| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| 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 This is a straightforward integration question requiring reverse chain rule for part (b) and solving a simple equation in part (a). Part (a) involves setting f'(a) = -16/27 and solving (a+2)^4 = 81/16, which is routine algebra. Part (b) is standard integration of (x+2)^{-4} followed by using a boundary condition. Both parts are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{-3}{(a+2)^4} = -\frac{16}{27} \rightarrow\) e.g. \(16(a+2)^4 = 81\) | M1 | Equate first derivative and \(-\frac{16}{27}\) and move term in \(a\) (or \(x\)) into the numerator. |
| \(\rightarrow (a+2)^2 = \frac{9}{4} \rightarrow a + 2 = [\pm]\frac{3}{2}\) | M1 | Solve for \((a+2)\) or \((x+2)\). |
| \(a = -\frac{1}{2}\) or \(-\frac{7}{2}\) | A1 A1 | Allow '\(x =\)' |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\text{f}(x)\right] = \frac{1}{(x+2)^3}\ [+c]\) | B1 | Allow unsimplified form and '\(y =\)' |
| \(5 = 1 + c\) | M1 | Sub \(x = -1\), \(y = 5\) into an integral. |
| \(\left[\text{f}(x)\right] = \frac{1}{(x+2)^3} + 4\) | A1 | Allow '\(y =\)' |
## Question 7:
### Part 7(a):
$\frac{-3}{(a+2)^4} = -\frac{16}{27} \rightarrow$ e.g. $16(a+2)^4 = 81$ | **M1** | Equate first derivative and $-\frac{16}{27}$ and move term in $a$ (or $x$) into the numerator.
$\rightarrow (a+2)^2 = \frac{9}{4} \rightarrow a + 2 = [\pm]\frac{3}{2}$ | **M1** | Solve for $(a+2)$ or $(x+2)$.
$a = -\frac{1}{2}$ or $-\frac{7}{2}$ | **A1 A1** | Allow '$x =$'
### Part 7(b):
$\left[\text{f}(x)\right] = \frac{1}{(x+2)^3}\ [+c]$ | **B1** | Allow unsimplified form and '$y =$'
$5 = 1 + c$ | **M1** | Sub $x = -1$, $y = 5$ into an integral.
$\left[\text{f}(x)\right] = \frac{1}{(x+2)^3} + 4$ | **A1** | Allow '$y =$'
---7 The curve $y = \mathrm { f } ( x )$ is such that $\mathrm { f } ^ { \prime } ( x ) = \frac { - 3 } { ( x + 2 ) ^ { 4 } }$.
\begin{enumerate}[label=(\alph*)]
\item The tangent at a point on the curve where $x = a$ has gradient $- \frac { 16 } { 27 }$.
Find the possible values of $a$.
\item Find $\mathrm { f } ( x )$ given that the curve passes through the point $( - 1,5 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q7 [7]}}