| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | January |
| 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 manipulation of powers (rewriting √x as x^(1/2)), application of standard power rule integration, and using a boundary condition to find the constant. Part (a) is simple substitution into the derivative. While it requires multiple steps, all techniques are routine P1/C1 material with no problem-solving insight needed, 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x=4\) into \(\frac{dy}{dx} = 3x\sqrt{x} - 10x^{-\frac{1}{2}} = 3\times4\times2 - \frac{10}{2} = 19\) | M1A1 | Do not award M1 if they attempt to differentiate the expression first (look for \(-10x^{-\frac{1}{2}}\) power decreasing as evidence) |
| Attempts \((y-(-2)) = \text{"19"}\times(x-4) \Rightarrow y = 19x - 78\) | M1A1 cao | At least one bracket correct; if \(y=mx+c\) used must find \(c\) with gradient and \((4,-2)\) substituted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 3x^{\frac{3}{2}} - 10x^{-\frac{1}{2}} \Rightarrow f(x) = \frac{6}{5}x^{\frac{5}{2}} - 20x^{\frac{1}{2}} + c\) | M1 A1 A1 | M1: raises power of any term by one e.g. \(x^{-\frac{1}{2}} \to x^{\frac{1}{2}}\); first A1: any term correct (unsimplified, with or without \(+c\)); second A1: both terms correct (with or without \(+c\)) |
| \(x=4,\ f(x)=-2 \Rightarrow -2 = 38.4 - 40 + c \Rightarrow c = ...(-0.4)\) | M1 | Substitutes \(x=4\), \(y=-2\) into \(f(x)\) containing \(+c\); condone errors in evaluating and rearranging |
| \(\left[f(x)=\right] \frac{6}{5}x^{\frac{5}{2}} - 20x^{\frac{1}{2}} - 0.4\) | A1 cso | Or equivalent including \((y=)...\); cso |
## Question 12(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x=4$ into $\frac{dy}{dx} = 3x\sqrt{x} - 10x^{-\frac{1}{2}} = 3\times4\times2 - \frac{10}{2} = 19$ | M1A1 | Do not award M1 if they attempt to differentiate the expression first (look for $-10x^{-\frac{1}{2}}$ power decreasing as evidence) |
| Attempts $(y-(-2)) = \text{"19"}\times(x-4) \Rightarrow y = 19x - 78$ | M1A1 cao | At least one bracket correct; if $y=mx+c$ used must find $c$ with gradient and $(4,-2)$ substituted | (4 marks) |
## Question 12(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 3x^{\frac{3}{2}} - 10x^{-\frac{1}{2}} \Rightarrow f(x) = \frac{6}{5}x^{\frac{5}{2}} - 20x^{\frac{1}{2}} + c$ | M1 A1 A1 | M1: raises power of any term by one e.g. $x^{-\frac{1}{2}} \to x^{\frac{1}{2}}$; first A1: any term correct (unsimplified, with or without $+c$); second A1: both terms correct (with or without $+c$) |
| $x=4,\ f(x)=-2 \Rightarrow -2 = 38.4 - 40 + c \Rightarrow c = ...(-0.4)$ | M1 | Substitutes $x=4$, $y=-2$ into $f(x)$ containing $+c$; condone errors in evaluating and rearranging |
| $\left[f(x)=\right] \frac{6}{5}x^{\frac{5}{2}} - 20x^{\frac{1}{2}} - 0.4$ | A1 cso | Or equivalent including $(y=)...$; cso | (5 marks) **(9 marks total)** |12. The curve with equation $y = \mathrm { f } ( x ) , x > 0$, passes through the point $P ( 4 , - 2 )$. Given that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x \sqrt { x } - 10 x ^ { - \frac { 1 } { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item find 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 integers to be found.
\item Find $\mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q12 [9]}}