| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| 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 question requiring expansion of the numerator, simplification to powers of x, term-by-term integration, and using a point to find the constant. Part (a) is routine substitution and tangent line work. While it has multiple parts and requires careful algebra, it follows standard P1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=4 \Rightarrow f'(4) = \frac{(4+3)^2}{4\sqrt{4}} = \frac{49}{8}\) (6.125) | B1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y - 20 = "\frac{49}{8}"(x-4)\) | M1A1ft | Correct straight line method using \(y=20\) and \(x=4\) with their \(f'(4)\); allow one sign error in brackets. If \(y=mx+c\) used must proceed to \(c=...\). Perpendicular gradient is M0. A1ft: correct equation following through on their \(f'(4) \) |
| \(49x - 8y - 36 = 0\) | A1 | Correct equation with all terms on one side (allow any integer multiple) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = \frac{(x+3)^2}{x\sqrt{x}} = \frac{x^2+6x+9}{x\sqrt{x}} = \frac{x^2}{x\sqrt{x}}+\frac{6x}{x\sqrt{x}}+\frac{9}{x\sqrt{x}}\) | M1 | Squares numerator (condone if only \(x^2+9\) terms found) and attempts to split. Score for one correct index from correct working: \(...x^{\frac{1}{2}}\) or \(...x^{-\frac{1}{2}}\) or \(...x^{-\frac{3}{2}}\) |
| \(= x^{\frac{1}{2}} + 6x^{-\frac{1}{2}} + 9x^{-\frac{3}{2}}\) | A1 | These terms may appear on different lines |
| \(f(x) = \frac{2}{3}x^{\frac{3}{2}} + 12x^{\frac{1}{2}} - 18x^{-\frac{1}{2}} + c\) | dM1 A1A1 | dM1: \(...x^{\frac{1}{2}} \to ...x^{\frac{3}{2}}\) or \(...x^{-\frac{1}{2}} \to ...x^{\frac{1}{2}}\) or \(...x^{-\frac{3}{2}} \to ...x^{-\frac{1}{2}}\); dependent on previous M. A1: any 2 correct terms; A1: all three terms \(+c\) (condone lack of \(+c\)) |
| \(20 = \frac{2}{3}(4)^{\frac{3}{2}} + 12(4)^{\frac{1}{2}} - 18(4)^{-\frac{1}{2}} + c \Rightarrow c = ...\) | M1 | Uses \(y=20\) and \(x=4\) in integrated function to find \(c\) (not differentiated or original function) |
| \(f(x) = \frac{2}{3}x^{\frac{3}{2}} + 12x^{\frac{1}{2}} - 18x^{-\frac{1}{2}} - \frac{1}{3}\) | A1 | Accept equivalent simplified forms with exact coefficients; condone \(y=\); isw after simplified correct answer unless they multiply through |
# Question 6:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=4 \Rightarrow f'(4) = \frac{(4+3)^2}{4\sqrt{4}} = \frac{49}{8}$ (6.125) | B1 | Correct value |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y - 20 = "\frac{49}{8}"(x-4)$ | M1A1ft | Correct straight line method using $y=20$ and $x=4$ with their $f'(4)$; allow one sign error in brackets. If $y=mx+c$ used must proceed to $c=...$. Perpendicular gradient is M0. A1ft: correct equation following through on their $f'(4) $ |
| $49x - 8y - 36 = 0$ | A1 | Correct equation with all terms on one side (allow any integer multiple) |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{(x+3)^2}{x\sqrt{x}} = \frac{x^2+6x+9}{x\sqrt{x}} = \frac{x^2}{x\sqrt{x}}+\frac{6x}{x\sqrt{x}}+\frac{9}{x\sqrt{x}}$ | M1 | Squares numerator (condone if only $x^2+9$ terms found) and attempts to split. Score for one correct index from correct working: $...x^{\frac{1}{2}}$ or $...x^{-\frac{1}{2}}$ or $...x^{-\frac{3}{2}}$ |
| $= x^{\frac{1}{2}} + 6x^{-\frac{1}{2}} + 9x^{-\frac{3}{2}}$ | A1 | These terms may appear on different lines |
| $f(x) = \frac{2}{3}x^{\frac{3}{2}} + 12x^{\frac{1}{2}} - 18x^{-\frac{1}{2}} + c$ | dM1 A1A1 | dM1: $...x^{\frac{1}{2}} \to ...x^{\frac{3}{2}}$ or $...x^{-\frac{1}{2}} \to ...x^{\frac{1}{2}}$ or $...x^{-\frac{3}{2}} \to ...x^{-\frac{1}{2}}$; dependent on previous M. A1: any 2 correct terms; A1: all three terms $+c$ (condone lack of $+c$) |
| $20 = \frac{2}{3}(4)^{\frac{3}{2}} + 12(4)^{\frac{1}{2}} - 18(4)^{-\frac{1}{2}} + c \Rightarrow c = ...$ | M1 | Uses $y=20$ and $x=4$ in integrated function to find $c$ (not differentiated or original function) |
| $f(x) = \frac{2}{3}x^{\frac{3}{2}} + 12x^{\frac{1}{2}} - 18x^{-\frac{1}{2}} - \frac{1}{3}$ | A1 | Accept equivalent simplified forms with exact coefficients; condone $y=$; isw after simplified correct answer unless they multiply through |
---6. The curve $C$ has equation $y = \mathrm { f } ( x )$ where $x > 0$
Given that
\begin{itemize}
\item $\mathrm { f } ^ { \prime } ( x ) = \frac { ( x + 3 ) ^ { 2 } } { x \sqrt { x } }$
\item the point $P ( 4,20 )$ lies on $C$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item find the value of the gradient at $P$
\item Hence find the equation of the tangent to $C$ at $P$, giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers to be found.
\end{enumerate}\item Find $\mathrm { f } ( x )$, simplifying your answer.\\
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q6 [11]}}