| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve from second derivative |
| Difficulty | Standard +0.3 This is a straightforward integration question requiring two successive integrations with constants of integration determined by given conditions. The second derivative simplifies to 6x^(-3/2) + x, both standard integral forms. Part (a) is routine gradient/normal calculation. Part (b) involves standard integration techniques taught early in C1/P1 with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| VIIIV SIHI NI IIIYM ION OC | VIIV SIHI NI JIHMM ION OO | VI4V SIHI NI JIIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of normal \(= \frac{1}{4}\) | B1 | Deduces gradient of normal is \(\frac{1}{4}\) |
| \((y+50) = \frac{1}{4}(x-4) \Rightarrow y = \frac{1}{4}x - 51\) | M1 A1 | M1: line through \(P(4,-50)\) with changed gradient; allow one sign slip on a coordinate; if \(y=mx+c\) used, at least one coordinate correctly substituted. A1: \(y=\frac{1}{4}x-51\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f''(x) = \frac{6}{\sqrt{x^3}} + x = 6x^{-\frac{3}{2}} + x \Rightarrow f'(x) = -12x^{-\frac{1}{2}} + \frac{1}{2}x^2 + k\) | M1 A1 | M1: integrate with one index correct (\(\ldots x^{-\frac{1}{2}}\) or \(\ldots x^2\)). A1: unsimplified, with or without \(+k\) |
| Substitute \(x=4\), \(f'(x)=-4 \Rightarrow k=-6\) | dM1 A1 | Dependent on first M1; substitute \(x=4\), \(f'(x)=-4\) into integrated form with \(+k\) |
| \(f'(x) = -12x^{-\frac{1}{2}} + \frac{1}{2}x^2 - 6 \Rightarrow f(x) = -24x^{\frac{1}{2}} + \frac{1}{6}x^3 - 6x + d\) | dM1 A1ft | dM1: dependent on first M; integrate again with one index correct (\(\ldots x^{\frac{1}{2}}\) or \(\ldots x^3\)). A1ft: follow through on their \(k\) only |
| Substitute \(x=4\), \(f(x)=-50 \Rightarrow d = \frac{34}{3}\) | dddM1 | Dependent on all three previous Ms; both \(k\) and \(d\) must have been added |
| \(f(x) = -24x^{\frac{1}{2}} + \frac{1}{6}x^3 - 6x + \frac{34}{3}\) | A1 | Exact equivalent expressions only; do not allow \(\frac{1}{6}\) written as \(0.167\); condone \(-6x^1\); all terms on one line including \(\frac{34}{3}\) |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of normal $= \frac{1}{4}$ | B1 | Deduces gradient of normal is $\frac{1}{4}$ |
| $(y+50) = \frac{1}{4}(x-4) \Rightarrow y = \frac{1}{4}x - 51$ | M1 A1 | M1: line through $P(4,-50)$ with **changed** gradient; allow one sign slip on a coordinate; if $y=mx+c$ used, at least one coordinate correctly substituted. A1: $y=\frac{1}{4}x-51$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f''(x) = \frac{6}{\sqrt{x^3}} + x = 6x^{-\frac{3}{2}} + x \Rightarrow f'(x) = -12x^{-\frac{1}{2}} + \frac{1}{2}x^2 + k$ | M1 A1 | M1: integrate with one index correct ($\ldots x^{-\frac{1}{2}}$ or $\ldots x^2$). A1: unsimplified, with or without $+k$ |
| Substitute $x=4$, $f'(x)=-4 \Rightarrow k=-6$ | dM1 A1 | Dependent on first M1; substitute $x=4$, $f'(x)=-4$ into integrated form with $+k$ |
| $f'(x) = -12x^{-\frac{1}{2}} + \frac{1}{2}x^2 - 6 \Rightarrow f(x) = -24x^{\frac{1}{2}} + \frac{1}{6}x^3 - 6x + d$ | dM1 A1ft | dM1: dependent on first M; integrate again with one index correct ($\ldots x^{\frac{1}{2}}$ or $\ldots x^3$). A1ft: follow through on their $k$ only |
| Substitute $x=4$, $f(x)=-50 \Rightarrow d = \frac{34}{3}$ | dddM1 | Dependent on all three previous Ms; both $k$ and $d$ must have been added |
| $f(x) = -24x^{\frac{1}{2}} + \frac{1}{6}x^3 - 6x + \frac{34}{3}$ | A1 | Exact equivalent expressions only; do not allow $\frac{1}{6}$ written as $0.167$; condone $-6x^1$; all terms on one line including $\frac{34}{3}$ |11. A curve has equation $y = \mathrm { f } ( x )$, where
$$f ^ { \prime \prime } ( x ) = \frac { 6 } { \sqrt { x ^ { 3 } } } + x \quad x > 0$$
The point $P ( 4 , - 50 )$ lies on the curve.\\
Given that $\mathrm { f } ^ { \prime } ( x ) = - 4$ at $P$,
\begin{enumerate}[label=(\alph*)]
\item find the equation of the normal at $P$, writing your answer in the form $y = m x + c$, where $m$ and $c$ are constants,\\
(3)
\item find $\mathrm { f } ( x )$.\\
(8)
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VIIIV SIHI NI IIIYM ION OC & VIIV SIHI NI JIHMM ION OO & VI4V SIHI NI JIIYM ION OO \\
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\hfill \mbox{\textit{Edexcel P1 2020 Q11 [11]}}