| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring two successive integrations of a polynomial and using given conditions to find constants. Part (a) requires recognizing that d²y/dx² > 0 everywhere implies a minimum. Part (b) is routine integration with boundary conditions. While multi-part, it involves only standard techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks |
|---|---|
| \(3x^2 + 2 > 0\) for all values of \(x\) therefore stationary point is a minimum | B1 (AO 2.4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y' = \int(3x^2 + 2)\,dx = x^3 + 2x + k\) | M1* (AO 2.1) | Attempt to integrate (at least one of the terms in \(x\) correct); condone with no \(+k\) |
| \(y' = 0\) at \(x = -1\); \(\Rightarrow (-1)^3 + 2(-1) + k = 0\) leading to \(k = \ldots\) | M1dep* (AO 1.1) | Uses correct conditions to find the value of \(k\) (candidates may use the fact that when \(x = 0\), \(y' = 3\)); if correct \(k = 3\) |
| \(y = \int(x^3 + 2x + {'}3{'}) \,dx = \tfrac{1}{4}x^4 + x^2 + {'}3{'}x + c\) | M1 (AO 1.1) | Integrates their \(y'\) correctly (allow with \(k = 0\)); condone with no \(+c\) (allow use of same letter for second constant) |
| \(\left(-1, \tfrac{1}{4}\right) \Rightarrow \tfrac{1}{4}(-1)^4 + (-1)^2 + {'}3{'}(-1) + c = \tfrac{1}{4}\) leading to \(c = \ldots\) | M1 (AO 1.1) | Uses correct conditions to find the value of \(c\) |
| \(y = \tfrac{1}{4}x^4 + x^2 + 3x + 2\) | A1 (AO 2.5) | cao (must include \(y =\)) |
## Question 6(a):
$3x^2 + 2 > 0$ for all values of $x$ therefore stationary point is a minimum | B1 (AO 2.4) |
**Total: [1]**
## Question 6(b):
$y' = \int(3x^2 + 2)\,dx = x^3 + 2x + k$ | M1* (AO 2.1) | Attempt to integrate (at least one of the terms in $x$ correct); condone with no $+k$
$y' = 0$ at $x = -1$; $\Rightarrow (-1)^3 + 2(-1) + k = 0$ leading to $k = \ldots$ | M1dep* (AO 1.1) | Uses correct conditions to find the value of $k$ (candidates may use the fact that when $x = 0$, $y' = 3$); if correct $k = 3$
$y = \int(x^3 + 2x + {'}3{'}) \,dx = \tfrac{1}{4}x^4 + x^2 + {'}3{'}x + c$ | M1 (AO 1.1) | Integrates their $y'$ correctly (allow with $k = 0$); condone with no $+c$ (allow use of same letter for second constant)
$\left(-1, \tfrac{1}{4}\right) \Rightarrow \tfrac{1}{4}(-1)^4 + (-1)^2 + {'}3{'}(-1) + c = \tfrac{1}{4}$ leading to $c = \ldots$ | M1 (AO 1.1) | Uses correct conditions to find the value of $c$
$y = \tfrac{1}{4}x^4 + x^2 + 3x + 2$ | A1 (AO 2.5) | cao (must include $y =$)
**Total: [5]**
---6 A curve $C$ has an equation which satisfies $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 3 x ^ { 2 } + 2$, for all values of $x$.
\begin{enumerate}[label=(\alph*)]
\item It is given that $C$ has a single stationary point. Determine the nature of this stationary point.
The diagram shows the graph of the gradient function for $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-04_702_442_1672_242}
\item Given that $C$ passes through the point $\left( - 1 , \frac { 1 } { 4 } \right)$, find the equation of $C$ in the form $y = \mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q6 [6]}}