| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (straightforward integration + point) |
| Difficulty | Moderate -0.8 This is a straightforward C1 integration question requiring basic algebraic expansion of a binomial, followed by term-by-term integration of powers of x and finding a constant using given coordinates. All steps are routine with no problem-solving insight needed, making it easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \((x^2+3)^2 = x^4 + 3x^2 + 3x^2 + 3^2\) | M1 | For attempting to expand \((x^2+3)^2\) and having at least 3 (out of the 4) correct terms |
| \(\frac{(x^2+3)^2}{x^2} = \frac{x^4+6x^2+9}{x^2} = x^2 + 6 + 9x^{-2}\) (*) | A1cso | (2) At least this should be seen and no incorrect working seen. If they never write \(\frac{9}{x^2}\) as \(9x^{-2}\) they score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(y = \frac{x^3}{3} + 6x + \frac{9}{-1}x^{-1}(+c)\) | M1A1A1 | 1st M1 for some correct integration, one correct \(x\) term as printed or better. Trying \(\frac{\int u}{\int v}\) loses the first M mark but could pick up the second. 1st A1 for two correct \(x\) terms, un-simplified, as printed or better. 2nd A1 for a fully correct expression. Terms need not be simplified and \(+c\) is not required. No \(+c\) loses the next 3 marks |
| \(20 = \frac{27}{3} + 6\times3 - \frac{9}{3} + c\) | M1 | 2nd M1 for using \(x=3\) and \(y=20\) in their expression for \(f(x)\) \(\left[\neq\frac{dy}{dx}\right]\) to form a linear equation for \(c\) |
| \(c = -4\) | A1 | 3rd A1 for \(c = -4\) |
| \([y=]\frac{x^3}{3} + 6x - 9x^{-1} - 4\) | A1ft | (6) 4th A1ft for an expression for \(y\) with simplified \(x\) terms: \(\frac{9}{x}\) for \(9x^{-1}\) is OK. Condone missing "\(y=\)". Follow through their numerical value of \(c\) only |
| Total: 8 |
## Question 11:
### Part (a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $(x^2+3)^2 = x^4 + 3x^2 + 3x^2 + 3^2$ | M1 | For attempting to expand $(x^2+3)^2$ and having at least 3 (out of the 4) correct terms |
| $\frac{(x^2+3)^2}{x^2} = \frac{x^4+6x^2+9}{x^2} = x^2 + 6 + 9x^{-2}$ (*) | A1cso | (2) At least this should be seen and no incorrect working seen. If they never write $\frac{9}{x^2}$ as $9x^{-2}$ they score A0 |
### Part (b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $y = \frac{x^3}{3} + 6x + \frac{9}{-1}x^{-1}(+c)$ | M1A1A1 | 1st M1 for some correct integration, one correct $x$ term as printed or better. Trying $\frac{\int u}{\int v}$ loses the first M mark but could pick up the second. 1st A1 for two correct $x$ terms, un-simplified, as printed or better. 2nd A1 for a fully correct expression. Terms need not be simplified and $+c$ is not required. No $+c$ loses the next 3 marks |
| $20 = \frac{27}{3} + 6\times3 - \frac{9}{3} + c$ | M1 | 2nd M1 for using $x=3$ and $y=20$ in their expression for $f(x)$ $\left[\neq\frac{dy}{dx}\right]$ to form a linear equation for $c$ |
| $c = -4$ | A1 | 3rd A1 for $c = -4$ |
| $[y=]\frac{x^3}{3} + 6x - 9x^{-1} - 4$ | A1ft | (6) 4th A1ft for an expression for $y$ with simplified $x$ terms: $\frac{9}{x}$ for $9x^{-1}$ is OK. Condone missing "$y=$". Follow through their numerical value of $c$ only |
| | | **Total: 8** |\begin{enumerate}
\item The gradient of a curve $C$ is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( x ^ { 2 } + 3 \right) ^ { 2 } } { x ^ { 2 } } , x \neq 0$.\\
(a) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } + 6 + 9 x ^ { - 2 }$.
\end{enumerate}
The point $( 3,20 )$ lies on $C$.\\
(b) Find an equation for the curve $C$ in the form $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel C1 2008 Q11 [8]}}