| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find constant using stationary point |
| Difficulty | Standard +0.3 This is a straightforward multi-part calculus question requiring chain rule differentiation, solving f'(2)=0 for k, using f''(x) to determine sign constraints, integration to find f(x), and finding/classifying another stationary point. All techniques are standard P1/AS-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f''(x) = -\left(\frac{1}{2}x + k\right)^{-3}\) | B1 | |
| \(f''(2) > 0 \Rightarrow -(1+k)^{-3} > 0\) | M1 | Allow for solving *their* \(f''(2) > 0\) |
| \(k < -1\) | A1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[f(x) = \int\left(\left(\frac{1}{2}x-3\right)^{-2} - (-2)^{-2}\right)dx =\right] \left\{\frac{\left(\frac{1}{2}x-3\right)^{-1}}{-1\times\frac{1}{2}}\right\}\left\{-\frac{x}{4}\right\}\) | B1 B1 | Allow \(-2\left(\frac{1}{2}x+k\right)^{-1}\) OE for 1st B1 and \(-(1+k)^{-2}x\) OE for 2nd B1 |
| \(3\frac{1}{2} = 1 - \frac{1}{2} + c\) | M1 | Substitute \(x = 2\), \(y = 3\frac{1}{2}\) into *their* integral with \(c\) present |
| \(f(x) = \frac{-2}{\left(\frac{1}{2}x-3\right)} - \frac{x}{4} + 3\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{1}{2}x-3\right)^{-2} - (-2)^{-2} = 0\) | M1 | Substitute \(k = -3\) and set to zero |
| leading to \(\left(\frac{1}{2}x-3\right)^2 = 4\ \left[\frac{1}{2}x-3 = (\pm)2\right]\) leading to \(x = 10\) | A1 | |
| \(\left(10, -\frac{1}{2}\right)\) | A1 | Or when \(x = 10\), \(y = -1 - 2\frac{1}{2} + 3 = -\frac{1}{2}\) |
| \(f''(10)\left[= -(5-3)^{-3} \to\right] < 0 \to\) MAXIMUM | A1 | WWW |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f''(x) = -\left(\frac{1}{2}x + k\right)^{-3}$ | B1 | |
| $f''(2) > 0 \Rightarrow -(1+k)^{-3} > 0$ | M1 | Allow for solving *their* $f''(2) > 0$ |
| $k < -1$ | A1 | WWW |
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[f(x) = \int\left(\left(\frac{1}{2}x-3\right)^{-2} - (-2)^{-2}\right)dx =\right] \left\{\frac{\left(\frac{1}{2}x-3\right)^{-1}}{-1\times\frac{1}{2}}\right\}\left\{-\frac{x}{4}\right\}$ | B1 B1 | Allow $-2\left(\frac{1}{2}x+k\right)^{-1}$ OE for 1st B1 and $-(1+k)^{-2}x$ OE for 2nd B1 |
| $3\frac{1}{2} = 1 - \frac{1}{2} + c$ | M1 | Substitute $x = 2$, $y = 3\frac{1}{2}$ into *their* integral with $c$ present |
| $f(x) = \frac{-2}{\left(\frac{1}{2}x-3\right)} - \frac{x}{4} + 3$ | A1 | OE |
---
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{1}{2}x-3\right)^{-2} - (-2)^{-2} = 0$ | M1 | Substitute $k = -3$ and set to zero |
| leading to $\left(\frac{1}{2}x-3\right)^2 = 4\ \left[\frac{1}{2}x-3 = (\pm)2\right]$ leading to $x = 10$ | A1 | |
| $\left(10, -\frac{1}{2}\right)$ | A1 | Or when $x = 10$, $y = -1 - 2\frac{1}{2} + 3 = -\frac{1}{2}$ |
| $f''(10)\left[= -(5-3)^{-3} \to\right] < 0 \to$ MAXIMUM | A1 | WWW |10 A curve has equation $y = \mathrm { f } ( x )$ and it is given that
$$\mathrm { f } ^ { \prime } ( x ) = \left( \frac { 1 } { 2 } x + k \right) ^ { - 2 } - ( 1 + k ) ^ { - 2 }$$
where $k$ is a constant. The curve has a minimum point at $x = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { f } ^ { \prime \prime } ( x )$ in terms of $k$ and $x$, and hence find the set of possible values of $k$.\\
It is now given that $k = - 3$ and the minimum point is at $\left( 2,3 \frac { 1 } { 2 } \right)$.
\item Find $\mathrm { f } ( x )$.
\item Find the coordinates of the other stationary point and determine its nature.\\
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 2021 Q10 [11]}}