| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find range using calculus |
| Difficulty | Standard +0.3 This is a straightforward calculus optimization problem requiring differentiation of a simple function (power rule only, despite the topic label), solving for a parameter, using the second derivative test, and interpreting the result to find a range. All steps are standard textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{f}'(x) =]\ 2x - \frac{k}{x^2}\) | B1 | |
| \(\text{f}'(2) = 0 \left[2\times2 - \frac{k}{2^2} = 0\right] \Rightarrow k = \ldots\) | M1 | Setting *their* 2-term \(\text{f}'(2) = 0\), at least one term correct and attempting to solve as far as \(k =\) |
| \(k = 16\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{f}''(2) =\) e.g. \(2 + \frac{2k}{2^3}\) | M1 | Evaluate a two term \(\text{f}''(2)\) with at least one term correct. Or other valid method |
| \(\left[2 + \frac{2k}{2^3}\right] > 0 \Rightarrow\) minimum or \(= 6 \Rightarrow\) minimum | A1 FT | WWW. FT on positive \(k\) value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(x = 2\), \(\text{f}(x) = 14\) | B1 | SOI |
| [Range is or \(y\) or \(\text{f}(x)\)] \(\geqslant\) *their* \(\text{f}(2)\) | B1 FT | Not \(x \geqslant\) *their* \(\text{f}(2)\) |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{f}'(x) =]\ 2x - \frac{k}{x^2}$ | B1 | |
| $\text{f}'(2) = 0 \left[2\times2 - \frac{k}{2^2} = 0\right] \Rightarrow k = \ldots$ | M1 | Setting *their* 2-term $\text{f}'(2) = 0$, at least one term correct and attempting to solve as far as $k =$ |
| $k = 16$ | A1 | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{f}''(2) =$ e.g. $2 + \frac{2k}{2^3}$ | M1 | Evaluate a two term $\text{f}''(2)$ with at least one term correct. Or other valid method |
| $\left[2 + \frac{2k}{2^3}\right] > 0 \Rightarrow$ minimum or $= 6 \Rightarrow$ minimum | A1 FT | WWW. FT on positive $k$ value |
---
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $x = 2$, $\text{f}(x) = 14$ | B1 | SOI |
| [Range is or $y$ or $\text{f}(x)$] $\geqslant$ *their* $\text{f}(2)$ | B1 FT | Not $x \geqslant$ *their* $\text{f}(2)$ |10 The function f is defined by $\mathrm { f } ( x ) = x ^ { 2 } + \frac { k } { x } + 2$ for $x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Given that the curve with equation $y = \mathrm { f } ( x )$ has a stationary point when $x = 2$, find $k$.
\item Determine the nature of the stationary point.
\item Given that this is the only stationary point of the curve, find the range of f .
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q10 [7]}}