| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (reverse chain rule / composite functions) |
| Difficulty | Moderate -0.3 Part (a) requires recognizing a reverse chain rule integral and evaluating an improper integral with straightforward limit calculation. Part (b) is a standard integration problem with boundary condition substitution. Both parts use routine techniques with no conceptual challenges beyond basic A-level integration skills. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{-2}{x+2}\) | B1 | Integrate \(f(x)\). Accept \(-2(x+2)^{-1}\). Can be unsimplified. |
| \(0 - \left(-\dfrac{2}{3}\right) = \dfrac{2}{3}\) | M1 A1 | Apply limit(s) to an integrated expansion. CAO for A1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1 = -2 + c\) | M1 | Substitute \(x = -1,\ y = -1\) into *their* integrated expression (\(c\) present) |
| \(y = \dfrac{-2}{x+2} + 1\) | A1 | Accept \(y = -2(x+2)^{-1} + 1\). \(-2\) must be resolved. |
## Question 2:
**Part (a):**
$\dfrac{-2}{x+2}$ | B1 | Integrate $f(x)$. Accept $-2(x+2)^{-1}$. Can be unsimplified.
$0 - \left(-\dfrac{2}{3}\right) = \dfrac{2}{3}$ | M1 A1 | Apply limit(s) to an integrated expansion. CAO for A1.
*Total: 3 marks*
**Part (b):**
$-1 = -2 + c$ | M1 | Substitute $x = -1,\ y = -1$ into *their* integrated expression ($c$ present)
$y = \dfrac{-2}{x+2} + 1$ | A1 | Accept $y = -2(x+2)^{-1} + 1$. $-2$ must be resolved.
*Total: 2 marks*2 The function f is defined by $\mathrm { f } ( x ) = \frac { 2 } { ( x + 2 ) ^ { 2 } }$ for $x > - 2$.
\begin{enumerate}[label=(\alph*)]
\item Find $\int _ { 1 } ^ { \infty } \mathrm { f } ( x ) \mathrm { d } x$.
\item The equation of a curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x )$. It is given that the point $( - 1 , - 1 )$ lies on the curve.
Find the equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q2 [5]}}