| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.3 This is a straightforward integration problem requiring students to differentiate f'(x) to find A using the condition f''(4)=0, then integrate f'(x) and use the boundary condition to find the constant. While it involves multiple steps and careful algebraic manipulation of powers, it follows standard A-level procedures without requiring problem-solving insight or novel approaches—slightly easier than average due to its routine nature. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 2x^{-\frac{1}{2}} + Ax^{-2} + 3 \Rightarrow f''(x) = ...x^{-\frac{3}{2}} + ...x^{-3}\) | M1 | Correct method of differentiation; at least one correct power reduced by 1 |
| \(f''(x) = -x^{-\frac{3}{2}} - 2Ax^{-3}\) | A1 | Correct differentiation; indices and coefficients correctly processed |
| \(f''(4) = 0 \Rightarrow -4^{-\frac{3}{2}} - 2A \times 4^{-3} = 0 \Rightarrow A = ...\) | dM1 | Sets \(f''(4) = 0\); proceeds to find \(A\); dependent on previous M1 |
| \(-\frac{1}{8} - \frac{2A}{64} = 0 \Rightarrow A = -4\) | A1 | \(A = -4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \int 2x^{-\frac{1}{2}} + Ax^{-2} + 3\, dx = \frac{2x^{\frac{1}{2}}}{\frac{1}{2}} + \frac{Ax^{-1}}{-1} + 3x (+c)\) | M1 | Attempts integration; at least one correct power increased by 1 |
| \(= 4x^{\frac{1}{2}} - \frac{A}{x} + 3x (+c)\) | A1ft | Correct integration with \(A\) or follow-through value of \(A\) |
| \(f(12) = 8\sqrt{3} \Rightarrow 4\sqrt{12} - \frac{A}{12} + 36 + c = 8\sqrt{3} \Rightarrow c = ...\) | dM1 | Sets \(f(12) = 8\sqrt{3}\); proceeds to find \(c\); dependent on previous M1 |
| \(c = 8\sqrt{3} - 4\sqrt{12} - 36 - \frac{4}{12} = -\frac{109}{3}\) or follow through \(c = \frac{A}{12} - 36\) | A1ft | Correct value of \(c\); follow through their \(A\); accept awrt 1 dp |
| \(f(x) = 4x^{\frac{1}{2}} + \frac{4}{x} + 3x - \frac{109}{3}\) | A1 | Correct final answer; not follow through; if multiplied by 3 etc, A0 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 2x^{-\frac{1}{2}} + Ax^{-2} + 3 \Rightarrow f''(x) = ...x^{-\frac{3}{2}} + ...x^{-3}$ | M1 | Correct method of differentiation; at least one correct power reduced by 1 |
| $f''(x) = -x^{-\frac{3}{2}} - 2Ax^{-3}$ | A1 | Correct differentiation; indices and coefficients correctly processed |
| $f''(4) = 0 \Rightarrow -4^{-\frac{3}{2}} - 2A \times 4^{-3} = 0 \Rightarrow A = ...$ | dM1 | Sets $f''(4) = 0$; proceeds to find $A$; dependent on previous M1 |
| $-\frac{1}{8} - \frac{2A}{64} = 0 \Rightarrow A = -4$ | A1 | $A = -4$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \int 2x^{-\frac{1}{2}} + Ax^{-2} + 3\, dx = \frac{2x^{\frac{1}{2}}}{\frac{1}{2}} + \frac{Ax^{-1}}{-1} + 3x (+c)$ | M1 | Attempts integration; at least one correct power increased by 1 |
| $= 4x^{\frac{1}{2}} - \frac{A}{x} + 3x (+c)$ | A1ft | Correct integration with $A$ or follow-through value of $A$ |
| $f(12) = 8\sqrt{3} \Rightarrow 4\sqrt{12} - \frac{A}{12} + 36 + c = 8\sqrt{3} \Rightarrow c = ...$ | dM1 | Sets $f(12) = 8\sqrt{3}$; proceeds to find $c$; dependent on previous M1 |
| $c = 8\sqrt{3} - 4\sqrt{12} - 36 - \frac{4}{12} = -\frac{109}{3}$ or follow through $c = \frac{A}{12} - 36$ | A1ft | Correct value of $c$; follow through their $A$; accept awrt 1 dp |
| $f(x) = 4x^{\frac{1}{2}} + \frac{4}{x} + 3x - \frac{109}{3}$ | A1 | Correct final answer; not follow through; if multiplied by 3 etc, A0 |7. The curve $C$ has equation $y = \mathrm { f } ( x ) , x > 0$
Given that
\begin{itemize}
\item $\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \sqrt { x } } + \frac { A } { x ^ { 2 } } + 3$, where $A$ is a constant
\item $\mathrm { f } ^ { \prime \prime } ( x ) = 0$ when $x = 4$
\begin{enumerate}[label=(\alph*)]
\item find the value of $A$.
\end{itemize}
Given also that
\begin{itemize}
\item $\mathrm { f } ( x ) = 8 \sqrt { 3 }$, when $x = 12$
\item find $\mathrm { f } ( x )$, giving each term in simplest form.\\
\end{itemize}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q7 [9]}}