A1 [4] | Use of $y = mx + c$ etc. gets 0/4
**(ii)** $(1 + 2x)^{(>)}9$ or $4x^2 + 4x - 8(>)0$ OE | M1 |
1, −2 | A1 |
$x > 1, x < -2$ ISW | A1 [3] |
---7 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { ( 1 + 2 x ) ^ { 2 } }$ and the point $\left( 1 , \frac { 1 } { 2 } \right)$ lies on the curve.\\
(i) Find the equation of the curve.\\
(ii) Find the set of values of $x$ for which the gradient of the curve is less than $\frac { 1 } { 3 }$.
\hfill \mbox{\textit{CAIE P1 2011 Q7 [7]}}