| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Tangent with given gradient/property |
| Difficulty | Standard +0.8 This question requires finding tangency conditions by equating curve and line equations with their derivatives, then solving a cubic equation for the normal intersection. It involves multiple conceptual steps (tangency condition, normal gradient, solving higher-order equations) beyond routine differentiation, making it moderately challenging but still within standard A-level scope. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mx + c = -\frac{m}{x} \Rightarrow mx^2 + cx + m = 0\) | M1 | All \(x\) terms in the numerator. OE e.g. \(mx^2 + cx = -m\). |
| \(b^2 - 4ac = 0 \Rightarrow c^2 - 4m^2 = 0\) | M1 | OE \(b^2-4ac=0\) is implied by \(c^2-4m^2=0\). |
| \(c = [\pm]\ 2m\) | A1 | SOI. Allow \(\pm\) at this stage. |
| \(mx^2\ [\pm]2mx + m = 0 \Rightarrow x^2[\pm]2x+1=0\) | M1 | Sub \(c=+2m\). Ignore substitution of \(-2m\). |
| \((x+1)^2 = 0 \Rightarrow x = -1\) only | A1 | |
| \(y = m\) only or \((-1,\ m)\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{m}{x^2}\) | M1 | As this is a method mark a sign error is allowed. |
| \(\frac{m}{x^2} = m \Rightarrow x^2 = 1\) | M1 A1 | Equating *their* \(\frac{dy}{dx}\) and \(m\) and attempt to solve. |
| \(x = \pm 1\) or \(x = -1\) | A1 | If \(x=-1\) and \(y=m\) are the only answers offered here award the final M1 A1. |
| Selecting \(x=-1\) as the only answer and attempt to find \(y\) | M1 | |
| \(y = m\) or \((-1,\ m)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equation of normal is \(y - m = \frac{-1}{m}(x+1)\) | *M1 | Through *their* \(P\) with gradient \(\frac{-1}{m}\). OE e.g. \(y = \frac{-1}{m}x + \frac{m^2-1}{m}\). Allow use of gradient of curve as \(-\dfrac{1}{\left[\frac{m}{(their\ x)^2}\right]}\) with *their* P. Coordinates of P must be in terms of \(m\) only. |
| \(\frac{-x}{m} - \frac{1}{m} + m = \frac{-m}{x} \Rightarrow x^2 + x(1-m^2) - m^2\ [=0]\) | DM1 | OE. Equating *their* normal equation to the equation of the curve and removing \(x\) from the denominator. |
| \((x+1)(x-m^2)\ [=0] \Rightarrow x = m^2\) | A1 | or \(x = \frac{m^2-1 \pm \sqrt{1-2m^2+m^4+4m^2}}{2} = \frac{m^2-1\pm(m^2+1)}{2} = m^2\) |
| \(y = \frac{-m}{m^2} = \frac{-1}{m}\) | A1 | or \(\left(m^2,\ \frac{-1}{m}\right)\), ignore the coordinates of P. |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mx + c = -\frac{m}{x} \Rightarrow mx^2 + cx + m = 0$ | M1 | All $x$ terms in the numerator. OE e.g. $mx^2 + cx = -m$. |
| $b^2 - 4ac = 0 \Rightarrow c^2 - 4m^2 = 0$ | M1 | OE $b^2-4ac=0$ is implied by $c^2-4m^2=0$. |
| $c = [\pm]\ 2m$ | A1 | SOI. Allow $\pm$ at this stage. |
| $mx^2\ [\pm]2mx + m = 0 \Rightarrow x^2[\pm]2x+1=0$ | M1 | Sub $c=+2m$. Ignore substitution of $-2m$. |
| $(x+1)^2 = 0 \Rightarrow x = -1$ only | A1 | |
| $y = m$ only or $(-1,\ m)$ only | A1 | |
**Alternative method to question 11(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{m}{x^2}$ | M1 | As this is a method mark a sign error is allowed. |
| $\frac{m}{x^2} = m \Rightarrow x^2 = 1$ | M1 A1 | Equating *their* $\frac{dy}{dx}$ and $m$ and attempt to solve. |
| $x = \pm 1$ or $x = -1$ | A1 | If $x=-1$ and $y=m$ are the only answers offered here award the final M1 A1. |
| Selecting $x=-1$ as the only answer and attempt to find $y$ | M1 | |
| $y = m$ or $(-1,\ m)$ | A1 | |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equation of normal is $y - m = \frac{-1}{m}(x+1)$ | *M1 | Through *their* $P$ with gradient $\frac{-1}{m}$. OE e.g. $y = \frac{-1}{m}x + \frac{m^2-1}{m}$. Allow use of gradient of curve as $-\dfrac{1}{\left[\frac{m}{(their\ x)^2}\right]}$ with *their* P. Coordinates of P must be in terms of $m$ only. |
| $\frac{-x}{m} - \frac{1}{m} + m = \frac{-m}{x} \Rightarrow x^2 + x(1-m^2) - m^2\ [=0]$ | DM1 | OE. Equating *their* normal equation to the equation of the curve and removing $x$ from the denominator. |
| $(x+1)(x-m^2)\ [=0] \Rightarrow x = m^2$ | A1 | or $x = \frac{m^2-1 \pm \sqrt{1-2m^2+m^4+4m^2}}{2} = \frac{m^2-1\pm(m^2+1)}{2} = m^2$ |
| $y = \frac{-m}{m^2} = \frac{-1}{m}$ | A1 | or $\left(m^2,\ \frac{-1}{m}\right)$, ignore the coordinates of P. |11 The point $P$ lies on the line with equation $y = m x + c$, where $m$ and $c$ are positive constants. A curve has equation $y = - \frac { m } { x }$. There is a single point $P$ on the curve such that the straight line is a tangent to the curve at $P$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $P$, giving the $y$-coordinate in terms of $m$.\\
The normal to the curve at $P$ intersects the curve again at the point $Q$.
\item Find the coordinates of $Q$ in terms of $m$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q11 [10]}}