| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| 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 | Standard +0.3 This is a straightforward integration question requiring reverse chain rule for the term -2/(x-1)^3, finding a constant using initial conditions, then basic tangent line work and algebraic verification. All techniques are standard P1 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([y =] \{x\}\{+(x-1)^{-2}\}\{+c\}\) | B1 B1 | May be unsimplified |
| Sub \(x=0\), \(y=3\) leading to \(3 = 0 + 1 + c\) | M1 | Substitution into an integral, expect \(c = 2\) |
| \(y = x + (x-1)^{-2} + 2\) or \(f(x) = x + (x-1)^{-2} + 2\) | A1 | \(\dfrac{-2}{(-2)(x-1)^2}\) or \(\dfrac{-2(x-1)^{-2}}{-2}\) must be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Gradient of tangent \(=\)] \(f'(0) = 3\) | B1 | |
| Equation of tangent is \(y - 3 =\) *their* gradient at \(x=0(x-0)\) | M1* | Expect \(y = 3x+3\), normal gets M0 |
| Intersection given by \(3x + 3 = x + (x-1)^{-2} + 2\) | DM1 | FT *their* equation from part (a) |
| \(2x+1 = \dfrac{1}{(x-1)^2} \rightarrow (2x+1)(x-1)^2 - 1 = 0\) or solve equation before given form reached and show solution \(x = 3/2\) satisfies given result | A1 | WWW AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(x = \dfrac{3}{2}\) leading to \((2x+1)(x-1)^2 - 1\) leading to \(4 \times \frac{1}{4} - 1 = 0\). Hence \(x = \dfrac{3}{2}\). If shown in (b) must be referenced here | B1 | Evaluation of each bracket must be shown. Allow \(\left(\dfrac{1}{2}\right)^2\) for second bracket. Solution of \((2x+1)(x-1)^2 - 1 = 0\) is acceptable |
| When \(x = \dfrac{3}{2}\), \(y = 7\frac{1}{2}\) | B1 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[y =] \{x\}\{+(x-1)^{-2}\}\{+c\}$ | B1 B1 | May be unsimplified |
| Sub $x=0$, $y=3$ leading to $3 = 0 + 1 + c$ | M1 | Substitution into an integral, expect $c = 2$ |
| $y = x + (x-1)^{-2} + 2$ or $f(x) = x + (x-1)^{-2} + 2$ | A1 | $\dfrac{-2}{(-2)(x-1)^2}$ or $\dfrac{-2(x-1)^{-2}}{-2}$ must be simplified |
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Gradient of tangent $=$] $f'(0) = 3$ | B1 | |
| Equation of tangent is $y - 3 =$ *their* gradient at $x=0(x-0)$ | M1* | Expect $y = 3x+3$, normal gets M0 |
| Intersection given by $3x + 3 = x + (x-1)^{-2} + 2$ | DM1 | FT *their* equation from part (a) |
| $2x+1 = \dfrac{1}{(x-1)^2} \rightarrow (2x+1)(x-1)^2 - 1 = 0$ or solve equation before given form reached and show solution $x = 3/2$ satisfies given result | A1 | WWW AG |
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = \dfrac{3}{2}$ leading to $(2x+1)(x-1)^2 - 1$ leading to $4 \times \frac{1}{4} - 1 = 0$. Hence $x = \dfrac{3}{2}$. If shown in **(b)** must be referenced here | B1 | Evaluation of each bracket must be shown. Allow $\left(\dfrac{1}{2}\right)^2$ for second bracket. Solution of $(2x+1)(x-1)^2 - 1 = 0$ is acceptable |
| When $x = \dfrac{3}{2}$, $y = 7\frac{1}{2}$ | B1 | |9 A curve which passes through $( 0,3 )$ has equation $y = \mathrm { f } ( x )$. It is given that $\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 2 } { ( x - 1 ) ^ { 3 } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the curve.\\
The tangent to the curve at $( 0,3 )$ intersects the curve again at one other point, $P$.
\item Show that the $x$-coordinate of $P$ satisfies the equation $( 2 x + 1 ) ( x - 1 ) ^ { 2 } - 1 = 0$.
\item Verify that $x = \frac { 3 } { 2 }$ satisfies this equation and hence find the $y$-coordinate of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q9 [10]}}