| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (extended problem with normals, stationary points, or further geometry) |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring standard techniques (power rule for x^{1/2}, constant, and x^{-2}) plus finding a normal line. Part (a) involves evaluating the derivative at x=4, finding the negative reciprocal, and writing in the required form. Part (b) is routine integration with a constant found using the given point. All techniques are standard P1/C1 material with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(f'(4) = 4\times2 - 2 - \frac{8}{3\times4^2} = \frac{35}{6}\) | M1 | Award if two of three terms correct e.g. two of \(8-2-``0.166``\); or award for sight of \(\frac{35}{6}\) or awrt 5.83 |
| Gradient of perpendicular \(= -\frac{6}{35}\) | dM1 | Using correct perpendicular gradient rule; M1 must be awarded |
| \(y - 1 = -\frac{6}{35}(x-4) \Rightarrow 6x + 35y - 59 = 0\) | M1A1 | Using point \((4,1)\) with changed gradient; accept \(\pm k(6x+35y-59)=0\) where \(k \in \mathbb{N}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(f'(x) = 4x^{\frac{1}{2}} - 2 - \frac{8}{3x^2} \Rightarrow f(x) = \frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x}(+c)\) | M1 A1 A1 | Integration of each term |
| \(x=4, f(x)=1 \Rightarrow 1 = \frac{8}{3}\times8 - 8 + \frac{2}{3} + c \Rightarrow c = \ldots(-13)\) | dM1 | Substituting \((4,1)\) to find \(c\) |
| \(f(x) = \frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x} - 13\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Raises any correct index by one: \(x^{\frac{1}{2}} \to x^{\frac{3}{2}}\), \(2 \to 2x\), \(\frac{1}{x^2} \to \frac{1}{x}\) | M1 | The indices must be processed |
| Any two terms correct (may be un-simplified) with or without \(+c\) | A1 | Do not allow indices or coefficients left unprocessed e.g. \(4x^{\frac{1}{2}} \to \frac{4}{1+\frac{1}{2}}x^{\frac{1}{2}+1}\) |
| \(\frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x}\) or exact simplified equivalent e.g. \(\frac{8x^{\frac{3}{2}}}{3} - 2x + \frac{8}{3}x^{-1}\) | A1 | All three terms correct and simplified with or without \(+c\). Condone spurious notation. May be implied by final line where \(c\) has been found. |
| Substitute \(x=4\), \(y=1\) into \(f(x)\) containing \(+c\) to obtain \(c\) | dM1 | |
| \(f(x) = \frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x} - 13\) or e.g. \(f(x) = \frac{8}{3}x\sqrt{x} - 2x + \frac{8}{3}x^{-1} - 13\) | A1 | Condone \(f(x)=y\) or allow if no left-hand side. Isw after correct answer. |
## Question 8:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $f'(4) = 4\times2 - 2 - \frac{8}{3\times4^2} = \frac{35}{6}$ | M1 | Award if two of three terms correct e.g. two of $8-2-``0.166``$; or award for sight of $\frac{35}{6}$ or awrt 5.83 |
| Gradient of perpendicular $= -\frac{6}{35}$ | dM1 | Using correct perpendicular gradient rule; M1 must be awarded |
| $y - 1 = -\frac{6}{35}(x-4) \Rightarrow 6x + 35y - 59 = 0$ | M1A1 | Using point $(4,1)$ with changed gradient; accept $\pm k(6x+35y-59)=0$ where $k \in \mathbb{N}$ |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $f'(x) = 4x^{\frac{1}{2}} - 2 - \frac{8}{3x^2} \Rightarrow f(x) = \frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x}(+c)$ | M1 A1 A1 | Integration of each term |
| $x=4, f(x)=1 \Rightarrow 1 = \frac{8}{3}\times8 - 8 + \frac{2}{3} + c \Rightarrow c = \ldots(-13)$ | dM1 | Substituting $(4,1)$ to find $c$ |
| $f(x) = \frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x} - 13$ | A1 | |
## Question (b) - Integration with constant
| Answer/Working | Mark | Guidance |
|---|---|---|
| Raises any correct index by one: $x^{\frac{1}{2}} \to x^{\frac{3}{2}}$, $2 \to 2x$, $\frac{1}{x^2} \to \frac{1}{x}$ | M1 | The indices must be processed |
| Any two terms correct (may be un-simplified) with or without $+c$ | A1 | Do not allow indices or coefficients left unprocessed e.g. $4x^{\frac{1}{2}} \to \frac{4}{1+\frac{1}{2}}x^{\frac{1}{2}+1}$ |
| $\frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x}$ or exact simplified equivalent e.g. $\frac{8x^{\frac{3}{2}}}{3} - 2x + \frac{8}{3}x^{-1}$ | A1 | All three terms correct and simplified with or without $+c$. Condone spurious notation. May be implied by final line where $c$ has been found. |
| Substitute $x=4$, $y=1$ into $f(x)$ containing $+c$ to obtain $c$ | dM1 | |
| $f(x) = \frac{8}{3}x^{\frac{3}{2}} - 2x + \frac{8}{3x} - 13$ or e.g. $f(x) = \frac{8}{3}x\sqrt{x} - 2x + \frac{8}{3}x^{-1} - 13$ | A1 | Condone $f(x)=y$ or allow if no left-hand side. Isw after correct answer. |
---\begin{enumerate}
\item The curve $C$ with equation $y = \mathrm { f } ( x ) , \quad x > 0$, passes through the point $P ( 4,1 )$.
\end{enumerate}
Given that $\mathrm { f } ^ { \prime } ( x ) = 4 \sqrt { x } - 2 - \frac { 8 } { 3 x ^ { 2 } }$\\
(a) find the equation of the normal to $C$ at $P$. Write your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.\\
(4)\\
(b) Find $\mathrm { f } ( x )$.\\
(5)\\
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\hfill \mbox{\textit{Edexcel P1 2019 Q8 [9]}}