| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve from second derivative |
| Difficulty | Moderate -0.8 This is a straightforward double integration problem with given boundary conditions. Students integrate f''(x) = 4x^(-1/2) - 3 twice using standard power rules, then apply two conditions to find constants. Part (a) is trivial given f'(4)=5. The only mild challenge is careful arithmetic with the constants, but the method is completely routine for P1/C1 level. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts \(y - \frac{32}{3} = 5(x-4) \Rightarrow y = 5x - \frac{28}{3}\) | M1 | Uses gradient of 5 at point \(P\left(4, \frac{32}{3}\right)\) to form tangent equation |
| \(y = 5x - \frac{28}{3}\) | A1 | Accept recurring decimal, but \(y = 5x - 9.33\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(f''(x) = \frac{4}{\sqrt{x}} - 3 \Rightarrow f'(x) = 8x^{\frac{1}{2}} - 3x + k\) | M1 | Attempts to integrate \(\frac{4}{\sqrt{x}} - 3\) with one index correct |
| \(f'(x) = \frac{4}{\sqrt{x}} - 3 \rightarrow 8x^{\frac{1}{2}} - 3x + k\) | A1 | With or without the \(+k\) |
| Substitutes \(x = 4,\ f'(x) = 5 \Rightarrow k = 1\) | dM1 | Substitutes \(x=4,\ f'(x)=5\) into integrated form (with \(+k\)) to find value of \(k\) |
| \(f'(x) = 8x^{\frac{1}{2}} - 3x + 1\) | A1 | Which may be implied (allow if \(k=1\) found following correct integral with \(k\)) |
| \(f'(x) = 8x^{\frac{1}{2}} - 3x + 1 \Rightarrow f(x) = \frac{16}{3}x^{\frac{3}{2}} - \frac{3}{2}x^2 + x + d\) | dM1 | Dependent on first M; integrating again with one term correct |
| \(f(x) = \frac{16}{3}x^{\frac{3}{2}} - \frac{3}{2}x^2 + kx + d\) | A1 | Following through on \(k \neq 0\); with or without \(d\) |
| Substitutes \(x = 4,\ f(x) = \frac{32}{3} \Rightarrow d = -12\) | ddM1 | Dependent on 1st and 3rd M's (integrated twice) and both \(k\) and \(d\) added and processed correctly. Note "\(cx + c\)" scores M0 |
| \(f(x) = \frac{16}{3}x^{\frac{3}{2}} - \frac{3}{2}x^2 + x - 12\) | A1 | \(x\sqrt{x}\) instead of \(x^{\frac{3}{2}}\) is fine |
## Question 11:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts $y - \frac{32}{3} = 5(x-4) \Rightarrow y = 5x - \frac{28}{3}$ | M1 | Uses gradient of 5 at point $P\left(4, \frac{32}{3}\right)$ to form tangent equation |
| $y = 5x - \frac{28}{3}$ | A1 | Accept recurring decimal, but $y = 5x - 9.33$ is A0 |
**(2 marks)**
---
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $f''(x) = \frac{4}{\sqrt{x}} - 3 \Rightarrow f'(x) = 8x^{\frac{1}{2}} - 3x + k$ | M1 | Attempts to integrate $\frac{4}{\sqrt{x}} - 3$ with one **index** correct |
| $f'(x) = \frac{4}{\sqrt{x}} - 3 \rightarrow 8x^{\frac{1}{2}} - 3x + k$ | A1 | With or without the $+k$ |
| Substitutes $x = 4,\ f'(x) = 5 \Rightarrow k = 1$ | dM1 | Substitutes $x=4,\ f'(x)=5$ into integrated form (with $+k$) to find value of $k$ |
| $f'(x) = 8x^{\frac{1}{2}} - 3x + 1$ | A1 | Which may be implied (allow if $k=1$ found following correct integral with $k$) |
| $f'(x) = 8x^{\frac{1}{2}} - 3x + 1 \Rightarrow f(x) = \frac{16}{3}x^{\frac{3}{2}} - \frac{3}{2}x^2 + x + d$ | dM1 | Dependent on first M; integrating again with one **term** correct |
| $f(x) = \frac{16}{3}x^{\frac{3}{2}} - \frac{3}{2}x^2 + kx + d$ | A1 | Following through on $k \neq 0$; with or without $d$ |
| Substitutes $x = 4,\ f(x) = \frac{32}{3} \Rightarrow d = -12$ | ddM1 | Dependent on 1st and 3rd M's (integrated twice) **and** both $k$ and $d$ added and processed correctly. Note "$cx + c$" scores M0 |
| $f(x) = \frac{16}{3}x^{\frac{3}{2}} - \frac{3}{2}x^2 + x - 12$ | A1 | $x\sqrt{x}$ instead of $x^{\frac{3}{2}}$ is fine |
**(8 marks)**
**Total: 10 marks**\begin{enumerate}
\item A curve has equation $y = \mathrm { f } ( x )$.
\end{enumerate}
The point $P \left( 4 , \frac { 32 } { 3 } \right)$ lies on the curve.\\
Given that
\begin{itemize}
\item $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 4 } { \sqrt { x } } - 3$
\item $\quad \mathrm { f } ^ { \prime } ( x ) = 5$ at $P$\\
find\\
(a) the equation of the tangent to the curve at $P$, writing your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found,\\
(b) $\mathrm { f } ( x )$.\\
\end{itemize}
\hfill \mbox{\textit{Edexcel P1 2019 Q11 [10]}}